Find other two sides of a right angle triangle
Given one side of right angle triangle, check if there exists a right angle triangle possible with any other two sides of the triangle. If possible print length of the other two sides.
Else print -1
Note: All the sides of triangle must be positive integers
Example 1:
Input : a = 3 Output : b = 4, c = 5 Explanation : a = 3, b = 4 and c = 5 form right angle triangle because 32 + 42 = 9 + 16 = 25 = 52 => 32 + 42 = 52
Example 2:
Input : a = 11 Output : b = 60, c = 61 Explanation : a = 11, b = 60 and c = 61 form right angle triangle because 112 + 602 = 121 + 3600 = 3721 = 612 => 112 + 602 = 612
To solve this problem we first observe the Pythagoras equation. If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
This relationship is represented by the formula:
a2 + b2 = c2
- Case 1 – a is an odd number: Given a, find b and c
c2 - b2 = a2 OR c = (a2 + 1)/2; b = (a2 - 1)/2;
- Above solution works only for case when a is odd, because a2 + 1 is divisible by 2 only for odd a.
- Case 2 – a is an even number: When c-b is 2 & c+b is (a2)/2
c-b = 2 & c+b = (a2)/2 Hence, c = (a2)/4 + 1; b = (a2)/4 - 1;
This works when a is even.
This solution doesn’t work for case of a = 1 and a = 2 because there isn’t any right angle triangle with side 1 or 2 with all integer sides.
C++
// C++ program to print other two sides of right // angle triangle given one side #include <bits/stdc++.h> using namespace std; // Finds two sides of a right angle triangle // if it exist. void printOtherSides( int n) { // if n is odd if (n & 1) { // case of n = 1 handled separately if (n == 1) cout << -1 << endl; else { int b = (n*n-1)/2; int c = (n*n+1)/2; cout << "b = " << b << ", c = " << c << endl; } } else { // case of n = 2 handled separately if (n == 2) cout << -1 << endl; else { int b = n*n/4-1; int c = n*n/4+1; cout << "b = " << b << ", c = " << c << endl; } } } // Driver program to test above function int main() { int a = 3; printOtherSides(a); return 0; } |
Java
// Java program to print other two // sides of right angle triangle // given one side class GFG { // Finds two sides of a right angle // triangle if it they exist. static void printOtherSides( int n) { // if n is odd if (n % 2 != 0 ) { // case of n = 1 handled separately if (n == 1 ) System.out.println( "-1" ); else { int b = (n * n - 1 ) / 2 ; int c = (n *n + 1 ) / 2 ; System.out.println( "b = " + b + ", c = " +c); } } else { // case of n = 2 handled separately if (n == 2 ) System.out.println( "-1" ); else { int b = n * n / 4 - 1 ; int c = n * n / 4 + 1 ; System.out.println( "b = " + b + ", c = " +c); } } } // Driver code public static void main (String[] args) { int a = 3 ; printOtherSides(a); } } // This code is contributed by Anant Agarwal. |
Python3
# Python program to print other # two sides of right angle # triangle given one side # Finds two sides of a right angle # triangle if it they exist. def printOtherSides(n): # if n is odd if (n & 1 ): # case of n = 1 handled # separately if (n = = 1 ): print ( - 1 ) else : b = (n * n - 1 ) / / 2 c = (n * n + 1 ) / / 2 print ( "b =" , b, ", c =" , c) else : # case of n = 2 handled # separately if (n = = 2 ): print ( - 1 ) else : b = n * n / / 4 - 1 c = n * n / / 4 + 1 print ( "b =" , b ", c =" , c) # Driver Code a = 3 printOtherSides(a) # This code is contributed by # Sanjit_Prasad |
C#
// C# program to print other two // sides of right angle triangle // given one side using System; class GFG { // Finds two sides of a right angle // triangle if it they exist. static void printOtherSides( int n) { // if n is odd if (n % 2 != 0) { // case of n = 1 handled // separately if (n == 1) Console.WriteLine( "-1" ); else { int b = (n * n - 1) / 2; int c = (n * n + 1) / 2; Console.Write( "b = " + b + ", c = " + c); } } else { // case of n = 2 handled // separately if (n == 2) Console.Write( "-1" ); else { int b = n * n / 4 - 1; int c = n * n / 4 + 1; Console.Write( "b = " + b + ", c = " + c); } } } // Driver code public static void Main () { int a = 3; printOtherSides(a); } } // This code is contributed by Nitin Mittal. |
PHP
<?php // PHP program to print other two // sides of right angle triangle // given one side // Finds two sides of a right angle // triangle if it they exist. function printOtherSides( $n ) { // if n is odd if ( $n & 1) { // case of n = 1 // handled separately if ( $n == 1) echo -1 ; else { $b = ( $n * $n - 1) / 2; $c = ( $n * $n + 1) / 2; echo "b = " , $b , ", c = " , $c ; } } else { // case of n = 2 // handled separately if ( $n == 2) echo -1 ; else { $b = $n * $n / 4 - 1; $c = $n * $n / 4 + 1; echo "b = " , $b , ", c = " , $c ; } } } // Driver Code $a = 3; printOtherSides( $a ); return 0; // This code is contributed by nitin mittal. ?> |
Javascript
<script> // javascript program to print other two // sides of right angle triangle // given one side // Finds two sides of a right angle // triangle if it they exist. function printOtherSides(n) { // if n is odd if (n % 2 != 0) { // case of n = 1 handled separately if (n == 1) document.write( "-1" ); else { var b = (n * n -1) / 2; var c = (n *n +1) / 2; document.write( "b = " + b + ", c = " +c); } } else { // case of n = 2 handled separately if (n == 2) document.write( "-1" ); else { var b = n * n / 4 - 1; var c = n * n / 4 + 1; document.write( "b = " + b + ", c = " +c); } } } // Driver code var a = 3; printOtherSides(a); // This code is contributed by 29AjayKumar </script> |
Output:
b = 4, c = 5 This article is contributed by Pratik Chhajer . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Time complexity: O(1)
Auxiliary Space: O(1)
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