Given six integers representing the vertices of a triangle, say A(x1, y1), B(x2, y2), and C(x3, y3), the task is to find the coordinates of the excenters of the given triangle.
Excenter is a point where the bisector of one interior angle and bisectors of two external angle bisectors of the opposite side of the triangle, intersect. There are a total of three excenters in a triangle.
Examples:
Input: x1 = 0, y1 = 0, x2 = 3, y2 = 0, x3 = 0, y3 = 4
Output:
6 6
-3 3
2 -2
Explanation: The coordinates of the Excenters of the triangle are: (6, 6), (-3, 3), (2, -2)Input: x1 = 0, y1 = 0, x2 = 12, y2 = 0, x3 = 0, y3 = 5
Output:
15 15
-3 3
10 -10
Approach: The given problem can be solved by using the formula for finding the excenter of the triangles. Follow the steps below to solve the problem:
- Suppose the vertices of the triangle are A(x1, y1), B(x2, y2), and C(x3, y3).
- Let the length of the sides be AB, BC and AC be c, a and b respectively.
Therefore, the formula to find the coordinates of the Excenters of the triangle is given by:
I1 = { (-a*x1 + b*x2 + c*x3) / (-a + b + c ), (-a*y1 +b*y2 + c*y3 ) / (-a + b + c) }
I2 = { ( a*x1 – b*x2 + c*x3) / ( a – b + c ), ( a*y1 -b*y2 + c*y3 ) / (a – b + c) }
I3 = { ( a*x1 + b*x2 – c*x3 / (a + b – c ), ( a*y1 +b*y2 – c*y3) / (a + b – c) }
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to calculate the // distance between a pair of points float distance( int m, int n, int p, int q) { return sqrt ( pow (n - m, 2) + pow (q - p, 2) * 1.0); } // Function to calculate the coordinates // of the excenters of a triangle void Excenters( int x1, int y1, int x2, int y2, int x3, int y3) { // Length of the sides of the triangle float a = distance(x2, x3, y2, y3); float b = distance(x3, x1, y3, y1); float c = distance(x1, x2, y1, y2); // Stores the coordinates of the // excenters of the triangle vector<pair< float , float > > excenter(4); // Applying formula to find the // excenters of the triangle // For I1 excenter[1].first = (-(a * x1) + (b * x2) + (c * x3)) / (-a + b + c); excenter[1].second = (-(a * y1) + (b * y2) + (c * y3)) / (-a + b + c); // For I2 excenter[2].first = ((a * x1) - (b * x2) + (c * x3)) / (a - b + c); excenter[2].second = ((a * y1) - (b * y2) + (c * y3)) / (a - b + c); // For I3 excenter[3].first = ((a * x1) + (b * x2) - (c * x3)) / (a + b - c); excenter[3].second = ((a * y1) + (b * y2) - (c * y3)) / (a + b - c); // Print the excenters of the triangle for ( int i = 1; i <= 3; i++) { cout << excenter[i].first << " " << excenter[i].second << endl; } } // Driver Code int main() { float x1, x2, x3, y1, y2, y3; x1 = 0; x2 = 3; x3 = 0; y1 = 0; y2 = 0; y3 = 4; Excenters(x1, y1, x2, y2, x3, y3); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ static class pair { float first, second; pair( float first, float second) { this .first = first; this .second = second; } } // Function to calculate the // distance between a pair of points static float distance( int m, int n, int p, int q) { return ( float )Math.sqrt(Math.pow(n - m, 2 ) + Math.pow(q - p, 2 ) * 1.0 ); } // Function to calculate the coordinates // of the excenters of a triangle static void Excenters( int x1, int y1, int x2, int y2, int x3, int y3) { // Length of the sides of the triangle float a = distance(x2, x3, y2, y3); float b = distance(x3, x1, y3, y1); float c = distance(x1, x2, y1, y2); // Stores the coordinates of the // excenters of the triangle pair[] excenter = new pair[ 4 ]; // Applying formula to find the // excenters of the triangle // For I1 excenter[ 1 ] = new pair((-(a * x1) + (b * x2) + (c * x3)) / (-a + b + c), (-(a * y1) + (b * y2) + (c * y3)) / (-a + b + c)); // For I2 excenter[ 2 ] = new pair(((a * x1) - (b * x2) + (c * x3)) / (a - b + c), ((a * y1) - (b * y2) + (c * y3)) / (a - b + c)); // For I3 excenter[ 3 ] = new pair(((a * x1) + (b * x2) - (c * x3)) / (a + b - c), ((a * y1) + (b * y2) - (c * y3)) / (a + b - c)); // Print the excenters of the triangle for ( int i = 1 ; i <= 3 ; i++) { System.out.println(( int )excenter[i].first + " " + ( int )excenter[i].second); } } // Driver code public static void main(String[] args) { int x1, x2, x3, y1, y2, y3; x1 = 0 ; x2 = 3 ; x3 = 0 ; y1 = 0 ; y2 = 0 ; y3 = 4 ; Excenters(x1, y1, x2, y2, x3, y3); } } // This code is contributed by offbeat |
Python3
