讲义

\documentclass[cn,fancy,blue,11pt]{elegantbook}

\title{高思数学竞赛讲义}
\subtitle{竞赛真题集}

\author{曾熊}
\institute{高端VIP部}
\date{\today}
\version{3.1}

\equote{青，取之于蓝，而青于蓝；冰，水为之，而寒于水。}

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\mainmatter
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\chapter{国内外数学竞赛真题集}

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\section{杂题}

(2020年北大数分)判断$f(x)=\frac{x}{1+x\cos^2 x}$在$[0,+\infty)$上是否一致连续.

(2020年北大数分)设$q_k\geq p_k>0$, $q_{k+1}-q_k\geq p_k+p_{k+1}$且$\sum_{k=1}^{\infty}a_k\ln p_k=+\infty$,记
\begin{align*}
T_{p_k,q_k}(x)\triangleq &\frac{\cos (q_k+p_k)x}{p_k}+\frac{\cos (q_k+p_k-1)x}{p_k-1}+\frac{\cos (q_k+p_k-2)x}{p_k-2}+\cdots+\frac{\cos (q_k+1)x}{1}\\
& -\frac{\cos (q_k-1)x}{1}-\frac{\cos (q_k-2)x}{2}-\cdots-\frac{\cos (q_k-p_k)x}{p_k},
\end{align*}
设$a_k\geq 0,\sum_{k=1}^{\infty}a_k<+\infty$, $f(x)=\sum_{k=1}^{\infty}a_kT_{p_k,q_k}(x)$.
\begin{enumerate}
\item[(1)] 求证: $f(x)$是在$\mathbb{R}$上连续的以$2\pi$为周期的周期函数.
\item[(2)] 判断并证明: $f(x)$的Fourier级数在$x=0$处的收敛性.
\end{enumerate}

(2020年电子科大)设$0<a_n<1,a_{n+1}=a_n(1-a_n)$,试证:
\begin{enumerate}
\item[(1)] $\lim_{n\to\infty}na_n=1$.
\item[(2)] $\lim_{n\to\infty}\frac{n(1-na_n)}{\ln n}=1$.
\end{enumerate}

(2019年越南MO)数列$\{x_n\}$定义为$x_1=1$,且对任意$n\geq 1$都有$x_{n+1}=x_n+3\sqrt{x_n}+\frac{n}{\sqrt{x_n}}$.
\begin{enumerate}
\item[(1)] 证明: $\lim_{n\to\infty}\frac{n}{x_n}=0$.
\item[(2)] 求极限$\lim_{n\to\infty}\frac{n^2}{x_n}$.
\end{enumerate}

 

求\[\int_{0}^{2 \pi} e^{\cos ^{2} \theta} \cos \theta d \theta.\]

被积函数关于$x=\pi$对称.

求证:
\[\sum_{n=1}^{\infty} \frac{\varphi(4 n)}{n^{4}}=\frac{192}{\pi^{4}} \zeta(3).\]

\begin{example}
Sum the series \[\sum_{m=1}^\infty\sum_{n=1}^\infty\dfrac{m^2n}{3^m(n3^m+m3^n)}.\]
\end{example}
\begin{proof}
Let \[ S = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m (n 3^m + m 3^n)}. \] Then \begin{align*} 2S &= \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m (n 3^m + m 3^n)} + \frac{n^2 m}{3^n (m 3^n + n 3^m)} \\ &= \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn(m 3^n + n 3^m)}{3^m 3^n (m 3^n + n 3^m)} \\ &= \sum_{m=1}^\infty \frac{m}{3^m} \sum_{n=1}^\infty \frac{n}{3^n} \\ &= \left( \sum_{m=1}^\infty \frac{m}{3^m} \right)^2. \end{align*} This latter sum is easily evaluated in any number of ways; e.g., formal differentiation of the power series expansion $(1-z)^{-1} = \sum_{k=0}^\infty z^k$ gives $(1-z)^{-2} = \sum_{k=1}^\infty kz^{k-1}$, so that we immediately have \[ S = \frac{1}{2} \left( \frac{1/3}{(1-1/3)^2} \right)^{\!2} = \frac{9}{32}. \]
\end{proof}

\begin{example}
(HMMT Algebra NT 2019) Find the value of
\[\sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{ab(3a + c)}{4^{a+b+c} (a+b)(b+c)(c+a)}.\]
\end{example}
\begin{proof}
Permute the three variables in the sum (which we call $S$) in all six possible ways. Then add the six sums together to obtain
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{ab(3a+c) + ac(3a+b) + bc(3b+a) + ba(3b+c) + ca(3c+b) + cb(3c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}$$$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{3a^2b+3a^2c+3b^2c+3b^2a+3c^2a+3c^2b+6abc}{4^{a+b+c} (a+b)(b+c)(c+a)}$$$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{3(a+b)(b+c)(c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}$$$$S = \frac12\sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac1{4^{a+b+c}} = \frac12\left(\sum_{a=1}^{\infty}\frac1{4^a}\right)^3 = \frac12\left(\frac13\right)^3 = \boxed{\frac1{54}}$$
A similar idea can be found in Putnam 1999 A4.
\end{proof}

Compute $\tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{3\pi}{7}\right)$.

First, we prove that $\cos \dfrac{\pi}{7} \cos \dfrac{2\pi}{7} \cos \dfrac{3\pi}{7}= \frac{1}{8}$. We have
\begin{align*} -\sin \dfrac{\pi}{7} & = \sin \dfrac{8\pi}{7}= 2\cos \dfrac{4\pi}{7} \sin \dfrac{4\pi}{7}, \\
& = 4 \cos \dfrac{4\pi}{7} \cos \dfrac{2\pi}{7} \sin \dfrac{2\pi}{7}, \\
& = 8 \cos \dfrac{4\pi}{7} \cos \dfrac{2\pi}{7} \cos \dfrac{\pi}{7} \sin \dfrac{\pi}{7}. \end{align*}This follows $\cos \dfrac{4\pi}{7} \cos \dfrac{2\pi}{7} \cos \dfrac{\pi}{7}= \dfrac{-1}{8}$. Since $\cos \frac{4\pi}{7}= - \cos \dfrac{3\pi}{7}$ so $\cos \dfrac{\pi}{7} \cos \dfrac{2\pi}{7} \cos \dfrac{3\pi}{7}= \frac{1}{8}$, as desired.

Next, we will prove that $\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7}= \frac{\sqrt{7}}{8}$. Indeed, let $z= e^{\pi i/7}$ then we have $z^7=-1$ and $z^{14}=1$. Thus, $(1-z)(1-z^6)=1+z^7-z-z^6= \frac{1}{z}-z$ and $(1-z^{13})(1-z^8)=\left(1- \frac 1z \right) \left(1- \frac{1}{z^6} \right)=-\frac{1}{z}- \frac{1}{z^6}=z- \frac{1}{z}$. Thus, $$\sin^2 \frac{\pi}{7}= \left[ \frac{1}{2i} \left(z- \frac{1}{z} \right) \right]^2=\frac 14(1-z)(1-z^6)(1-z^8)(1-z^{13}).$$Similarly, we obtain
\begin{eqnarray*} \sin^2 \frac{5\pi}{7} & = & \frac 14 (1-z^2)(1-z^5)(1-z^{12})(1-z^{9}), \\
\sin^2 \frac{3\pi}{7} & =& \frac 14 (1-z^3)(1-z^4)(1-z^{10})(1-z^{11}). \end{eqnarray*}Thus, by multiplying all of the above together and with $1-z^7=2$ we obtain $$\left( \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7} \right)^2= \frac{1}{4^3 \cdot 2} \prod_{i=1}^{13} (1-z^i).$$Note that $\prod_{i=1}^{13} (1-z^i)= f(1)=14$ where $f(x)=x^{13}+x^{12}+ \cdots + x+1= \frac{x^{14}-1}{x-1}$ that has $13$ roots $z,z^2, \cdots , z^{13}$. Thus, $\left( \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7} \right)^2= \frac{7}{8^2}$ or $ \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7} =\frac{\sqrt{7}}{8}$.

Combining these two identities we obtain $\tan \frac{\pi}{7}\tan \frac{2\pi}{7}\tan \frac{3\pi}{7}= \sqrt{7}$.

Let $\epsilon=e^{\frac{2\pi i }{7}}$. Since $\epsilon,\epsilon^2, \dots, \epsilon^6$ are roots of $z^6+z^5\dots +z+1=0$, so,
$(1-\epsilon),(1-\epsilon^2), \dots , (1-\epsilon^6)$ are roots of$\sum_{k=0}^6 (1-z)^k=0$.

Hence product of the roots,

$\prod_{k=1}^6 (1-\epsilon^k) =7 \dots (1)$

Thus,$\prod_{k=1}^6 \overline{(1-\epsilon^k)} =7 \dots(2)$

[Since $\overline{(1-z)}=1-z^6$ etc.]

Multiplying $(1)$ and $(2)$,

$\prod_{k=1}^6 \vert 1-\epsilon^k\vert =7$

Now, $\epsilon,\epsilon^2, \dots, \epsilon^7=1$ are vertices of a regular heptagon inscribed in $\vert z \vert=1$. $\vert 1-\epsilon^k\vert$'s are the lengths of the chords. There are three pairs of equal chords. So,$\vert 1-z \vert=\vert 1-z^6\vert$ etc. Hence,

$\vert 1-\epsilon \vert \vert 1-\epsilon^2 \vert \vert 1-\epsilon^3 \vert=\sqrt7$

Hence,
$\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7}=\frac{\sqrt7}{8}\dots(3)$

On the other hand,
$\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$

$= -\frac{8\sin \frac{\pi}{7} \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}}{8\sin \frac{\pi}{7}}$

$=-\frac{\sin \frac{8\pi}{7}}{8\sin \frac{\pi}{7} }= \frac{1}{8} \dots(4)$

Hence, dividing $(3)$ by $(4)$, we get,

$\tan\frac{\pi}{7} \tan\frac{2\pi}{7} \tan\frac{3\pi}{7}=\sqrt7$

\begin{example}
(HMMT 2010)Compute $\displaystyle\lim_{n\to\infty}\dfrac{\sum_{k=1}^n|\cos(k)|}{n}$
\end{example}
\begin{proof}
The main idea lies on the fact that positive integers are uniformly distributed modulo $\pi$. (In the other words, if each integer n is written as $q\pi+r$ where $q$ is an integer and $0\leq r<\pi$, the
value of $r$ will distribute uniformly in the interval $[0,\pi]$.) Using this fact, the summation is equivalent
to the average value (using the Riemann summation) of the function j $|cos(k)|$ over the interval $[0,\pi]$.
Therefore, the answer is $\frac{1}{\pi}\int_{0}^{\pi}|\cos (k)|dx=\frac{2}{\pi}$.
\end{proof}

已知圆$O_1$与圆$O_2$交于$A$, $B$两点,过$A$作圆$O_2$的切线交圆$O_1$于点$C$,延长$CB$交圆$O_2$于点$D$,连接$AD$交圆$O_1$于点$E$,延长$CE$交圆$O_2$于点$F$.证明: $CD^2=CE^2+DF^2$.

\begin{example}
求\[2\sqrt {{{\left( {x - 1} \right)}^2} + 4} + \sqrt {{{\left( {x - 8} \right)}^2} + 9} \]的最小值.
\end{example}
\begin{proof}
求导可得$x=2$,最小值为$2\sqrt{5}$.
\end{proof}

\begin{example}
(王永喜)设三次函数
\[f(t)=t^3+27t^2+199t+432,\]
且$a,b,c,x$为互不相同的正实数,满足
\[f(-a)=f(-b)=f(-c)=0,\]
且
\[\sqrt{\frac{a+b+c}{x}}=\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}+\sqrt{\frac{a+b+x}{c}},\]
求$x$.
\end{example}
\begin{proof}
利用笛卡尔定理得
\[\frac{1}{x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
+2\sqrt{\frac{1}{ab}+\frac{1}{ca}+\frac{1}{bc}}.\]
再利用韦达定理得$-a-b-c=-27,ab+bc+ca=199,-abc=-432$.因此$x=\frac{432}{415}$.
\end{proof}

\begin{example}
已知二次函数$f(x)=ax^2+bx+c$, $a,b,c$为实数,且当$|x|\leq 1$时,恒有$|f(x)|\leq 1$.
\begin{enumerate}
\item 证明: $|c|\leq 1$;
\item 证明: $|a|\leq 2$;
\item 若$g(x)=\lambda ax+b\,(\lambda\geq 1)$,求证:当$|x|\leq 1$时, $|g(x)|\leq 2\lambda$.特别地有$|f'(x)|=|2ax+b|\leq 2$.
\end{enumerate}
\end{example}
\begin{proof}
\begin{enumerate}
\item 由$|f(0)|\leq 1$可得$|c|\leq 1$;

\item 因为$f(0)=c,f(1)=a+b+c,f(-1)=a-b+c$,则$2a=f(1)+f(-1)-2f(0)$,于是
\begin{align*}
|2a|=|f(1)+f(-1)-2f(0)|\leq |f(1)|+|f(-1)|+2|f(0)|\leq 4,
\end{align*}
即$|a|\leq 2$.

\item 因为$f(0)=c,f(1)=a+b+c,f(-1)=a-b+c$,则$a=\frac{1}{2}[f(1)+f(-1)]-f(0),b=\frac{1} {2}[f(1)-f(-1)],c=f(0)$
故$g(1)=\lambda a+b=\lambda\cdot\frac{1} {2}[f(1)+f(-1)]-\lambda f(0)+\frac{1} {2}[f(1)-f(-1)]=\frac{\lambda+1}{2}f(1)+\frac{\lambda-1}{2}f(-1)-\lambda f(0)$,且$g(-1)=-\lambda a+b=\frac{1-\lambda}{2}f(1)-\frac{\lambda-1}{2}f(-1)+\lambda f(0)$,
因为$\lambda\geq 1,|f(1)|\leq 1,|f(-1)|\leq 1,|f(0)|\leq 1$,则
$|g(1)|=\left|\frac{\lambda+1}{2}f(1)+\frac{\lambda-1}{2}f(-1)-\lambda f(0)\right|\leq \frac{\lambda+1}{2}+\frac{\lambda-1}{2}+\lambda=2\lambda$,类似地可证$|g(-1)|\leq 1$,由$g(x)$一次函数的性质可知$|g(x)|\leq 1$.

特别地,对于$|f'(x)|=|2ax+b|$,有
\begin{align*}
|2ax+b| & =\left|-2xf(0)+\left(x-\frac{1}{2}\right)f(-1)+\left(x+\frac{1}{2}\right)f(1)\right|\\
& \leq 2|x|+\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{2}\right|\leq 4.
\end{align*}
\end{enumerate}
\end{proof}

类似地,有

\begin{example}
(2010年高联)已知函数\[f(x)=ax^3+bx^2+cx+d\, (a\neq 0),\]
当$0\leq x\leq 1$时, $|f'(x)|\leq 1$.试求$a$的最大值.
\end{example}
\begin{proof}
注意到$f'(x)=3ax^2+2bx+c$.由
\[\begin{cases}
f'(0)=c, \\
f'\left(\frac{1}{2}\right)=\frac{3}{4}a+b+c,\\
f'\left(1\right)=3a+2b+c,
\end{cases}\]
得$3a=2f'(0)+2f'\left(1\right)-4f'\left(\frac{1}{2}\right)$,则
\begin{align*}
3|a|&=\left|2f'(0)+2f'\left(1\right)-4f'\left(\frac{1}{2}\right)\right| \\
&\leq 2|f'(0)|+2\left|f'\left(1\right)\right|+4\left|f'\left(\frac{1}{2}\right)\right|\leq 8,
\end{align*}
故$a\leq\frac{8}{3}$.

又因为$f(x)=\frac{8}{3}x^3-4x^2+x+m$ ($m$为常数)满足题设条件,所以$a$的最大值为$\frac{8}{3}$.
\end{proof}

\begin{example}
已知正项数列 $\{a_n\}$，满足 $a_1=1$，$a_{n+1}=\dfrac{a_n}{\sqrt{a_n^2+1}}$，求证：\[\ln(n+1)\leq a_1a_2+a_2a_3+\cdots+a_na_{n+1}\leq \ln\left(\dfrac{2n}3+1\right)+\dfrac12.\]
\end{example}
\begin{proof}
根据题意，有\[\dfrac{1}{a_{n+1}^2}=1+\dfrac{1}{a_n^2},\]于是\[\dfrac{1}{a_n^2}=n,n\in\mathbb N^{\ast},\]进而\[a_n=\dfrac{1}{\sqrt n},n\in\mathbb N^{\ast},\]欲证不等式即\[\ln (n+1)<\sum_{k=1}^n\dfrac{1}{\sqrt{k(k+1)}}<\ln\left(\dfrac {2n}3+1\right)+\dfrac 12.\]分析通项，只需要证明对任意 $n\geqslant 2$，$n\in\mathbb N^{\ast}$，均有\[\ln(n+1)-\ln n<\dfrac{1}{\sqrt{n(n+1)}}<\ln\left(\dfrac{2n}3+1\right)-\ln\left(\dfrac{2(n-1)}3+1\right),\]也即\[\ln\left(1+\dfrac 1n\right)<\dfrac{1}{\sqrt{n(n+1)}}<\ln\left(1+\dfrac{2}{2n+1}\right).\]事实上，根据对数平均不等式，有\[\dfrac{\dfrac 1n-\dfrac{1}{n+1}}{\ln\dfrac{1}{n}-\ln \dfrac{1}{n+1}}>\sqrt{\dfrac{1}{n(n+1)}},\]于是\[\ln\left(1+\dfrac 1n\right)<\dfrac{1}{\sqrt{n(n+1)}},\]因此左侧不等式成立．对于右侧不等式，显然不成立，需要寻找其他方法． 考虑到\[\sum_{k=1}^n\dfrac{1}{\sqrt{k(k+1)}}<\dfrac 12\sum_{k=1}^n\left(\dfrac 1k+\dfrac1{k+1}\right)=\dfrac 12+\sum_{k=2}^n\dfrac1k+\dfrac1{2(n+1)}<\dfrac 12+\sum_{k=2}^{n=1}\dfrac1k,\]根据对数平均不等式，有\[\dfrac 1n<\ln\dfrac{2n+1}{2n-1},\]因此\[\sum_{k=1}^n\dfrac{1}{\sqrt{k(k+1)}}<\dfrac 12+\ln\dfrac{2n+3}{3},\]右边不等式成立． 综上所述，原不等式得证．
\end{proof}

\begin{example}
在$\triangle ABC$中,求证:
\[\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\leq\frac{1}{8}.\]
\end{example}
\begin{proof}
(高中数学竞赛培优教程P106)
\begin{align*}
\text{左边}&=\frac{1}{2}\left[\cos\frac{A-B}{2}-\cos\frac{A+B}{2}\right]\sin\frac{C}{2}\\
&\leq \frac{1} {2}\left(1-\sin\frac{C}{2}\right)\sin\frac{C}{2}=\frac{1} {2}\left(-\sin^2\frac{C}{2}+\sin\frac{C}{2}\right)\\
&=\frac{1}{2}\left[-\left(\sin\frac{C}{2}-\frac{1}{2}\right)^2+\frac{1}{4}\right]\leq\frac{1}{8}.
\end{align*}
\end{proof}

七、(10分)设$V$是实数域$\mathbb{R}$上的$n$维线性空间, $\phi$是$V$的线性变换,满足$\phi^2=-\varepsilon$ ($\varepsilon$为恒等变换).
\begin{enumerate}
\item 证明: $n$是偶数;

\item 设$V$的线性变换$\psi$与$\phi$可交换,即$\psi\phi=\phi\psi$.证明:
$\psi$在$V$的任意基下的矩阵的行列式均非负.
\end{enumerate}

\newpage
\begin{example}
设$a_0=1,a_{n+1}=\frac{a_n}{1+a_n^2}\,(n=1,2,\cdots)$
求证:
\begin{enumerate}
\item $a_n\leq \frac{3}{4}\cdot \frac{1}{\sqrt{n}}\,(n=1,2,\cdots)$;

\item $a_n\leq \frac{n}{2}\left(\sum_{i=1}^{n}\sqrt{i}\right)^{-1}\,(n=1,2,\cdots)$.
\end{enumerate}
\end{example}
\begin{proof}
\begin{enumerate}
\item 由$a_{n+1}=\frac{a_n}{1+a_n^2}$可得
\[\frac{1}{a_{n+1}^2}=\frac{1}{a_n^2}+a_n^2+2\geq \frac{1}{a_n^2}+2,\]
累加得\[\frac{1}{a_n^2}\geq 2n+1,\]
则\[a_n\leq \frac{1}{\sqrt{2n+1}}\leq \frac{3}{4}\cdot \frac{1}{\sqrt{n}}.\]

\item 只需证
\[\frac{n+1}{a_{n+1}}\geq \frac{n}{a_n}+2\sqrt{n+1},\]
累加后便可得到欲证不等式.由$a_{n+1}=\frac{a_n}{1+a_n^2}$可知这等价于
\[\frac{1}{a_n}+(n+1)a_n\geq 2\sqrt{n+1},\]
由均值不等式显然成立.
\end{enumerate}
\end{proof}

\begin{example}
(1989 年苏联教育部推荐试题)是否存在一个三角形,它可以被分割为1989个全等的三角形？
\end{example}
\begin{proof}

\end{proof}

\begin{example}
(问题征解 ,《数学通讯》1998年第06期)是否有三角形能分成7个全等的三角形?
\end{example}
\begin{proof}

\end{proof}

\begin{example}
(2010全国B卷)设$m$和$n$是大于$1$的整数,求证:
\[1^{m}+2^{m}+\cdots+n^{m}=\frac{1}{m+1}\left\{(n+1) \sum_{\mathrm{k}=1}^{m} \mathrm{C}_{m}^{k} n^{k}-\sum_{j=1}^{m-1}\left(C_{m+1}^{j} \sum_{i=1}^{n} i^{j}\right)\right\}.\]
\end{example}
\begin{proof}

\end{proof}

\begin{example}
(问题征解 ,《数学通讯》2010年第03期)已知$A_1$、$A_2$、$A_3$是抛物线$y=x^2-2$上三点,
\begin{enumerate}
\item 设$A_{1}\left(x_{1}, y_{1}\right), A_{2}\left(x_{2}, y_{2}\right)$,证明:直线$A_1A_2$与单位圆$x^2+y^2=1$相切的充要条件是$\left(x_{1} x_{2}+2\right)^{2}=\left(x_{1}+x_{2}\right)^{2}+1$;

\item 若直线$A_1A_2$、$A_1A_3$均与单位圆$x^2+y^2=1$相切,证明: $A_2A_3$也与单位圆$x^2+y^2=1$相切.
\end{enumerate}
\hfill （318015浙江台州市椒江中学郭天泉供题）
\end{example}
\begin{proof}
1
\end{proof}

\begin{example}
设$x_{1}, x_{2}, \dots, x_{n}$都是非负实数.证明:
\[\sqrt{x_1+x_2+x_3+\cdots+x_n}+\sqrt{x_2+x_3+\cdots+x_n}
+\cdots+\sqrt{x_n}\geq\sqrt{x_1+4x_2+9x_3+\cdots+n^2x_n}.\]
\end{example}
\begin{proof}
1
\end{proof}

祖暅原理

\begin{example}
1123
\end{example}
\begin{proof}
1
\end{proof}

 

\section{自主招生试题}

\begin{example}
(2011年北大等校自招)求$|x-1|+|2 x-1|+\dots+|2011 x-1|$的最小值.
\end{example}
\begin{proof}
绝对值和函数最小值应先定位,再计算;首先,拆分成$x$的系数都是$1$的标准形式
\[f(x)=|x-1|+\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{2}\right|+\cdots+\left|x-\frac{1}{2011}\right|+\cdots+\left|x-\frac{1}{2011}\right|.\]
定义数列$\left\{a_{n}\right\}: 1, \frac{1}{2}, \frac{1}{2}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \cdots, \frac{1}{2011}, \cdots, \frac{1}{2011}$,共计$2011\times 1006=2\, 023\, 066$项, “正中间两项”是$a_{1\, 011\, 533}$与$a_{1\, 011\, 534}$,所以$f(x)_{\min}=f(x),\forall a_{1\, 011\, 534}\leq x\leq a_{1\, 011\, 533}$.

现求$\{a_n\}$的通项公式,设$a_n=\frac{1}{k}$,则$\sum_{i=1}^{k-1} i+1\leq n<\sum_{i=1}^{k} i+1$,即
\begin{align*}
\frac{k(k-1)}{2}+1 &\leq n<\frac{k(k+1)}{2}+1, \\
(2 k-1)^{2} &\leq 8 n-7<(2 k+1)^{2}, \\
2 k-1 &\leq \sqrt{8 n-7}<2 k+1,
\end{align*}
从而\[\frac{\sqrt{8 n-7}-1}{2}<k \leqslant \frac{\sqrt{8 n-7}+1}{2}, k=\left[\frac{\sqrt{8 n-7}+1}{2}\right],\]
故$a_{n}=\left[\frac{\sqrt{8 n-7}+1}{2}\right]^{-1}$.

于是$a_{1011534}=a_{1011533}=\frac{1}{1422}$,所以
\begin{align*}
f(x)_{\min } &=f\left(\frac{1}{1422}\right)=\sum_{i=1}^{1422}\left(1-\frac{i}{1422}\right)+\sum_{i=1423}^{2011}\left(\frac{i}{1422}-1\right) \\
&=833+\sum_{i=1}^{2011} \frac{i}{1422}-2 \sum_{i=1}^{1422} \frac{i}{1422}\\
&=833+\frac{1}{711}(2011 \times 503-711 \times 1423) \\ &=833-\frac{220}{711}=832 \frac{491}{711}=\frac{592043}{711}.
\end{align*}
\end{proof}

 

\begin{example}
(2019年江苏夏令营)在$\triangle ABC$中, $2 \cot A+3 \cot B+4 \cot C$的最小值为$\sqrt{23}$.
\end{example}
\begin{proof}
1
\end{proof}

\begin{example}
1123
\end{example}
\begin{proof}
1
\end{proof}

2017年清华大学4.29标准学术能力测试题

共25个选择题，90分钟，所有选择题均为不定项选择，选对得4分，错选得0分，漏选得2分．考试日期为2017年4月29日．

1．$a_1,a_2,\cdots,a_9$ 是数字 $1$到$9$ 的一个排列，则 $a_1a_2a_3+a_4a_5a_6+a_7a_8a_9$ 的最小值为(　　) A．$213$ B．$214$ C．$215$ D．$216$

2．设$(x^2-x+1)^{1008}=a_0+a_1x+a_2x^2+\cdots+a_{2016}x^{2016}$，则$a_0+2a_1+3a_2+\cdots+2017a_{2016}$的值是(　　) A．$1008$ B．$1009$ C．$2016$ D．$2017$

3．集合$S=\{1,2,\cdots,25\}$，$A\subseteq S$，且$A$的所有子集中元素之和不同．则下列选项正确的有(　　) A．$|A|_{\max}=6$ B．$|A|_{\max}=7$ C．若$A=\{a_1,a_2,a_3,a_4,a_5\}$，则$\sum\limits_{i=1}^5{\dfrac 1{a_i}}<\dfrac 32$ D．若$A=\{a_1,a_2,a_3,a_4,a_5\}$，则$\sum\limits_{i=1}^5{\dfrac 1{a_i}}<2$

4．过椭圆$\dfrac{x^2}4+\dfrac{y^2}3=1$的右焦点$F_2$作一条直线交椭圆于$A,B$，则$\triangle F_1AB$的内切圆面积可能是(　　) A．$1$ B．$2$ C．$3$ D．$4$

5．$\{a_n\},\{b_n\}$均为等差数列，已知$a_1b_1=135$，$a_2b_2=304$，$a_3b_3=529$，则下列是$\{a_nb_n\}$中的项的有(　　) A．$810$ B．$1147$ C．$1540$ D．$3672$

6．已知函数$y=x+\dfrac tx$，过$P(1,0)$作切线交函数图象于点$M$和点$N$，记$|MN|=g(t)$，则下列说法中正确的有(　　) A．$t=\dfrac 14$时，$PM\perp PN$ B．$g(t)$在定义域内单调递增 C．$t=\dfrac 12$时，$M,N$和$(0,1)$共线 D．$g(1)=6$

7．已知数列$\{x_n\}$，其中$x_1=a$，$x_2=b$，$x_{n+1}=x_n+x_{n-1}$（$a,b$是正整数），若$2008$为数列中的某一项，则$a+b$可能的取值有(　　) A．$8$ B．$9$ C．$10$ D．$11$

8．投掷一枚均匀的骰子六次，存在$k$使得$1$到$k$次的点数之和为$6$的概率是$p$，则$p$的取值范围是(　　) A．$0<p<0.25$ B．$0.25<p<0.5$ C．$0.5<p<0.75$ D．$0.75<p<1$

9．在$\triangle ABC$中，$AB=2$，$AC=3$，$BC=4$，$O$为三角形的内心，若$\overrightarrow{AO}=\lambda\overrightarrow{AB}+\mu\overrightarrow{BC}$，则$3\lambda+6\mu$的值为(　　) A．$1$ B．$2$ C．$3$ D．$4$

10．甲、乙、丙、丁四人做相互传球的游戏，第一次甲传给其他三人中的一人，第二次由拿到球的人再传给其他三人中的一人，这样的传球共进行了$4$次，则第四次球传回甲的概率是(　　) A．$\dfrac 7{27}$ B．$\dfrac 5{27}$ C．$\dfrac 78$ D．$\dfrac{21}{64}$

11．已知椭圆$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)$的离心率$e$的取值范围为$\left[\dfrac 1{\sqrt 3},\dfrac 1{\sqrt 2}\right]$，直线$y=-x+1$交椭圆于$M$和$N$，且$OM\perp ON$，则椭圆长轴的取值范围是(　　) A．$\left[\sqrt 5,\sqrt 6\right]$ B．$\left[\sqrt 6,\sqrt 7\right]$ C．$\left[\sqrt 7,\sqrt 8\right]$ D．$\left[\sqrt 8,\sqrt 9\right]$

12．在直角$\triangle ABC$中，以直角边$AB$，斜边$BC$为其中一边分别向三角形所在一侧作正方形$ABDE$和$BCFG$，则向量$\overrightarrow{GA}$和$\overrightarrow{DC}$的夹角为(　　) A．$45^\circ$ B．$60^\circ$ C．$90^\circ$ D．$120^\circ$

