_LNHTD_'s blog

By _LNHTD_, history, 5 years ago, In English

The problem is : We are given perimeter (P) of a triangle. We need to find the number of triplet edges (a, b, c) of a triangle, so that three edges are all integer, the area and the length of the radius of the incircle and circumcircle is also an integer.

In the solution, they have an observation that : In order to exist at least a triplet satisfy the problem, 4 must be a divisor of P (perimeter) and a, b, c (three edges) must be all even.

I have proofed all a, b, c are even. But I can't figure out how 4 is a divisor of P.

Could someone help me to proof this ? Thanks in advance!!

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5 years ago, # |
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Very interesting! I need a proof too! Thanks ! <3 <3

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5 years ago, # |
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Edit: Successfully hacked by _LNHTD_

I keep this comment just in case I have some new ideas to prove it.

Given $$$P, r, R, k, x, y, z$$$ are integer, $$$a = 2x, b = 2y, c = 2z$$$, prove that $$$p = x + y + z$$$ is even.

$$$P$$$ is the perimeter of the triangle, $$$r$$$ is the radius of the incircle, $$$R$$$ is the radius of the circumcircle, $$$k$$$ is the area of the triangle, $$$a, b, c$$$ are the length of the edges, $$$p$$$ is the semiperimeter.

We have:

$$$k = pr \Rightarrow r = \frac{k}{p} = \frac{\sqrt{p(p-a)(p-b)(p-c)}}{p}$$$ and is an integer.

$$$r^2 = \frac{(p-a)(p-b)(p-c)}{p}$$$ is an integer.

$$$r^2 = \frac{(p-2x)(p-2y)(p-2z)}{p}$$$ is an integer.

Suppose that $$$p$$$ is odd.

We have

$$$r^2 = \frac{(p - 2 x) (p - 2 y)p}{p} - \frac{2z (p - 2 x) (p - 2 y) }{p} = (p-2x)(p-2y) - \frac{2z (p - 2 x) (p - 2 y) }{p}$$$

But since $$$p$$$ is odd, $$$\frac{2z(p - 2 x) (p - 2 y)}{p}$$$ is not an integer $$$\Rightarrow r^2$$$ is not an integer (contradiction).

Therefore, $$$p$$$ must be even $$$\Rightarrow$$$ $$$P$$$ is divisible by 4.

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    5 years ago, # ^ |
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    Thanks but how to ensure that $$$2z(p-2x)(p-2y)$$$ can't be divided by $$$p$$$?

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      5 years ago, # ^ |
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      Thank you for pointing out that flaw. I thought even cannot be divided by odd, which is clearly not true.

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5 years ago, # |
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How did you prove that all $$$a, b, c$$$ are even?

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    5 years ago, # ^ |
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    Never mind ^^

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      5 years ago, # ^ |
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      Your proof is wrong, in case 3rd sinC can be say 2/3.

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      5 years ago, # ^ |
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      Your proof for case 1 and 2 too is wrong. For example, if $$$p = 19/2, r = 6$$$ then $$$S$$$ is still integer.

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5 years ago, # |
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Hint
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    5 years ago, # ^ |
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    Actually I have thought a lot about those formulas. But still ended up stucked :<. Could you show us more hints or something like that? Thanks a lot.

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    5 years ago, # ^ |
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    I just can prove that 'a' must divide 8 and the area must be an even number.But I don't know if it was efficient:< (trying to prove that the case having one edge is even and the 2 others are odd, and 'a' is the even one)

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    5 years ago, # ^ |
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    Can u send me your proof :< It takes me my whole yesterday evening

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    5 years ago, # ^ |
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    Could you help us :< ?

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      5 years ago, # ^ |
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      s is an integer

      If $$$s$$$ is even then we're done, otherwise let's consider the case when $$$s$$$ is odd.

      all of a, b, c can't be even
      a can't be even if both b and c are odd
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        5 years ago, # ^ |
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        If s is even, how, i can prove that s is even why you're trying to consider the case in which s is odd, bro?

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          5 years ago, # ^ |
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          We assume that $$$s$$$ is not even, i.e, $$$s$$$ is odd, and arrive at contradiction.

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        5 years ago, # ^ |
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        Have you remembered it back yet?