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Right Triangle (Posted on 2003-08-05) Difficulty: 3 of 5

Given a right triangle with lengths that are reciprocals of integers, what is the smallest possible sum of these the integers?

In other words, given a right triangle with lengths 1/a, 1/b, and 1/c, where a, b, and c are all integers, what is the lowest value of a+b+c? Also, prove it.

Taken from CAML, which did not ask for a proof.

See The Solution Submitted by np_rt    
Rating: 3.4000 (5 votes)

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Question re: I have a proof now! :) | Comment 4 of 12 |
(In reply to I have a proof now! :) by Gamer)

Nice solve Gamer. I kind of figured it was going to begin with a pythagorean triple, but I didn't get much past 3,4,5. LOL.

I have been out of the math game for a few years and I'm afraid my brain has atrophied. I have to get it in gear again. I'm afraid I'm going to sound dumb, but I'm just going to jump right in. May I ask some possibly obvious questions ...

You state, "Since the left side is a perfect square, the right side must be a perfect square". How do we know this? Are you referring to aČ + bČ? How do we know it is a perfect square? And how do we know that means that the right side must also be one?

I have one or two more questions, but I don't think I will display all of my ignorance in one post! And perhaps your response will clear up the rest for me.
  Posted by Mathcop on 2003-08-05 16:26:22

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