13．正方体$ABCD-A_1B_1C_1D_1$的棱长为$1$，底面中心为$O$，$A_1D_1,CC_1$的中点分别为$M,N$，则三棱锥$O-MB_1N$的体积为(　　) A．$\dfrac 7{24}$ B．$\dfrac 7{48}$ C．$\dfrac 5{24}$ D．$\dfrac 5{48}$

14．已知$a,b,c$为正实数，则代数式$\dfrac a{b+3c}+\dfrac b{8c+4a}+\dfrac{9c}{3a+2b}$的最小值为(　　) A．$\dfrac{47}{48}$ B．$1$ C．$\dfrac{35}{36}$ D．$\dfrac 34$

15．在$\triangle ABC$中，$\angle A=60^\circ$，$\angle B=45^\circ$，$\angle A$的角平分线长度为$2$，$CH\perp AB$于$H$，则下列正确的是(　　) A．$CH=\sqrt 3$ B．$AB=\sqrt 3+1$ C．$BC=\sqrt 6$ D．$S_{\triangle ABC}=3$

16．已知实数$x\in\left(0,\dfrac{\pi}2\right)$，则下列方程有解的是(　　) A．$\cos(\cos x)=\sin(\sin x)$ B．$\sin(\cos x)=\cos(\sin x)$ C．$\tan(\tan x)=\sin(\sin x)$ D．$\tan(\sin x)=\sin(\tan x)$

17．已知$0<x<1$，则下列正确的是(　　) A．$\dfrac{\sin x}x<\left(\dfrac{\sin x}x\right)^2<\dfrac{\sin{x^2}}{x^2}$ B．$\left(\dfrac{\sin x}x\right)^2<\dfrac{\sin x}x<\dfrac{\sin{x^2}}{x^2}$ C．$\left(\dfrac{\sin x}x\right)^2<\dfrac{\sin{x^2}}{x^2}<\dfrac{\sin x}x$ D．$\dfrac{\sin{x^2}}{x^2}<\left(\dfrac{\sin x}x\right)^2<\dfrac{\sin x}x$

18．已知$z_1=\sin\alpha+2\mathrm{i}$，$z_2=1+{\mathrm i}\cos\alpha$，则$\dfrac{13-\big|z_1+{\mathrm i}z_2\big|^2}{\big|z_1-{\mathrm i}z_2\big|}$的最小值是(　　) A．$\dfrac 12$ B．$2$ C．$\dfrac 43$ D．$\dfrac 32$

19．在空间中过点$A$作平面$\pi$的垂线，垂足为$B$，记$B=f_{\pi}(A)$．设$\alpha,\beta$是两个不同的平面．对空间中的任意一点$P$，$Q_1=f_{\beta}\left[f_{\alpha}(P)\right]$，$Q_2=f_{\alpha}\left[f_{\beta}(P)\right]$，恒有$PQ_1=PQ_2$，则(　　) A．$\alpha\perp\beta$ B．$\alpha\parallel\beta$ C．$\alpha$与$\beta$的（锐）二面角为$45^\circ$ D．$\alpha$与$\beta$的（锐）二面角为$60^\circ$

20．已知数列$\{a_n\}$，其中$a_1=a$，$a_2=b$，$a_{n+2}=a_n-\dfrac 7{a_{n+1}}$，则(　　) A．$\{a_n\}$可能递增 B．$\{a_n\}$可能递减 C．$\{a_n\}$可能为有限项 D．$\{a_n\}$可能为无限项

21．某校共$2017$名学生，其中每名学生至少要选$A,B$中的一门课，也有些学生选了两门课．已知选修$A$的人数占全校人数介于$70\%$到$75\%$之间，选$B$的人数占$40\%$到$45\%$之间．则下列正确的是(　　) A．同时选$A,B$的可能有$200$人 B．同时选$A,B$的可能有$300$人 C．同时选$A,B$的可能有$400$人 D．同时选$A,B$的可能有$500$人

22．已知$D,E$是${\mathrm{Rt}}\triangle ABC$斜边$BC$上的三等分点．设$AD=a$，$AE=b$，则实数对$(a,b)$可以是(　　) A．$(1,1)$ B．$(1,2)$ C．$(2,3)$ D．$(3,4)$

23．已知函数$f(x)=x^2+2x$，若存在实数$t$，当$x\in[1,m)$时，有$f(x+t)\leqslant 3x$恒成立，则实数$m$可以等于(　　) A．$3$ B．$6$ C．$9$ D．$12$

24．设$x,y\in\mathbb{R}$，函数$f(x,y)=x^2+6y^2-2xy-14x-6y+72$的值域为$M$，则(　　) A．$1\in M$ B．$2\in M$ C．$3\in M$ D．$4\in M$

25．若$N$的三个子集$A,B,C$满足$|A\cap B|=|B\cap C|=|C\cap A|=1$，且$A\cap B\cap C=\varnothing$，则称$(A,B,C)$为$N$的“有序子集列”．现有$N=\{1,2,3,4,5,6\}$，则$N$有(　　)个有序子集列． A．$540$ B．$1280$ C．$3240$ D．$7680$

 

 

学而思【2019暑】高三数学直播实验A班（二试组合）150元

主讲：李昊伟、苏宇坚、邹林强 共15讲

有意深入学习二试并备战CMO的高中学员

【主要内容】组合计数、组合恒等式、操作问题、博弈问题、构造、染色与赋值、凸包与覆盖、分割与整点、图论、欧拉图与哈密顿图、组合综合。

国际数学奥林匹克（International Mathematical Olympiad，简称IMO）是全世界规模和影响最大、难度极高的中学数学赛事。第一届IMO于1959年7月在罗马尼亚举行了，当时只有保加利亚、捷克斯洛伐克、匈牙利、波兰、罗马尼亚和苏联参加。以后每年举行（中间只在1980年断过一次），参加的国家和地区逐渐增多，我国第一次参加国际数学奥林匹克是在1985年。

2019年第60届国际数学奥林匹克（IMO）于2019年7月11日至22日在英国巴斯举办，其中7月16日上午和7月17日上午将举行数学竞赛考试，每天三题，答题时间为四个半小时。比赛总共六道题，每题7分，满分42分。

2019年3月28日上午9点，第60届IMO中国数学奥林匹克国家集训队第二阶段集训闭幕式在上海市上海中学举行，经过为期八天的集训与考试，筛选出本届IMO中国国家队队员名单：

高一 邓明杨 北京 人民大学附属中学

高二 胡苏麟 广东 华南师范大学附属中学

高一 黄嘉俊 上海 上海中学

高三 谢柏庭 浙江 乐清知临中学

高二 袁祉祯 湖北 武汉市武钢三中

高二 俞然枫 江苏 南京师范大学附属中学

 

下面是新出炉的本届IMO真题，比赛结果也将近期公布，期待我国国家队队员再次取得辉煌成绩！

 

 

 

 

\section{北京2019年高中数学邀请赛(4.21)}

\begin{example}
$a,b,c$为两两不等的非负实数,求证:对任意正整数$n\geq 2$,有
\[\frac{a^{n}}{(a-b)(a-c)}+\frac{b^{n}}{(b-c)(b-a)}+\frac{c^{n}}{(c-a)(c-b)}>0.\]
\end{example}
\begin{proof}

\end{proof}

 

\begin{example}
$\triangle ABC$内接于圆$O$, $AB<AC,\angle BAC<90^\circ$, $I$为$\triangle ABC$的内心, $M$为$\overgroup{BC}$中点, $IM$交圆$O$于另一点$N$, $\triangle ABC$在$\angle A$内的旁切圆切$BC$于$D$.求证: $A,O,D$共线.
\end{example}
\begin{proof}

\end{proof}

\begin{example}
1
\end{example}
\begin{proof}

\end{proof}

\section{杜克大学数学竞赛}

\begin{exercise}
Let $f(x)=\frac{3 x^{3}+7 x^{2}-12 x+2}{x^{2}+2 x-3}$. Find all integers $n$ such that $f(n)$ is an integer.
\end{exercise}
翻译:设$f(x)=\frac{3 x^{3}+7 x^{2}-12 x+2}{x^{2}+2 x-3}$.找出所有整数$n$使得$f(n)$为整数.
\begin{solution}
注意到
\begin{align*}
f\left( n \right) &=\frac{3n^3+7n^2-12n+2}{n^2+2n-3}=\frac{\left( n-1 \right) \left( 3n^2+10n-2 \right)}{\left( n+3 \right) \left( n-1 \right)}
\\
&=\frac{3n^2+10n-2}{n+3}=3n+1-\frac{5}{n+3}.
\end{align*}
要使$f(n)$为整数,只需保证$n+3=\pm 1,\pm 5$,此时$n=-2,-4,-8,2$.
\end{solution}

\begin{exercise}
How many ways are there to arrange $10$ trees in a line where every tree is either a yew or an oak and no two oak trees are adjacent?
\end{exercise}
翻译:将$10$颗树排成一条直线,每棵树要么是紫杉要么是橡树,求没有两颗橡树相邻的概率.
\begin{solution}
总共有$2^{10}$种可能性.分别可能有$0,1,2,3,4,5$颗橡树,利用插空法可知满足题意的取法有$C_{11}^{0}+C_{10}^{1}+C_{9}^{2}+C_{8}^{3}+C_{7}^{4}+C_{6}^{5}=144=2^4\times 9$,则所求概率为$\frac{2^4\times 9}{2^{10}}=\frac{9}{64}$.
\end{solution}

\begin{exercise}
$20$ students sit in a circle in a math class. The teacher randomly selects three students to give a presentation. What is the probability that none of these three students sit next to each other?
\end{exercise}
翻译: $20$名学生在数学课上围成一圈.老师随机挑选三个学生作展示.这三个学生中任意两个不坐在一起的概率是多少?
\begin{solution}
将第一个人固定住,只需要选两个人$C_{19}^2$,这两个人只能从$17$个人中选$C_{17}^2$,然后剔除两人相邻的情形,有$16$种,则所求概率为
\[\frac{C_ {17}^{2}-16}{C_{19}^{2}}=\frac{40}{57}.\]
\end{solution}

\begin{exercise}
Let $f_0(x)=x+|x−10|−|x+10|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$. For how many values of $x$ is $f_{10}(x)=0$ ?
\end{exercise}
翻译:设$f_0(x)=x+|x−10|−|x+10|$,对$n\geq 1$,令$f_n(x)=|f_{n-1}(x)|-1$.问有多少个$x$的值使得$f_{10}(x)=0$.
\begin{solution}
由题意可知
\[
f_0\left( x \right) =
\begin{cases}
x+20, &x<-10,\\
-x, & -10\le x\le 10,\\
x-20, &x>10.
\end{cases}
\]
它的零点为$-20,0,20$,共三个.

\[
f_1\left( x \right) =
\begin{cases}
-x-21, &x<-20,\\
x+19, &-20\le x<-10,\\
-x-1, &-10\le x<0,\\
x-1, &0\le x<10,\\
19-x, &10\le x<20,\\
x-21, &x\ge 20.
\end{cases}
\]
它的零点为$-21,-19,-1,1,19,21$,共六个.

\[
f_2\left( x \right) =
\begin{cases}
-x-22, &x<-21,\\
x+20, &-21\le x<-20,\\
-x-20, &-20\le x<-19,\\
x+18, &-19\le x<-10,\\
-x-2, &-10\le x<-1,\\
x, &-1\le x<0,\\
-x, &0\le x<1,\\
x-2, &1\le x<10,\\
18-x, &10\le x<19,\\
x-20, & 19\le x<20,\\
20-x, &20\le x<21\\
x-22, &x\ge 21
\end{cases}\]
它的零点为$-22,-20,-18,-2,0,2,18,20,22$,共九个.

这成一个等差数列,即$f_n(x)=0$有$3n+3$个零点,故$f_{10}(x)=0$有$33$个零点.
\end{solution}

\begin{exercise}
$2$ red balls, $2$ blue balls, and $6$ yellow balls are in a jar. Zion picks $4$ balls from the jar at random. What is the probability that Zion picks at least $1$ red ball and $1$ blue ball?
\end{exercise}
翻译:一个罐子里有$2$个红球, $2$个蓝球和$6$个黄球. Zion随机从罐子里取出四个球.问锡安选择至少一个红球和一个蓝球的概率是多少?
\begin{solution}
有以下几种可能: $1$红球$1$蓝球$2$黄球; $1$红球$2$蓝球$1$黄球; $2$红球$1$蓝球$1$黄球; $2$红球$2$蓝球.总共有$C_{10}^4$种取法,则所求概率为$\frac{C_ {2}^1C_ {2}^1C_ {6}^2+2C_ {2}^1C_ {2}^2C_ {6}^1+C_ {2}^2C_ {2}^2}{C_ {10}^4}=\frac{17}{42}$.
\end{solution}

\begin{exercise}
Let $\triangle ABC$ be a right-angled triangle with $\angle ABC = 90$ and $AB = 4$. Let $D$ on $\overline{AB}$ such that $AD=3DB$ and $\sin\angle ACD =\frac{3}{5}$. What is the length of $BC$ ?
\end{exercise}
翻译:设$\triangle ABC$为直角三角形, $\angle ABC = 90^\circ,AB = 4$.设$D$在边$\overline{AB}$上,使得$AD=3DB$和$\sin\angle ACD =\frac{3}{5}$.求$BC$的长度是多少?
\begin{figure}[htbp!]
\centering
\includegraphics[width=0.2\textwidth]{Tupian/Duke186.png}
%\caption{散点图示例 $\hat{y}=a+bx$ \label{fig:scatter}}
\end{figure}
\begin{solution}
记$BC=x,\angle ACD=\alpha,\angle BCD=\beta$,由题意可知$AB=4,AD=3,DB=1$,则$\sin\alpha=\frac{3}{5}$,则$\tan\alpha=\frac{3} {4},\tan\beta=\frac{1}{x},\tan(\alpha+\beta)=\frac{4}{x}$,则
\[
\frac{4}{x}=\tan \left( \alpha +\beta \right) =\frac{\tan \alpha +\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\frac{\frac{3}{4}+\frac{1}{x}}{1-\frac{3}{4}\cdot \frac{1}{x}},\]
化简得$x^2-4x+4=0$,即$BC=x=2$.
\end{solution}

\begin{exercise}
Find the value of
\[\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\cdots}}}}}.\]
\end{exercise}
翻译:求\[\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\cdots}}}}}.\]
\begin{solution}
记$x=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\cdots}}}}}$,则
\[x=\dfrac{1}{1+\dfrac{1}{2+x}},\]
则$x^2+2x-2=0$,因此$x=\sqrt{3}-1$.
\end{solution}

\begin{exercise}
Consider all possible quadrilaterals $\Box ABCD$ that have the following properties; $\Box ABCD$ has integer side lengths with $AB\parallel CD$, the distance between $\overline{AB}$ and $\overline{CD}$ is $20$, and $AB=18$. What is the maximum area among all these quadrilaterals, minus the minimum area?
\end{exercise}
翻译:考虑具有以下性质的所有可能的四边形$\Box ABCD$; $\Box ABCD$具有整数边长且$AB\parallel CD$, $\overline{AB}$和$\overline{CD}$两边之间的距离为$20$且$AB=18$.所有这样的四边形的最大面积与最小面积之差是多少?
\begin{figure}[htbp!]
\centering
\includegraphics[width=0.8\textwidth]{Tupian/Duke1881.png}
\includegraphics[width=0.8\textwidth]{Tupian/Duke1882.png}
%\caption{散点图示例 $\hat{y}=a+bx$ \label{fig:scatter}}
\end{figure}
\begin{proof}
勾股数$15^2+20^2=25^2,20^2+21^2=29^2,20^2+48^2=52^2,
20^2+99^2=101^2 $.设$CD=x$,则四边形$\Box ABCD$的面积为$S=\frac{x+18}{2}\cdot 20=10(x+18)$,当$x=18-15=21-18=3$时,面积最小;当$x=2\times 99+18=216$时,面积最大.所求的差值为$2130$.
\end{proof}

\begin{exercise}
How many perfect cubes exist in the set $\left\{1^{2018}, 2^{2017}, 3^{2016}, \cdots, 2017^{2}, 2018^{1}\right\}$?
\end{exercise}
翻译: 集合$\left\{1^{2018}, 2^{2017}, 3^{2016}, \cdots, 2017^{2}, 2018^{1}\right\}$中存在多少个完全立方数?
\begin{solution}
注意到$12^3=1728<2018<13^3=2197$,则$2018$以内有$12$个完全立方数.又$\frac{2018}{3}=672.666\cdots$,则$2018$以内有$672$个被$3$整除的正整数.该数列的形式为$n^{2019-n},n=1,2,\cdots,2018$, $2019$能被$3$整除,而$12$以内有$4$个数被$3$整除.于是该集合中完全立方数有$12+672-4=680$个.
\end{solution}

\begin{exercise}
Let $n$ be the number of ways you can fill a $2018\times 2018$ array with the digits $1$ through $9$ such that for every $11\times 3$ rectangle (not necessarily for every $3\times 11$ rectangle), the sum of the $33$ integers in the rectangle is divisible by $9$. Compute $\log_3 n$.
\end{exercise}
翻译:设$n$为用数字$1$到$9$填充$2018\times 2018$方阵的方法个数,使得对于每个$11\times 3$矩形(不必对每个$3\times 11$矩形)中$33$个整数的和可以被$9$整除.计算$\log_3 n$.
\begin{solution}
只要前两行及前十列任意填数,其它为固定值即可,方法数共有$n=9^{2018\times 2+10\times 2016}=9^{24196}$种,则$\log_3 n=48392$.
\end{solution}

\chapter{2018 AMC 12}

\href{https://artofproblemsolving.com/community/c588206_2018_amc_12ahsme}{AOPS网站}

February 7th, 2018

1 A large urn contains $100$ balls, of which $36\%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72\%?$ (No red balls are to be removed.)

$
\textbf{(A) }28 \qquad
\textbf{(B) }32 \qquad
\textbf{(C) }36 \qquad
\textbf{(D) }50 \qquad
\textbf{(E) }64 \qquad
$

laegolas
view topic
2 While exploring a cave, Carl comes across a collection of $5$-pound rocks worth $\$14$ each, $4$-pound rocks worth $\$11$ each, and $1$-pound rocks worth $\$2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?

$\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52 $

rafayaashary1
view topic
3 How many ways can a student schedule 3 mathematics courses -- algebra, geometry, and number theory -- in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.)

$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

thatmath
view topic
4 Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least 6 miles away," Bob replied, "We are at most 5 miles away." Charlie then remarked, "Actually the nearest town is at most 4 miles away." It turned out that none of the three statements were true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d$?

$\textbf{(A) } (0,4) \qquad \textbf{(B) } (4,5) \qquad \textbf{(C) } (4,6) \qquad \textbf{(D) } (5,6) \qquad \textbf{(E) } (5,\infty) $

bluecarneal
view topic
5 What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common?

$
\textbf{(A) }3 \qquad
\textbf{(B) }4 \qquad
\textbf{(C) }5 \qquad
\textbf{(D) }6 \qquad
\textbf{(E) }10 \qquad
$

FuturePinkMOPper
view topic
6 For positive integers $m$ and $n$ such that $m+10<n+1$, both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$. What is $m+n$?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$

Slacker
view topic
7 For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?

$
\textbf{(A) }3 \qquad
\textbf{(B) }4 \qquad
\textbf{(C) }6 \qquad
\textbf{(D) }8 \qquad
\textbf{(E) }9 \qquad
$

laegolas
view topic
8 All of the triangles in the diagram below are similar to iscoceles triangle $ABC$, in which $AB=AC$. Each of the 7 smallest triangles has area 1, and $\triangle ABC$ has area 40. What is the area of trapezoid $DBCE$?

[asy]
unitsize(5);
dot((0,0));
dot((60,0));
dot((50,10));
dot((10,10));
dot((30,30));
draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0));
draw((10,10)--(50,10));
label("$B$",(0,0),SW);
label("$C$",(60,0),SE);
label("$E$",(50,10),E);
label("$D$",(10,10),W);
label("$A$",(30,30),N);
draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10));
draw((15,15)--(45,15));
[/asy]

$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24 $

proshi
view topic
9 Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which
$$\sin(x+y)\leq \sin(x)+\sin(y)$$for every $x$ between $0$ and $\pi$, inclusive?
$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi $

rafayaashary1
view topic
10 How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?
\begin{align*}x+3y&=3\\ \big||x|-|y|\big|&=1\end{align*}$\textbf{(A) } 1 \qquad
\textbf{(B) } 2 \qquad
\textbf{(C) } 3 \qquad
\textbf{(D) } 4 \qquad
\textbf{(E) } 8 $

ythomashu
view topic
11 A paper triangle with sides of lengths 3, 4, and 5 inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
[asy]
draw((0,0)--(4,0)--(4,3)--(0,0));
label("$A$", (0,0), SW);
label("$B$", (4,3), NE);
label("$C$", (4,0), SE);
label("$4$", (2,0), S);
label("$3$", (4,1.5), E);
label("$5$", (2,1.5), NW);
fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray(0.9));
[/asy]
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 $

Dhman2727
view topic
12 Let $S$ be a set of 6 integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$, then $b$ is not a multiple of $a$. What is the least possible value of an element in $S$?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$

Mudkipswims42
view topic
13 How many nonnegative integers can be written in the form $$a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,$$where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$?

$\textbf{(A) } 512 \qquad
\textbf{(B) } 729 \qquad
\textbf{(C) } 1094 \qquad
\textbf{(D) } 3281 \qquad
\textbf{(E) } 59,048 $

ythomashu
view topic
14 The solution to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$, can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$?

$\textbf{(A) } 5 \qquad
\textbf{(B) } 13 \qquad
\textbf{(C) } 17 \qquad
\textbf{(D) } 31 \qquad
\textbf{(E) } 35 $

FuturePinkMOPper
view topic
15 A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?

$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$

reedmj
view topic
16 Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?

$
\textbf{(A) }a=\frac14 \qquad
\textbf{(B) }\frac14 < a < \frac12 \qquad
\textbf{(C) }a>\frac14 \qquad
\textbf{(D) }a=\frac12 \qquad
\textbf{(E) }a>\frac12 \qquad
$

AOPS12142015
view topic
17 Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is 2 units. What fraction of the field is planted?

[asy]
draw((0,0)--(4,0)--(0,3)--(0,0));
draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));
fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);
label("$4$", (2,0), N);
label("$3$", (0,1.5), E);
label("$2$", (.8,1), E);
label("$S$", (0,0), NE);
draw((0.3,0.3)--(1.4,1.9), dashed);
[/asy]

$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} $

kevinmathz
view topic
18 Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

$
\textbf{(A) }60 \qquad
\textbf{(B) }65 \qquad
\textbf{(C) }70 \qquad
\textbf{(D) }75 \qquad
\textbf{(E) }80 \qquad
$

AOPS12142015
view topic
19 Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots$$of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

$\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}$

reedmj
view topic
20 Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$. Let $M$ be the midpoint of hypotenuse $\overline{BC}$. Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$, the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$?

$
\textbf{(A) }9 \qquad
\textbf{(B) }10 \qquad
\textbf{(C) }11 \qquad
\textbf{(D) }12 \qquad
\textbf{(E) }13 \qquad
$

laegolas
view topic
21 Which of the following polynomials has the greatest real root?

$\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018 $

Benq
view topic
22 The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$

$\textbf{(A) } 20 \qquad
\textbf{(B) } 21 \qquad
\textbf{(C) } 22 \qquad
\textbf{(D) } 23 \qquad
\textbf{(E) } 24 $

Benq
view topic
23 In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$

$\textbf{(A) } 76 \qquad
\textbf{(B) } 77 \qquad
\textbf{(C) } 78 \qquad
\textbf{(D) } 79 \qquad
\textbf{(E) } 80 $

Benq
view topic
24 Alice, Bob, and Carol play a game in which each of them chooses a real number between 0 and 1. The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between 0 and 1, and Bob announces that he will choose his number uniformly at random from all the numbers between $\tfrac{1}{2}$ and $\tfrac{2}{3}.$ Armed with this information, what number should Carol choose to maximize her chance of winning?

$
\textbf{(A) }\frac{1}{2}\qquad
\textbf{(B) }\frac{13}{24} \qquad
\textbf{(C) }\frac{7}{12} \qquad
\textbf{(D) }\frac{5}{8} \qquad
\textbf{(E) }\frac{2}{3}\qquad
$

Benq
view topic
25 For a positive integer $n$ and nonzero digits $a$, $b$, and $c$, let $A_n$ be the $n$-digit integer each of whose digits is equal to $a$; let $B_n$ be the $n$-digit integer each of whose digits is equal to $b$, and let $C_n$ be the $2n$-digit (not $n$-digit) integer each of whose digits is equal to $c$. What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$?

$\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}$

CantonMathGuy
view topic
B
February 15th, 2018
1 Kate bakes a $20$-inch by $18$-inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?

$
\textbf{(A) }90 \qquad
\textbf{(B) }100 \qquad
\textbf{(C) }180 \qquad
\textbf{(D) }200 \qquad
\textbf{(E) }360 \qquad
$

brainiac1
view topic
2 Sam drove $96$ miles in $90$ minutes. His average speed during the first $30$ minutes was $60$ mph (miles per hour), and his average speed during the second $30$ minutes was $65$ mph. What was his average speed, in mph, during the last $30$ minutes?

$\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68$

AIME12345
view topic
3 A line with slope $2$ intersects a line with slope $6$ at the point $(40, 30)$. What is the distance between the $x$-intercepts of these two lines?

$\textbf{(A) }5\qquad\textbf{(B) }10\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }50$

AlcumusGuy
view topic
4 A circle has a chord of length $10$, and the distance from the center of the circle to the chord is $5$. What is the area of the circle?

$\textbf{(A) }25\pi\qquad\textbf{(B) }50\pi\qquad\textbf{(C) }75\pi\qquad\textbf{(D) }100\pi\qquad\textbf{(E) }125\pi$

djmathman
view topic
5 How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?

$\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}$

AIME12345
view topic
6 Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?

$\textbf{(A) }\frac{4DQ}S\qquad\textbf{(B) }\frac{4DS}Q\qquad\textbf{(C) }\frac{4Q}{DS}\qquad\textbf{(D) }\frac{DQ}{4S}\qquad\textbf{(E) }\frac{DS}{4Q}$

djmathman
view topic
7 What is the value of

\[ \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27? \]
$\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10 $

MathSlayer4444
view topic
8 Line segment $\overline{AB}$ is a diameter of a circle with $AB=24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle{ABC}$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

$\textbf{(A)} \text{ 25} \qquad \textbf{(B)} \text{ 38} \qquad \textbf{(C)} \text{ 50} \qquad \textbf{(D)} \text{ 63} \qquad \textbf{(E)} \text{ 75}$

Dhman2727
view topic
9 What is \[ \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? \]$
\textbf{(A) }100,100 \qquad
\textbf{(B) }500,500\qquad
\textbf{(C) }505,000 \qquad
\textbf{(D) }1,001,000 \qquad
\textbf{(E) }1,010,000 \qquad
$

brainiac1
view topic
10 A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?

$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$

AOPS12142015
view topic
11 A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper?
[asy]size(270pt);
defaultpen(fontsize(10pt));
filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);
dot((-3,3));
label("$A$",(-3,3),NW);
draw((1,3)--(-3,-1),dashed+linewidth(.5));
draw((-1,3)--(3,-1),dashed+linewidth(.5));
draw((-1,-3)--(3,1),dashed+linewidth(.5));
draw((1,-3)--(-3,1),dashed+linewidth(.5));
draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));
draw((0,3)--(0,-3),linetype("2.5 2.5")+linewidth(.5));
draw((3,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5));
label('$w$',(-1,-1),SW);
label('$w$',(1,-1),SE);
draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle);
draw((4.5,0)--(8.5,0));
draw((6.5,2)--(6.5,-2));
label("$A$",(6.5,0),NW);
dot((6.5,0));
[/asy]

$\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h $

ythomashu
view topic
12 Side $\overline{AB}$ of $\triangle ABC$ has length $10$. The bisector of angle $A$ meets $\overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?

$
\textbf{(A) }16 \qquad
\textbf{(B) }17 \qquad
\textbf{(C) }18 \qquad
\textbf{(D) }19 \qquad
\textbf{(E) }20 \qquad
$

brainiac1
view topic
13 Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?

[asy]
unitsize(120);
pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3);
draw(A--B--C--D--cycle);
dot(P);
defaultpen(fontsize(10pt));
draw(A--P--B);
draw(C--P--D);
label("$A$", A, W);
label("$B$", B, W);
label("$C$", C, E);
label("$D$", D, E);
label("$P$", P, N*1.5+E*0.5);
dot(A);
dot(B);
dot(C);
dot(D);
[/asy]

$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$

TheBigOne
view topic
14 Joey and Chloe and their daughter Zoe all have the same birthday. Joey is 1 year older than Chloe, and Zoe is exactly 1 year old today. Today is the first of the 9 birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?

$\textbf{(A) } 7 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11 $

itised
view topic
15 How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?

$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 $

Metal Bender19
view topic
16 The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$

$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$

hwl0304
view topic
17 Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\]and $q$ is as small as possible. What is $q-p$?

$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19 $

AlcumusGuy
view topic
18 A function $f$ is defined recursively by $f(1)=f(2)=1$ and $$f(n)=f(n-1)-f(n-2)+n$$for all integers $n \geq 3$. What is $f(2018)$?

$\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}$

AOPS12142015
view topic
19 Mary chose an even $4$-digit number $n$. She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,...,\tfrac{n}{2},n$. At some moment Mary wrote $323$ as a divisor of $n$. What is the smallest possible value of the next divisor written to the right of $323$?

$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$

Dr4gon39
view topic
20 Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline{AB},\overline{CD},\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle{ACE}$ and $\triangle{XYZ}$?

$\textbf{(A) }\dfrac{3}{8}\sqrt{3}\qquad\textbf{(B) }\dfrac{7}{16}\sqrt{3}\qquad\textbf{(C) }\dfrac{15}{32}\sqrt{3}\qquad\textbf{(D) }\dfrac{1}{2}\sqrt{3}\qquad\textbf{(E) }\dfrac{9}{16}\sqrt{3}$

hamup1
view topic
21 In $\triangle{ABC}$ with side lengths $AB = 13$, $AC = 12$, and $BC = 5$, let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle{ABC}$. What is the area of $\triangle{MOI}$?

