可内接于立方体内的球体内接的最大直圆锥

给定一个边长为a 的立方体，它内接一个球体，该球体又内接一个直圆锥。任务是找到这个圆锥体的最大可能体积。
例子：

Input:  a = 5
Output: 58.1481

Input: a = 8
Output: 238.175

方法：
让，正圆锥的高度 = h 。
圆锥半径 = r
球体的半径 = R
我们知道立方体内球体的半径， r = a/2 。请参考（可内切立方体的最大球体）。
此外，球体内锥体的高度h = 4r/3 。
球内圆锥的半径， r = 2√2r/3 。请参阅（可内接在球体内的最大正圆锥）。
因此，球体内部锥体的高度又内接在立方体中， h = 2a/3 。
球体内锥体的半径，而球体又内接在立方体内， r = √2a/3 。
下面是上述方法的实现：

C++

// C++ Program to find the biggest right circular cone
// that can be inscribed within a right circular cone
// which in turn is inscribed within a cube
 
#include 
using namespace std;
 
// Function to find the biggest right circular cone
float cone(float a)
{
 
    // side cannot be negative
    if (a < 0)
        return -1;
 
    // radius of right circular cone
    float r = (a * sqrt(2)) / 3;
 
    // height of right circular cone
    float h = (2 * a) / 3;
 
    // volume of right circular cone
    float V = 3.14 * pow(r, 2) * h;
 
    return V;
}
 
// Driver code
int main()
{
    float a = 5;
    cout << cone(a) << endl;
 
    return 0;
}

Java

// Java Program to find the biggest right circular cone
// that can be inscribed within a right circular cone
// which in turn is inscribed within a cube
import java.io.*;
 
class GFG
{
     
// Function to find the biggest right circular cone
static float cone(float a)
{
 
    // side cannot be negative
    if (a < 0)
        return -1;
 
    // radius of right circular cone
    float r = (float) (a * Math.sqrt(2)) / 3;
 
    // height of right circular cone
    float h = (2 * a) / 3;
 
    // volume of right circular cone
    float V = (float)(3.14 *Math. pow(r, 2) * h);
 
    return V;
}
 
// Driver code
public static void main (String[] args)
{
    float a = 5;
    System.out.println( cone(a));
}
}
 
// This code is contributed by anuj_67..

Python3

# Python3 Program to find the biggest right
# circular cone that can be inscribed within
# a right circular cone which in turn is
# inscribed within a cube
import math
 
# Function to find the biggest
# right circular cone
def cone(a):
 
    # side cannot be negative
    if (a < 0):
        return -1;
 
    # radius of right circular cone
    r = (a * math.sqrt(2)) / 3;
 
    # height of right circular cone
    h = (2 * a) / 3;
 
    # volume of right circular cone
    V = 3.14 * math.pow(r, 2) * h;
 
    return V;
 
# Driver code
a = 5;
print(cone(a));
 
# This code is contributed by
# Shivi_Aggarwal

C#

// C# Program to find the biggest
// right circular cone that can be
// inscribed within a right circular cone
// which in turn is inscribed within a cube
using System;
 
class GFG
{
     
// Function to find the biggest
// right circular cone
static double cone(double a)
{
 
    // side cannot be negative
    if (a < 0)
        return -1;
 
    // radius of right circular cone
    double r = (double) (a * Math.Sqrt(2)) / 3;
 
    // height of right circular cone
    double h = (2 * a) / 3;
 
    // volume of right circular cone
    double V = (double)(3.14 * Math.Pow(r, 2) * h);
 
    return Math.Round(V,4);
}
 
// Driver code
static void Main ()
{
    double a = 5;
    Console.WriteLine(cone(a));
}
}
 
// This code is contributed by chandan_jnu

Javascript

 
// javascript Program to find the biggest right circular cone
// that can be inscribed within a right circular cone
// which in turn is inscribed within a cube
 
// Function to find the biggest right circular cone
function cone(a)
{
 
    // side cannot be negative
    if (a < 0)
        return -1;
 
    // radius of right circular cone
    var r =  (a * Math.sqrt(2)) / 3;
 
    // height of right circular cone
    var h = (2 * a) / 3;
 
    // volume of right circular cone
    var V = (3.14 *Math. pow(r, 2) * h);
 
    return V;
}
 
// Driver code
var a = 5;
document.write( cone(a).toFixed(5));
 
 
// This code is contributed by Amit Katiyar
 

输出：

58.1481

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