# Python3 program for the above approach from math import sqrt # Function to calculate the # distance between a pair of points def distance(m, n, p, q): return (sqrt( pow (n - m, 2 ) + pow (q - p, 2 ) * 1.0 )) # Function to calculate the coordinates # of the excenters of a triangle def Excenters(x1, y1, x2, y2, x3, y3): # Length of the sides of the triangle a = distance(x2, x3, y2, y3) b = distance(x3, x1, y3, y1) c = distance(x1, x2, y1, y2) # Stores the coordinates of the # excenters of the triangle excenter = [[ 0 , 0 ] for i in range ( 4 )] # Applying formula to find the # excenters of the triangle # For I1 excenter[ 1 ][ 0 ] = (( - (a * x1) + (b * x2) + (c * x3)) / / ( - a + b + c)) excenter[ 1 ][ 1 ] = (( - (a * y1) + (b * y2) + (c * y3)) / / ( - a + b + c)) # For I2 excenter[ 2 ][ 0 ] = (((a * x1) - (b * x2) + (c * x3)) / / (a - b + c)) excenter[ 2 ][ 1 ] = (((a * y1) - (b * y2) + (c * y3)) / / (a - b + c)) # For I3 excenter[ 3 ][ 0 ] = (((a * x1) + (b * x2) - (c * x3)) / / (a + b - c)) excenter[ 3 ][ 1 ] = (((a * y1) + (b * y2) - (c * y3)) / / (a + b - c)) # Print the excenters of the triangle for i in range ( 1 , 4 ): print ( int (excenter[i][ 0 ]), int (excenter[i][ 1 ])) # Driver Code if __name__ = = '__main__' : x1 = 0 x2 = 3 x3 = 0 y1 = 0 y2 = 0 y3 = 4 Excenters(x1, y1, x2, y2, x3, y3) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approach using System; class GFG{ class pair { public float first, second; public pair( float first, float second) { this .first = first; this .second = second; } } // Function to calculate the // distance between a pair of points static float distance( int m, int n, int p, int q) { return ( float )Math.Sqrt(Math.Pow(n - m, 2) + Math.Pow(q - p, 2) * 1.0); } // Function to calculate the coordinates // of the excenters of a triangle static void Excenters( int x1, int y1, int x2, int y2, int x3, int y3) { // Length of the sides of the triangle float a = distance(x2, x3, y2, y3); float b = distance(x3, x1, y3, y1); float c = distance(x1, x2, y1, y2); // Stores the coordinates of the // excenters of the triangle pair[] excenter = new pair[4]; // Applying formula to find the // excenters of the triangle // For I1 excenter[1] = new pair((-(a * x1) + (b * x2) + (c * x3)) / (-a + b + c), (-(a * y1) + (b * y2) + (c * y3)) / (-a + b + c)); // For I2 excenter[2] = new pair(((a * x1) - (b * x2) + (c * x3)) / (a - b + c), ((a * y1) - (b * y2) + (c * y3)) / (a - b + c)); // For I3 excenter[3] = new pair(((a * x1) + (b * x2) - (c * x3)) / (a + b - c), ((a * y1) + (b * y2) - (c * y3)) / (a + b - c)); // Print the excenters of the triangle for ( int i = 1; i <= 3; i++) { Console.WriteLine(( int )excenter[i].first + " " + ( int )excenter[i].second); } } // Driver code static void Main() { int x1, x2, x3, y1, y2, y3; x1 = 0; x2 = 3; x3 = 0; y1 = 0; y2 = 0; y3 = 4; Excenters(x1, y1, x2, y2, x3, y3); } } // This code is contributed by abhinavjain194 |
Javascript
<script> // Javascript implementation for the above approach // Function to calculate the // distance between a pair of points function distance( m, n, p, q) { return Math.sqrt(Math.pow(n - m, 2) + Math.pow(q - p, 2) * 1.0); } // Function to calculate the coordinates // of the excenters of a triangle function Excenters( x1, y1, x2, y2, x3, y3) { // Length of the sides of the triangle var a = distance(x2, x3, y2, y3); var b = distance(x3, x1, y3, y1); var c = distance(x1, x2, y1, y2); // Stores the coordinates of the // excenters of the triangle var excenter = new Array(4); for ( var i= 0; i<4;i++) excenter[i] = new Array(2); // Applying formula to find the // excenters of the triangle // For I1 excenter[1][0] = (-(a * x1) + (b * x2) + (c * x3)) / (-a + b + c); excenter[1][1] = (-(a * y1) + (b * y2) + (c * y3)) / (-a + b + c); // For I2 excenter[2][0] = ((a * x1) - (b * x2) + (c * x3)) / (a - b + c); excenter[2][1] = ((a * y1) - (b * y2) + (c * y3)) / (a - b + c); // For I3 excenter[3][0] = ((a * x1) + (b * x2) - (c * x3)) / (a + b - c); excenter[3][1] = ((a * y1) + (b * y2) - (c * y3)) / (a + b - c); // Print the excenters of the triangle for ( var i = 1; i <= 3; i++) { document.write(excenter[i][0] + " " + excenter[i][1] + "<br>" ); } } // Driver Code var x1, x2, x3, y1, y2, y3; x1 = 0; x2 = 3; x3 = 0; y1 = 0; y2 = 0; y3 = 4; Excenters(x1, y1, x2, y2, x3, y3); // This code is contributed by Shubham Singh </script> |
6 6 -3 3 2 -2
Time Complexity: O(logn) as using inbuilt sqrt function
Auxiliary Space: O(1)
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