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++ 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 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 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# 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 |
<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)