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 pointsfloat 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 trianglevoid 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 Codeint 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 approachimport 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 pointsstatic 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 trianglestatic 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 codepublic 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 approachfrom math import sqrt# Function to calculate the# distance between a pair of pointsdef 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 triangledef 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 Codeif __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 approachusing 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 pointsstatic 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 trianglestatic 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 codestatic 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 pointsfunction 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 trianglefunction 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 Codevar 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> |
Output:
6 6 -3 3 2 -2
Time Complexity: O(logn) as using inbuilt sqrt function
Auxiliary Space: O(1)
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