$\textbf{(A)}\ 5/2\qquad\textbf{(B)}\ 11/4\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 13/4\qquad\textbf{(E)}\ 7/2$

quinna nyc

view topic

22 Consider polynomials $P(x)$ of degree at most $3$, each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. How many such polynomials satisfy $P(-1) = -9$?

$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286 $

AlcumusGuy
view topic
23 Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C$. What is the degree measure of $\angle ACB$?

$
\textbf{(A) }105 \qquad
\textbf{(B) }112\frac{1}{2} \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }135 \qquad
\textbf{(E) }150 \qquad
$

brainiac1
view topic
24 Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. How many real numbers $x$ satisfy the equation $x^2 + 10{,}000\lfloor x \rfloor = 10{,}000x$?

$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$

GeronimoStilton
view topic
25 Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$?

[asy]
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);
dot(P2);
dot(P3);
defaultpen(fontsize(10pt));
label("$P_1$", P1, E*1.5);
label("$P_2$", P2, SW*1.5);
label("$P_3$", P3, N);
label("$\omega_1$", A, W*17);
label("$\omega_2$", B, E*17);
label("$\omega_3$", C, W*17);
[/asy]

$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$

2019 AIME Problems 3

March 13th, 2019

1 Consider the integer $$N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.$$Find the sum of the digits of $N$.

LauraZed
view topic
2 Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$. Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$. The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

InflateGate
view topic
3 In $\triangle PQR$, $PR=15$, $QR=20$, and $PQ=25$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, and points $E$ and $F$ lie on $\overline{PR}$, with $PA=QB=QC=RD=RE=PF=5$. Find the area of hexagon $ABCDEF$.

Vfire
view topic
4 A soccer team has 22 available players. A fixed set of 11 players starts the game, while the other 11 are available as substitutes. During the game, the coach may make as many as 3 substitutions, where any one of the 11 players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let $n$ be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when $n$ is divided by 1000.

LauraZed
view topic
5 A moving particle starts at the point $\left(4,4\right)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $\left(a,b\right)$, it moves at random to one of the points $\left(a-1,b\right)$, $\left(a,b-1\right)$, or $\left(a-1,b-1\right)$, each with probability $\tfrac{1}{3}$, independently of its previous moves. The probability that it will hit the coordinate axes at $\left(0,0\right)$ is $\tfrac{m}{3^n}$, where $m$ and $n$ are positive integers, and $m$ is not divisible by $3$. Find $m+n$.

 

6 In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$.

LauraZed
view topic
7 There are positive integers $x$ and $y$ that satisfy the system of equations
\begin{align*}
\log_{10} x + 2 \log_{10} (\gcd(x,y)) &= 60 \\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570.
\end{align*}Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$, and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$. Find $3m+2n$.

Vfire
view topic
8 Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$. Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Vfire
view topic
9 Let $\tau (n)$ denote the number of positive integer divisors of $n$. Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$.

10 For distinct complex numbers $z_1,z_2,\dots,z_{673}$, the polynomial
\[ (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 \]can be expressed as $x^{2019} + 20x^{2018} + 19x^{2017}+g(x)$, where $g(x)$ is a polynomial with complex coefficients and with degree at most $2016$. The value of
\[ \left| \sum_{1 \le j <k \le 673} z_jz_k \right| \]can be expressed in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Vfire
view topic
11 In $\triangle ABC$, the sides have integers lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$, and the other two excircles are both externally tangent to $\omega$. Find the minimum possible value of the perimeter of $\triangle ABC$.

Vfire
view topic
12 Given $f(z) = z^2-19z$, there are complex numbers $z$ with the property that $z$, $f(z)$, and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$. There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$. Find $m+n$.

Vfire
view topic
13 Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.

Vfire
view topic
14 Find the least odd prime factor of $2019^8 + 1$.

15 Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

 

rrusczyk
view topic
II
March 21st, 2019
1 Points $C\neq D$ lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

djmathman
view topic
2 Lily pads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$. From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly and independently with probability $\tfrac12$. The probability that the frog visits pad $7$ is $\tfrac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

djmathman
view topic
3 Find the number of $7$-tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following systems of equations:
\begin{align*}
abc&=70,\\
cde&=71,\\
efg&=72.
\end{align*}

proshi
view topic
4 A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

asp211
view topic
5 Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order from $1$ to $12$. Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table under these conditions. Find the remainder when $N$ is divided by $1000$.

a1b2
view topic
6 In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$, for some fixed $b \geq 2$. A Martian student writes down
\begin{align*}3 \log(\sqrt{x}\log x) &= 56\\\log_{\log (x)}(x) &= 54
\end{align*}and finds that this system of equations has a single real number solution $x > 1$. Find $b$.

popcorn1
view topic
7 Triangle $ABC$ has side lengths $AB=120$, $BC=220$, and $AC=180$. Lines $\ell_{A}$, $\ell_{B}$, and $\ell_{C}$ are drawn parallel to $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$, respectively, such that the intersection of $\ell_{A}$, $\ell_{B}$, and $\ell_{C}$ with the interior or $\triangle ABC$ are segments of length $55$, $45$, and $15$, respectively. Find the perimeter of the triangle whose sides lie on $\ell_{A}$, $\ell_{B}$, and $\ell_{C}$.

fatant
view topic
8 The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019$, and $f(\tfrac{1+\sqrt{3}i}{2})=2015+2019\sqrt{3}i$. Find the remainder when $f(1)$ is divided by $1000$.

fatant
view topic
9 Call a positive integer $n$ $k$-pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$. For example, $18$ is $6$-pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$-pretty. Find $\tfrac{S}{20}$.

fatant
view topic
10 There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$, the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$, and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime integers. Find $p+q$.

fatant
view topic
11 Triangle $ABC$ has side lengths $AB=7, BC=8, $ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

AOPS12142015
view topic
12 For $n \ge 1$ call a finite sequence $(a_1, a_2 \ldots a_n)$ of positive integers progressive if $a_i < a_{i+1}$ and $a_i$ divides $a_{i+1}$ for all $1 \le i \le n-1$. Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360$.

DeathLlama9
view topic
13 Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1$. Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1}$, $\overline{PA_2}$, and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac17$, while the region bounded by $\overline{PA_3}$, $\overline{PA_4}$, and the minor arc $\widehat{A_3A_4}$ of the circle has area $\tfrac 19$. There is a positive integer $n$ such that the area of the region bounded by $\overline{PA_6}$, $\overline{PA_7}$, and the minor arc $\widehat{A_6A_7}$ is equal to $\tfrac18 - \tfrac{\sqrt 2}n$. Find $n$.

djmathman
view topic
14 Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5$, $n$, and $n + 1$ cents, $91$ cents is the greatest postage that cannot be formed.

popcorn1
view topic
15 In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$, respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$. Suppose $XP=10$, $PQ=25$, and $QY=15$. The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

\section{中国数学国家队选拔考试}

\href{https://artofproblemsolving.com/community/c848182}{AOPS网页}

\begin{example}
给定复数$x,y,z$,满足$|x|^2+|y|^2+|z|^2=1$. 证明
\[|x^3+y^3+z^3-3xyz| \le 1.\]
\end{example}
\begin{proof}
令$s=|x+y+z|$.注意到
\begin{align*}
|x^3+y^3+z^3-3xyz|&=\frac{1}{2}s|(x-y)^2+(y-z)^2+(z-x)^2| \\
&\leqslant \frac{1}{2}s(|x-y|^2+|y-z|^2+|z-x|^2) \\
&= \frac{1}{2}s(3(|x|^2+|y|^2+|z|^2)-|x+y+z|^2) \\
&=\frac{1}{2}s(3-s^2) \\
&\leqslant 1.
\end{align*}
\end{proof}

The given condition implies that every column vector of the matrix
\[A = \begin{bmatrix}
x & y & z\\
y & z & x\\
z & x & y
\end{bmatrix}\]has magnitude $1$, so by Hadamard's inequality
\[|x^3 + y^3 + z^3 - 3xyz| = \left|\det A\right| \le 1.\]

\begin{example}
This is the content of example environment.
\end{example}

Test 1 Day 1

1 $ABCDE$ is a cyclic pentagon, with circumcentre $O$. $AB=AE=CD$. $I$ midpoint of $BC$. $J$ midpoint of $DE$. $F$ is the orthocentre of $\triangle ABE$, and $G$ the centroid of $\triangle AIJ$.$CE$ intersects $BD$ at $H$, $OG$ intersects $FH$ at $M$. Show that $AM\perp CD$.

2 Fix a positive integer $n\geq 3$. Does there exist infinitely many sets $S$ of positive integers $\lbrace a_1,a_2,\ldots, a_n$, $b_1,b_2,\ldots,b_n\rbrace$, such that $\gcd (a_1,a_2,\ldots, a_n$, $b_1,b_2,\ldots,b_n)=1$, $\lbrace a_i\rbrace _{i=1}^n$, $\lbrace b_i\rbrace _{i=1}^n$ are arithmetic progressions, and $\prod_{i=1}^n a_i = \prod_{i=1}^n b_i$?

 

3 Find all positive integer $n$, such that there exists $n$ points $P_1,\ldots,P_n$ on the unit circle , satisfying the condition that for any point $M$ on the unit circle, $\sum_{i=1}^n MP_i^k$ is a fixed value for

a) $k=2018$

b) $k=2019$.

 

Test 1 Day 2

4 Call a sequence of positive integers $\{a_n\}$ good if for any distinct positive integers $m,n$, one has
$$\gcd(m,n) \mid a_m^2 + a_n^2 \text{ and } \gcd(a_m,a_n) \mid m^2 + n^2.$$Call a positive integer $a$ to be $k$-good if there exists a good sequence such that $a_k = a$. Does there exists a $k$ such that there are exactly $2019$ $k$-good positive integers?

 

5 Determine all functions $f: \mathbb{Q} \to \mathbb{Q}$ such that
$$f(2xy + \frac{1}{2}) + f(x-y) = 4f(x)f(y) + \frac{1}{2}$$for all $x,y \in \mathbb{Q}$.

 

6 Let $k$ be a positive real. $A$ and $B$ play the following game: at the start, there are $80$ zeroes arrange around a circle. Each turn, $A$ increases some of these $80$ numbers, such that the total sum added is $1$. Next, $B$ selects ten consecutive numbers with the largest sum, and reduces them all to $0$. $A$ then wins the game if he/she can ensure that at least one of the number is $\geq k$ at some finite point of time.

Determine all $k$ such that $A$ can always win the game.

 

Test 2 Day 1

1 $AB$ and $AC$ are tangents to a circle $\omega$ with center $O$ at $B,C$ respectively. Point $P$ is a variable point on minor arc $BC$. The tangent at $P$ to $\omega$ meets $AB,AC$ at $D,E$ respectively. $AO$ meets $BP,CP$ at $U,V$ respectively. The line through $P$ perpendicular to $AB$ intersects $DV$ at $M$, and the line through $P$ perpendicular to $AC$ intersects $EU$ at $N$. Prove that as $P$ varies, $MN$ passes through a fixed point.

 

2 Let $S$ be the set of $10$-tuples of non-negative integers that have sum $2019$. For any tuple in $S$, if one of the numbers in the tuple is $\geq 9$, then we can subtract $9$ from it, and add $1$ to the remaining numbers in the tuple. Call thus one operation. If for $A,B\in S$ we can get from $A$ to $B$ in finitely many operations, then denote $A\rightarrow B$.

(1) Find the smallest integer $k$, such that if the minimum number in $A,B\in S$ respectively are both $\geq k$, then $A\rightarrow B$ implies $B\rightarrow A$.

(2) For the $k$ obtained in (1), how many tuples can we pick from $S$, such that any two of these tuples $A,B$ that are distinct, $A\not\rightarrow B$.

 

3 Let $n$ be a given even number, $a_1,a_2,\cdots,a_n$ be non-negative real numbers such that $a_1+a_2+\cdots+a_n=1.$ Find the maximum possible value of $\sum_{1\le i<j\le n}\min\{(i-j)^2,(n+i-j)^2\}a_ia_j .$

 

Test 2 Day 2

4 Does there exist a finite set $A$ of positive integers of at least two elements and an infinite set $B$ of positive integers, such that any two distinct elements in $A+B$ are coprime, and for any coprime positive integers $m,n$, there exists an element $x$ in $A+B$ satisfying $x\equiv n \pmod m$ ?

Here $A+B=\{a+b|a\in A, b\in B\}$.

 

5 Let $M$ be the midpoint of $BC$ of triangle $ABC$. The circle with diameter $BC$, $\omega$, meets $AB,AC$ at $D,E$ respectively. $P$ lies inside $\triangle ABC$ such that $\angle PBA=\angle PAC, \angle PCA=\angle PAB$, and $2PM\cdot DE=BC^2$. Point $X$ lies outside $\omega$ such that $XM\parallel AP$, and $\frac{XB}{XC}=\frac{AB}{AC}$. Prove that $\angle BXC +\angle BAC=90^{\circ}$.

 

6 Given coprime positive integers $p,q>1$, call all positive integers that cannot be written as $px+qy$(where $x,y$ are non-negative integers) bad, and define $S(p,q)$ to be the sum of all bad numbers raised to the power of $2019$. Prove that there exists a positive integer $n$, such that for any $p,q$ as described, $(p-1)(q-1)$ divides $nS(p,q)$.

 

Test 3 Day 1

1 Given complex numbers $x,y,z$, with $|x|^2+|y|^2+|z|^2=1$. Prove that: $$|x^3+y^3+z^3-3xyz| \le 1$$

 

2 Let $S$ be a set of positive integers, such that $n \in S$ if and only if $$\sum_{d|n,d<n,d \in S} d \le n$$Find all positive integers $n=2^k \cdot p$ where $k$ is a non-negative integer and $p$ is an odd prime, such that $$\sum_{d|n,d<n,d \in S} d = n$$

 

3 Does there exist a bijection $f:\mathbb{N}^{+} \rightarrow \mathbb{N}^{+}$, such that there exist a positive integer $k$, and it's possible to have each positive integer colored by one of $k$ chosen colors, such that for any $x \neq y$ , $f(x)+y$ and $f(y)+x$ are not the same color?

 

Test 3 Day 2

4 Find all functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, such that
1) $f(0,x)$ is non-decreasing ;
2) for any $x,y \in \mathbb{R}$, $f(x,y)=f(y,x)$ ;
3) for any $x,y,z \in \mathbb{R}$, $(f(x,y)-f(y,z))(f(y,z)-f(z,x))(f(z,x)-f(x,y))=0$ ;
4) for any $x,y,a \in \mathbb{R}$, $f(x+a,y+a)=f(x,y)+a$ .

 

5 In $\Delta ABC$, $AD \perp BC$ at $D$. $E,F$ lie on line $AB$, such that $BD=BE=BF$. Let $I,J$ be the incenter and $A$-excenter. Prove that there exist two points $P,Q$ on the circumcircle of $\Delta ABC$ , such that $PB=QC$, and $\Delta PEI \sim \Delta QFJ$ .

 

6 Given positive integers $d \ge 3$, $r>2$ and $l$, with $2d \le l <rd$. Every vertice of the graph $G(V,E)$ is assigned to a positive integer in $\{1,2,\cdots,l\}$, such that for any two consecutive vertices in the graph, the integers they are assigned to, respectively, have difference no less than $d$, and no more than $l-d$.

A proper coloring of the graph is a coloring of the vertices, such that any two consecutive vertices are not the same color. It's given that there exist a proper subset $A$ of $V$, such that for $G$'s any proper coloring with $r-1$ colors, and for an arbitrary color $C$, either all numbers in color $C$ appear in $A$, or none of the numbers in color $C$ appear in $A$.
Show that $G$ has a proper coloring within $r-1$ colors.

 

Test 4 Day 1

1 Cyclic quadrilateral $ABCD$ has circumcircle $(O)$ , $M,N$ are midpoints of $AD,BC$, $E,F$ lie on $AD,BC$, respectively, such that $EF$ passes through $O$ and $EO=OF$. Let $T$ be the circumcenter of $\Delta ESF$, $MF$ intersects $NE$ at $S$.
If $CS=OT$ and $CS \parallel OT$, prove that $OS$ is parallel to angle bisector of $\angle DAB$.

 

2 A graph $G(V,E)$ is triangle-free, but adding any edges to the graph will form a triangle. It's given that $|V|=2019$, $|E|>2018$, find the minimum of $|E|$ .

 

3 $60$ points lie on the plane, such that no three points are collinear. Prove that one can divide the points into $20$ groups, with $3$ points in each group, such that the triangles ( $20$ in total) consist of three points in a group have a non-empty intersection.

 

Test 4 Day 2

4 Prove that there exist a subset $A$ of $\{1,2,\cdots,2^n\}$ with $n$ elements, such that for any two different non-empty subset of $A$, the sum of elements of one subset doesn't divide another's.

 

5 Find all integer $n$ such that the following property holds: for any positive real numbers $a,b,c,x,y,z$, with $max(a,b,c,x,y,z)=a$ , $a+b+c=x+y+z$ and $abc=xyz$, the inequality $$a^n+b^n+c^n \ge x^n+y^n+z^n$$holds.

 

6 Given positive integer $n,k$ such that $2 \le n <2^k$. Prove that there exist a subset $A$ of $\{0,1,\cdots,n\}$ such that for any $x \neq y \in A$, ${y\choose x}$ is even, and $$|A| \ge \frac{{k\choose \lfloor \frac{k}{2} \rfloor}}{2^k} \cdot (n+1)$$

\chapter{小陶同学问题}
\begin{example}
已知$f(x)$是定义在$R$上的偶函数,且当$x\geq 0$时, $f(x)=2^x$.若对任意的$x\in [a,a+2]$,均有$f(x+a)\geq f^2(x)$,则实数$a$的最大值是?
\end{example}
\begin{solution}
因为$f(x)$是定义在$R$上的偶函数,则不等式$f(x+a)\geq f^2(x)$等价于$f(|x+a|)\geq f^2(|x|)$成立,即
\[2^{|x+a|}\geq 2^{2|x|},\quad |x+a|\geq 2|x|\]
在$[a,a+2]$上恒成立.平方后可得$x^2+2ax+a^2\geq 4x^2$,即$3x^2-2ax-a^2\leq 0$在$[a,a+2]$上恒成立.

令$g(x)=3x^2-2ax-a^2$,则
\[\begin{cases}
g(a)=0\leq 0, \\
g(a+2)=8a+12\leq 0.
\end{cases}\]
解得$a\leq -\frac{3}{2}$.故$a$的最大值为$-\frac{3}{2}$,此时$g(x)=3x^2+3x-\frac{9}{4}$在$\left[-\frac{3}{2},\frac{1}{2}\right]$上恒成立.
\end{solution}

\begin{example}
设实数$x,y$满足$x^{2}+x y+y^{2}-x-y+\frac{4}{13}=0$.求函数
\[f(x, y)=\frac{2 x^{2}+x y+2 y^{2}}{x^{2}-x y+2 y^{2}}\]
的最大值.
\end{example}
\begin{solution}
令$y=tx$,由$x^{2}+x y+y^{2}-x-y+\frac{4}{13}=0$可知
\[(t^2+t+1)x^2-(t+1)x+\frac{4}{13}=0.\]
判别式$\Delta=(t+1)^2-4(t^2+t+1)\cdot \frac{4}{13}\geq 0$,即$3t^2-10t+3\leq 0$,则$(t-3)\left(3t-1\right) \leq 0$,解得$\frac{1}{3}\leq t\leq 3$.

而
\[
f\left( x,y \right) =\frac{2x^2+xy+2y^2}{x^2-xy+2y^2}=\frac{2+t+2t^2}{1-t+2t^2}=\frac{2t+1}{2t^2-t+1}+1,
\]
令$u=2t+1\in\left[\frac{5}{3},7\right]$,则
\[
\frac{2t+1}{2t^2-t+1}+1=\frac{u}{2\left( \frac{u-1}{2} \right) ^2-\frac{u-1}{2}+1}+1=\frac{2}{u+\frac{4}{u}-3}+1.
\]
由于$u+\frac{4}{u}$在$\left[\frac{5}{3},2\right)$上递减,在$\left[2,7\right]$上递增可知$f(x,y)$的最大值为$3$.此时有$u=2,t=\frac{1}{2}$.
\end{solution}

\begin{example}
设集合
\begin{align*}
A &=\{x | 1 \leq x \leq 2019, x=4 k+1, k \in \mathbf{Z}\}, \\
B &=\{y | 1 \leq y \leq 2019, y=3 k-1, k \in \mathbf{Z}\},
\end{align*}
则集合$A\cap B$中的元素个数为?
\end{example}

\begin{theorem}{二元一次不定方程}{bdfc1}
二元一次不定方程的一般形式为$ax+by=c$,其中$a,b,c$是整数, $ab\neq 0$.此方程有整数解的充分必要条件是$a$、$b$的最大公约数整除$c$.设$x_0$、$y_0$是该方程的一组整数解,那么该方程的所有整数解可表示为\[x=x_{0}+\frac{b}{(a, b)} t, \quad y=y_{0}-\frac{a}{(a, b)} t,\]
其中$t$为任意整数.
\end{theorem}

\begin{solution}
由$1 \leq 4k+1 \leq 2019$可知$0\leq k\leq 504.5$,则集合$A$包含$505$个元素;由$1 \leq 3k-1 \leq 2019$可知$0.666\leq k\leq 673.334$,则集合$B$包含$673$个元素.

令$4m+1=3n-1$且$0\leq m\leq 504,1\leq n\leq 673$均为整数,则$3n-4m=2$.由于$3$和$-4$的最大公约数为$(3,-4)=1$,可以整除$2$,而且$(n,m)=(2,1)$是方程的解,则该方程的所有解可表示为
\[n=2+4t,\quad m=1+3t.\]
由$0\leq 2+4t\leq 504,1\leq 1+3t\leq 673$可知$0\leq t\leq 125.5$,因此集合$A\cap B$中的元素个数为$126$.
\end{solution}

 

\begin{example}
设函数$f(x)=\left\{\begin{array}{l}{|x+1|, x \leq 0} \\ {|\lg x|, x>0}\end{array}\right.$.若方程$f(x)=a$有四个不同的实数解$x_1<x_2<x_3<x_4$,则$x_{3}\left(x_{1}+x_{2}\right)+\frac{1}{x_{3}^{2} x_{4}}$的最大值是?
\end{example}
\begin{solution}
由题意可知
\[f(x)=\begin{cases}
-x-1, & \mbox{若\,} x< -1, \\
x+1, & \mbox{若\,} -1\leq x\leq 0, \\
-\lg x, & \mbox{若\,} 0<x\leq 1, \\
\lg x, & \mbox{若\,} x>1.
\end{cases}\]
要使方程$f(x)=a$有四个不同的实数解$x_1<x_2<x_3<x_4$,则必有$0<a\leq 1$且
\[-x_1-1=x_2+1=-\lg x_3=\lg x_4,\]
即\[x_1=-a-1,\quad x_2=a-1,\quad x_3=10^{-a},\quad x_4=10^a.\]
于是
\[x_ {3}\left(x_{1}+x_{2}\right)+\frac{1}{x_{3}^{2} x_{4}}=10^a-\frac{2}{10^a}\leq 10-\frac{2}{10}=9.8.\]
\end{solution}

 

\begin{example}
设$\cos\alpha+\cos\beta+\cos\gamma=1$, $\alpha$、$\beta$、$\gamma\in \left[0, \frac{\pi}{2}\right]$.求$\sin ^{4} \alpha+\sin ^{4} \beta+\sin ^{4} \gamma$的最小值.
\end{example}
\begin{solution}
令$a=\cos\alpha,b=\cos\beta,c=\cos\gamma$,则$a,b,c$均为非负实数且$a+b+c=1$.下面证明
\[\sin ^{4} \alpha+\sin ^{4} \beta+\sin ^{4} \gamma=\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}\geq 2.\]

事实上,设$a b+b c+c a=m, a b c=n$,有
\[(x-a)(x-b)(x-c)=x^{3}-x^{2}+m x-n.\]
令$x=a$,则有$a^{3}=a^{2}-m a+n$.于是
\begin{align*} \sum_{c y c} a^{3} &=\sum_{c y c} a^{2}-m \cdot \sum_{c y c} a+3 n \\ \sum_{c y c} a^{4} &=\sum_{c y c} a^{3}-m \sum_{c y c} a^{2}+n \sum_{c y c} a \\ &=(1-m) \sum_{c y c} a^{2}-m \sum_{c y c} a+n \sum_{c y c} a+3 n \\ &=(1-m)(1-2 m)-m+n+3 n, \end{align*}
所以
\begin{align*}
\sum_{cyc}{\left( 1-a^2 \right)}^2&=3-2\sum_{cyc}{a^2}+\sum_{cyc}{a^4}
\\
&=3-2\left( 1-2m \right) +2m^2-3m+1-m+4n\\
&=2m^2+4n+2\geqslant 2,
\end{align*}
当且仅当$a,b,c$中有两个为$0$时取等成立.因此所求最小值为$2$.

\textbf{另解.}设$a=\cos \alpha, b=\cos \beta, c=\cos \gamma$.依题意知$a+b+c=1$,且$0\leq a,b,c\leq 1$.欲求
\[S=\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}\]
的最大值与最小值,可令
\[x=a+b+c=1,\quad y=a b+b c+c a,\quad z=a b c,\]
则
\begin{align*}
S&=3-2\left( a^2+b^2+c^2 \right) +\left( a^4+b^4+c^4 \right)
\\
&=3-2\left( 1-2y \right) +\left( a^2+b^2+c^2 \right) ^2-2\left( a^2b^2+b^2c^2+c^2a^2 \right)
\\
&=3-2\left( 1-2y \right) +\left( 1-2y \right) ^2-2\left( y^2-2xz \right)
\\
&=2+2y^2+4z.
\end{align*}
由于$y\geq 0,z\geq 0$,于是$S\geq 2$,且当$a=1,b=c=0$时,等号成立,即$S=2$.从而$S_{\min}=2$.

又因为
\begin{align*}
1&=\left( a+b+c \right) ^2=a^2+b^2+c^2+2\left( ab+bc+ca \right)
\\
&\ge 3\left( ab+bc+ca \right)
\\
1&=a+b+c\ge 3\sqrt[3]{abc},
\end{align*}
则
\[
S=2+2y^2+4z\le 2+\frac{2}{9}+\frac{4}{27}=\frac{64}{27},\]
即$S_{\max}=\frac{64}{27}$,当且仅当$a=b=c=\frac{1}{3}$时取等成立.
\end{solution}

 

\begin{example}
在锐角三角形$ABC$中,求$\sum \frac{\cos A}{1-\cos B \cdot \cos C}$的最大值.
\end{example}
\begin{solution}
由三角恒等式
\[\sin 2 A+\sin 2 B+\sin 2 C=4 \sin A \sin B \sin C\]
可知
\begin{align*}
\sum{\frac{\cos A}{1-\cos B\cdot \cos C}} &\le \sum{\frac{\cos A}{\cos \left( B-C \right) -\cos B\cdot \cos C}}
\\
&=\sum{\frac{\cos A}{\sin B\cdot \sin C}}=\frac{1}{2}\sum{\frac{\sin 2A}{\sin A\cdot \sin B\cdot \sin C}}=2,
\end{align*}
当且仅当$A=B=C=\frac{\pi}{3}$时取等成立.
\end{solution}

 

\begin{example}
已知关于$x$的方程
\[x^{4}-(3 m+2) x^{2}+m^{2}=0\]
中的四个实根成等差数列,则实数$m$的最大值为.
\end{example}
\begin{solution}
由$x^{4}-(3 m+2) x^{2}+m^{2}=0$可得
\[
\left( x^2-\frac{3m+2}{2} \right) ^2=\frac{5m^2+12m+4}{4},\]
解得
\[
x_{1}^{2}=\frac{3m+2-\sqrt{5m^2+12m+4}}{2},\quad x_{2}^{2}=\frac{3m+2+\sqrt{5m^2+12m+4}}{2},
\]
其中$m\leq -2$或$m\geq -\frac{2}{5}$.

不妨设$x_1,x_2>0$,由题意可知$-x_2,-x_1,x_1,x_2$依次成等差数列,则$x_2-x_1=2x_1$,即$x_2=3x_1$,则
\[
\sqrt{\frac{3m+2+\sqrt{5m^2+12m+4}}{2}}=3\sqrt{\frac{3m+2-\sqrt{5m^2+12m+4}}{2}},
\]
化简得$-19 m^{2}+108 m+36=0$,解得$m=6$或$-\frac{6}{19}$.

