Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Count number of right triangles possible with a given perimeter

  • Difficulty Level : Easy
  • Last Updated : 27 Jan, 2021

Given a perimeter P, the task is to find the number of right triangles possible with perimeter equal to p.
Examples: 
 

Input: P = 12
Output: number of right triangles = 1 
The only right angle possible is with sides 
hypotenuse = 5, perpendicular = 4 and base = 3. 

Input: p = 840
Output: number of right triangles = 8

 

So the aim is to find the number of solutions which satisfy equations a + b + c = p and a2 + b2 = c2.
A naive approach is to run two loops for a(1 to p/2) and b(a+1 to p/3) then make c=p-a-b and increase count by one if a*a + b*b == c*c   . This will take O(p^{2})   time.
An efficient approach can be found by little algebraic manipulation :
 

a^{2}+b^{2}=c^{2} or, (a+b)^{2}-2ab = c^{2} or, (p-c)^{2}-2ab = c^{2} or, p^{2}-2cp-2ab = 0 or, 2ab = p^{2}-2cp or, 2ab = p^{2}-2p(p-a-b) or, 2a(p-b) = p(p-2b) or, a = (p/2) * ((p-2b)/(p-b))



Since a + c > b or, p – b > b or, b < p/2. Thus iterating b from 1 to p/2, calculating a and storing only the whole number a would give all solutions for a given p. There are no right triangles are possible for odd p as right angle triangles follow the Pythagoras theorem. Use a list of pairs to store the values of a and band return the count at the end. 
Below is the implementation of the above approach. 
 

C++




// C++ program to find the number of
// right triangles with given perimeter
#include<bits/stdc++.h>
using namespace std;
 
// Function to return the count
int countTriangles(int p)
{
  // making a list to store (a, b) pairs
  vector<pair<int,int>> store;
 
  // no triangle if p is odd
  if (p % 2 != 0)
    return 0;
  else
  {
    int count = 1;
 
    for(int b = 1; b < p / 2; b++)
    {
      float a = (float)p / 2.0f * ((float)((float)p -
                                           2.0 * (float)b) /
                                   ((float)p - (float)b));
 
      int inta = (int)(a);
 
      if (a == inta)
      {
        // make (a, b) pair in sorted order
        pair<int,int> ab;
 
        if(inta<b)
        {
          ab = {inta, b};
        }
        else
        {
          ab = {b, inta};
        }
 
        // check to avoid duplicates
        if(find(store.begin(), store.end(), ab) == store.end())
        {
          count += 1;
 
          // store the new pair
          store.push_back(ab);
        }
      }
 
    }
    return count;
  }
}
 
// Driver Code
int main()
{
  int p = 840;
  cout << "number of right triangles = " << countTriangles(p);
  return 0;
}
 
// This code is contributed by rutvik_56.

Python3




# python program to find the number of
# right triangles with given perimeter
 
# Function to return the count
def countTriangles(p):
     
    # making a list to store (a, b) pairs
    store =[]
 
    # no triangle if p is odd
    if p % 2 != 0 : return 0
    else :
        count = 0
        for b in range(1, p // 2):
 
            a = p / 2 * ((p - 2 * b) / (p - b))
            inta = int(a)
            if (a == inta ):
 
                # make (a, b) pair in sorted order
                ab = tuple(sorted((inta, b)))
 
                # check to avoid duplicates
                if ab not in store :
                    count += 1
                    # store the new pair
                    store.append(ab)
        return count
 
# Driver Code
p = 840
print("number of right triangles = "+str(countTriangles(p)))
Output: 
number of right triangles = 8

 

Time complexity: O(P)
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :