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
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:
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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