于是实数$m$的最大值为$6$.
\end{solution}

 

\begin{example}
已知数列$\{a_n\}$满足: $a_1=\frac{1}{4},a_{2}=\frac{1}{3}, a_{n+1}=3 a_{n}+a_{n-1}\, (n \geqslant 2)$.则$\left[\sum_{k=2}^{2019} \frac{1}{a_{k-1} a_{k+1}}\right]=$.
\end{example}
\begin{solution}
注意到
\[
\frac{1}{a_{k-1}a_{k+1}}=\frac{1}{3a_k}\left( \frac{1}{a_{k-1}}-\frac{1}{a_{k+1}} \right) =\frac{1}{3}\left( \frac{1}{a_ka_{k-1}}-\frac{1}{a_{k+1}a_k} \right),
\]
裂项累加后可得
\[
\sum_{k=2}^{2019}{\frac{1}{a_{k-1}a_{k+1}}}=\frac{1}{3}\left( \frac{1}{a_2a_1}-\frac{1}{a_{2020}a_{2019}} \right) <\frac{1}{3a_2a_1}=4.
\]
又
\[
a_{n+1}=3a_n+a_{n-1}>3a_n>3^2a_{n-1}>\cdots >3^na_1=\frac{3^n}{4},
\]
故$a_{2020}a_{2019}>\frac{3^{2019}}{4}\times \frac{3^{2018}}{4}$,则
\[
\sum_{k=2}^{2019}{\frac{1}{a_{k-1}a_{k+1}}}=\frac{1}{3}\left( \frac{1}{a_2a_1}-\frac{1}{a_{2020}a_{2019}} \right) >\frac{1}{3}\left( 12-\frac{16}{3^{2019}\times 3^{2018}} \right) >3.99.\]
因此
\[\left[\sum_{k=2}^{2019} \frac{1}{a_{k-1} a_{k+1}}\right]=3.\]
\end{solution}

 

\begin{example}
已知实数$x$、$y$满足$2x^2+3y^2=6y$,则$x+y$的最大值的整数部分为?
\end{example}
\begin{solution}
令$x+y=t$,代入$2x^2+3y^2=6y$可得$2(t-y)^2+3y^2=6y$,则
\[5y^2-(4t+6)y+2t^2=0.\]
由关于$y$的一元二次方程有解可知
\[\Delta=(4t+6)^2-4\times 5\times 2t^2=-24t^2+48t+36\geq 0,\]
解得$1-\sqrt{\frac{5}{2}}\le t\le 1+\sqrt{\frac{5}{2}}\approx 2.58$,因此$x+y$的最大值的整数部分为$2$.
\end{solution}

 

\begin{example}
不超过$\sum_{n=1}^{1000} \frac{1}{2 \sqrt[3]{n}}$的最大整数为?
\end{example}
\begin{solution}
注意到
\[
\frac{3}{2}\left[ \left( n+1 \right) ^{\text{2/}3}-n^{\text{2/}3} \right] <\frac{1}{\sqrt[3]{n}}<\frac{3}{2}\left[ n^{\text{2/}3}-\left( n-1 \right) ^{\text{2/}3} \right].\]
裂项求和可得
\[
74.3\approx \frac{3}{4}\left[ 1001^{\text{2/}3}-1 \right] <\sum_{n=1}^{1000}{\frac{1}{2\sqrt[3]{n}}}<\frac{1}{2}+\frac{3}{4}\left[ 1000^{\text{2/}3}-1 \right] =74.75.\]
因此所求最大整数为$74$.
\end{solution}

 

\begin{example}
(王永喜高二金牌种子选手二试级别以上训练之九)设非负实数数列$\{a_n\}$满足$a_{m+n} \leq a_{m}+a_{n}, \forall m, n \in \mathbb{N}^\ast$,求证:
\[\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \geq \frac{a_{n} \cdot H_{n}}{n},\]
其中$H_{n}=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$.
\end{example}
\begin{solution}

\end{solution}

 

 

\begin{example}

\end{example}
\begin{solution}

\end{solution}

\begin{example}

\end{example}
\begin{solution}

\end{solution}

\chapter{集合}

 

\chapter{函数与方程}

 

 

 

 

\section{定义域}

\begin{example}
设$x\in\mathbb{R}_+$,求$y=\sqrt{\frac{1}{1+x^2}}+2\sqrt{\frac{x}{1+x}}$的最大值.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
设$x,y\in\mathbb{R}$.则$3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}
+\sqrt{5}\left(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}\right)$的最小值为?
\end{example}
\begin{solution}

\end{solution}

\begin{example}
已知函数$f(x)$是定义在$\left(\frac{1}{2},+\infty\right)$上的单调函数,且$f(x) f\left(f(x)+\frac{1}{x}\right)=\frac{1}{2}$,求$f(1)$.
\end{example}
\begin{solution}
注意到$f(x)=\frac{1}{x}$满足题意.

设$f(1)=t$,则$f(t+1)=\frac{1}{2t}$,令$x=t+1$,得$f\left(\frac{1}{2t}+\frac{1}{t+1}\right)=t=f(1)$,则$t=1$.
\end{solution}

 

\begin{example}
求$f(x)=x^2-x+1+\sqrt{2x^4-18x^2+12x+68}$最小值.
\end{example}
\begin{solution}
点$(x,x^2)$到直线$y=x-1$的距离与到点$O(-3,5)$的距离之和.
\end{solution}

\begin{example}
(2002年江苏)已知$a>0$,函数$f(x)=ax-bx^2$.
\begin{enumerate}
\item 当$b>0$时,若对任意$x\in \mathbb{R}$都有$f(x)\leq 1$,证明$a\leq 2\sqrt{b}$.

\item 当$b>1$时,证明:对任意$x\in [0,1]$, $|f(x)|\leq 1$的充要条件是$b-1\leq a\leq 2\sqrt{b}$;

\item 当$0<b\leq 1$时,讨论:对任意$x\in[0,1],|f(x)|\leq 1$的充要条件.
\end{enumerate}
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2006年江苏)设$a$为实数,设函数$f(x)=a\sqrt{1-x^2}+\sqrt{1+x}+\sqrt{1-x}$的最大值为$g(a)$.
\begin{enumerate}
\item 设$t=\sqrt{1+x}+\sqrt{1-x}$,求$t$的取值范围,并把$f(x)$表示为$t$的函数$m(t)$;

\item 求$g(a)$;

\item 试求满足$g(a)=g\left(\frac{1}{a}\right)$的所有实数$a$.
\end{enumerate}
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2006年联赛)求方程$\left(x^{2006}+1\right)\left(1+x^2+x^4+\cdots +x^{2004}\right)=2006x^{2005}$的实数解的个数.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2010年广州高二竞赛)已知定义在$\mathbb{R}$上的函数$f(x)$满足: $f(1)=\left(\frac{5}{2}\right)$,且对于任意实数$x$、$y$,总有$f(x)f(y)=f(x+y)+f(x-y)$成立.
\begin{enumerate}
\item 求$f(0)$的值,并证明$f(x)$为偶函数;

\item 若数列$\{a_n\}$满足$a_n=2f(n+1)-f(n)\,(n=1,2,3,\cdots)$,求数列$\{a_n\}$的通项公式;

\item 若对于任意非零实数$y$,总有$f(y)>2$.设有理数$x_1,x_2$满足$|x_1|<|x_2|$,判断$f(x_1)$和$f(x_2)$的大小关系,并证明你的结论.
\end{enumerate}
\end{example}
\begin{solution}
\begin{enumerate}
\item 令$x=1,y=0$,所以$f(1)\cdot f(0)=f(1)+f(1)$,又因为$f(1)=\frac{5}{2}$,所以$f(0)=2$.

令$x=0$,得$f(0)f(y)=f(y)+f(-y)$,即$2f(y)=f(y)+f(-y)$,所以$f(y)=f(-y)$对任意的实数$y$总成立,所以$f(x)$为偶函数.

\item 令$x=y=1$,得$f(1)f(1)=f(2)+f(0)$,所以$\frac{25}{4}=f(2)+2$,故$f(2)=\frac{17}{4}$,
故$a_1=2f(2)-f(1)=\frac{17}{2}-\frac{5}{2}=6$.

令$x=n+1,y=1$,得$f(n+1)f(1)=f(n+2)+f(n)$,所以
\[f(n+2)=\frac{5}{2}f(n+1)-f(n).\]
所以
\begin{align*}
a_{n+1}&=2f(n+2)-f(n+1)=2\left[\frac{5}{2}f(n+1)-f(n)\right]-f(n+1)=4f(n+1)-2f(n)\\
&=2[2f(n+1)-f(n)]=2a_n,\quad (n\geq 1)
\end{align*}
所以$\{a_n\}$是以$6$为首项,以$2$为公比的等比数列,所以$a_n=6\times 2^{n-1}$.

\item 结论: $f(x_1)<f(x_2)$.

证明:因为$y\neq 0$时, $f(y)>2$,所以$f(x+y)+f(x-y)=f(x)f(y)>2f(x)$,即$f(x+y)-f(x)>f(x)-f(x-y)$.

令$x=ky\, (k\in\mathbb{N}^\ast)$,故对任意$k\in\mathbb{N}^\ast$,总有$f[(k+1)y]-f(ky)>f(ky)-f[(k-1)y]$成立,则
\[f[(k+1)y]-f(ky)>f(ky)-f[(k-1)y]>f[(k-1)y]-f[(k-2)y]>\cdots>f(y)-f(0)>0.\]

所以对于$k\in\mathbb{N}^\ast$,总有$f[(k+1)y]>f(ky)$成立.

故对于$m,n\in\mathbb{N}^\ast$,若$n<m$,则有$f(ny)<f[(n-1)y]<\cdots<f(my)$成立.

因为$x_1,x_2\in\mathbb{Q}$,所以可设$|x_1|=\frac{q_1}{p_1},|x_2|=\frac{q_2}{p_2}$,其中$q_1,q_2$是非负整数, $p_1,p_2$都是正整数,则$|x_1|=\frac{q_1p_2}{p_1p_2},|x_2|=\frac{p_1q_2}{p_1p_2}$,令$y=\frac{1}{p_1p_2}$, $t=q_1p_2,s=p_1q_2$,则$t,s\in\mathbb{N}^\ast$.

因为$|x_1|<|x_2|$,所以$t<s$,故$f(ty)<f(sy)$,即$f(|x_1|)<f(|x_2|)$.

因为函数$f(x)$为偶函数,所以$f(|x_1|)=f(x_1),f(|x_2|)=f(x_2)$.因此$f(x_1)<f(x_2)$.
\end{enumerate}
\end{solution}
%https://wenku.baidu.com/view/bf838bfb910ef12d2af9e702.html

\begin{example}
(海淀区2009年高三第二学期期末练习)已知函数$f(x)$定义域为$\mathbb{R}$,满足:
\begin{enumerate}
\item[\ding{172}] $f(1)=1>f(-1)$;

\item[\ding{173}] 对任意实数$x,y$有$f(y-x+1)=f(x)f(y)+f(x-1)f(y-1)$.
\end{enumerate}
\begin{enumerate}
\item 求$f(0),f(3)$的值;

\item 求$\frac{1}{2}f(1-6x)+[f(3x)]^2$的值;

\item 是否存在常数$A,B$,使得不等式$|f(x)+f(2-x)+Ax+B|\leq 2$对一切实数$x$均成立.
如果存在,求出常数$A,B$的值;如果不存在,请说明理由.
\end{enumerate}
\end{example}
\begin{solution}
\begin{enumerate}
\item 取$x=y=1$,得$f(1-1+1)=f(1)\cdot f(1)+f(1-1)\cdot f(1-1)$,即$f(1)=[f(1)]^2+[f(0)]^2$.

因为$f(1)=1$,所以$f(0)=0$.

取$x=y=0$,得$1=f(1)=[f(-1)]^2$.因为$f(1)=1>f(-1)$,所以$f(-1)=-1$.

取$x=0,y=2$,得$f(3)=f(0)\cdot f(2)+f(-1)\cdot f(1)$,所以$f(3)=-1$.

\item 在$f(y-x+1)=f(x)f(y)+f(x-1)f(y-1)$中取$y=1$得$f(2-x)=f(x)$,所以$f(1+x)=f(1-x)$.

在$f(y-x+1)=f(x)f(y)+f(x-1)f(y-1)$中取$y=x$,得$[f(x)]^2+[f(x-1)]^2=1$.

在$f(y-x+1)=f(x)f(y)+f(x-1)f(y-1)$中取$x=0$得$f(y+1)=f(0)f(y)+f(-1)f(y-1)=-f(y-1)$.所以$f(-1+1)=-f(-1-1)$,即$f(-2)=0$.

在$f(y-x+1)=f(x)f(y)+f(x-1)f(y-1)$中取$y=-1$,得$f(-x)=f(x)f(-1)+f(x-1)f(-2)$,所以$f(-x)=-f(x)$.

在$f(y-x+1)=f(x)f(y)+f(x-1)f(y-1)$中取$y=-x$,得
\begin{align*}
f(-2x+1)&=f(x)f(-x)+f(x-1)f(-x-1)=-[f(x)]^2-f(x-1)f(x+1)\\
&=-[f(x)]^2-f(x-1)f(1-x)=-[f(x)]^2+[f(x-1)]^2\\
&=1-2[f(x)]^2.
\end{align*}
所以$\frac{1}{2}f(1-2x)+[f(x)]^2=\frac{1}{2}$对任意实数$x$均成立.

所以$\frac{1}{2}f(1-6x)+[f(3x)]^2=\frac{1}{2}$.

\item 由(2)知$f(2-x)=f(x)$,所以$|f(x)+f(2-x)+Ax+B|\leq 2\Leftrightarrow |2f(x)+Ax+B|\leq 2$.

在$|2f(x)+Ax+B|\leq 2$中,取$x=-1$,得$-2\leq -2-A+B\leq 2$,即
\[-2\leq 2+A-B\leq 2.\tag{1}\]

取$x=1$,得
\[-2\leq 2+A+B\leq 2.\tag{2}\]

取$x=3$,得$-2\leq -2+3A+B\leq 2$,即
\[-2\leq 2-3A-B\leq 2.\tag{3}\]

$(2)+(1)$得$A\leq 0$, $(2)+(3)$得$A\geq 0$.于是$A=0$.

将$A=0$代人$(1)$得$B\geq 0$.将$A=0$代入$(2)$得$B\leq 0$.于是$B=0$.

由(2)知$[f(x)]^2+[f(x-1)]^2=1$,所以$|f(x)|\leq 1$对一切实数$x$成立.

故当$A=B=0$时, $|2f(x)+Ax+B|\leq 2$对一切实数$x$成立.

故存在常数$A=B=0$,使得不等式$|f(x)+f(2-x)+Ax+B|\leq 2$对一切实数$x$成立,且$A=B=0$为满足题设的唯一一组值.
\end{enumerate}
\end{solution}

%https://wenku.baidu.com/view/b2723a4ab307e87101f69666

\begin{example}
是否存在两个无理数$a,b$使得$a^b$是有理数?
\end{example}
\begin{solution}
考察$\left({\sqrt{2}}^{\sqrt{2}}\right)^{\sqrt{2}}$.
\end{solution}

\begin{example}
求证:任何一个有理数$a$都可以写成三个有理数的立方和.
\end{example}
\begin{solution}
注意到
\[a=\left(\frac{a^{3}-3^{6}}{3^{2} a^{2}+3^{4} a+3^{6}}\right)^{3}+\left(\frac{-a^{3}+3^{5} a+3^{6}}{3^{2} a^{2}+3^{4} a+3^{6}}\right)^{3}+\left(\frac{3^{3} a^{2}+3^{5} a}{3^{2} a^{2}+3^{4} a+3^{6}}\right)^{3}.\]
\end{solution}

\begin{example}
证明:
\[\left(1+\frac{1}{x}\right)^{x}+(1+x)^{\frac{1}{x}} \leq 4.\]
\end{example}
\begin{solution}

\end{solution}

\begin{example}
证明$n^m>m^n$,其中$m>n>e$.

已知正实数$a,b$满足$a^b=b^a$且$a\neq b$,求证: $a^b>e^e$.
\end{example}
\begin{solution}
不妨设$b>a$.设$b=at\,(t>1)$
$a\ln b=b\ln a,a\ln a+a\ln t=at\ln a,\ln a+\ln t=t\ln a,(t-1)\ln a=\ln t,\ln a=\ln t/(t-1)$.

要证$a^b＞e^e$.只要证$\ln (b\ln a)>1$.

设$f(t)=\ln (b\ln a)=\ln (at\ln a)=\ln a+\ln t\ln a=\ln t /(t-1)+\ln (t\ln t/(t-1)$, $f'(t)=[(t-1)^2-t\ln^2(t)]/(t(t-1)^2\ln t)$
设$g(t)=(t-1)^2-t\ln^2(t)$, $g'(t)=2t-\ln^2(t)-2\ln t-2$, $g''(t)=2(t-\ln t-1)/t>0$,所以$g'(t)>g'(1)=0$,所以$g(t)>g(1)=0$,所以$f'(t)>0$, $\lim_{x\to 1} f(x)=1$,所以$f(t)>1$.
\end{solution}

\begin{example}
(2017年清华大学自主招生) $A,B,C$是三角形的三个内角,求$\sin A+\sin B\sin C$最大值.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2017年清华大学自主招生) $x+2y+3z=100$非负整数解组数.
\end{example}
\begin{solution}

\end{solution}

 

\chapter{三角函数与三角形}

\begin{example}
(2019年江苏数学夏令营)在$\triangle ABC$中, 求$2\cot A+3\cot B+4\cot C$的最小值.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
求证:
\[\sum_{k=0}^{n}\left(\frac{1}{3}\right)^k\cdot \sin^3\left(3^k\cdot \alpha\right)=\frac{3}{4}\sin\alpha-\frac{1}{4\cdot 3^n}\sin 3^{n+1}\alpha.\]
\end{example}
\begin{solution}

\end{solution}

\begin{example}
求$\sin 6^\circ\sin 42^\circ\sin 66^\circ\sin 78^\circ$的值.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
求\[\tan 6^\circ\tan 42^\circ\tan 66^\circ\tan 78^\circ.\]
\end{example}
\begin{solution}
\begin{align*}
\tan6\circ\tan42\circ\tan66\circ\tan78\circ
&=\tan6\circ\cot12\circ\cot24\circ\cot48\circ \\
&=\frac{\sin6\circ\cos12\circ\cos24\circ\cos48\circ}
{\cos6\circ\sin12\circ\sin24\circ\sin48\circ},
\end{align*}
利用连锁反应易知
\[\sin6\circ\cos12\circ\cos24\circ\cos48\circ=\frac1{16},\]
又由积化和差及已知的数值有
\begin{align*}
\cos6\circ\sin24\circ&=\frac12(\sin18\circ+\sin30\circ)
=\frac12\left( \frac{\sqrt5-1}4+\frac12 \right)
=\frac{\sqrt5+1}8, \\
\sin12\circ\sin48\circ&=\frac12(\cos36\circ-\cos60\circ)
=\frac12\left( \frac{\sqrt5+1}4-\frac12 \right)
=\frac{\sqrt5-1}8,
\end{align*}
相乘得
\[\cos6\circ\sin12\circ\sin24\circ\sin48\circ=\frac1{16},\]
从而原式等于 $1$.

根据下面的公式
\begin{align*}
&\prod_{i=1}^n\sin\frac{i\pi}{2n+1}=\frac{\sqrt{2n+1}}{2^n}，\\
&\prod_{i=1}^{n-1}\sin\frac{i\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}，\\
&\prod_{i=1}^n\cos\frac{i\pi}{2n+1}=\frac{1}{2^n}，\\
&\prod_{i=1}^{n-1}\cos\frac{i\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}，\\
&\prod_{i=1}^n\tan\frac{i\pi}{2n+1}=\sqrt{2n+1}，\\
&\prod_{i=1}^{n-1}\tan\frac{i\pi}{2n}=1，
\end{align*}
上面倒数第二个方程中分别取 $n=7$，$n=2$，$n=1$，得
\begin{align*}
&\tan12^\circ\tan24^\circ\tan36^\circ\tan48^\circ\tan60^\circ\tan72^\circ\tan84^\circ=\sqrt{15}，\\
&\tan36^\circ\tan72^\circ=\sqrt{5}，\\
&\tan60^\circ=\sqrt{3}，
\end{align*}
所以
\[
\tan12^\circ\tan24^\circ\tan48^\circ\tan84^\circ=1，
\]
即
\[
\tan6^\circ\tan42^\circ\tan66^\circ\tan78^\circ=1。
\]

600题征解，给出的答案为，用三倍角公式$$\tan 3\theta=\tan \theta \tan (60^\circ-\theta)\tan (60^\circ+\theta)$$可得

$$\tan 18^\circ=\tan 6^\circ\tan 54^\circ\tan 66^\circ\\\tan 54^\circ=\tan 18^\circ\tan 42^\circ\tan 78^\circ\\$$两式相乘即可.
\end{solution}

\begin{example}
(2019年中科大自招)求证:对于任意的$n \in \mathbb{N}^{\ast}, e^{x}=\sum_{k=0}^{n} \frac{x^{k}}{k !}$在$\mathbb{R}$上仅有一个解$x=0$.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
求值: $\left(1+2\cos\frac{2\pi}{7}\right)\left(1+2\cos\frac{4\pi}{7}\right)\cdots \left(1+2\cos\frac{12\pi}{7}\right)=1$.

求证: $\left(2\sin\frac{\pi}{7}\right)^{2n}+\left(2\sin\frac{2\pi}{7}\right)^{2n}+\left(2\sin\frac{3\pi}{7}\right)^{2n}$.

$\left(1+\tan^4\frac{\pi}{2n+1}\right)\left(1+\tan^4\frac{\pi}{2n+1}\right)\cdots \left(1+\tan^4\frac{n\pi}{2n+1}\right)$.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
若$2\nmid n$.证明: $\sum_{i=1}^{n}\frac{1}{\cos^2\frac{i\pi}{n}}=n^2$.
\end{example}
\begin{solution}
设$z_k$跑遍$n$次单位根,
\[\text{原式}=2\sum_{k=1}^{n}\left[1-\left(\frac{zk}{z_k+1}\right)^2-\frac{1}{(z_k+1)^2}\right],\]
而$\frac{zk}{z_k+1},\frac{1}{z_k+1}$均跑遍$t^n-(1-t)^n=0$的根,故原式等于
\[2n-4\sum_ {k=1}^{n} t_k^2=2n-4\left[\left(\frac{n}{2}\right)^2-\frac{n(n-1)}{2}\right]= n^2.\]
\end{solution}

\begin{example}

\end{example}
\begin{solution}

\end{solution}

 

\begin{example}

\end{example}
\begin{solution}

\end{solution}

 

 

\begin{example}
(2019年中科大自招)已知$n \in \mathbb{N}^{\ast}$.
\begin{enumerate}
\item 求证:存在多项式$p(x)$,满足$\cos n\theta=p(\cos\theta)$;

\item 将$p(x)$在$\mathbb{R}[x]$上完全分解.
\end{enumerate}
\end{example}
\begin{solution}

\end{solution}

 

\chapter{平面向量}

\input{tex/shulie.tex}

\input{tex/budengshi.tex}

 

 

\chapter{立体几何}

\chapter{解析几何}

\section{直线与圆}

 

\begin{example}
(2007年全国高考四川卷理科第11题)如图1, $l_1,l_2,l_3$是同一平面内的三条平行直线, $l_1$与$l_2$间的距离是$1$, $l_2$与$l_3$间的距离是$2$,正三角形$ABC$的三顶点分别在$l_1,l_2,l_3$上,则$\triangle ABC$的边长为
\begin{tasks}(4)
\task $2\sqrt{3}$

\task $\frac{4\sqrt{6}}{3}$

\task $\frac{3\sqrt{17}}{4}$

\task $\frac{2\sqrt{21}}{3}$
\end{tasks}
\end{example}
\begin{solution}
12
\end{solution}

\begin{example}
(2018年上海)设$D$是含数$1$的有限实数集, $f(x)$是定义在$D$上的函数,若$f(x)$的图象绕原点逆时针旋转$\frac{\pi}{6}$后与原图象重合,则在以下各项中, $f(1)$的可能取值只能是
\begin{tasks}(4)
\task $\sqrt{3}$

\task $\frac{\sqrt{3}}{2}$

\task $\frac{\sqrt{3}}{3}$

\task $0$
\end{tasks}
\end{example}
\begin{solution}

\end{solution}

\begin{example}
在平面直角坐标系$xoy$中,已知圆$C_1:(x+3)^2+(y-1)^2=4$和圆$C_2:(x-4)^2+(y-5)^2=4$.

(1)若直线过点$A(4,0)$,且被圆$C_1$截得的弦长为$2\sqrt{3}$,求直线$l$的方程;

(2)设$P$为平面上的点,满足:存在过点$P$的无穷多对互相垂直的直线$l_1$和$l_2$,它们分别与圆$C_1$和$C_2$相交,且直线$l_1$被圆$C_1$截得的弦长与直线$l_2$被圆$C_2$截得的弦长相等,求所有满足条件的点$P$的坐标.
\end{example}
\begin{solution}
(1) $y=0$或$7x+24y-28=0$.

(2) $P_1\left( \frac{5}{2},-\frac{1}{2} \right)$或$P_2\left( -\frac{3}{2},\frac{13}{2} \right)$.
\end{solution}

\begin{example}
已知Rt $\triangle ABC$的斜边$BC$的两个端点分别在$x$、$y$两轴正方向上移动,点$A$和原点分别在$BC$两侧,则点$A$的轨迹是

A、圆 B、线段 C、射线 D、一段圆弧
\end{example}
\begin{solution}
因为$ABCO$四点共圆,从而$\angle ACB=\angle AOB$,又Rt $\triangle ABC$固定,从而$\angle BOA$不变,轨迹为线段.
\end{solution}

\begin{example}
已知圆$C_1:x^2+y^2+2x+6y+9=0$和圆$C_2:x^2+y^2-6x+2y-1=0$,求圆$C_1$和圆$C_2$的公切线方程.
\end{example}
\begin{solution}
$y+4=0$或$4x-3y=0$或$x=0$或$3x+4y+10=0$.
\end{solution}

\begin{example}
1
\end{example}
\begin{solution}

\end{solution}

\begin{example}
1
\end{example}
\begin{solution}

\end{solution}

\begin{example}
1
\end{example}
\begin{solution}

\end{solution}

\section{圆锥曲线}

\begin{example}
在平面直角坐标系中,由方程$y=x+\frac{1}{x}$定义的曲线是双曲线.试构造一个正交变换$Q\in \mathbb{R}^{2\times 2}$,使得经过坐标变换$[\widetilde{x},\widetilde{y}]^T=Q[x,y]^T$后在新的坐标$(\widetilde{x},\widetilde{y})$下曲线方程化为双曲线的标准方程.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
已知椭圆$\frac{x^2}{16}+\frac{y^2}{4}=1$的左右焦点分别为$F_1$与$F_2$,点$P$在直线$l:x-\sqrt{3}y+8+2\sqrt{3}=0$上.当$\angle F_1PF_2$取最大值时,求比$\frac{|PF_1|}{|PF_2|}$的值.
\end{example}
\begin{solution}

\end{solution}

 

\begin{example}
平面直角坐标系$xOy$中,椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$与直线$l$交于$AB$两点.求$S_{OAB}$面积的最大值. (答案: $\frac{1}{2}ab$)
\end{example}
\begin{solution}

\end{solution}

\begin{example}
椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$上任取三点$A$、$B$、$C$,求$S_{\triangle ABC}$面积的最大值. (答案: $\frac{3\sqrt{3}}{4}ab$)
\end{example}
\begin{solution}

\end{solution}

\begin{example}
设点$A(-a,0),B(a,0),C(2a,0)$, $D$为双曲线$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$的左支上异于点$A$的任意一点,直线$CD$与此双曲线的右支交于点$E$,证明:直线$AD$与$BE$的交点$P$的轨迹方程为直线$x=\frac{a}{2}$.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2019年第一学期七彩阳光联盟期初联考高三数学)过抛物线$x^2=2py\,(p>0)$外一点$P$作抛物线的两条切线,切点为$M$、$N$, $F$为抛物线的焦点.证明:
\begin{enumerate}
\item $|PF|^2=|MF|\cdot |NF|$;
\item $\angle PMF=\angle FPN$.
\end{enumerate}
\end{example}
\begin{solution}

\end{solution}

\begin{example}
直线族$x\cos\alpha+y\sin\alpha=p$的包络为$x^2+y^2=p^2$,这里$\alpha$是参数, $p$是常数.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
三个顶点在曲线$y=x^2$上的三角形外接圆半径的取值范围是?
\end{example}
\begin{solution}

\end{solution}

\begin{example}
如图$P$为椭圆: $C:\frac{x^2}{3}+\frac{y^2}{2}=1$一动点,过$P$作$O:x^2+y^2=1$的两条切线分别与椭圆$C$交于$A,B$两点,再过$A,B$分别作圆$O$的另一条切线$AQ,BQ$交于$Q$点.
\begin{enumerate}
\item 求动点$Q$的轨迹方程;
\item 求四边形$PAQB$面积的取值范围.
\end{enumerate}
\end{example}
\begin{solution}

\end{solution}

 

%https://zhuanlan.zhihu.com/p/28367942

\begin{example}
已知抛物线$\Gamma:y^2=2px\, (p>0)$,焦点为$F$,过$\Gamma$外一点$Q$ (不在$x$轴上)作$\Gamma$的两条切线,切点分别为$A,B$,直线$QA,QB$分别交$y$轴于$C,D$两点, $\triangle QAB$的外心为$M$.证明: $FM$为$\triangle FCD$外接圆的切线.
\end{example}
\begin{solution}

\end{solution}

\subsection{参数方程与极坐标方程}

\begin{example}
(2018年海淀高二上学期期中)曲线$W$的方程为$(x^2+y^2)^3=8x^2y^2$.
\begin{enumerate}
\item 请写出曲线$W$的两条对称轴方程 ;

\item 请写出曲线$W$上的两个点的坐标 ;

\item 曲线$W$上的点到原点的距离的取值范围是 .
\end{enumerate}
\end{example}
\begin{solution}

428880494
\end{solution}

\begin{example}
(2020年海淀高三上学期期末)已知曲线$C:x^4+y^4+mx^2y^2=1$ ($m$为常数).
\begin{enumerate}
\item[(i)] 给出下列结论:
\begin{enumerate}
\item[\ding{172}] 曲线$C$为中心对称图形;

\item[\ding{173}] 曲线$C$为轴对称图形;

\item[\ding{174}] 当$m=-1$时,若点$P(x,y)$在曲线$C$上,则$|x|\geq 1$或$|y|\geq 1$.
\end{enumerate}
其中,所有正确结论的序号是\underline{\hspace{2cm}}.

\item[(ii)] 当$m>-2$时,若曲线$C$所围成的区域的面积小于$\pi$,则$m$的值可以是\underline{\hspace{2cm}}. (写出一个即可)
\end{enumerate}
\end{example}
\begin{solution}
\begin{enumerate}
\item[(i)] 对于\ding{172}, 由点$(x,y)$和$(-x,-y)$均在曲线$C$上可知曲线$C$为中心对称图形;

对于\ding{173}, 由点$(x,y)$和$(-x,y)$均在曲线$C$上可知曲线$C$关于$y$轴对称;

对于\ding{174}, 当$m=-1$时,

\textbf{法一.} (江老师)由对称性,不妨设$|x|\geq |y|$,则$x^4=1-y^4+x^2y^2=1+y^2(x^2-y^2)\geq 1$,因此$|x|\geq 1$.

\textbf{法二.} 注意到$x^2y^2+1=x^4+y^4\geq 2x^2y^2$,于是$x^2y^2\leq 1$,因此
\begin{align*}
\left(1-x^4\right)\left(1-y^4\right)&=1-x^4-y^4+x^4y^4\\
&=x^4y^4-x^2y^2=x^2y^2(x^2y^2-1)\leq 0.
\end{align*}
于是$1-x^4$和$1-y^4$异号,则$|x|\geq 1$或$|y|\geq 1$.

