When will the area of and perimeter of a right triangle be numerically equal?


Your question has infinitely many answers as stated, I suspect that you had in mind integer valued sides for the right triangle. It is not too hard to find examples of such triangles -- the area and perimeter of a right triangle will be numerically the same when one side is 6, another 8 and the hypotenuse 10.

This suggests one approach: use a table and start with the Pythagorean triple 3,4,5; then try 6, 8, 10 etc.; you might then 5, 12, 13 and then the multiples of these and so on.

Algebraically, if we let the sides have lengths a, b and c (where c is the hypotenuese) then we have to solve a + b + c = ab/2. If we use our knowledge about Pythagoras' Theorem to eliminate c, and with some not too difficult algebra we are led to requiring that ab = 4a + 4b - 8, equivalently, for b 4, a = 4(b-2)/(b-4). We now need a little number theory to determine the integer solutions.

If b is odd so is b - 4 thus unless b = 5, b - 4 needs to divide b - 2 for a to be an integer; since a is to be positive we find the only solution is b = 5 and thus a = 12 and c = 13.

If b is even, say b = 2t, then we have a = 4(t-1)/(t-2). The integers t - 2 and t - 1 are relatively prime so unless t = 3 we have t - 2 must divide 4; i.e. t - 2 = 1, 2 or 4. We end up with b = 6, 8 or 12 and correspondingly a = 8, 6 and 5.

We have thus shown that all the integer solutions are found in the pythagorean triples (6, 8, 10) and (5, 12, 13).

One curiosity. Note that if you only required a and b to be integers you still end up with the same solutions - Pythagorous at work again!

Penny and Maxine

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