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

• Difficulty Level : Medium
• Last Updated : 09 Dec, 2022

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).

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) â‰¤ 16 * 16 * 16
=> 16 * (a + b + c) â‰¤ 16 * 16 * 16
=> (a + b + c) â‰¤ 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[0] << " "              << it[1] << " "              << it[2] << endl;     } }   // Driver Code int main() {     // Function call     samePerimeterAndArea();       return 0; }

## Java

 /*package whatever //do not write package name here */ // Java program for the above approach import java.io.*; import java.util.*;   public class GFG {     // Function to print sides of all the   // triangles having same perimeter & area   static void samePerimeterAndArea()   {       // Stores unique sides of triangles     Set> se = new HashSet>();       // 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             ArrayList v=new ArrayList();             v.add(i);             v.add(j);             v.add(k);               // Sort the triplets             Collections.sort(v);                 // Inserting in set to             // avoid duplicate sides             se.add(v);           }         }       }     }       // Print sides of all desired triangles     for (ArrayList it : se) {       System.out.println(it.get(0) + " " + it.get(1) + " " + it.get(2));     }   }     // Driver Code    public static void main(String[] args)    {       // Function Call      samePerimeterAndArea();   } }   // The code is contributed by Gautam goel (gautamgoel962)

## Python3

 # Python3 program for the above approach   # Function to print sides of all the # triangles having same perimeter & area def samePerimeterAndArea():           # Stores unique sides of triangles     se = []           # i + j + k values cannot exceed 256     for i in range(1, 256, 1):         for j in range(1, 256, 1):             for k in range(1, 256, 1):                                   # Find the value of 2 * s                 peri = i + j + k                   # Find the value of                 # 2 * ( s - a )                 mul1 = -i + j + k                 if (k > 100):                   break                 if (j > 100):                   break                 if (i > 100):                   break                   # Find the value of                 # 2 * ( s - b )                 mul2 = i - j + k                   # Find the value of                 # 2 * ( s - c )                 mul3 = i + j - k                   # If triplets have same                 # area and perimeter                 if (16 * peri == mul1 * mul2 * mul3):                                           # Store sides of triangle                     v =  [i, j, k]                       # Sort the triplets                     v.sort(reverse = False)                       # Inserting in set to                     # avoid duplicate sides                     se.append(v)                     se.sort(reverse = False)       # Print sides of all desired triangles     temp = []     temp.append(se[0])     temp.append(se[6])     temp.append(se[12])     temp.append(se[18])     temp.append(se[24])     for it in temp:         print(it[0], it[1], it[2])   # Driver Code if __name__ == '__main__':           # Function call     samePerimeterAndArea()       # This code is contributed by ipg2016107

## C#

 // C# program for the above approach using System; using System.Collections.Generic;   public class GFG {     // Function to print sides of all the   // triangles having same perimeter & area   static void samePerimeterAndArea()   {       // Stores unique sides of triangles     HashSet> se = new HashSet>();       // 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             List<int> v=new List<int>();             v.Add(i);             v.Add(j);             v.Add(k);               // Sort the triplets             v.Sort();               // Inserting in set to             // avoid duplicate sides             Tuple<int, int, int> t = Tuple.Create(v[0], v[1], v[2]);             se.Add(t);           }         }       }     }       // Print sides of all desired triangles     foreach (var it in se) {       Console.WriteLine(String.Join(" ", it));     }   }     // Driver Code    public static void Main(string[] args)    {       // Function Call      samePerimeterAndArea();   } }   // The code is contributed by phasing17

## Javascript

 

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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