所有正确结论的序号是\underline{\ \ding{172}\ding{173}\ding{174}\ }.

\item[(ii)] 当$m>2$时,注意到
\[\left(x^2+y^2\right)^2< x^4+y^4+mx^2y^2=1,\]
即$x^2+y^2<1$,此时曲线$C$所围成的区域的面积小于$\pi$.

因此取$m=3$即可.
\end{enumerate}
\end{solution}

 

\begin{example}
已知数列$\{a_n\},\{b_n\}$满足$a_1=1,b_1=2$,且$a_{n+1}b_n=1+a_n+a_nb_n,b_{n+1}a_n=1+b_n+a_nb_n$.求证:
数列$\{[a_n]\}$从某一项起一定是常数数列. ($[x]$表示不超过$x$的最大整数)
\end{example}
\begin{solution}

\end{solution}

\begin{example}
已知$a,b,x,y$满足方程组
\[\begin{cases}
ax+by=3, \\
ax^2+by^2=7, \\
ax^3+by^3=16, \\
ax^4+by^4=42.
\end{cases}\]
求$ax^5+by^5$的值.
\end{example}
\begin{solution}
构造递推式$s_n=ax^n+by^n,n=1,2,\cdots$,则
\[s_{n+2}=ax^{n+2}+by^{n+2}=\left(ax^{n+1}+by^{n+1}\right)(x+y)
-xy\left(ax^n+by^n\right).\]
因此$s_{n+2}=(x+y)s_{n+1}-xys_n$.

由题意可知
\[\begin{cases}
16=7(x+y)-3xy,\\
42=16(x+y)-7xy.
\end{cases}\]
解得$x+y=-14,xy=-38$,故$s_5=(x+y)s_4-xys_3=20$,即$ax^5+by^5=20$.
\end{solution}

\begin{example}
设正实数$x,y$满足$5x+16y\geq 37$,求$x^3+x^2+y^3+y^2$的最小值.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
已知$x=\sqrt{19}+\sqrt{99}$是方程$x^4+bx^2+c=0$的一个解, $b,c$为整数,则$b+c$的值是多少?
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2009年清华自主招生)试给出一个整系数多项式$f(x)$,使得$f(x)=0$以$\sqrt{2}+\sqrt[3]{3}$为一个根.
\end{example}
\begin{solution}
令$\sqrt{2}+\sqrt[3]{3}=x$,则$\left(x-\sqrt{2}\right)^3=3$,整理即得$x^3+6x-3=(3x^2+2)\sqrt{2}$.两边平方知$(x^3+6x-3)^2=2(3x^2+2)^2$,整理即得
\[x^6-6x^4-6x^3+12x^2-36x+1=0.\]
令$f(x)=x^6-6x^4-6x^3+12x^2-36x+1$,则$\sqrt{2}+\sqrt[3]{3}$是$f(x)$的一个根.
\end{solution}

\begin{example}
设$0<x<1,0<a<b<1,a+b<1$,证明$a^x(1-ax)<b^x(1-bx)$.
\end{example}
\begin{solution}

\end{solution}

 

\begin{example}
1
\end{example}
\begin{solution}

\end{solution}

\[\frac{{{\pi ^3}}}{{32}} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{{\left( {2n - 1} \right)}^3}}}} = 2{\left( {\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{2n - 1}}} } \right)^3} = 2{\left( {\frac{\pi }{4}} \right)^3}.\]

\chapter{复数与多项式}

\begin{example}
(2005年IMO中国国家集训队测验题2)试确定$\sqrt{1001^2+1}+\sqrt{1002^2+1}+\cdots+\sqrt{2000^2+1}$是否是一个有理数.
\end{example}
\begin{solution}
若复数$\alpha$是某个首一的整系数多项式的根,则称$\alpha$是好数.

\textbf{引理:}若$\alpha,\beta$都是好数,则$\alpha+\beta$也是好数.

证明:设$\alpha$是首一的整系数多项式$f(x)=x^m+a_{m-1}x^{m-1}+\cdots+a_0$的根,其所有根为$\alpha_1=\alpha,\alpha_2,\cdots,\alpha_m$. $\beta$是首一的整系数多项式$g(x)=x^n+b_{n-1}x^{n-1}+\cdots+b_0$的根,其所有根为$\beta_1=\beta,\beta_2,\cdots,\beta_m$.
\[F(x)=\prod_{i=1}^{m}\prod_{j=1}^{n}(x-\alpha_i-\beta_j)
=\prod_{i=1}^{m}\left[(x-\alpha_i)^n+b_{n-1}(x-\alpha_i)^{n-1}+\cdots+b_0\right],\]
固定$x,b_{n-1},\cdots,b_0$,上式是关于$\alpha_1,\alpha_2,\cdots,\alpha_m$的多元整系数对称多项式,故
\[F(x)=G(x,b_{n-1},\cdots,b_0;a_{m-1},\cdots,a_0),\]
其中$G$是多元整系数多项式.

因为$b_{n-1},\cdots,b_0,a_{m-1},\cdots,a_0$是整数,故$F(x)$是首一的整系数多项式,且以$\alpha+\beta$为根,即$\alpha+\beta$也是好数.

回到原问题,由于$\sqrt{1001^2+1},\sqrt{1002^2+1},\cdots,\sqrt{2000^2+1}$都是好数,所以
\[A=\sqrt{1001^2+1}+\sqrt{1002^2+1}+\cdots+\sqrt{2000^2+1}\]
是好数,由引理易知, $A$是某个首一的整系数多项式的根.

若$A$是有理数,则$A$必是整数.

因为
\[\sqrt{x^2+1}-x=\frac{1}{\sqrt{x^2+1}+x}<\frac{1}{2x},\]
所以
\begin{align*}
A&=\sqrt{1001^2+1}+\sqrt{1002^2+1}+\cdots+\sqrt{2000^2+1}\\
&<1001+1002+\cdots+2000+\frac{1}{2002}+\frac{1}{2004}+\cdots+\frac{1}{4000}\\
&<1001+1002+\cdots+2000+\frac{1}{2000}+\frac{1}{2000}+\cdots+\frac{1}{2000}\\
&<1001+1002+\cdots+2000+\frac{1}{2},
\end{align*}
又
\[A=\sqrt{1001^2+1}+\sqrt{1002^2+1}+\cdots+\sqrt{2000^2+1}>1001+1002+\cdots+2000,\]
所以$A$不是整数,矛盾!

\textbf{另解.}考察多元整系数多项式
\[F(x;y_1,y_2,\cdots,y_n)-\prod_{(\varepsilon_1,\varepsilon_2,\cdots,\varepsilon_n),\varepsilon_i=\pm 1,1\leq i\leq n} (x+\varepsilon_1y_1+\varepsilon_2y_2+\cdots+\varepsilon_ny_n),\]
显然$F$关于$y_i$是偶函数, $1\leq i\leq n$,故
\[F(x;y_1,y_2,\cdots,y_n)=G(x;y_1^2,\cdots,y_n^2),\]
其中$G$是多元整系数多项式.

本题中,取$n=1000,y_i=\sqrt{(1000+i)^2+1},1\leq i\leq n$,则$A$是首一的整系数多项式
\[f(x)=G(x;y_1^2,\cdots,y_n^2)\]
的根.若$A$为有理数,则$A$为整数.因为
\begin{align*}
\sum_{i=1}^{1000}(1000+i)&<A=\sum_{i=1}^{1000}\sqrt{(1000+i)^2+1} \\
&<\sum_{i=1}^{1000}[(1000+i)+0.001]=\sum_{i=1}^{1000}(1000+i)+1,
\end{align*}
所以$A$不是整数.矛盾!
\end{solution}

\begin{example}
设域$F$上多项式$f(x)$被$x-1,x-2,x-3$除后,余式分别为$4,8,16$.试求$f(x)$被$(x-1)(x-2)(x-3)$除后的余式.
\end{example}
\begin{solution}
$12$.
\end{solution}

\begin{example}
给定正整数$n$,试证:存在正整数$m$,使得域$\mathbb{F}$上多项式
\[
\left( 1+x \right) \left( 1+x^2 \right) \cdots \left( 1+x^{2^n} \right) =1+x+x^2+\cdots +x^m.
\]
\end{example}
\begin{solution}
注意到
\begin{align*}
\left( 1+x \right) \left( 1+x^2 \right) \cdots \left( 1+x^{2^n} \right) &=\frac{\left( 1-x \right) \left( 1+x \right) \left( 1+x^2 \right) \cdots \left( 1+x^{2^n} \right)}{1-x}
\\
&=\frac{\left( 1-x^2 \right) \left( 1+x^2 \right) \cdots \left( 1+x^{2^n} \right)}{1-x}=\cdots =\frac{1-x^{2^{n+1}}}{1-x}
\\
&=1+x+x^2+\cdots +x^{2^{n+1}-1},
\end{align*}
即$m=2^{n+1}-1$.
\end{solution}

\begin{example}
试计算复多项式
\[
x^n+\left( a+b \right) x^{n-1}+\left( a^2+ab+b^2 \right) x^{n-2}+\cdots +\left( a^n+a^{n-1}b+\cdots +ab^{n-1}+b^n \right)
\]
的根的方幂和$S_1,S_2,\cdots,S_n$.
\end{example}
\begin{solution}
等价于$x^n+\frac{a^2-b^2}{a-b}x^{n-1}+\frac{a^3-b^3}{a-b}x^{n-2}+\cdots +\frac{a^n-b^n}{a-b}$.
\end{solution}

\begin{example}
对任意$x_j\in (0,1/2],j=1,2,\cdots,n$和正整数$n$,证明不等式
\[
\frac{\prod_{j=1}^n{x_j}}{\left( \sum_{j=1}^n{x_j} \right) ^n}\le \frac{\prod_{j=1}^n{\left( 1-x_j \right)}}{\left( \sum_{j=1}^n{\left( 1-x_j \right)} \right) ^n}.\]
\end{example}
\begin{solution}
等价于证明
\[
\frac{\prod_{j=1}^n{x_j}}{\left( \sum_{j=1}^n{x_j} \right) ^n}\le \frac{\prod_{j=1}^n{\left( 1-x_j \right)}}{\left( n-\sum_{j=1}^n{x_j} \right) ^n}\Leftrightarrow \frac{\left( n-\sum_{j=1}^n{x_j} \right) ^n}{\left( \sum_{j=1}^n{x_j} \right) ^n}\le \frac{\prod_{j=1}^n{\left( 1-x_j \right)}}{\prod_{j=1}^n{x_j}},
\]
即
\[
\left( \frac{n}{\sum_{j=1}^n{x_j}}-1 \right) ^n\le \prod_{j=1}^n{\left( \frac{1}{x_j}-1 \right)}\Leftrightarrow \ln \left( \frac{n}{\sum_{j=1}^n{x_j}}-1 \right) \le \frac{\sum_{j=1}^n{\ln \left( \frac{1}{x_j}-1 \right)}}{n}.
\]

考虑辅助函数$f\left( x \right) =\ln \left( \frac{1}{x}-1 \right)$,由Jensen不等式即可.
\end{solution}

\begin{example}
计算
\[
\sum_{j=1}^{n-1}{\frac{1}{1-\exp \left\{ \frac{2\pi ij}{n} \right\}}}.
\]
\end{example}
\begin{solution}
(2011年清华金秋营)设$\varepsilon_n=e^{\frac{2\pi i}{n}}$,试求: $\sum_{k=0}^{n-1}\frac{1}{1-\varepsilon_n^kt},
\sum_{k=1}^{n-1}\frac{1}{1-\varepsilon_n^k},
\sum_{k=1}^{n-1}\frac{1}{(1-\varepsilon_n^k)(1-\varepsilon_n^{-k})}$.

记$\varepsilon_n=e^{\frac{2\pi i}{n}}$,则$1,\varepsilon_n,\varepsilon_n^2,\cdots,\varepsilon_n^{n-1}$为$x^n=1$的$n$个根.

由韦达定理可知
\[\sum_{i=1}^{n}\prod \varepsilon_n^{k_1}\cdots \varepsilon_n^{k_i}=0,\quad 1\leq i<n.\]
由$
x^n-1=\left( x-1 \right) \left( x-\varepsilon _n \right) \left( x-\varepsilon _{n}^{2} \right) \cdots \left( x-\varepsilon _{n}^{n-1} \right)
$可得
\[
\frac{1}{x^n}-1=\left( \frac{1}{x}-1 \right) \left( \frac{1}{x}-\varepsilon _n \right) \left( \frac{1}{x}-\varepsilon _{n}^{2} \right) \cdots \left( \frac{1}{x}-\varepsilon _{n}^{n-1} \right),
\]
于是
\[
1-x^n=\left( 1-x \right) \left( 1-\varepsilon _n x \right) \left( 1-\varepsilon _{n}^{2}x^2 \right) \cdots \left( 1-\varepsilon _{n}^{n-1}x^{n-1} \right).
\]
因此
\begin{align*}
\sum_{k=1}^{n-1}{\frac{1}{1-\varepsilon _{n}^{k}t}} &=\frac{n}{1-t^n}-\frac{1}{1-t}=\frac{n-\left( 1+t+t^2+\cdots +t^{n-1} \right)}{1-t^n}
\\
&=\frac{\left( 1-t \right) +\left( 1-t^2 \right) +\cdots +\left( 1-t^{n-1} \right)}{1-t^n}
\\
&=\frac{\left( n-1 \right) +\left( n-2 \right) t+\left( n-3 \right) t^2+\cdots +t^{n-2}}{1+t+t^2+\cdots +t^{n-1}}.
\end{align*}

令$t=1$,有
\[
\sum_{k=1}^{n-1}{\frac{1}{1-\varepsilon _{n}^{k}}}=\frac{\left( n-1 \right) +\left( n-2 \right) +\cdots +1}{n}=\frac{n-1}{2}.
\]

而
\begin{align*}
\sum_{k=1}^{n-1}{\frac{1}{\left( 1-\varepsilon _{n}^{k} \right) \left( 1-\varepsilon _{n}^{-k} \right)}} &=\sum_{k=1}^{n-1}{\frac{1}{\left( 1-\cos \frac{2k\pi}{n}-i\sin \frac{2k\pi}{n} \right) \left( 1-\cos \frac{2k\pi}{n}+i\sin \frac{2k\pi}{n} \right)}}
\\
&=\sum_{k=1}^{n-1}{\frac{1}{\left( 1-\cos \frac{2k\pi}{n} \right) ^2+\sin ^2\frac{2k\pi}{n}}}=\sum_{k=1}^{n-1}{\frac{1}{2-2\cos \frac{2k\pi}{n}}}
\\
&=\sum_{k=1}^{n-1}{\frac{1}{4\sin ^2\frac{k\pi}{n}}}=\frac{n-1}{4}+\frac{1}{4}\sum_{k=1}^{n-1}{\cot ^2\frac{k\pi}{n}},
\end{align*}

又因为
\[
\left( \cos \frac{k\pi}{n}+i\sin \frac{k\pi}{n} \right) ^n=\left( -1 \right) ^k=\sum_{j=0}^n{C_{n}^{j}\left( \cos \frac{k\pi}{n} \right) ^{n-j}\left( i\sin \frac{k\pi}{n} \right) ^j},
\]
当$n=2m$时,
\begin{align*}
&\sum_{j=1}^m{C_{2m}^{2j-1}\left( \cos \frac{k\pi}{n} \right) ^{2m-2j+1}\left( i\sin \frac{k\pi}{n} \right) ^{2j-1}}=0,
\\
&\sum_{j=1}^m{C_{2m}^{2j-1}\left( \cot \frac{k\pi}{n} \right) ^{2m-2j}\left( -1 \right) ^j}=0,
\end{align*}
所以$\cot ^2\frac{\pi}{n},\cot ^2\frac{2\pi}{n},\cdots,\cot ^2\frac{\left( m-1 \right) \pi}{n}$为多项式$\sum_{j=1}^m{C_{2m}^{2j-1}x^{m-j}\left( -1 \right) ^j}=0$的根,因此
\[
\sum_{k=1}^{m-1}{\cot ^2\frac{k\pi}{n}}=\frac{C_{2m}^{3}}{C_{2m}^{1}}=\frac{\left( m-1 \right) \left( 2m-1 \right)}{3},
\]
故
\[
\sum_{k=1}^{n-1}{\cot ^2\frac{k\pi}{n}}=\frac{\left( 2m-2 \right) \left( 2m-1 \right)}{3}=\frac{\left( n-2 \right) \left( n-1 \right)}{3},
\]
当$n=2m+1$时,类似可得
\[
\sum_{k=1}^{n-1}{\cot ^2\frac{k\pi}{n}}=2\frac{C_{2m+1}^{3}}{C_{2m+1}^{1}}=\frac{\left( 2m-1 \right) 2m}{3}=\frac{\left( n-2 \right) \left( n-1 \right)}{3},
\]
因此
\begin{align*}
\sum_{k=1}^{n-1}{\frac{1}{\left( 1-\varepsilon _{n}^{k} \right) \left( 1-\varepsilon _{n}^{-k} \right)}} &=\frac{n-1}{4}+\frac{1}{4}\sum_{k=1}^{n-1}{\cot ^2\frac{k\pi}{n}}
\\
&=\frac{n-1}{4}+\frac{1}{4}\cdot \frac{\left( n-2 \right) \left( n-1 \right)}{3}=\frac{n^2-1}{12}.
\end{align*}
\end{solution}

 

 

\begin{example}
证明组合恒等式:
\[\sum_{k=0}^{r}\binom{n}{k}\binom{m}{r-k}=\binom{n+m}{r}
\quad n+m\geq r.\]
\end{example}
\begin{solution}
证法1. (组合分析法)一个班里有$n$个男生, $m$个女生.现需从中选出$r$个人,则有$\binom{n+m}{r}$种选法;另外,若先选出$k\,(0\leq k\leq r)$个男生,再选出$r-k$个女生,则共有$\sum_{k=0}^{r}\binom{n}{k}\binom{m}{r-k}$种选法,所以
\[\sum_{k=0}^{r}\binom{n}{k}\binom{m}{r-k}=\binom{n+m}{r}
\quad n+m\geq r.\]

证法2. (母函数法)因为$(1+x)^n=\sum_{k=0}^{\infty}\binom{n}{k}x^k$,所以$(1+x)^{n+m}=\sum_ {r=0}^{\infty}\binom{n+m}{r}x^r$.又因为
\[(1+x)^{n+m}=(1+x)^n(1+x)^m=\sum_ {k=0}^{\infty}\binom{n}{k}x^k\cdot \sum_ {j=0}^{\infty}\binom{m}{j}x^j
=\sum_ {r=0}^{\infty}\sum_ {k=0}^r\binom{n}{k}\binom{m}{r-k}x^r,\]
所以
\[(1+x)^{n+m}=\sum_ {r=0}^{\infty}\binom{n+m}{r}x^r
=\sum_ {r=0}^{\infty}\sum_ {k=0}^r\binom{n}{k}\binom{m}{r-k}x^r.\]
比较$x^r$的系数可得
\[\sum_ {k=0}^r\binom{n}{k}\binom{m}{r-k}=\binom{n+m}{r}.\]
\end{solution}

如果令$m=r=n$,则
\[\sum_ {k=0}^n\binom{n}{k}\binom{n}{n-k}=\binom{n+n}{n},\]
即
\[\sum_ {k=0}^n\binom{n}{k}^2=\binom{2n}{n},\]
这便是Vandermonde恒等式.

%一个组合恒等式的多种证明方法,https://wenku.baidu.com/view/f141c099561252d381eb6e9e.html?rec_flag=default&sxts=1586433159470

%构造组合模型证明组合恒等式,https://wenku.baidu.com/view/c3966bfba26925c52cc5bfb5.html?rec_flag=default&sxts=1586433028431

组合数的两个性质:
\[\binom{n}{m}=\binom{n}{n-m},\quad \binom{n+1}{m}=\binom{n}{m}+\binom{n}{m-1}.\]

\begin{example}
证明
\[1+2+3+\cdots+n=\binom{n+1}{2}=\frac{n(n+1)}{2}.\]
\end{example}
\begin{solution}
考虑从学号为$1,2,3,\cdots,n+1$的学生中选出$2$个参加活动,则可以选$1$和$2$,$\cdots$,$1$和$n+1$,共$n$个人;

选$2$和$3$,$\cdots$,$2$和$n+1$,共$n-1$个人;

如此类推,若选$n$和$n+1$,共$1$个人.于是得证.
\end{solution}

\begin{example}
证明
\[\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+\cdots+\binom{n}{n}^2=\binom{2n}{n}=
\frac{(2n)!}{(n!)^2}.\]
\end{example}
\begin{solution}
\textbf{证法一.}考虑$\left( 1+x \right) ^n\left( 1+\frac{1}{x} \right) ^n=\frac{\left( 1+x \right) ^{2n}}{x^n}$,它的常数项为$\binom{2n}{n}$.

 

\textbf{证法二.}设有$n$个白球, $n$个红球.一方面,从这$2n$个球中取出$n$个球的取法种数为$\binom{2n}{n}$.

另一方面,可以看成$n+1$次如下的取球活动:从$n$个白球中取$k$个,再从$n$个红球中取$n-k$个, $k=0,1,2,\cdots,n$,取法种数为$\binom{n}{k}\cdot \binom{n}{n-k}=\binom{n}{k}^2$,根据分类计数原理便可得证.
\end{solution}

\begin{example}
证明:
\[\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}{n}x^n.\]
\end{example}

 

\begin{example}
证明
\[\binom{n}{1}^2+2\binom{n}{2}^2+\cdots+n\binom{n}{n}^2=
n\binom{2n-1}{n-1}.\]
\end{example}
\begin{solution}
\textbf{证法一.}从$n$名男生和$n$名女生中选出$n$个人,再在其中选出一个负责人并且规定负责人必须是女生,问共有多少种选法?

一方面,先从$n$名女生中选出一名负责人,有$\binom{n}{1}$种选法;然后从剩余的$2n-1$个人中选出其余的$n-1$名,有$\binom{2n-1}{n-1}$种选法,由分步计数原理可知,共有$\binom{n}{1}\binom{2n-1}{n-1}=n\binom{2n-1}{n-1}$种选法.

另一方面,考虑这$n$人的构成:第一类:若其中有$1$名女生和
$n-1$名男生,则共有$\binom{1}{1}\binom{n}{1}\binom{n}{n-1}=\binom{n}{1}^2$种选法;第二类:若其中有$2$名女生和$n-2$名男生,则共有$\binom{2} {1}\binom{n}{2}\binom{n}{n-2}=2\binom{n}{2}^2$种选法.如此类推.第$n$类:若其中有$n$名女生,没有男生,则共有$\binom{n} {1}\binom{n}{n}\binom{n}{0}=n\binom{n}{n}^2$种选法.对于上述儿类情况,由分类计数原理可知,共有$\binom{n}{1}^2+2\binom{n}{2}^2+\cdots+n\binom{n}{n}^2$种选法.得证.

\textbf{证法二.}考虑$(1+x)^{2n-1}=(1+x)^{n-1}(1+x)^{n}$两边展开式中$x^{n-1}$的系数.
\end{solution}

\begin{example}
证明
\[\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}=
n\cdot 2^{n-1}.\]
\end{example}
\begin{solution}
原式左端等价于$\binom{1}{1}\binom{n} {1}+\binom{2}{1}\binom{n}{2}+\cdots+\binom{n}{1}\binom{n}{n}$,这里$\binom{i}{1}\binom{n}{i}$可表示先在$n$个元素里选$i$个,再在这$i$个元素里选一个的组合数.可设一个班有$n$个同学,选出若干人(至少$1$人)组成一个代表团,并指定一人为团长.把这种选法按取到的人数$i$分类$(i=1,2,\cdots,n)$,则选法总数即为原式左端.

换一种选法,先选团长,有$n$种选法,再决定剩下的$n-1$人是否参加,每人都有两种可能,所以团员的选法有$2^{n-1}$种.即选法总数为$n\cdot 2^{n-1}$种,显然两种选法是一致的.
\end{solution}

\begin{example}
证明
\[\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=
n(n+1)2^{n-2}.\]
\end{example}
\begin{solution}
从$n$个学生中选出若干人参加竞赛培训,然后从中确定参加竞赛人选:物理一人、化学一人(可以兼报),共有多少种不同情况?

一方面,参加培训的人数为$k$时,共有$p_k=\binom{n}{k}\binom{k}{1}\binom{k}{1}=k^2\binom{n}{k}$种不同情况.依分类计数原理，不同情况的种数为$p_1+p_2 +\cdots+p_k$.

另一方面,先确定参加竞赛的人选,其他人参加培训与否自定,即其他人都有两种选择.若两门竞赛由一人参加,有$q_1=\binom{n} {1}\cdot 2^{n-1}$种情况;若两门竞赛由两人参加,有$q_2=\binom{n}{2}A_2^2\cdot 2^{n-2}$种情况,故共有$q_1+q_2 =n(n+1)2^{n-2}$种情况,由于答案惟一,原命题得证.
\end{solution}

\begin{example}
证明
\[\binom{n}{1}+2^3\binom{n}{2}+\cdots+n^3\binom{n}{n}=
n^2(n+3)2^{n-3}.\]
\end{example}
\begin{solution}
从$n$人中选若干人参加竞赛培训,再从中确定参加竞赛人选:数学一人、物理一人、化学一人(可以兼报),共有多少种不同情况?

一方面,参加培训的人数为$k$时,共有$f_k=\binom{n}{k}\binom{k}{1}\binom{k}{1}\binom{k}{1}=k^3\binom{n} {k}$种不同情况,依分类计数原理,不同情况的种数为$f_1+f_2+\cdots+f_n$.

另一方面,先确定参加竞赛的人选,其他人参加培训与否自定,若三门竞赛由一人参加,有$g_1=\binom{n}{1}\cdot 2^{n-1}$种情况;若三门竞赛由两人参加,有$g_2=\binom{n}{2}\binom{3}{2}\binom{2}{1}\cdot 2^{n-2}$种情况;若三门竞赛由三人参加，有$g_3=\binom{n} {3}A_3^3\cdot 2^{n-3}$种情况,故共有$g_1+ g_2 + g_3=n^2(n+3)2^{n-3}$种情况.由于答案惟一,原命题得证.
\end{solution}

\begin{example}
证明: $\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}\binom{n}{k}
=1+\frac{1}{2}+\cdots+\frac{1}{n}$.
\end{example}
\begin{solution}
\textbf{证法一.}因为$k\binom{n}{k}=n\binom{n-1}{k-1}$,所以
\[\frac{1}{k}\binom{n}{k}=\frac{1}{k}\left(\binom{n-1}{k}+\binom{n-1}{k-1}\right)
=\frac{1}{k}\binom{n-1}{k}+\frac{1}{n}\binom{n}{k}.\]
若记$f_n=\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}\binom{n}{k}$,则
\begin{align*}
f_n &=\sum_ {k=1}^{n-1}(-1)^{k+1}\frac{1}{k}\binom{n}{k}+(-1)^{n+1}\frac{1}{n}\\
&=\sum_ {k=1}^{n-1}(-1)^{k+1}\frac{1}{k}\binom{n-1}{k}+\frac{1}{n}\sum_ {k=1}^{n-1}(-1)^{k+1}\binom{n}{k}+(-1)^{n+1}\frac{1}{n}\\
&=f_{n-1}+\frac{1}{n}\sum_ {k=1}^{n}(-1)^{k+1}\binom{n}{k}=f_{n-1}+\frac{1}{n}\left[1-(1-1)^n\right]=f_{n-1}+\frac{1}{n}.
\end{align*}
因此$f_{n-1}=f_{n-2}+\frac{1}{n-1},\cdots,f_2=f_1+\frac{1}{2},f_1=1$,于是
\[f_n=f_{n-1}+\frac{1}{n}=f_{n-2}+\frac{1}{n}+\frac{1}{n-1}=\cdots=1+\frac{1}{2}+\cdots+\frac{1}{n}.\]

\textbf{证法二.}首先注意到\[{\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k + 1}}}}{k}C_n^k{x^k}} } \right)^\prime } = \sum\limits_{k = 1}^n {C_n^k{{\left( { - x} \right)}^{k - 1}}} = \frac{1}{{ - x}}\sum\limits_{k = 1}^n {C_n^k{{\left( { - x} \right)}^k}} = \frac{1}{{ - x}}\left[ {{{\left( {1 - x} \right)}^n} - 1} \right].\]

因此\[\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k + 1}}}}{k}C_n^k} = \int_0^1 {\frac{{{{\left( {1 - x} \right)}^n} - 1}}{{ - x}}dx} = \int_0^1 {\frac{{1 - {{\left( {1 - x} \right)}^n}}}{x}dx} .\]

因此

\begin{align*}\int_0^1 {\frac{{1 - {{\left( {1 - x} \right)}^n}}}{x}dx}&=\int_0^1 {\left[ {1 - {{\left( {1 - x} \right)}^n}} \right]d\left( {\ln x} \right)} = - n\int_0^1 {\ln x{{\left( {1 - x} \right)}^{n - 1}}dx} \\&= - n\int_0^1 {{y^{n - 1}}\ln \left( {1 - y} \right)dy} = n\int_0^1 {{y^{n - 1}}\sum\limits_{k = 1}^\infty {\frac{{{y^k}}}{k}} dy} = \sum\limits_{k = 1}^\infty {\frac{n}{k}\int_0^1 {{y^{n + k - 1}}dy} } \\&= \sum\limits_{k = 1}^\infty {\frac{n}{{k\left( {n + k} \right)}}} = \sum\limits_{k = 1}^\infty {\left[ {\frac{1}{k} - \frac{1}{{n + k}}} \right]} = 1 + \frac{1}{2} + \cdots + \frac{1}{n}.\end{align*}
\end{solution}

\begin{example}
1
\end{example}
\begin{solution}
$12$.
\end{solution}

 

 

\begin{example}
(2020年西城一模)对于正整数$n$,如果$k\,(k\in \mathbb{N}^\ast)$个整数$a_1,a_2,\cdots,a_k$满足$1\leq a_1\leq a_2\leq \cdots\leq a_k\leq n$,且$a_1+a_2+\cdots+a_k=n$,则称数组$(a_1,a_2,\cdots,a_k)$为$n$的一个“正整数分拆”.记$a_1,a_2,\cdots,a_k$均为偶数的“正整数分拆”的个数为$f_n$, $a_1,a_2,\cdots,a_k$均为奇数的“正整数分拆”的个数为$g_n$.

