# Program to find all possible triangles having same Area and Perimeter

The task is to find all possible triangles having the same perimeter and area.

Examples:

The triangle having sides (6, 8, 10) have the same perimeter (= (6 + 8 + 10) = 24) and area (= 0.5 * 6 * 8 = 24).

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

Approach: The idea is based on the observation from Heron’s Formula. Below are the observations:

Let the sides of the triangle be a, b, c.
Perimeter(P) = a + b + c
Area(A) using Heron’s Formula: where s = (a + b + c) / 2

Experimental Observation:

We know that:
4 * s2 = s * (s – a) * (s – b) * (s – c)
=> 4 * s = (s – a) * (s – b) * (s – c)
=> 2 * 2 * 2 * 4 * s = 2 * (s – a) * 2 * (s -b) * 2 * (s – c)
=> 16 * (a + b + c) = (- a + b + c) * (a – b + c) * (a + b – c)

Due to this condition:
Max value of (- a + b + c), (a – b + c), (a + b – c) is as follows:
(- a + b + c) * (a – b + c) * (a + b – c) ≤ 16 * 16 * 16
=> 16 * (a + b + c) ≤ 16 * 16 * 16
=> (a + b + c) ≤ 256

From the above equation, the sum of sides of the triangle doesn’t exceed 256 whose perimeter of triangle and area of the triangle can be the same. Therefore, the idea is to iterate three nested loops over the range [1, 256] and print those triplets of sides having the same area and perimeter.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach     #include  using namespace std;     // Function to print sides of all the  // triangles having same perimeter & area  void samePerimeterAndArea()  {      // Stores unique sides of triangles      set > se;         // i + j + k values cannot exceed 256      for (int i = 1; i <= 256; ++i) {             for (int j = 1; j <= 256; ++j) {                 for (int k = 1; k <= 256; ++k) {                     // Find the value of 2 * s                  int peri = i + j + k;                     // Find the value of                  // 2 * ( s - a )                  int mul1 = -i + j + k;                     // Find the value of                  // 2 * ( s - b )                  int mul2 = i - j + k;                     // Find the value of                  // 2 * ( s - c )                  int mul3 = i + j - k;                     // If triplets have same                  // area and perimeter                  if (16 * peri == mul1 * mul2 * mul3) {                         // Store sides of triangle                      vector<int> v = { i, j, k };                         // Sort the triplets                      sort(v.begin(), v.end());                         // Inserting in set to                      // avoid duplicate sides                      se.insert(v);                  }              }          }      }         // Print sides of all desired triangles      for (auto it : se) {          cout << it << " "              << it << " "              << it << endl;      }  }     // Driver Code  int main()  {      // Function call      samePerimeterAndArea();         return 0;  }

Output:

5 12 13
6 8 10
6 25 29
7 15 20
9 10 17


Time Complexity: O(2563)
Auxiliary Space: O(2563)

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