Count number of right triangles possible with a given perimeter

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

# 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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