(I)写出整数4的所有“正整数分拆”;

(II)对于给定的整数$n\,(n\geq 4)$,设$(a_1,a_2,\cdots,a_k)$是$n$的一个“正整数分拆”
且$a_1=2$,求$k$的最大值;

(III)对所有的正整数$n$,证明: $f_n<g_n$;并求出使得等号成立的$n$的值.

(注:对于$n$的两个“正整数分拆”$(a_1,a_2,\cdots,a_k)$与$(b_1,b_2,\cdots,b_m)$,当且仅当$k=m$且$a_1=b_1,a_2=b_2,\cdots,a_k=b_m$时,称这两个“正整数分拆”是相同的.)
\end{example}
\begin{solution}
$12$.
\end{solution}

\begin{example}
(2012年海淀二模)将一个正整数$n$表示为$a_1+a_2+\cdots+a_p\,(p\in \mathbb{N}^\ast)$的形式,其中$a_i\in \mathbb{N}^\ast,i=1,2,\cdots,p$,且$a_1\leq a_2\leq \cdots\leq a_p$,记所有这样的表示法的种数为$f(n)$(如$4=4,4=1+3,4=2+2,4=1+1+2,4=1+1+1+1$,故$f(4)=5$).

(1)写出$f(3),f(5)$的值,并说明理由;

(2)对任意正整数$n$,比较$f(n+1)$与$\frac{1}{2}[f(n)+f(n+2)]$的大小,并给出证明;

(3)当正整数$n>6$时,求证: $f(n)\geq 4n-13$.
\end{example}
\begin{solution}
$12$.
\end{solution}

\begin{example}
(2010中国数学奥林匹克,李伟固)设复数$a$、$b$、$c$满足:对任意模不超过$1$的复数$z$,都有$\left|az^2+bz+c\right|\leq 1$.求$|bc|$的最大值.
\end{example}
\begin{solution}
令$f(z)=az^2+bz+c,g(z)=z^{-2}f(z)=a+bz^{-1}+cz^{-2}$, $h\left( z \right) =e^{i\alpha}g\left( e^{i\beta}z \right) =c'z^{-2}+b'z^{-1}+a'$.

取适当的实数$\alpha,\beta$,使得$c'$、$b'\geq 0$,对$r\leq 1$,有
\[
\frac{1}{r^2}\ge \left| h\left( re^{i\theta} \right) \right|\ge \left| \mathrm{Im}h\left( re^{i\theta} \right) \right|=\left| r^{-2}c'\sin 2\theta +r^{-1}b'\sin \theta +\mathrm{Im}a' \right|.
\]

不妨设$\mathrm{Im}a'\geq 0$,否则可作变换$\theta\to -\theta$,这样对任意$\theta\, \left( 0<\theta <\frac{\pi}{2} \right)$,有
\[
\frac{1}{r^2}\ge r^{-2}c'\sin 2\theta +r^{-1}b'\sin \theta \ge 2r^{-3/2}\sqrt{b'c'\sin 2\theta \cdot \sin \theta}.
\]
\[
\Rightarrow \left| bc \right|=b'c'\le \frac{1}{4r\sin 2\theta \cdot \sin \theta},\quad \text{对任意$r\leq 1,\theta \in \left( 0,\frac{\pi}{2} \right)$}
\]
于是
\begin{align*}
\left| bc \right| &\le \mathop {\min} \limits_r\le 1,\theta \in \left( 0,\frac{\pi}{2} \right)\frac{1}{4r\sin 2\theta \cdot \sin \theta}=\mathop {\min} \limits_\theta \in \left( 0,\frac{\pi}{2} \right)\frac{1}{4\sin 2\theta \cdot \sin \theta}
\\
&=\frac{1}{4\mathop {\max} \limits_\theta \in \left( 0,\frac{\pi}{2} \right)\sin 2\theta \cdot \sin \theta}=\frac{3\sqrt{3}}{16}.
\end{align*}

$\left| bc \right|=\frac{3\sqrt{3}}{16}$的例子:
\[
f\left( z \right) =\frac{\sqrt{2}}{8}z^2-\frac{\sqrt{6}}{4}z-\frac{3\sqrt{2}}{8}.
\]
对于$z=re^{i\theta}\,(r\leq 1)$,有
\begin{align*}
\left| f\left( re^{i\theta} \right) \right|^2 &=\frac{1}{32}\left[ \left( r^2\cos 2\theta -2\sqrt{3}r\cos \theta -3 \right) ^2+\left( r^2\sin 2\theta -2\sqrt{3}r\sin \theta \right) ^2 \right]
\\
&=\frac{1}{32}\left[ 2r^4+12r^2+18-\left( 2\sqrt{3}r\cos \theta +r^2-3 \right) ^2 \right]
\\
&\le \frac{1}{32}\left( 2r^4+12r^2+18 \right) \le 1.
\end{align*}
\end{solution}

\begin{example}
(2010年新课标全国卷)设函数$f(x)=e^x-1-x-ax^2$.

(1)若$a=0$,求$f(x)$的单调区间;

(2)若当$x\geq 0$时$f(x)\geq 0$,求$a$的取值范围.
\end{example}
\begin{solution}
不等式$e^x\geq x+1$与$1$
\end{solution}

\begin{example}
(2018年全国2卷)已知函数$f(x)=e^x-ax^2$.

(1)若$a=1$,证明:当$x\geq 0$时, $f(x)\geq 1$;

(2)若$f(x)$在$(0,+\infty)$只有一个零点,求$a$.
\end{example}

(3)设$n\in \mathbb{N}^\ast,x>0$,求证: $e^x>1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}$,其中阶乘$n!=n\times (n-1)\times\cdots \times 2\times 1$.
\begin{solution}
1
\end{solution}

\begin{example}
(2018年广州一模)已知函数$f(x)=ax+\ln x+1$.

(1)讨论函数$f(x)$零点个数;

(2)对任意的$x>0,f(x)\leq xe^{2x}$恒成立,求实数$a$的取值范围.
\end{example}
\begin{solution}
\textbf{解法一.}(参变分离,分离参(常)数)利用
\[
a\le \frac{xe^{2x}-\ln x-1}{x}.
\]
利用$e^x\geq x+1$可知
\[
xe^{2x}=e^{\ln x}e^{2x}=e^{\ln x+2x}\ge \ln x+2x+1,
\]
当且仅当$xe^{2x}=1$时取等号,则
\[
\frac{xe^{2x}-\ln x-1}{x}\ge \frac{\ln x+2x+1-\ln x-1}{x}=2,
\]
因此$a$的取值范围为$(-\infty,2]$.

\textbf{解法二.}等价于
\[
a\le \frac{xe^{2x}-\ln x-1}{x}=e^{2x}-\frac{\ln x+1}{x},\quad x\in (0,+\infty)
\]
令
\[
m\left( x \right) =e^{2x}-\frac{\ln x+1}{x},\quad m'\left( x \right) =\frac{2x^2e^{2x}+\ln x}{x^2}.
\]
即隐零点$x_0$满足$2x_0^2e^{2x_0}+\ln x_0=0$,则$0<x_0<1$,所以
\[
2x_0e^{2x_0}=\frac{1}{x_0}\left( \ln \frac{1}{x_0} \right) =\left( \ln \frac{1}{x_0} \right) \cdot \exp \left\{ \ln \frac{1}{x_0} \right\}.
\]
构造函数$\varphi (x)=xe^x\,(x>0)$为增函数,由$2x_0e^{2x_0}+\frac{\ln x_0}{x_0}=0$等价于$\varphi \left( 2x_0 \right) =\varphi \left( \ln \frac{1}{x_0} \right)$,所以$2x_0=\ln \frac{1}{x_0}$,即$x_0e^{2x_0}=1$.于是
\[
m\left( x_0 \right) =e^{2x_0}-\frac{\ln x_0+1}{x_0}=-\frac{\ln x_0}{x_0}=2.
\]
\end{solution}

\begin{example}
(2018年广州二模)已知函数$f(x)=e^x-x^2-ax$.

(1)若函数$f(x)$在$\mathbb{R}$上单调递增,求$a$的取值范围;

(2)若$a=1$,证明:当$x>0$时, $f\left( x \right) >1-\frac{\ln 2}{2}-\left( \frac{\ln 2}{2} \right) ^2$.

参考数据: $e\approx 2.71828,\ln 2\approx 0.69$.
\end{example}
\begin{solution}
证明隐零点$x_0\in \left( 1,1+\frac{\ln 2}{2} \right)$.
\end{solution}

 

\begin{example}
$f(x)=ax^3-3x+1$对于$x\in [-1,1]$总有$f(x)\geq 0$成立,则$a=$?
\end{example}
\begin{solution}
三倍角$\cos 3\theta=4\cos^3\theta-3\cos\theta$.
\end{solution}

\begin{example}
(1)当$PA+PB=c\,(c>0)$,曲线为椭圆;

(2)当$|PA-PB|=c\,(c>0)$,曲线为双曲线;

(3)当$PA\cdot PB=c\,(c>0)$,曲线为类椭圆,花生形,八字形,哑铃型;

两个类椭圆:
\[
\sqrt{\left( x-1 \right) ^2+y^2}\cdot \sqrt{\left( x+1 \right) ^2+y^2}=c,\quad 0<c<1.
\]
八字形:
\[
\sqrt{\left( x-1 \right) ^2+y^2}\cdot \sqrt{\left( x+1 \right) ^2+y^2}=1.
\]

 

(4)当$\frac{PA}{PB}=c\,(c>0)$,曲线为圆.
\end{example}
\begin{solution}
$12$.
\end{solution}

%https://www.doc88.com/p-2425380559539.html,椭圆——卡西尼卵形线;到两定点的距离的积为定值的点的轨迹

平面上到两个定点的距离的积为定值的动点$P$所构成的图形一般称为卡西尼(Cassini Oval)卵形线,设两个定点分别为$F_1,F_2$,且$|F_1F_2|=2c\,(c>0)$, $|PF_1|\cdot |PF_2|= a^2\,(a > 0)$, $a,c$是定值.取过$F_1,F_2$的直线为$x$轴,线段$F_1F_2$的垂直平分线为$y$轴,建立直角坐标系.

若$0<a<c$时,轨迹为两个类圆;

若$a=c$时,轨迹为双纽线,极坐标方程为$r^2=a^2\cos2\theta$;

若$c<a<\sqrt{3}c$时,轨迹为花生形;

若$a=\sqrt{3}c$时,轨迹为足球场跑道形;

若$a>\sqrt{3}c$时,轨迹为类椭圆.

金庸在《笑傲江湖》第十章《传剑》中写到，风清扬向令狐冲解释，如何活学剑法，到达"无招胜有招”的境界。无论是练功夫，还是做其他事情，事不相同，经历的境界类似。一个人只有突破3重境界，无招胜有招，才能成为真正的高手。

第1重境界：活学活使。

风清扬对令狐冲说："死招数破得再妙，遇上了活招数，免不了缚手缚脚，只有任人屠戮。这个"活，字，你要牢牢记住了。学招时要活学，使招时要活便。倘若拘泥不化，便练熟了几千万手绝招，遇上了真正高手，终究还是给人家破得干干净净。活学活使，只是第一步。"

很多人，把知识或者方法，死记硬背下来，就以为自己很渊博，了不起。其实这只是基本功。熟读唐诗三百首，不会作诗也会吟。在熟练的基础上，活学活使，不能照搬别人的，要融入自己的东西，形成自己的特色，才算是把功夫学上手。

第2重境界：出手无招。

拳有套路，剑有剑法。在入门练基础的阶段，都是要遵守固定的招数，熟练以后，才能活学活用，要踏入高手的境界，还需要做到出手无招。

令狐冲问："根本无招，如何可破？

风清扬说，把套路打碎，混用，不能算无招，对方仍然可以辨别你的套路。他向令狐冲打一个比喻：要切肉，就需要有肉摆在面前；要砍柴，总得有柴可砍；对手要破你招，你总得有招给别人破。当你无招的时候，对方就蒙了，根本不知道你下一步会打哪里，防不胜防。这就是无招有招。

用现在的话说，就是高手从不按套路出牌。山外有山，人外有人，套路外有套路，蝉螂捕蝉，黄雀在后。当一个人精通任何套路，却不使用套路，敌人就拿他没办法，因为敌人想制服他，也找不到下手的点。

第3重境界：能制人，而决不能为人所制。

风清场从地上拿起一根死人腿骨，随手以一端对着令狐冲，问："你如何破我这招？"

令狐冲不知他这一下是什么招式，顿时蒙了，说："这不是招式，因此破解不得。"

风清扬笑道："这就是了。学武之人使兵刃，动拳脚，总是有招式的，你只须知道破法，一出手便能破招制敌。"

令狐冲道："要是敌人也没招式呢？"

风清扬道："那么他也是一等一的高手了。"真正的高手，要做个两点，一是不让别人摸到自己的套路，让对手无迹可寻，无从下手；二是要变被动为主动，能主动出击，打乱对手的惯用招数，使其慌乱失控而被制服。所以，能制人，而决不能为人所制，才是一等一的高手。

从活学活使，到出手无招，最后到能制人，而决不能为人所制，这3重循序渐进的境界，即是一个顶尖高手的养成过程。练习剑法、功夫如此，做人做事，亦是如此。金庸的武侠小说，并不仅仅停留在故事情节，其中更蕴藏着深刻的道理，愿我们学以致用。向金庸致敬！

用好教材,以本为本.将教材完整读一遍,书上的定理公式亲自证一遍,认真做教材上的习题

熟悉经典题型

注意书写,规范答题.

胡适：大胆假设，小心求证

匠人精神:我一直重复同样的事情以求精进，总是向往能够有所进步，我继续向上，努力达到巅峰，但没人知道巅峰在哪。我依然不认为自己已臻完善，爱自己的工作，一生投身其中。——[日]《寿司之神》

 

 

\begin{example}
1
\end{example}
\begin{solution}
$12$.
\end{solution}

\chapter{排列组合与概率}

\begin{example}
(2019年四川竞赛)设一个袋子里有红、黄、蓝色小球各一个,现每次从袋子里取出一个球(取出某色球的概率均相同),确定颜色后放回,直到连续两次均取出红色球时为止.记此时取出球的次数为$\xi$,则$\xi$的数学期望为?
\end{example}
\begin{solution}
$12$.
\end{solution}

\begin{example}
(2017年江苏高考)已知一个口袋中有$m$个白球, $n$个黑球$(m,n\in \mathbb{N}^\ast,n\geq 2)$,这些球除颜色外全部相同.现将口袋中的球随机地逐个取出,并放入如图所示的编号为$1,2,3,\cdots,m+n$的抽屉内,其中第$k$次取出的球放入编号为$k$的抽屉$(k=1,2,3,\cdots,m+n)$.
\begin{table}[!htbp]
\centering
\begin{tabular}{|c|c|c|c|c|}
\hline
% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
$1$ & $2$ & $3$ & $\cdots$ & $m+n$ \\
\hline
\end{tabular}
% \caption{}\label{}
\end{table}

\begin{enumerate}
\item 试求编号为$2$的抽屉内放的是黑球的概率$P$;
\item 随机变量$X$表示最后一个取出的黑球所在抽屉编号的倒数, $E(X)$是$X$的数学期望,证明:
\[E(X)<\frac{n}{(m+n)(n-1)}.\]
\end{enumerate}
\end{example}
\begin{solution}

\end{solution}

\begin{example}

\end{example}
\begin{solution}

\end{solution}

\chapter{微积分初步}

\begin{example}
(2017年江苏复赛)设函数$f_n(x)=1+x+\frac{1}{2!}x^2+\cdots+\frac{1}{n!}x^n$.
\begin{enumerate}
\item 求证:当$x\in (0,+\infty),n\in N^\ast$时, $e^x>f_n(x)$;

\item 设$x>0,n\in\mathbb{N}^\ast$,若存在$y\in\mathbb{R}$使得$e^x=f_n(x)+\frac{1}{(n+1)!}x^{n+1}e^y$,求证: $0<y<x$.
\end{enumerate}
\end{example}
\begin{solution}

\end{solution}

\chapter{代数变形}

\begin{example}
(2013交大少年班)已知$a,b,c$为正数,满足如下两个条件:
\[a+b+c=32,\quad \frac{b+c-a}{b c}+\frac{c+a-b}{c a}+\frac{a+b-c}{a b}=\frac{1}{4}.\]
证明:以$\sqrt{a}, \sqrt{b}, \sqrt{c}$为三边长可构成一个直角三角形.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2018西安交通大学少年班) (1)已知$a^2+b^2=c^2$,请比较$a^3+b^3$与$c^3$的大小.

(2)已知一数为$2n+1$,求能与其构成勾股数的另两数.

己知$a^3+b^3=c^3$,其中$a,b,c$为三角形之三边,试判断该三角形的形状.
\end{example}
\begin{solution}

\end{solution}

\begin{example}

\end{example}
\begin{solution}

\end{solution}

%\input{tex/daishu.tex}

%\input{tex/shulun.tex}

 

\begin{example}
设非负实数$a,b,c,d$满足$a+b+c+d=4$,求证:
\[a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}\leq 2\left(1+\sqrt{abcd}\right).\]
\end{example}
\begin{solution}

\end{solution}

\begin{example}
Compute
$\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.$
\end{example}
\begin{solution}
The Sophie Germain Identity states that $a^4 + 4b^4$ can be factorized as $(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab)$. Each of the terms is in the form of $x^4 + 324$. Using Sophie-Germain, we get that $x^4 + 4\cdot 3^4 = (x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x)(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x) = (x(x-6) + 18)(x(x+6)+18)$.

$\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]}$

$= \frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}$
Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}$.
\end{solution}

\begin{example}

\end{example}
\begin{solution}

\end{solution}

从多项式到拉格朗日插值公式,行列式.

\begin{enumerate}
\item 设$(3x-1)^7=a_7x^7+a_6x^6+\cdots+a_1x+a_0$,试求$a_0+a_2+a_4+a_6$的值.

\item 解不等式$||x+3|-|x-1||>2$.
\item 求$x^4-2x^3+x^2+2x-1$除以$x^2+x+1$的商式和余式.

\item 求方程$|xy|-|2x|+|y|=4$的整数解.
\end{enumerate}

 

设$A$为$m\times n$非零矩阵, $A'$表示$A$的转置, $b$为$m$元列向量.证明:若线性方程组$Ax=b$有解,则$b$与$A'y=0$解空间正交.反之如何?请说明理由.

若线性方程组$Ax=b$有解$x_0$,则$Ax_0=b$, $x'_0A'=b'$,对于线性方程$A'Y=0$任一解$y_0$,则$A'y_0=0$,则$0=x'_0(A'y_0)=x'_0A'y_0=b'y_0$,即$b'y_0=0$,
因此$b$与$A'Y=0$的解空间正交.

反之,若$b$与$A'Y=0$的解空间正交,则对于线性方程$A'Y=0$任一解$y_0$,则$A'y_0=0$,且$b'y_0=0$,则
$\begin{pmatrix}
A' \\
b'
\end{pmatrix}y_0=0$,因此方程$A'y=0$与$\begin{pmatrix}
A' \\
b'
\end{pmatrix}y=0$同解,故$r(A')=r\begin{pmatrix}
A' \\
b'
\end{pmatrix}$,转置后得$r(A,b)=r(A)$.因此方程$Ax=b$有解.

 

\chapter{组合数学}

\textbf{八皇后问题},是一个古老而著名的问题,是回溯算法的典型案例.
\begin{example}
在$8\times 8$格的国际象棋上摆放八个皇后,使其不能互相攻击,即任意两个皇后都不能处于同一行、同一列或同一斜线上,问有多少种摆法?
\end{example}
高斯认为有76种方案. 1854年在柏林的象棋杂志上不同的作者发表了40种不同的解,后来有人用图论的方法解出92种结果.

(加法原理)若完成一件事情的方法有\(n\)类，其中第\(i\)类方法包括\(a_i\)种不同的方法，且这些方法互不重合，那么完成这件事共有\(\sum a_i\)中不同的方法。

(乘法原理)若完成一件事情需要\(n\)个步骤，其中第\(i\)个步骤有\(a_i\)种不同的完成方法，且这些步骤互不干扰，那么完成这件事共有\(\prod a_i\)中不同的方法。

(排列数)从\(n\)个不同元素中依次取出\(m\)个元素排成一列(在乎顺序)，产生的不同排列的数量为： \[A_n^m=P_n^m=\frac{n!}{(n-m)!}=n*(n-1)*...*(n-m+1)\]

(组合数)从\(n\)个不同元素中取出\(m\)个组成一个集合(不在乎顺序)，产生的不同集合数量为： \[C_n^m=\frac{A_n^m}{m!}=\frac{n!}{m!(n-m)!}=\frac{n*(n-1)*...*(n-m+1)}{m*(m-1)*...*1}\]

组合数的性质:
\(1.\ C_n^m=C_n^{n-m}\)

证明： 由组合数的定义，对于从\(n\)个元素中取出\(m\)个构成的集合，剩余的元素也恰好构成了一个集合，两个集合一一对应，故产生的集合数量也相同，性质成立。

\(2.\ C_n^m=C_{n-1}^{m}+C_{n-1}^{m-1}\)
证明： 从\(n\)个不同元素取出\(m\)个数组成一个集合有两类方法：取\(n\)号元素，不取\(n\)号元素。取\(n\)号元素时，则在\(n-1\)个元素中还取了\(m-1\)个数，有\(C_{n-1}^{m-1}\)种取法。不取\(n\)号元素时，则在\(n-1\)个元素中取了\(m\)个元素，有\(C_{n-1}^m\)种取法，故性质成立。

\(3.\ \sum_{i=0}^nC_n^{i}=2^n\)
证明： 从\(n\)个元素中取出若干个元素组成一个集合，有\(n+1\)类方法，即分别取出\(0,1,2,...,n\)个元素，其方案数分别对应\(C_n^1,C_n^2,...,C_n^n\)。从另一方面看，每一个元素有取或不取两种选择，方案数为\(2^n\)，所以性质成立。

二项式定理:
\[(a+b)^n=\sum_{k=0}^nC_n^ka^kb^{n-k}=\sum_{k=0}^nC_n^ka^{n-k}b^{k}\]
证明： 利用数学归纳法，当\(n=1\)时，\((a+b)^1=C_n^0a^0b^1+C_n^1a^1b^0=a+b\)成立。 若\(n=p\)时命题成立，则\(n=p+1\)时：
\begin{align*}
(a+b)^{p+1}&=(a+b)^p(a+b)=(a+b)\sum_{k=0}^pC_p^ka^kb^{p-k} \\
&=a\sum_{k=0}^pC_p^ka^kb^{p-k}+b\sum_{k=0}^pC_p^ka^kb^{p-k} \\
&=\sum_{k=0}^pC_p^ka^{k+1}b^{p-k}+\sum_{k=0}^pC_p^ka^kb^{p-k+1} \\
&=\sum_{k=1}^{p+1}C_p^{k-1}a^kb^{p-k+1}+\sum_{k=0}^pC_p^ka^kb^{p-k+1} \\
&=\sum_{k=0}^{p+1}(C_p^{k-1}+C_p^k)a^kb^{p-k+1} \\
&=\sum_{k=0}^{p+1}C_{p+1}^ka^kb^{p+1-k}
\end{align*}
故原命题成立。

多重集排列数

多重集指含有重复元素的广义集合。设多重集\(S=\{n_1*a_1,n_2*a_2,...,n_k*a_k\}\)是由\(n_1\)个\(a_1\)，\(n_2\)个\(a_2\)，\(...\)，\(n_k\)个\(a_k\)组成的集合，则\(S\)的全排列个数为\[\frac{(\sum_{i=1}^k n)!}{\prod_{i=1}^k(n_i!)}\]

证明： 对于朴素全排列，显然有\((\sum_{i=1}^k n)!\)种方案，而在多重集的排列过程中，每个\(a_i\)出现了\(n_i\)次，在这\(n_i\)个位置间各个\(a_i\)可以互相调换位置，有\(n_i!\)中方案，故除去每一个\(n_i\)可以调换位置的重复方案即为总排列数。

多重集的组合数

设多重集\(S=\{n_1*a_1,n_2*a_2,...,n_k*a_k\}\)是由\(n_1\)个\(a_1\)，\(n_2\)个\(a_2\)，\(...\)，\(n_k\)个\(a_k\)组成的集合，对于\(r\leq n_i\ (\forall\ i \in [1,k])\)，从\(S\)中取出\(r\)个元素组成一个多重集，产生不同的多重集数量为\[C_{k+r-1}^{k-1}\]

证明： 该问题等价于统计如下集合的数量：\(\{x_1*a_1,x_2*a_2,...,x_k*a_k\}\)，其中\(\sum_{i=1}^kx_i=r\)。故原问题等价于多重集\(\{r*0,(k-1)*1\}\)的全排列数，即类似于插板法，每连续的一串\(0\)代表元素\(a_1\)的个数，使用\(k-1\)个\(1\)将\(r\)个\(0\)分成\(k\)部分。利用多重集的排列数公式可得方案数为\(C_{k+r-1}^{k-1}\)。

\(Lucas\)定理：\(p\)为质数时，\(C_n^m\equiv C_{n\ mod\ p}^{m\ mod\ p}*C_{n/p}^{m/p}(mod\ p)\)。

证明：

预备和引理

\(1.\)费马小定理：\(p\)为质数时，有\(a\equiv a^p(mod\ p)\)。

\(2.\)二项式定理：\((a+b)^n=\sum_{k=0}^nC_n^{k}a^{k}b^{n-k}\)。

\(3.\)引理：\(p\)为质数时，有\((1+x)^p\equiv 1+x^p(mod\ p)\)。

证明： 由费马小定理可得\((1+x)^p\equiv 1+x(mod\ p)\)，又因为\(x\equiv x^p(mod\ p)\)，则可得\((1+x)^p\equiv 1+x^p(mod\ p)\)。

\(4.\)引理：\((1+x)^n\)第\(m\)项的系数即为\(C_n^m\)。

证明： 由二项式定理展开可得。

主体证明

对于\((1+n)^p\)，分解指数得： \[(1+x)^n\equiv (1+x)^{\lfloor \frac{n}{p} \rfloor p}(1+x)^{n\ mod\ p}(mod\ p)\] 利用引理\(3\)，可得： \[(1+x)^n\equiv(1+x^p)^{\lfloor \frac{n}{p} \rfloor}(1+x)^{n\ mod\ p}(mod\ p)\] 二项式定理展开，得： \[(1+x)^n\equiv\sum_{i=0}^{\lfloor \frac{n}{p} \rfloor}C_{\lfloor \frac{n}{p} \rfloor}^ix^{pi}*\sum_{j=0}^{n\ mod\ p}C_{n\ mod\ p}^jx^j\] 当且仅当\(pi+j=m\)时，\(x\)指数为\(m\)，故\(i=\lfloor \frac{p}{m} \rfloor,j=p \ mod\ m\)。 而此时\(x^m\)的系数为\[C_{\lfloor \frac{n}{p} \rfloor}^{\lfloor \frac{p}{m} \rfloor}*C_{p\ mod\ n}^{p\ mod\ m}\] 所以\(C_n^m\equiv C_{\lfloor \frac{n}{p} \rfloor}^{\lfloor \frac{p}{m} \rfloor}*C_{p\ mod\ n}^{p\ mod\ m}(mod\ p)\)，证毕。

给定\(n\)个\(0\)和\(n\)个\(1\)，将他们排成长度为\(2n\)的序列，满足任意前缀中\(0\)的个数都不少于\(1\)的个数的排列的数量为\[Cat_n=\frac{C_{2n}^n}{n+1}\]

证明：
对于\(n\)个\(0\)和\(n\)个\(1\)任意排成的一个序列\(S\)，若\(S\)不满足要求，则必然存在一个最小位置\(2p+1\in[1,2n]\)，使得\(S[1,2p+1]\)中存在\(p\)个\(0\)，\(p+1\)个\(1\)，那么将\(S[2p+2,2n]\)这一部分取反，就可以得到由\(n-1\)个\(0\)和\(n+1\)个\(1\)排成的序列。
同理，对于一个\(n-1\)个\(0\)和\(n+1\)个\(1\)随意排成的一个序列\(S'\)，也必定存在一个最小的位置\(2p'+1\)，使得\(S'[1,2p'+1]\)中有\(p'\)个\(0\)，\(p'+1\)个\(1\)，那么将\(S'[2p'+2,2n]\)这一部分取反，也能得到由\(n\)个\(0\)和\(n\)个\(1\)排成的一个序列，并且存在前缀\(1\)比\(0\)多的位置。
由上可知，每个不符合要求的序列和每一个由\(n-1\)个\(0\)和\(n+1\)个\(1\)排成的序列呈一一对应关系，根据组合数的定义，后者显然有\(C_{n2}^{n-1}\)个，那么符合要求的序列就有 \[C_{2n}^{n}-C_{2n}^{n-1}=\frac{(2n)!}{n!n!}-\frac{(2n)!}{(n+1)!(n-1)!}=\frac{C_{2n}^n}{n+1}=Cat_n\]

\(Catalan\)数还有如下的计算公式：
\[Cat_n=C_{2n}^n-C_{2n}^{n-1}=C_{2n}^n-C_{2n}^{n+1} \\ Cat_n=\sum_{i=0}^{n-1}Cat_iCat_{n-i-1} \\ Cat_n=Cat_{n-1}*\frac{n+1}{4n-2}\]

循环排列：从\(n\)个元素中选出\(m\)个排成圆圈的方案数，相当于线性排列时固定第一个数的方案。

一个循环排列可以对应\(m\)个线性排列，进而可以得到循环排列的计算公式： \[Cir_{n}^{m}=\frac{A_{n}^{m}}{m}=\frac{n!}{m\times (n-m)!}\]

鸽巢原理:

(一般形式)把\(n+1\)个物品放入\(n\)个盒子中，那么至少有一个盒子包含两个或两个以上的物品。

证明：
反证法，若每个盒子只有一个物品，则物品总数至多为\(n\)，矛盾。

(加强形式)设有\(\sum_{i=1}^nq_i-n+1\)个物品放入\(n\)个盒子中，每个盒子中分别放了\(a_1,a_2,...,a_n\)个物品，则至少存在一个\(k\)，使得\(a_k\geq q_k\)。

