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 program for the above approach #include <bits/stdc++.h> using namespace std;
// Function returns the required distance double 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 program for the above approach import java.util.*;
class GFG{
// Function returns the required distance static double distance( int r, int R)
{ double d = Math.sqrt(Math.pow(R, 2 ) -
( 2 * r * R));
return d;
} // Driver code public 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 program for the above approach import math
# Function returns the required distance def distance(r,R):
d = math.sqrt( (R * * 2 ) - ( 2 * r * R))
return d
# Driver Code # Length of Inradius r = 2
# Length of Circumradius R = 5
print ( round (distance(r,R), 2 ))
|
// C# program for the above approach using System;
class GFG{
// Function returns the required distance static double distance( int r, int R)
{ double d = Math.Sqrt(Math.Pow(R, 2) -
(2 * r * R));
return d;
} // Driver code public 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 |
<script> // Javascript program for // the above approach // Function returns the required distance function 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> |
Output:
2.24
Time Complexity: O(logn) since time complexity of sqrt is O(logn)
Auxiliary Space: O(1)