Given two integers r and R representing the length of Inradius and Circumradius respectively, the task is to calculate the distance d between Incenter and Circumcenter.
Inradius The inradius( r ) of a regular triangle( ABC ) is the radius of the incircle (having center as l), which is the largest circle that will fit inside the triangle.
Circumradius: The circumradius( R ) of a triangle is the radius of the circumscribed circle (having center as O) of that triangle.
Examples:
Input: r = 2, R = 5
Output: 2.24Input: r = 5, R = 12
Output: 4.9
Approach:
The problem can be solved using Euler’s Theorem in geometry, which states that the distance between the incenter and circumcenter of a triangle can be calculated by the equation:
Below is the implementation of the above approach:
C++14
// C++14 program for the above approach#include <bits/stdc++.h> using namespace std;// Function returns the required distancedouble distance(int r, int R) { double d = sqrt(pow(R, 2) - (2 * r * R)); return d; } // Driver code int main() { // Length of Inradius int r = 2; // Length of Circumradius int R = 5; cout << (round(distance(r, R) * 100.0) / 100.0); } // This code is contributed by sanjoy_62 |
Java
// Java program for the above approach import java.util.*;class GFG{ // Function returns the required distancestatic double distance(int r,int R){ double d = Math.sqrt(Math.pow(R, 2) - (2 * r * R)); return d;}// Driver codepublic static void main(String[] args){ // Length of Inradius int r = 2; // Length of Circumradius int R = 5; System.out.println(Math.round( distance(r, R) * 100.0) / 100.0);}}// This code is contributed by offbeat |
Python3
# Python3 program for the above approachimport math# Function returns the required distancedef distance(r,R): d = math.sqrt( (R**2) - (2 * r * R)) return d# Driver Code# Length of Inradiusr = 2# Length of CircumradiusR = 5print(round(distance(r,R),2)) |
C#
// C# program for the above approach using System;class GFG{ // Function returns the required distancestatic double distance(int r, int R){ double d = Math.Sqrt(Math.Pow(R, 2) - (2 * r * R)); return d;}// Driver codepublic static void Main(string[] args){ // Length of Inradius int r = 2; // Length of Circumradius int R = 5; Console.Write(Math.Round( distance(r, R) * 100.0) / 100.0);}}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript program for// the above approach// Function returns the required distancefunction distance(r, R){ let d = Math.sqrt(Math.pow(R, 2) - (2 * r * R)); return d;}// Driver code // Length of Inradius let r = 2; // Length of Circumradius let R = 5; document.write(Math.round( distance(r, R) * 100.0) / 100.0); // This code is contributed by susmitakundugoaldanga.</script> |
2.24
Time Complexity: O(logn) since time complexity of sqrt is O(logn)
Auxiliary Space: O(1)
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