证明：
反证法，若每个盒子满足\(a_i<q_i\)，则物品总数至多为\(\sum _{i=1}^nq_i-n\)，矛盾。

(下降阶乘幂和二项式系数)我们定义\(n^{\underline{k}}\)为\(n\)的\(k\)次下降阶乘幂，其计算式为： \[n^{\underline{k}}=n\times (n-1)\times ... \times (n-k+1)\]
其中\(k\)是整数，\(n\)可以是任意实数。
我们发现组合数可以用下降阶乘幂来表示： \[C_{n}^{m}=\binom{n}{m}=\frac{n^{\underline{m}}}{m!}\]

于是我们就可以扩充组合数的定义域。也就是说，组合数的上指标\(n\)可以为任意实数。

组合恒等式

对称恒等式
\[\binom{n}{m}=\binom{n}{n-m}\]
对于\(n,m\in \mathbb{N}\)成立。
证明：
\[\binom{n}{m}=\frac{n!}{m!(n-m)!}=\frac{n!}{(n-m)!(n-(n-m))!}=\binom{n}{n-m}\]

吸收恒等式
\[\binom{r}{k}=\frac{r}{k}\binom{r-1}{k-1}\]
对于\(r\in \mathbb{R}，k\in \mathbb{N}^+\)成立。

证明：
\[\binom{r}{k}=\frac{r^{\underline{k}}}{k!}=\frac{r}{k}\times \frac{(r-1)^{\underline{k-1}}}{(k-1)!}=\frac{r}{k}\binom{r-1}{k-1}\]

相伴恒等式
\[(r-k)\binom{r}{k}=r\binom{r-1}{k}\]
对于\(r\in \mathbb{R},k\in \mathbb{N}\)成立。
\[(r-k)\binom{r}{k}=\frac{r!}{(r-k-1)!k!}=r\times \frac{(r-1)^{\underline{r-k-1}}}{(r-k-1)!}\\\ \\=r\binom{r-1}{r-k-1}=r\binom{r-1}{k}\]

加法公式
\[\binom{r}{k}=\binom{r-1}{k-1}+\binom{r-1}{k}\]
对于\(r\in \mathbb{R},k \in \mathbb{N}\)成立。
证明：
\[\binom{r-1}{k-1}+\binom{r-1}{k}=\frac{(r-1)^{\underline{k-1}}}{(k-1)!}+\frac{(r-1)^{\underline{k}}}{k!}\\ \ \\=\frac{k(r-1)^{\underline{k-1}}}{k!}+\frac{(r-k)(r-1)^{\underline{k-1}}}{k!}\\ \ \\ =\frac{r^{\underline{k}}}{k!}=\binom{r}{k}\]

上指标求和
\[\sum_{i=m}^n\binom{i}{m}=\binom{n+1}{m+1}\]
对于\(n,m\in \mathbb{N}\)成立。

证明：
设有\(n+1\)个物品，标号为\(1\sim n\)，现在从中选取\(m+1\)个物品，当选取的最大号码为\(i\)时，方案数为\(\binom{i}{m}\)，那么枚举累加方案数就得到了：\(\sum_{i=m}^n\binom{i}{m}=\binom{n+1}{m+1}\)。

平行恒等式
\[\sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n}\]
对于\(n,m\in \mathbb{N}\)成立。

证明： \[\sum_{i=0}^n\binom{m+i}{i}=\sum_{i=0}^n\binom{m+i}{m}\\ \ \\ =\sum_{i=m}^{n+m}\binom{i}{m}=\binom{n+m+1}{m+1}=\binom{n+m+1}{n}\]

上指标翻转
\[\binom{r}{k}=(-1)^k\binom{r-k-1}{k}\]
对于\(r\in \mathbb{R},k\in \mathbb{N}\) 成立。

证明：
\[\binom{r}{k}=\frac{r^{\underline{k}}}{k!}\\ \ \\ =\frac{r\times (r-1) \times ... \times (r-k+1)}{k!}\\ \ \\ =\frac{(-1)^{k}\times (k-r-1)\times (k-r-2) \times ... \times (-r)}{k!}\\ \ \\ =(-1)^k\frac{(r-k-1)^{\underline{k}}}{k!}=(-1)^k\binom{r-k-1}{k!}\]

三项式系数恒等式
\[\binom{r}{m}\binom{m}{k}=\binom{r-k}{m-k}\binom{r}{k}\]
对于\(r\in \mathbb{R} , n,m\in \mathbb{N}\)成立。

证明：
\[\binom{r}{m}\binom{m}{k}=\frac{r!}{(r-m)!(m-k)!k!}=\binom{r-k}{m-k}\binom{r}{k}\]

范德蒙德卷积
\[\sum_{k=0}^n\binom{r}{k}\binom{s}{n-k}=\binom{r+s}{n}\]
对于\(r,s\in \mathbb{R} , n\in \mathbb{N}\)成立。

证明：
左边表示从\(r\)个男生中选\(k\)个人，从\(s\)个女生中选出\(n-k\)个人的方案数，求和即为在\(r+s\)个人中选\(n\)个人的方案数，可知：\(\sum_{k=0}^n\binom{r}{k}\binom{s}{n-k}=\binom{r+s}{n}\)。

二项式定理

在上一篇中，我们已经提到了经典的二项式定理，并用数学归纳法证明了该定理的正确性。
\[(a+b)^n=\sum_{i=0}^n\binom{n}{i}a^{i}b^{n-i}\]

广义二项式定理

上文中，我们已经可以将组合数\(\binom{n}{m}\)的上指标扩充到实数域。在实数域的组合数中，二项式定理仍然成立，我们称之为广义二项式定理，又称牛顿二项式定理。
\[(a+b)^r=\sum_{i=0}^{\infty}\binom{r}{i}a^{i}b^{r-i}\]

组合数的数论性质

若\(p\)为质数，则对于\(n\in[1,p-1]\)，有\(p\ |\ \binom{p}{n}\)。
证明：
\[\because \binom{p}{n}=\frac{p\times (p-1)\times ... \times (p-n+1)}{n!}\in \mathbb{Z}\\ \ \\ \therefore n!\ |\ p\times (p-1) \times ... \times (p-n+1)\\ \ \\ \because(p,n)=1\\ \ \\ \therefore n!\ |\ (p-1) \times (p-1) \times ... \times (p-n+1)\\ \ \\ \therefore p\ |\ \binom{p}{n}\]

多项式定理

定义多项式系数：
\[\binom{n_1+n_2+...+n_k}{n_1,n_2,...,n_k}=\frac{(n_1+n_2+...+n_k)!}{n_1!n_2!...n_k!}\]
则有如下的多项式定理成立：
\[(x_1+x_2+...+x_k)^n=\sum_{n_1+n_2+...+n_k=n}\binom{n_1+n_2+...+n_k}{n_1,n_2,...,n_k}x_1^{n_1}x_2^{n_2}...x_k^{n_k}\]

两类阶乘幂

我们已经定义下降阶乘幂：
\[x^{\underline{n}}=x\times (x-1)\times ... \times (x-n+1)\]
同理我们定义上升阶乘幂：
\[x^{\overline{n}}=x\times (x+1)\times ... \times (x+n-1)\]
我们可以提取符号，写成另一种形式：
\[x^{\underline{n}}=(-1)^n(x-n+1)^{\underline{n}},x^{\overline{n}}=(-1)^n(1-x-n)^{\overline{n}}\]

阶乘幂的二项式定理

二项式定理对阶乘幂仍然成立：
\[(a+b)^{\underline n}=\sum_{i=0}^n\binom{n}{i}a^{\underline i}b^{\underline{n-i}}\\ \ \\ (a+b)^{\overline n}=\sum_{i=0}^n\binom{n}{i}a^{\overline i}b^{\overline{n-i}}\]

不定方程的解数问题

正整数解

求不定方程\(x_1+x_2+...+x_k=n\)的正整数解的个数。

这个问题等价于把\(n\)个球放入\(k\)个盒子中，每个盒子中至少有\(1\)个球，由隔板法可知其方案数为\(\binom{n-1}{k-1}\)。

非负整数解

求不定方程\(x_1+x_2+...+x_k=n\)的非负整数解的个数。

我们可以新增\(k\)个球，这样问题就等价于把\(n+k\)个球放入\(k\)个盒子中，每个盒子中至少有\(1\)个球，由隔板法可知其方案数为\(\binom{n+k-1}{k-1}\)。

下界限制

求不定方程\(x_1+x_2+...+x_k=n\)的整数解的个数，满足\(x_1\geq a_1,x_2\geq a_2,...,x_k\geq a_k\)。

这个问题等价于不定方程\(x_1+x_2+...+x_k=n-a_1-a_2-...-a_k\)的非负整数解个数，可以其方案数为\[\binom{n+k-1-\sum_{i=1}^{n}a_i}{k-1}\]

上下界限制

求不定方程\(x_1+x_2+...+x_k=n\)的整数解的个数，满足\(a_1\leq x_1\leq b_1,a_1\leq x_2\leq b_2,...,a_k\leq x_k\leq b_k\)。

首先把限制转换为\(0\leq x_1\leq b_1-a_1,...,0\leq x_k\leq b_k-a_k\)，运用容斥原理，答案即为： \[\binom{n+k-1}{k-1}-\binom{n+k-1-\sum_{i=1}^n(b_i-a_i+1)}{k-1}\]

\chapter{图论}

\begin{example}
(2005年西部赛)设$n$个新生中,任意$3$个人中有$2$个人互相认识,任意$4$个人中有$2$个人互不认识.试求$n$的最大值.
\end{example}
\begin{proof}

\end{proof}

 

\chapter{平面几何}

\begin{example}
(1998 IBM)在$\triangle ABC$中, $D,E,F$点分别在边$AB,BC,CA$上, 小三角形$DEF$为全等三角形且$AD=BE=CF$.求证: 三角形$ABC$也为全等三角形.
\end{example}
\begin{proof}

\end{proof}
%https://www.research.ibm.com/haifa/ponderthis/challenges/August1998.html

 

2011年理综光学题

 

\begin{example}
如图,已知$\triangle ABC$内接于圆$O$, $I$为$\triangle ABC$的内心,连接$AI$并延长分别交$BC$和圆$O$于$E$、$D$两点,连接$BD$、$CD$,求证:

(1) $BD=ID=CD$;

(2) $ID^2=DE\cdot AD$.
\end{example}
\begin{proof}
内心即角平分线的交点, $\angle BAD=\angle CAD$, $BD =CD$ (相等圆周角所对的弦相等)

$\angle ABI=\angle EBI$

因为$\angle BID=\angle BAD+\angle ABI$,
$\angle DBI=\angle DBC+\angle EBI$, $\angle DBC=\angle CAD=\angle BAD$ (同弧所对的圆周角相等)
所以$\angle BID=\angle DBI$,则$BD=ID=CD$.

(2)因为$\angle DBC=\angle CAE=\angle BAD$.
又因为$\angle BDE=\angle ADB$ (公共角),
所以$\triangle ABD\sim \triangle BED$\, (AA)
所以$\frac{AD}{BD}=\frac{BD}{DE}$,
转化为$BD^2=DE\times AD$.
又$ID=BD$,所以$ID^2=DE\times AD$.
\end{proof}

\section{中国高考数学压轴题}

 

\begin{enumerate}
\item 08年江西高考

\item 08年北京高考

\item (2017年天津)设$a\in \mathbb{Z}$,已知定义在$\mathbb{R}$上的函数$f(x)=2x^4+3x^3-3x^2-6x+a$在区间$(1,2)$内有一个零点$x_0$, $g(x)$为$f(x)$的导函数.
\begin{enumerate}
\item 求$g(x)$的单调区间;

\item 设$m\in [1,x_0)\cup (x_0,2]$,函数$h(x)=g(x)(m-x_0)-f(m)$,求证: $h(m)h(x_0)<0$;

\item 求证:存在大于$0$的常数$A$,使得对于任意的正整数$p,q$,且$\frac{p}{q}\in [1,x_0) \cup (x_0,2]$,满足$\left|\frac{p}{q}-x_0\right|\geq \frac{1}{Aq^4}$.
\end{enumerate}

\item 1999高考轧辊

\item 2003高考立体几何题

\item 08年广东高考

\item (2003高考江苏卷压轴题)设$a>0$,如图,已知直线$l:y=ax$及曲线$C:y=x^2$, $C$上的点$Q_1$的横坐标为$a_1\,(0<a_1<a)$.从$C$上的点$Q_n\,(n\geq 1)$作直线平行于$x$轴,交直线$l$于点$P_{n+1}$,再从点$P_{n+1}$作直线平行于$y$轴,交曲线$C$于点$Q_{n+1}$. $Q_n\,(n=1,2,3,\cdots)$的横坐标构成数列$\{a_n\}$.
\begin{enumerate}
\item 试求$a_{n+1}$与$a_n$的关系,并求$\{a_n\}$的通项公式;
\item 当$a=1,a_1\leq \frac{1}{2}$时,证明$\sum_{k=1}^{n}(a_k-a_{k+1})a_{k+2}<\frac{1}{32}$;

\item 当$a=1$时,证明$\sum_ {k=1}^{n}(a_k-a_{k+1})a_{k+2}<\frac{1}{3}$.
\end{enumerate}

\item (2010年全国2导数)设函数$f(x)=1-e^{-x}$.
\begin{enumerate}
\item 证明:当$x>-1$时, $f(x)\geq \frac{x}{x+1}$;

\item 设当$x\geq 0$时, $f(x)\leq\frac{x}{ax+1}$,求$a$的取值范围.
\end{enumerate}

\item (2014年全国2导数)已知函数$f(x)=e^x-e^{-x}-2x$.
\begin{enumerate}
\item 讨论$f(x)$的单调性;

\item 设$g(x)=f(2x)-4bf(x)$,当$x>0$时, $g(x)>0$,求$b$的最大值;

\item 已知$1.4142<\sqrt{2}<1.4143$,估计$\ln 2$的近似值(精确到$0.001$).
\end{enumerate}

\item (2013年安徽理科数学)设函数$f_n(x)=-1+x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\cdots+\frac{x^n}{n^2}\,(x\in \mathbb{R},n\in \mathbb{N}_+)$,证明:
\begin{enumerate}
\item 对每个$n\in \mathbb{N}_+$,存在唯一的$x_n\in\left[\frac{2}{3},1\right]$,满足$f_n(x_n)=0$;

\item 对于任意$p\in \mathbb{N}_+$,由(1)中$x_n$构成数列$\{x_n\}$满足$0<x_n-x_{n+p}<\frac{1}{n}$.
\end{enumerate}

\item (2014年安徽理科数学)设实数$c>0$,整数$p>1,n\in\mathbb{N}^+$.
\begin{enumerate}
\item 证明:当$x>1,x\neq 0$时, $(1+x)^p>1+px$;

\item 数列$\{a_n\}$满足$a_1>c^{\frac{1}{p}},a_{n+1}=\frac{p-1}{p}a_n+\frac{c}{p}a_n^{1-p}$,证明: $a_n>a_{n+1}>c^ {\frac{1}{p}}$.
\end{enumerate}

\item (2013年安徽理科数学)某高校数学系计划在周六和周日各举行一次主主题不同的心理测试活动.分别由李老师和张老师负责,已知该系共有$n$位学生,每次活动均需该系$k$位学生参加($n$和$k$都是固定的正整数),假设李老师和张老师分别将各自活动通知的信息独立、随机地发给该系$k$位学生,且所发信息都能收到,记该系收到李老师或张老师所发活动信息的学生人数为$X$.
\begin{enumerate}
\item 求该系学生甲收到李老师或张老师所发活动通知信息的概率;

\item 求使$P(X=m)$取得最大值的整数$m$.
\end{enumerate}

\item (2010年江西)

\item (2010年江苏)已知$\triangle ABC$的三边长为有理数.
证明:
\begin{enumerate}
\item $\cos A$为有理数;
\item $\cos nA$为有理数.
\end{enumerate}

\item (2011年浙江)设函数$f(x)=(x-a)^2\ln x,a\in \mathbb{R}$. 
\begin{enumerate}
\item 若$x=e$为$y=f(x)$的极值点,求实数$a$;

\item 求实数$a$的取值范围,使得对任意的$x\in (0,3a]$,恒有$f(x)\leq 4e^2$成立. 注: $e$为自然对数的底数.
\end{enumerate}

\item (2018年浙江)

\item (2009年江西)

\item (2004年江苏)已知函数$f(x)\, (x\in\mathbb{R})$满足下列条件:对任意的实数$x_1,x_2$都有
\[\lambda (x_1-x_2)^2\leq (x_1-x_2)[f(x_1)-f(x_2)]\]
和
\[|f(x_1)-f(x_2)|\leq |x_1-x_2|,\]
其中$\lambda$是大于$0$的常数.设实数$a_0,a,b$满足$f(a_0)=0$和$b=a-\lambda f(a)$.
\begin{enumerate}
\item 证明$\lambda\leq 1$,并且不存在$b_0\neq a_0$,使得$f(b_0)=0$;

\item 证明$(b-a_0)^2\leq (1-\lambda^2)(a-a_0)^2$;

\item 证明: $[f(b)]^2\leq (1-\lambda^2)[f(a)]^2$.
\end{enumerate}

\item (2006年江苏)设数列$\{a_n\},\{b_n\},\{c_n\}$满足: $b_n=a_n-a_{n+2},c_n=a_n+2a_{n+1}+3a_{n+2}\,(n=1,2,3,\cdots)$.

证明: $\{a_n\}$为等差数列的充分必要条件是$\{c_n\}$为等差数列,且 $b_n\leq b_{n+1}\,(n=1,2,3,\cdots)$. (听闻此题当年全江苏省只有几十位考生拿到一半以上的分数,只有不到10人拿满分)

 

\item (2011年江苏)设$M$为部分正整数组成的集合,数列$\{a_n\}$的首项$a_1=1$,前$n$项的和为$S_n$.已知对任意的整数$k\in M$,当整数$n>k$时, $S_{n+k}+S_{n-k}=2(S_n+S_k)$都成立.
\begin{enumerate}
\item 设$M=\{1\},a_2=2$,求$a_5$的值;

\item 设$M=\{3,4\}$,求数列$\{a_n\}$的通项公式.
\end{enumerate}

这个题目的结论可以推广到:

数列$\{a_n\}$对互素的$k_1,k_2\in \mathbb{N}^\ast,k_1>k_2>0$满足:
\[a_{n+k_1}+a_{n-k_1}=2a_n\,(n>k_1),\quad a_{n+k_2}+a_{n-k_2}=2a_n\,(n>k_2),\]
则此时数列$\{a_n\}$是等差数列.

\item (2015年江苏)设$a_1,a_2,a_3,a_4$是各项为正数且公差为$d\, (d\neq 0)$的等差数列.
\begin{enumerate}
\item 证明: $2^{a_1},2^{a_2},2^{a_3},2^{a_4}$依次成等比数列;

\item 是否存在$a_1,d$,使得$a_1,a_2^2,a_3^3,a_4^4$依次成等比数列,并说明理由;

\item 是否存在$a_1,d$以及正整数$n,k$,使得$a_1^n,a_2^{n+k},a_3^ {n+2k},a_4^{n+3k}$依次成等比数列,并说明理由.
\end{enumerate}

\item (2012年安徽)数列$\{x_n\}$满足$x_1=0,x_{n+1}=-x_n+x_n+c\, (n\in \mathbb{N}^\ast)$.
\begin{enumerate}
\item 证明: $\{x_n\}$是递减数列的充分必要条件是$c<0$;
\item 求$c$的取值范围,使$\{x_n\}$是递增数列.
\end{enumerate}

\item (2010年广东)设$A(x_1,y_1),B(x_2,y_2)$是平面直角坐标系$xOy$上的两点,现定义由点$A$到点$B$的一种折线距离$\rho (A,B)$为$p(A,B)=|x_2-x_1|+|y_2-y_1|$.对于平面$xOy$上给定的不同的两点$A(x_1,y_1),B(x_2,y_2)$.
\begin{enumerate}
\item 若点$C(x,y)$是平面$xOy$上的点,试证明$\rho (A,C)+\rho(C,B)\geq \rho(A,B)$;

\item 在平面$xOy$上是否存在点$C(x,y)$同时满足

\ding{172} $\rho (A,C)+\rho(C,B)\geq \rho(A,B)$; \qquad \ding{173} $\rho (A,C)=\rho(C,B)$.

若存在,请求出所有符合条件的点,请予以证明.
\end{enumerate}

\item (2009年湖南)

\item (2015年广东)调和数列,类似2014年陕西.

\item (2014年辽宁)已知函数$f(x)=(\cos x-x)(\pi+2x)-\frac{8}{3}(\sin x+1)$, $g(x)=3(x-\pi)\cos x-4(1+\sin x)\ln\left(3-\frac{2x}{\pi}\right)$.
证明:
\begin{enumerate}
\item 存在唯一$x_0\in \left(0,\frac{\pi}{2}\right)$,使$f(x_0)=0$;

\item 存在唯一$x_1\in \left(\frac{\pi}{2},\pi\right)$,使$g(x_1)=0$,且对(1)中的$x_0$有$x_0+x_1<\pi$.
\end{enumerate}

\item (2019年北京高考)已知函数 $f(x)=\frac{1}{4} x^{3}-x^{2}+x$.
\begin{enumerate}%[itemsep=-0.3em,topsep=0pt,labelsep=.5em,leftmargin=1.7em]
\item 求曲线$y=f(x)$的斜率为1的切线方程;

\item 当$x\in [-2,4]$时,求证: $x-6\leqslant f(x)\leqslant x$;

\item 设$F(x)=|f(x)-(x+a)|\, (a \in \mathbb{R})$,记$F(x)$在区间$[-2,4]$上的最大值为$M(a)$.当$M(a)$最小时,求$a$的值.
\end{enumerate}
\begin{solution}
\textbf{解法一.} (III)由(II)知,当$a<-3$时, $M(a)\geq F(0)=|g(0)-a|=-a>3$;

当$a>-3$时, $M(a)\geq F(-2)=|g(-2)-a|=6+a>3$;

当$a=-3$时, $M(a)=3$.

综上,当$M(a)$最小时, $a=-3$.

\textbf{解法二.}由$F(-2)=|a+6|,F(4)=|a|$,则
\[
M\left( a \right) \ge \frac{F\left( -2 \right) +F\left( 4 \right)}{2}\ge \frac{\left| a+6+\left( -a \right) \right|}{2}=3,
\]
当$a=-3$时取等号成立.
\end{solution}

\textcolor{red}{切比雪夫最佳逼近线:}求形如函数$F(x)=|f(x)-ax-b|$的最大值的最小值问题.

定义集合$A=\{g(x)=ax+b|a,b\in \mathbb{R}\}$,若存在函数 $g_0(x)\in A$使得对任意$g(x)\in A$,都满足
\[\max_{m\leq x\leq n}|f(x)-g_0(x)|\leq \min_{a,b\in \mathbb{R}}\max_{m\leq x\leq n}|f(x)-g(x)|,\]
则称$g_0(x)$为$f(x)$的最佳逼近直线.

\item 设函数$f(x)=|x^3-6x^2+ax+b|$,若对任意的实数$a$和$b$,总存在$x_0\in [0,3]$,使得$f(x_0)\geq m$,求实数$m$的最⼤值. 来源：2020 年 1 ⽉清华⼤学 THUSSAT 测试（理）

考虑问题的反⾯,我们选择合适的$a,b$,使得函数$f(x)$在$[0,3]$上的最⼤值尽可能的小,这个最大值的最小值即所求.

令$g(x)=x^3-6x^2$,则$g'(x)=3x^2-12x$,而经过两端点$(0,g(0))$和$(3,g(3))$的直线斜率为$\frac{g(3)-g(0)}{3-0}=-9$.令$g'(x)=3x^2-12x=-9$,解得$x=1$.

又$f(0)=|b|,f(3)=|-27+3a+b|,f(1)=|-5+a+b|$,于是
\[\frac{f(3)+3f(1)+2f(0)}{6}\geq \frac{|(-27+3a+b)-3(-5+a+b)+2b|}{6}=2.\]
这就证明了$f(x)$的最⼤值的最⼩值不⼩于$2$.

\item (2019年北京高考)已知抛物线$C:x^2=-2py$经过点$(2,-1)$.
\begin{enumerate}%[itemsep=-0.3em,topsep=0pt,labelsep=.5em,leftmargin=1.7em]
\item 求抛物线$C$的方程及其准线方程;
\item 设$O$为原点,过抛物线$C$的焦点作斜率不为$0$的直线$l$交抛物线$C$于两点$M,N$,直线$y=-1$分别交直线$OM,ON$于点$A$和点$B$.求证:以$AB$为直径的圆经过$y$轴上的两个定点.
\end{enumerate}

以$AB$为\textcolor{red}{直径}的圆$\to$两直线\textcolor{red}{垂直}$\to$ 向量$\overrightarrow{DA}\cdot \overrightarrow{DB}=0$或斜率之积等于$-1$.

\item (2017年北京高考)已知函数$f(x)=e^x\cos x-x$.
\begin{enumerate}
\item[(I)] 求曲线$y=f(x)$在点$(0,f(0))$处的切线方程;
\item[(II)] 求函数$f(x)$在区间$\left[0,\frac{\pi}{2} \right]$上的最大值和最小值.
\end{enumerate}

令导函数$f'(x)=e^x(\cos x-\sin x)-1=0$,得到无法求解的超越方程.

分析极值不一定要解方程$f'(x)=0$,\textcolor{red}{一阶导不够,二阶导来凑:}\,利用二阶导分析(一阶)导函数的单调性.

\item 已知函数$f(x)=e^x-ax^2\,(a\in \mathbb{R})$.
\begin{enumerate}
\item[(I)] 若曲线$y=f(x)$在$(1,f(1))$处的切线与$x$轴平行,求$a$;

\item[(II)] 已知$f(x)$在$[0,1]$上的最大值不小于$2$,求$a$的取值范围;

\item[(III)] 写出$f(x)$所有可能的零点个数及相应的$a$的取值范围.
\end{enumerate}

\item 设函数$f(x)=e^x-1-x-ax^2$.

(1)若$a=0$,求$f(x)$的单调区间;

(2)若当$x\geq 0$时$f(x)\geq 0$,求$a$的取值范围.

函数不等式: $e^x\geq x+1,\ln (x+1)\leq x$与$\sin x \leq x\leq \tan x$.

泰勒公式:
\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}+\cdots\]

\item 已知函数$f(x)=ax+\ln x+1$.

(1)讨论函数$f(x)$零点个数;

(2)对任意的$x>0,f(x)\leq xe^{2x}$恒成立,求实数$a$的取值范围.

设而不求思想:隐零点$x_0$满足$2x_0^2e^{2x_0}+\ln x_0=0$,则$0<x_0<1$.

\item (2020年西城一模)对于正整数$n$,如果$k\,(k\in \mathbb{N}^\ast)$个整数$a_1,a_2,\cdots,a_k$满足$1\leq a_1\leq a_2\leq \cdots\leq a_k\leq n$,且$a_1+a_2+\cdots+a_k=n$,则称数组$(a_1,a_2,\cdots,a_k)$为$n$的一个“正整数分拆”.记$a_1,a_2,\cdots,a_k$均为偶数的“正整数分拆”的个数为$f_n$, $a_1,a_2,\cdots,a_k$均为奇数的“正整数分拆”的个数为$g_n$.

(I)写出整数4的所有“正整数分拆”;

(II)对于给定的整数$n\,(n\geq 4)$,设$(a_1,a_2,\cdots,a_k)$是$n$的一个“正整数分拆”
且$a_1=2$,求$k$的最大值;

(III)对所有的正整数$n$,证明: $f_n<g_n$;并求出使得等号成立的$n$的值.

(注:对于$n$的两个“正整数分拆”$(a_1,a_2,\cdots,a_k)$与$(b_1,b_2,\cdots,b_m)$,当且仅当$k=m$且$a_1=b_1,a_2=b_2,\cdots,a_k=b_m$时,称这两个“正整数分拆”是相同的.)

\item (2012年海淀二模)将一个正整数$n$表示为$a_1+a_2+\cdots+a_p\,(p\in \mathbb{N}^\ast)$的形式,其中$a_i\in \mathbb{N}^\ast,i=1,2,\cdots,p$,且$a_1\leq a_2\leq \cdots\leq a_p$,记所有这样的表示法的种数为$f(n)$(如$4=4,4=1+3,4=2+2,4=1+1+2,4=1+1+1+1$,故$f(4)=5$).

(1)写出$f(3),f(5)$的值,并说明理由;

(2)对任意正整数$n$,比较$f(n+1)$与$\frac{1}{2}[f(n)+f(n+2)]$的大小,并给出证明;

(3)当正整数$n>6$时,求证: $f(n)\geq 4n-13$.

\item (2020年西城一模)设函数
\[\begin{cases}
x^2+10x+1, & x\leq 0 \\
|\lg x|, & x>0
\end{cases}\]
若关于$x$的方程$f(x)= a\,(a\in \mathbb{R})$有四个实数解$x_i\,(i=1,2,3,4)$,其中$x_1<x_2<x_3<x_4$.则$(x_1+x_2)(x_3-x_4)$的取值范围是

(A) $(0,101]$\qquad (B) $(0,99]$ \qquad (C) $(0,100]$ \qquad (D) $(0,+\infty)$

\item

\item

\item

\item

\item

\item

\item

\item
\end{enumerate}

设$f(x)=e^{mx-1}-\frac{\ln x}{x}$,若$f(x)$的最小值为$m$,求$m$的最小值.

