Triángulo de Reuleaux más grande inscrito dentro de un cuadrado que está inscrito dentro de una elipse

Dada una elipse con longitud de eje mayor y eje menor 2a y 2b respectivamente que inscribe un cuadrado que a su vez inscribe un triángulo reuleaux. La tarea es encontrar el área máxima posible de este triángulo de Reuleaux.
Ejemplos: 
 

Input: a = 5, b = 4
Output: 0.0722389

Input: a = 7, b = 11
Output: 0.0202076

Planteamiento : Como, el lado del cuadrado inscrito dentro de una elipse es, x = √(a^2 + b^2)/ab. Consulte Área del cuadrado más grande que se puede inscribir en una elipse.  
Además, en el triángulo de Reuleaux, h = x = √(a^2 + b^2)/ab . 
Entonces, Área del triángulo de Reuleaux , A = 0.70477*h^2 = 0.70477*((a^2 + b^2)/a^2b^2) .
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the biggest reuleaux triangle
float Area(float a, float b)
{
 
    // length of the axes cannot be negative
    if (a < 0 && b < 0)
        return -1;
 
    // height of the reuleaux triangle
    float h = sqrt(((pow(a, 2) + pow(b, 2))
                    / (pow(a, 2) * pow(b, 2))));
 
    // area of the reuleaux triangle
    float A = 0.70477 * pow(h, 2);
 
    return A;
}
 
// Driver code
int main()
{
    float a = 5, b = 4;
    cout << Area(a, b) << endl;
 
    return 0;
}

// Java Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
import java.io.*;
 
class GFG
{
     
// Function to find the biggest reuleaux triangle
static float Area(float a, float b)
{
 
    // length of the axes cannot be negative
    if (a < 0 && b < 0)
        return -1;
 
    // height of the reuleaux triangle
    float h = (float)Math.sqrt(((Math.pow(a, 2) + Math.pow(b, 2))
                / (Math.pow(a, 2) * Math.pow(b, 2))));
 
    // area of the reuleaux triangle
    float A = (float)(0.70477 * Math.pow(h, 2));
 
    return A;
}
 
// Driver code
public static void main (String[] args)
{
    float a = 5, b = 4;
    System.out.println(Area(a, b));
}
}
 
// This code is contributed by anuj_67..

# Python3 Program to find the biggest Reuleaux
# triangle inscribed within in a square
# which in turn is inscribed within an ellipse
import math;
 
# Function to find the biggest
# reuleaux triangle
def Area(a, b):
 
    # length of the axes cannot
    # be negative
    if (a < 0 and b < 0):
        return -1;
 
    # height of the reuleaux triangle
    h = math.sqrt(((pow(a, 2) + pow(b, 2)) /
                   (pow(a, 2) * pow(b, 2))));
 
    # area of the reuleaux triangle
    A = 0.70477 * pow(h, 2);
 
    return A;
 
# Driver code
a = 5;
b = 4;
print(round(Area(a, b), 7));
 
# This code is contributed by chandan_jnu

// C# Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
using System;
 
class GFG
{
     
// Function to find the biggest reuleaux triangle
static double Area(double a, double b)
{
 
    // length of the axes cannot be negative
    if (a < 0 && b < 0)
        return -1;
 
    // height of the reuleaux triangle
    double h = (double)Math.Sqrt(((Math.Pow(a, 2) +
                                    Math.Pow(b, 2)) /
                                   (Math.Pow(a, 2) *
                                   Math.Pow(b, 2))));
 
    // area of the reuleaux triangle
    double A = (double)(0.70477 * Math.Pow(h, 2));
 
    return A;
}
 
// Driver code
static void Main()
{
    double a = 5, b = 4;
    Console.WriteLine(Math.Round(Area(a, b),7));
}
}
 
// This code is contributed by chandan_jnu

<?php
// PHP Program to find the biggest Reuleaux
// triangle inscribed within in a square
// which in turn is inscribed within an ellipse
 
// Function to find the biggest
// reuleaux triangle
function Area($a, $b)
{
 
    // length of the axes cannot
    // be negative
    if ($a < 0 && $b < 0)
        return -1;
 
    // height of the reuleaux triangle
    $h = sqrt(((pow($a, 2) + pow($b, 2)) /
               (pow($a, 2) * pow($b, 2))));
 
    // area of the reuleaux triangle
    $A = 0.70477 * pow($h, 2);
 
    return $A;
}
 
// Driver code
$a = 5;
$b = 4;
echo round(Area($a, $b), 7);
 
// This code is contributed by Ryuga
?>

<script>
 
// Javascript Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
 
// Function to find the biggest reuleaux triangle
function Area(a, b)
{
 
    // length of the axes cannot be negative
    if (a < 0 && b < 0)
        return -1;
 
    // height of the reuleaux triangle
    let h = Math.sqrt(((Math.pow(a, 2) + Math.pow(b, 2))
                    / (Math.pow(a, 2) * Math.pow(b, 2))));
 
    // area of the reuleaux triangle
    let A = 0.70477 * Math.pow(h, 2);
 
    return A;
}
 
// Driver code
    let a = 5, b = 4;
    document.write(Area(a, b) + "<br>");
     
// This code is contributed by Mayank Tyagi
 
</script>

0.0722389

Complejidad de tiempo: O (logn)

Espacio Auxiliar: O(1)