Problem D

Trouble with a Pentagon

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

 

 

You are asked to place the largest possible square inside a regular pentagon (whose internal angles are same and all the sides are same in length). You are given the information that one vertex of the square will be coincident with a vertex of the square as shown in the figure below. You will have to find the length of a side of the square when a side of the regular pentagon is given.

 

Fig: Square in a pentagon.

Input

The input file contains several lines of input. Each line contains a floating point number F (0<=F<=100000) which indicates the length of a side of the pentagon. Input is terminated by end of file.

 

Output

For each line of input produce one line of output containing a floating point number with ten digits after the decimal point. This number indicates the largest possible side of a square that fits in the pentagon. This output will be judged with a special correction program, so don’t worry about small precision errors.

 

Sample Input

0.0000001

0.0000002

0.0000003 

 

Sample Output

0.0000001067

0.0000002135

0.0000003202

(World Finals Warm-up Contest, Problem Setter: Shahriar Manzoor)

 

#include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;

const double PI = 3.14159265359;

class TroubleWithAPentagon{
private:
    double length;
    double side;
public:
    void initialize(double l){length = l;}
    void computing(){side = length * sin(108 * PI / 180) / sin(63 * PI / 180);}
    void outResult(){cout << setiosflags(ios::fixed) << setprecision(10) << side << endl;}
};

int main(){
    TroubleWithAPentagon twap;
    double l;
    while(cin >> l){
        twap.initialize(l);
        twap.computing();
        twap.outResult();
    }
    return 0;
}