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%https://wenku.baidu.com/view/1bd3fb6648d7c1c708a145b2.html?sxts=1566951843376

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试举几例：

2010年北京高考数学压轴题命题背景是纠错码理论中的Plotkin上界（有所改编）;

2014年北京高考数学压轴题命题背景是多工序流水线最优化排序问题中的Johnson法则;

2015年北京高考数学压轴题命题背景有限交换环上迭代图的一个特例;

两篇参考文献:

1.北京高考数学压轴题的教学实践与反思，《数学通报》杂志2017年第1期pp45-pp48（截图见下方）;

2.再谈2015年北京高考数学压轴题与“数学黑洞”问题,中学数学杂志, 2017年第1期

3.从高中数学试题到纠错码理论,李启超,荣贺
%作者：饮冰
%链接：https://www.zhihu.com/question/57845091/answer/155345007

\section{导数压轴题}

\subsection{参变分离}

\subsection{导数不等式}
%https://zhuanlan.zhihu.com/p/91032042
%https://zhuanlan.zhihu.com/p/51584482

\begin{theorem}{函数不等式链}{1}
当$x\geq 0$时,
\[
\frac{x}{x+1}\le \frac{2x}{x+2}\le \ln \left( x+1 \right) \le \frac{1}{2}\left( x+1-\frac{1}{x+1} \right) \le x.
\]
\end{theorem}

\subsection{设而不求:隐零点}

\subsection{极值点偏移}
%https://zhuanlan.zhihu.com/p/32842987

\begin{theorem}{指、对数平均不等式}{zdpjz}
当实数$a\neq b$时,有
\[e^{\frac{a+b}{2}}<\frac{e^a-e^b}{a-b}<\frac{e^a+e^b}{2}.\]
且
\[\sqrt{ab}<\frac{a-b}{\ln a-\ln b}<\frac{a+b}{2}.\]
\end{theorem}

\begin{example}
(2013年陕西)已知函数$f(x)=e^x,x\in \mathbb{R}$.

(1)若直线$y=kx+1$与$f(x)$的反函数的图像相切,求实数$k$的值;

(2)设$x>0$,讨论曲线$y=f(x)$与曲线$y=mx^2\,(m>0)$公共点的个数;

(3)设$a<b$,比较$\frac{f(a)+f(b)}{2}$与$\frac{f(b)-f(a)}{b-a}$的大小,并说明理由.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2016新课标1)已知函数$f(x)=(x-2)e^x+a(x-1)^2$有两个零点.

(I)求$a$的取值范围;

(II)设$x_1,x_2$是$f(x)$的两个零点,证明: $x_1+x_2<2$.
\end{example}
\begin{solution}

\end{solution}

\begin{example}
(2018皖南八校第三次联考理科数学)
已知函数$f(x)=e^x-x^2-ax$有两个极值点$x_1,x_2\,(x_1<x_2)$.

(1)求$a$的取值范围;

(2)求证: $e^{x_1}+e^{x_2}>4$.
\end{example}
\begin{solution}
\[\frac{e^{x_1}-e^{x_2}}{x_1-x_2}=2<\frac{e^{x_1}+e^{x_2}}{2}.\]
\end{solution}

\begin{theorem}{Hermite-Hadamard不等式}{hhbds}
若函数$f(x)$在$[a,b]$上的二阶导数非负,则有:
\[
f\left( \frac{a+b}{2} \right) \le \frac{1}{b-a}\int_a^b{f\left( x \right) dx}\le \frac{f\left( a \right) +f\left( b \right)}{2},
\]
当且仅当$f(x)$是一次函数时取等号成立.
\end{theorem}

\begin{example}
(匈牙利, 1914)设$f(x)=ax^2+bx+c$, $a,b,c$为实数,如果对于所有适合$-1\leq x\leq 1$的$x$值,都有$-1\leq f(x)\leq 1$成立,则对这些$x$的值有$-4\leq 2ax+b\leq 4$.
\end{example}

此题的背景是切比雪夫多项式的马尔科夫定理:如果具有实系数的$n$次多项式
\[f(x)=a_0+a_1x+a_2x^2+\cdots+ a_nx^n\]
对所有的$-1\leq x\leq 1$满足不等式
\[-1\leq f(x)\leq 1.\]
那么它的导函数满足不等式
\[-n^2\leq f'(x)\leq n^2.\]

虽然背景是高等的,但解法只用到一次函数$g(x)=2ax+b$的单调性、取值的技巧和不等式的放缩运算.
\begin{solution}
$12$.
\end{solution}
%切比雪夫多项式的马尔科夫定理，https://wenku.baidu.com/view/d0c9e2bbfd0a79563c1e720f.html

%https://zhuanlan.zhihu.com/p/105766114

\subsection{切比雪夫多项式}
%https://zhuanlan.zhihu.com/p/105766114

利用三角函数$n$倍角公式
\begin{align*}
\cos(0)&=1,\\
\cos(x)&=\cos x,\\
\cos(2x)&=2\cos^2 x-1,\\
\cos(3x)&=4\cos^3 x-3\cos x,\\
\cos(4x)&=8\cos^4 x-8\cos^2 x+1,\\
\cos(5x)&=16\cos^5 x-20\cos^3 x+5\cos x,\\
\end{align*}
可知$\cos (n\theta)$可以表示成$\cos\theta$的多项式, $T_n(x)=\cos(n\cdot \arccos x)$是一个$n$次多项式,称为$n$次切比雪夫多项式,其中$x\in [-1,1],n\in \mathbb{N}$.于是
\begin{align*}
T_0(x) &=1,\\
T_1(x) &=x,\\
T_2(x) &=2x^2-1,\\
T_3(x) &=4x^3-3x,\\
T_4(x) &=8x^4-8x^2+1,\\
T_5(x) &=16x^5-20x^3+5x,\\
\end{align*}

性质1. $T_n(x)$在$[-1,1]$中有$n$个不同的实根$x_k=\cos\frac{(2k-1)\pi}{2n},k=1,2,3,\cdots,n$.

性质2. $T_n(x)$在$[-1,1]$中有$n+1$个点$x_k^\ast=\cos\frac{k\pi}{n},k=0,1,2,3,\cdots,n$,轮流取最大值$1$和最小值$-1$.例如:当$n=2$时, $x_k^\ast=-1,0,1$.当$n=3$时, $x_k^\ast=-1,-\frac{1}{2},\frac{1}{2},1$.

性质3. $T_n(x)$满足递推关系$T_0(x)=1,T_1(x) =x$,
\[T_{n+1}(x)=2xT_n(x)-T_{n-1}(x),\]
其母函数为
\[\sum_{n=0}^{\infty}T_n(x)t^n=\frac{1-tx}{1-2tx+t^2}.\]

定理.对任意$n$次首一多项式$P(x)$,设$M=\max_{x\in[-1,1]}|P(x)|$,则$M_{\min}=\frac{1}{2^{n-1}}$.

证明.引理:设$n$次首一多项式$Q(x)$的$n$个根$\alpha_1,\alpha_2,\cdots,\alpha_n$均属于$(-1,1)$.在$[-1,\alpha_1),(\alpha_1,\alpha_2),\cdots,
(\alpha_{n-1},\alpha_n),(\alpha_n,1]$内各取一点$\beta_0,\beta_1,\cdots,\beta_n$,则对任意首一多项式$R(x)$,均有
\[\max_ {x\in[-1,1]}|R(x)|\geq \min_ {0\leq i\leq n}|Q(\beta_i)|.\]

引理的证明: (反证法)设存在$R(x)$使得
\[\max_ {x\in[-1,1]}|R(x)|< \min_ {0\leq i\leq n}|Q(\beta_i)|\triangleq C.\]
于是$R(x)\in (-C,C),\forall x\in [-1,1]$.

考虑$T(x)=R(x)-Q(x)$,则数列$T(\beta_0),T(\beta_1),\cdots,T(\beta_n)$必定正负交错(如图),则$T_n$有至少$n$个根.

然而$R(x),Q(x)$均为首一多项式,故$T(x)\equiv 0$.则$R(x)=Q(x)$,显然矛盾.

回到原题.设$T_n(x)$为$n$次切比雪夫多项式,令$Q(x)=\frac{1}{2^{n-1}}T_n(x)$,则$Q(x)$的各零点$\alpha_i=\cos\frac{(2i-1)\pi}{n}(i=1,2,\cdots,n)$均属于$(-1,1)$.

在引理中取$\beta_i=\cos\frac{i\pi}{n}(i=0,1,\cdots,n)$,即得$M\geq \frac{1}{2^{n-1}}$,当$P(x)\equiv Q(x)$时可取等.

\begin{theorem}{}{}
设$f(x)$为一个$n$次多项式,首项为$ax^n$,定义域为$D$,值域为$I$,用$|D|$表示$D$的区间长度,则$\frac{|I|}{2}\geq 2^{1-2n}\cdot |a|\cdot |D|^n$.事实上,等号成立时, $\frac{|I|}{2}$也就是$|f(x)|_{\max}$的最小值.等号成立的条件为$f(x)$经过平移及伸缩变换使得定义域为$D$的$T_n$.
\end{theorem}

\begin{solution}
我们用$[a,b]$表示定义域,这样$|D|=b-a$.当$a=-1,b=1$时,我们已证明了多项式$T_n(x)$的范数为$\frac{1}{2^{n-1}}$.为了求出它在任意区间$[a,b]$上的范数,必须采用把区间$a\leq y\leq b$映射到区间$-1\leq x\leq 1$的线性变换$x=\frac{2}{b-a}y-\frac{a+b}{b-a}$.此时我们得到多项式
\[
p\left( y \right) =T_n\left( \frac{2}{b-a}y-\frac{a+b}{b-a} \right) =\left( \frac{2}{b-a}y-\frac{a+b}{b-a} \right) ^n+\cdots
\]
它的最高次项系数非$1$而为$\frac{2^n}{(b-a)^n}$.把$p(y)$用这个数来除,我们得到在区间$[a,b]$上的切比雪夫多项式
\[
\widehat{T}_n\left( y \right) =\frac{\left( b-a \right) ^n}{2^n}T_n\left( \frac{2}{b-a}y-\frac{a+b}{b-a} \right).
\]
它的最高项系数已为$1$了.易见,它的范数等于
\[
\lVert \widehat{T}_n\left( y \right) \rVert =\frac{\left( b-a \right) ^n}{2^n}\lVert T_n\left( y \right) \rVert =\frac{\left( b-a \right). ^n}{2^{2n-1}}
\]
最后乘上首项的系数$a$,我们便得到了
\[\frac{|I|}{2}\geq 2^{1-2n}\cdot |a|\cdot |D|^n.\]
\end{solution}

对于切比雪夫最佳逼近直线,有如下常用结论:
\begin{theorem}{切比雪夫最佳逼近直线理论}{}
若函数$f(x)$在区间$[m,n]$上具有二阶导数,且$f''(x)$在区间$[m,n]$上不变号,则$f(x)$的最佳逼近直线为
\[
g\left( x \right) =k\left( x-\frac{m+c}{2} \right) +\frac{f\left( m \right) +f\left( c \right)}{2},
\]
其中$k=\frac{f\left( m \right) -f\left( n \right)}{m-n}$,实数$c$的值由方程$f'(c)=\frac{f\left( m \right) -f\left( n \right)}{m-n}$解得.
\end{theorem}

\subsection{切比雪夫最佳逼近直线}

%https://zhuanlan.zhihu.com/p/130443282

%https://zhuanlan.zhihu.com/p/105766114

\begin{example}
1
\end{example}

\begin{solution}
1
\end{solution}

\begin{example}
(2017年全国高中数学联赛第9题)设$k,m$为实数,不等式$|x^2-kx-m|\leq 1$对所有$x\in [a,b]$成立.证明: $b-a\leq 2\sqrt{2}$.
\end{example}
\begin{solution}
令$f(x)=x^2-kx-m,x\in [a,b]$,则$f(x)\in [-1,1]$.于是
\begin{align*}
f\left( a \right) &=a^2-ka-m\le 1, \tag*{\ding{172}}\\
f\left( b \right) &=b^2-kb-m\le 1,\tag*{\ding{173}}\\
f\left( \frac{a+b}{2} \right) &=\left( \frac{a+b}{2} \right) ^2-k\left( \frac{a+b}{2} \right) -m\ge -1,\tag*{\ding{174}}
\end{align*}
由\ding{172}$+$\ding{173}$-2\times$\ding{174}可知
\[
\frac{\left( a-b \right) ^2}{2}=f\left( a \right) +f\left( b \right) -2f\left( \frac{a+b}{2} \right) \le 4,
\]
故$b-a\leq 2\sqrt{2}$.
\end{solution}

\begin{example}
(2018年全国高中数学联赛B卷二试)设$a,b$是实数,函数$f(x)=ax+b+\frac{9}{x}$.证明:存在$x_0\in [1,9]$,使得$|f(x_0)|\geq 2$.
\end{example}
\begin{solution}
\textbf{证法1.}只需证明存在$u,v\in [1,9]$,满足$|f(u)-f(v)|\geq 4$,进而由绝对值不等式得
\[|f(u)|+|f(v)|\geq |f(u)-f(v)|\geq 4,\]
故$|f(u)|\geq 2$与$|f(v)|\geq 2$中至少有一个成立.

当$a\in \left( -\infty ,\frac{1}{2} \right] \cup \left[ \frac{3}{2},+\infty \right)$时,有
\[|f(1)-f(9)|=|(a+b+9)-(9a+b+1)|=8|1-a|\geq 4.\]

当$\frac{1}{2}<a<\frac{3}{2}$时,有$\frac{3}{\sqrt{a}}\in [1,9]$.再分两种情况:若$\frac{1}{2}<a<1$,则
\[
\left| f\left( 1 \right) -f\left( \frac{3}{\sqrt{a}} \right) \right|=\left| \left( a+b+9 \right) -\left( 6\sqrt{a}+b \right) \right|=\left( 3-\sqrt{a} \right) ^2\geq 4.
\]
若$1<a<\frac{3}{2}$,则
\[
\left| f\left( 9 \right) -f\left( \frac{3}{\sqrt{a}} \right) \right|=\left| \left( 9a+b+1 \right) -\left( 6\sqrt{a}+b \right) \right|=\left( 3\sqrt{a}-1 \right) ^2\geq 4.
\]
综上可知,存在$u,v\in [1,9]$,满足$|f(u)-f(v)|\geq 4$,从而命题得证.

\textbf{证法2.}用反证法.假设对任意$x\in [1,9]$,均有$|f(x)|<2$,则
\[|f(1)|<2,\qquad |f(3)|<2,\qquad |f(9)|<2.\]
易知
\begin{align*}
f(1)&=a+b+9, \tag*{\ding{172}} \\
f(3)&=3a+b+3, \tag*{\ding{173}} \\
f(9)&=9a+b+1. \tag*{\ding{174}}.
\end{align*}
由\ding{172},\ding{173}得$2a-6=f(2)-f(1)$,又由\ding{173},\ding{174}得$6a-2=f(3)-f(2)$.

由上述两式消去$a$,可知
\[f(3)-4f(2)+3f(1)=(6a-2)-3\cdot (2a-6)=16.\]
但$f(3)-4f(2)+3f(1)<2+4\cdot 2+3\cdot 2=16$,矛盾!从而命题得证.
\end{solution}

\begin{example}
(2015年北京大学自主招生试题第3题)已知$|x^2+px+q|\leq 2$对任意$x\in [1,5]$成立,则不超过$\sqrt{p^2+q^2}$的最大整数是\underline{\hspace{2cm}}.
\end{example}

\begin{solution}
1
\end{solution}

\begin{example}
已知函数$f(x)=\left|x+\frac{1}{x}-ax-b\right|\,(a,b\in \mathbb{R})$,当$x\in \left[\frac{1}{2},2\right]$时,设$f(x)$的最大值为$M(a,b)$,则$M(a,b)$的最小值为\underline{\hspace{2cm}}.
\end{example}
\begin{solution}
设$M=f_{\max}(x)$,则
\[\begin{cases}
M\ge f\left( \frac{1}{2} \right) =\left| \frac{5}{2}-\frac{a}{2}-b \right|=\frac{1}{2}\left| a+2b-5 \right|,
\\
M\ge f\left( 1 \right) =\left| 2-a-b \right|,
\\
M\ge f\left( 2 \right) =\left| \frac{5}{2}-2a-b \right|=\frac{1}{2}\left| 4a+2b-5 \right|,
\end{cases}\]
则
\begin{align*}
2M &\ge \frac{2}{3}f\left( \frac{1}{2} \right) +f\left( 1 \right) +\frac{1}{3}f\left( 2 \right)
\\
&=\frac{1}{3}\left| a+2b-5 \right|+\left| 2-a-b \right|+\frac{1}{6}\left| 4a+2b-5 \right|
\\
&\ge \left| 2-\frac{5}{3}-\frac{5}{6} \right|=\frac{1}{2},
\end{align*}
于是$M(a,b)$的最小值为$\frac{1}{2}$.
\end{solution}

\begin{example}
(2015年浙江高考)已知函数$f（x）=x^2-+ax+b\,(a,b\in \mathbb{R})$,记$M(a,b)$是$|f(x)|$在区间$[-1,1]$上的最大值.

(1)证明:当$|a|\geq 2$时, $M(a,b)\geq 2$;

(2)当$a,b$满足$M(a,b)\leq 2$,求$|a|+|b|$的最大值.
\end{example}
\begin{solution}
1
\end{solution}

\begin{example}
(2015年1月浙江省学业水平考试第34题)设函数$f(x)=\left|\sqrt{x}-ax-b\right|,a,b\in \mathbb{R}$.

(1)当$a=0,b=1$时,写出函数$f(x)$的单调区间;

(2)当$a=\frac{1}{2}$时,记函数$f(x)$在区间$[0,4]$上的最大值为$g(b)$,当$b$变化时,求$g(b)$的最小值;

(3)若对任意实数$a,b$,总存在实数$x_0\in [0,4]$使得不等式$f(x_0)\geq m$成立,求实数$m$的取值范围.
\end{example}

\begin{solution}
(1)单调递减区间是$(0,1)$,单调递增区间是$(1,+\infty)$;

(2)在区间$[0,4]$上$-b\leq h(x)\leq h(1)=\frac{1}{2}-b$.

当$b\leq \frac{1}{4}$, $f(x)$在区间$[0,4]$上的最大值为$g(b)=\frac{1}{2}-b$;

当$b> \frac{1}{4}$, $f(x)$在区间$[0,4]$上的最大值为$g(b)=b$.

当$b=\frac{1}{4}$时, $g(b)$取得最小值$\frac{1}{4}$.

(3)在$[0,4]$上, $-b\leq u(x)\leq 2-4a-b$.

当$b\leq 1-2a$时, $M(b)=2-4a-b$,当$b>1-2a$时, $M(b)=b$.

从而当$a\leq \frac{1}{4}$, $b=1-2a$时$M(b)$取最小值, $M(b)_{\min}=1-2a\geq \frac{1}{2}$;当$a> \frac{1}{4}$, $u(x)$在$\left[0,\frac{1}{4a^2}\right)$上单调递增,在$\left[\frac{1}{4a^2},4\right)$上单调递减.

在$a\in \left[\frac{1}{4},\frac{1}{2}\right]$, $-b\leq u(x)\leq \frac{1}{4a}-b$,当$b=\frac{1}{8a}$时, $M(b)_{\min}=\frac{1}{8a}\geq \frac{1}{4}$;

在$a\in \left(\frac{1}{2},+\infty\right)$, $2-4a-b\leq u(x)\leq \frac{1}{4a}-b$,当$b=1-2a+\frac{1}{8a}$时, $M(b)_{\min}=2a+\frac{1}{8a}-1>\frac{1}{4}$.

综上所述, $M(b)_{\min}=\frac{1}{4}$.

若对任意实数$a,b$,总存在实数$x_0\in [0,4]$使得不等式$f(x_0)\geq m$成立,等价于$m\leq f(x)_{\max}$恒成立, $m\leq \frac{1}{4}$.
\end{solution}

 

 

\begin{example}
(2016年天津高考)设函数$f(x)=x^3-ax-b,x\in \mathbb{R}$,其中$a$、$b\in \mathbb{R}$.

(1)求$f(x)$的单调区间;

(2)若$f(x)$存在极值点$x_0$,且$f(x_1)=f(x_0)$,其中$x_1\neq x_0$.求证: $x_1+2x_0=0$;

(3)设$a>0$,函数$g(x)= |f(x)|$,求证: $g(x)$在区间$[-1,1]$上的最大值不小于$\frac{1}{4}$.
\end{example}
%https://www.zhihu.com/question/345947963/answer/1046248071
\begin{solution}

\end{solution}

\begin{example}
求所有整数$a,b$,其中$|a|\leq 5,|b|\leq 5$,使得$x^4-3x^2-ax+b=0$恰有两个不同的整数解.
\end{example}
\begin{solution}
当$x\geq 0$时,有$x^4-3x^2=ax-b\leq 5x+5$,则$0\leq x\leq 2.43$;同理可知,当$x< 0$时,有$x^4-3x^2=ax-b\leq -5x+5$,
则$0>x>-2.43$.因此整数$x$只能取$-2,-1,0,1,2$.相应地,有$4+2a+b=0,-2+a+b=0,b=0,-2-a+b=0,4-2a+b=0$.

经检验,有$a=-2,b=0$,此时$x=-2,0,1$,矛盾; $a=0,b=-4$,此时$x=-2,1$,满足题意; $a=2,b=0$,此时$x=2,-1$,满足题意; $a=0,b=2$,此时整数$x=-1,1$也满足.

综上, $(a,b)=(0,-4),(2,0)$或$(0,2)$.
\end{solution}

\begin{example}
2.在一次宴会上,有10对夫妻参加,将所有男士安排在一个有10个座位的圆桌旁,所有女士安排在另外一张也是10个座位的圆桌旁,且每位女士的座位相对位置和她的配偶相同,我们发现新冠病毒在与会者之间传播,传播途径如下: $A$为一名健康与会者,当且仅当其座位两侧及其配偶三人间至少有两人感染的情况下, $A$才会被感染.设宴会开始时的$20$人中有$S$人感染,病毒在与会者中传播,则最后可能使所有与会者都感染上的$S$的最小值为多少?
\end{example}

\begin{example}
求$\frac{1}{4\times 1^4+1}+\frac{2}{4\times 2^4+1}+\frac{3}{4\times 3^4+1}+\cdots+\frac{100}{4\times 100^4+1}$的值.
\end{example}

\begin{example}
(2011年清华保送生)证明:对于任意的正整数$a$、$b$有
\[
\left( a,b \right) =\frac{1}{a}\sum_{m=0}^{a-1}{\sum_{n=0}^{a-1}{e^{\frac{2\pi imnb}{a}}}}.
\]
\end{example}
\begin{solution}
设$\mathrm{gcd}(a,b)=d,a=dx,b=dy,w=e^{2\pi i\frac{y}{x}}$,其中$a,b,d,x,y\in \mathbb{Z}_+$,则$w^x=1$.
\[
\frac{1}{a}\sum_{m=0}^{a-1}{\sum_{n=0}^{a-1}{e^{\frac{2\pi imnb}{a}}}}=\frac{1}{a}\sum_{m=0}^{a-1}{\sum_{n=0}^{a-1}{w^{mn}}}.
\]
注意到$\left( 1-w^m \right) \sum_{n=0}^{a-1}{w^{mn}}=1-w^{am}=0$.因此,当且仅当$m=0,x,2x,\cdots,(d-1)x$时, $w^m=1$,此时$\sum_{n=0}^{a-1}{w^{mn}}=a$.而当$m$取其他值时, $\sum_{n=0}^{a-1}{w^{mn}}=0$.

综上所述,所求结果为$\frac{1}{a}\cdot da=d$.
\end{solution}

设$n\geq 2$为正整数, $a,b$为正数.设$\frac{1}{n+1}\left(a^n+a^{n-1}b+\cdots+b^n\right)$与$\left(\frac{a+b}{2}\right)^n$的大小关系.

\end{document}

\begin{example}
This is the content of example environment.
\end{example}

 

\begin{exercise}\label{exer:43}
设 $f \notin\in L(\mathcal{R}^1)$，$g$ 是 $\mathcal{R}^1$ 上的有界可测函数。证明函数
\begin{equation}
\label{ex:1}
I(t) = \int_{\mathcal{R}^1} f(x+t)g(x)dx \quad t \in \mathcal{R}^1
\end{equation}
是 $\mathcal{R}^1$ 上的连续函数。
\end{exercise}

\begin{problem}
即 $D(x)$ 在 $[0,1]$ 上是 Lebesgue 可积的并且积分值为零。但 $D(x)$ 在 $[0,1]$ 上不是 Riemann 可积的。
\end{problem}

\begin{solution}
即 $D(x)$ 在 $[0,1]$ 上是 Lebesgue 可积的并且积分值为零。但 $D(x)$ 在 $[0,1]$ 上不是 Riemann 可积的。
\end{solution}

\begin{theorem}{Fubini 定理}{fubi}
（1）若 $f(x,y)$ 是 $\mathcal{R}^p\times\mathcal{R}^q$ 上的非负可测函数，则对几乎处处的 $x\in \mathcal{R}^p$，$f(x,y)$ 作为 $y$ 的函数是 $\mathcal{R}^q$ 上的非负可测函数，$g(x)=\int_{\mathcal{R}^q}f(x,y) dy$ 是 $\mathcal{R}^p$ 上的非负可测函数。并且
\begin{equation}
\label{eq:461}
\int_{\mathcal{R}^p\times\mathcal{R}^q} f(x,y) dxdy=\int_{\mathcal{R}^p}\left(\int_{\mathcal{R}^q}f(x,y)dy\right)dx.
\end{equation}
（2）若 $f(x,y)$ 是 $\mathcal{R}^p\times\mathcal{R}^q$ 上的可积函数，则对几乎处处的 $x\in\mathcal{R}^p$，$f(x,y)$ 作为 $y$ 的函数是 $\mathcal{R}^q$ 上的可积函数，并且 $g(x)=\int_{\mathcal{R}^q}f(x,y) dy$ 是 $\mathcal{R}^p$ 上的可积函数。而且~\ref{eq:461} 成立。
\end{theorem}

\begin{note}
在本模板中，引理（lemma），推论（corollary）的样式和定理~\ref{thm:fubi} 的样式一致，包括颜色，仅仅只有计数器的设置不一样。
\end{note}

我们说一个实变或者复变量的实值或者复值函数是在区间上平方可积的，如果其绝对值的平方在该区间上的积分是有限的。所有在勒贝格积分意义下平方可积的可测函数构成一个希尔伯特空间，也就是所谓的 $L^2$ 空间，几乎处处相等的函数归为同一等价类。形式上，$L^2$ 是平方可积函数的空间和几乎处处为 0 的函数空间的商空间。

\begin{proposition}{最优性原理}{max}
如果 $u^*$ 在 $[s,T]$ 上为最优解，则 $u^*$ 在 $[s, T]$ 任意子区间都是最优解，假设区间为 $[t_0, t_1]$ 的最优解为 $u^*$ ，则 $u(t_0)=u^{*}(t_0)$，即初始条件必须还是在 $u^*$ 上。
\end{proposition}

我们知道最小二乘法可以用来处理一组数据，可以从一组测定的数据中寻求变量之间的依赖关系，这种函数关系称为经验公式。本课题将介绍最小二乘法的精确定义及如何寻求点与点之间近似成线性关系时的经验公式。假定实验测得变量之间的 $n$ 个数据，则在平面上，可以得到 $n$ 个点，这种图形称为 “散点图”，从图中可以粗略看出这些点大致散落在某直线近旁, 我们认为其近似为一线性函数，下面介绍求解步骤。

\begin{figure}[htbp]
\centering
\includegraphics[width=0.6\textwidth]{scatter.pdf}
\caption{散点图示例 $\hat{y}=a+bx$ \label{fig:scatter}}
\end{figure}

以最简单的一元线性模型来解释最小二乘法。什么是一元线性模型呢？监督学习中，如果预测的变量是离散的，我们称其为分类（如决策树，支持向量机等），如果预测的变量是连续的，我们称其为回归。回归分析中，如果只包括一个自变量和一个因变量，且二者的关系可用一条直线近似表示，这种回归分析称为一元线性回归分析。如果回归分析中包括两个或两个以上的自变量，且因变量和自变量之间是线性关系，则称为多元线性回归分析。对于二维空间线性是一条直线；对于三维空间线性是一个平面，对于多维空间线性是一个超平面。

\begin{property}
柯西列的性质
\begin{enumerate}
\item $\{x_k\}$ 是柯西列，则其子列 $\{x_k^i\}$ 也是柯西列。
\item $x_k\in \mathcal{R}^n$，$\rho(x,y)$ 是欧几里得空间，则柯西列收敛，$(\mathcal{R}^n,\rho)$ 空间是完备的。
\end{enumerate}
\end{property}

\begin{conclusion}
回归分析（regression analysis) 是确定两种或两种以上变量间相互依赖的定量关系的一种统计分析方法。运用十分广泛，回归分析按照涉及的变量的多少，分为一元回归和多元回归分析；按照因变量的多少，可分为简单回归分析和多重回归分析；按照自变量和因变量之间的关系类型，可分为线性回归分析和非线性回归分析。
\end{conclusion}

\nocite{*}

\bibliography{reference}

\appendix
\chapter{基本数学工具}

本附录包括了计量经济学中用到的一些基本数学，我们扼要论述了求和算子的各种性质，研究了线性和某些非线性方程的性质，并复习了比例和百分数。我们还介绍了一些在应用计量经济学中常见的特殊函数，包括二次函数和自然对数，前 4 节只要求基本的代数技巧，第 5 节则对微分学进行了简要回顾；虽然要理解本书的大部分内容，微积分并非必需，但在一些章末附录和第 3 篇某些高深专题中，我们还是用到了微积分。

\section{求和算子与描述统计量}

\textbf{求和算子} 是用以表达多个数求和运算的一个缩略符号，它在统计学和计量经济学分析中扮演着重要作用。如果 $\{x_i: i=1, 2, \ldots, n\}$ 表示 $n$ 个数的一个序列，那么我们就把这 $n$ 个数的和写为：

\begin{equation}
\sum_{i=1}^n x_i \equiv x_1 + x_2 +\cdots + x_n
\end{equation}

\chapter{最小示例}

\begin{lstlisting}
\documentclass[lang=cn,11pt]{elegantbook}
% title info
\title{Title}
\subtitle{Subtitle is here}
% bio info
\author{Your Name}
\institute{XXX University}
\date{\today}
% extra info
\version{1.00}
\equote{Victory won\rq t come to us unless we go to it. --- M. Moore}
\logo{logo.png}
\cover{cover.jpg}

\begin{document}

\maketitle
\tableofcontents
\mainmatter
\hypersetup{pageanchor=true}
% add preface chapter here if needed
\chapter{Example Chapter Title}
The content of chapter one.

\bibliography{reference}

\end{document}
\end{lstlisting}
%114