连接一个正 N 边多边形的边中点形成的两个嵌套多边形的面积比

给定一个N边多边形，任务是找到第N^个与^第(N + 1)^个N边正则嵌套多边形的面积之比，这些多边形是通过连接原始多边形的边中点生成的。

例子 ：

Input: N = 3
Output: 4.000000
Explanation:

Nested Triangle

Ratio of the length of the sides formed by joining the mid-points of the triangle with the length of the side of the original triangle is 0.5. Hence, R = (Area of N^th triangle) / (Area of (N + 1)^th triangle) = 4

Input: N = 4
Output: 2.000000

方法：该问题可以基于以下观察来解决：

• 考虑如下图所示的N边正多边形。

N 边嵌套正多边形的表示。

• A = 2 * ℼ / N
  B = ℼ / N
  h = r * cos(B)
  b = h * cos(B)
  c = h((1 – cos(A)) / 2) ^1/2
• 黑色等腰三角形的面积：

• 红色等腰三角形的面积：

• r = s / (2 * [1 – cos(2B)]) ^1/2和b = r * [cos(B)] ²
• 将上述等式合并后：

• 最终得到的结果如下：

下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to calculate the ratio of
// area of N-th and (N + 1)-th nested
// polygons formed by connecting midpoints
void AreaFactor(int n)
{
    // Stores the value of PI
    double pi = 3.14159265;
 
    // Calculating area the factor
    double areaf = 1 / (cos(pi / n)
                        * cos(pi / n));
 
    // Printing the ratio
    // precise upto 6 decimal places
    cout << fixed << setprecision(6)
         << areaf << endl;
}
 
// Driver Code
int main()
{
    int n = 4;
    AreaFactor(n);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to calculate the ratio of
// area of N-th and (N + 1)-th nested
// polygons formed by connecting midpoints
static void AreaFactor(int n)
{
     
    // Stores the value of PI
    double pi = 3.14159265;
  
    // Calculating area the factor
    double areaf = 1 / (Math.cos(pi / n) *
                        Math.cos(pi / n));
  
    // Printing the ratio
    // precise upto 6 decimal places
    System.out.format("%.6f", areaf);
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 4;
     
    AreaFactor(n);
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 code for the above approach
import math
 
# Function to calculate the ratio of
# area of N-th and (N + 1)-th nested
# polygons formed by connecting midpoints
def AreaFactor(n):
 
    # Stores the value of PI
    pi = 3.14159265
 
    # Calculating area the factor
    areaf = 1 / (math.cos(pi / n) *
                 math.cos(pi / n))
 
    # Printing the ratio
    # precise upto 6 decimal places
    print('%.6f' % areaf)
     
# Driver Code
if __name__ == "__main__":
   
    n = 4
    AreaFactor(n)
 
# This code is contributed by ukasp

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
// Function to calculate the ratio of
// area of N-th and (N + 1)-th nested
// polygons formed by connecting midpoints
static void AreaFactor(int n)
{
      
    // Stores the value of PI
    double pi = 3.14159265;
   
    // Calculating area the factor
    double areaf = 1 / (Math.Cos(pi / n) *
                        Math.Cos(pi / n));
   
    // Printing the ratio
    // precise upto 6 decimal places
    Console.WriteLine(Math.Round(areaf));
 
}
    // Driver Code
    public static void Main(string[] args)
    {
        int n = 4;
      
        AreaFactor(n);
    }
}
 
// This code is contributed by susmitakundugoaldanga.

Javascript

 
// Javascript program implementation
// of the approach
 
// Function to calculate the ratio of
// area of N-th and (N + 1)-th nested
// polygons formed by connecting midpoints
function AreaFactor(n)
{
    // Stores the value of PI
    let pi = 3.14159265;
  
    // Calculating area the factor
    let areaf = (1 / (Math.cos(pi / n)
                        * Math.cos(pi / n)));
  
    // Printing the ratio
    // precise upto 6 decimal places
    document.write(areaf.toFixed(6));
}
 
// Driver Code
     
        let n = 4;
    AreaFactor(n);
 
// This code is contributed by splevel62.

输出：

2.000000

时间复杂度： O(1)
辅助空间： O(1)

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