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!!
Very interesting! I need a proof too! Thanks ! <3 <3
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.
Thanks but how to ensure that $$$2z(p-2x)(p-2y)$$$ can't be divided by $$$p$$$?
Thank you for pointing out that flaw. I thought even cannot be divided by odd, which is clearly not true.
How did you prove that all $$$a, b, c$$$ are even?
Never mind ^^
Your proof is wrong, in case 3rd sinC can be say 2/3.
Whoops ^^. Sorry for my mistake.
Your proof for case 1 and 2 too is wrong. For example, if $$$p = 19/2, r = 6$$$ then $$$S$$$ is still integer.
Play with, $$$A = \frac{abc}{4R} = \sqrt{s(s - a)(s - b)(s - c)} = rs$$$
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.
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)
Can u send me your proof :< It takes me my whole yesterday evening
Could you help us :< ?
We have $$$r^2 s = (s - a)(s - b)(s - c)$$$, putting $$$s = P/2$$$,
$$$4 * r^2 P = (P - 2a)(P - 2b)(P - 2c)$$$, if P is odd then RHS is odd but, LHS is even. Hence contradiction.
If $$$s$$$ is even then we're done, otherwise let's consider the case when $$$s$$$ is odd.
$$$r^2 s = s^3 - (a + b + c)s^2 + (ab + bc + ca)s - abc$$$
$$$r^2 s = s^3 - 2s^3 + \frac{( (a + b + c) ^2 - (a^2 + b^2 + c^2))}{2} s - 4Rrs$$$
$$$r^2s = s^3 -\frac{a^2 + b^2 + c^2}{2}s - 4Rrs$$$
$$$r^2 = s^2 -\frac{a^2 + b^2 + c^2}{2} - 4Rr$$$ -----equation 1.
If all $$$a, b, c$$$ are even then, put $$$a = 2x, b = 2y, c = 2z$$$
making $$$Rr = \frac{2xyz}{x + y + z}$$$, thus $$$Rr$$$ is even, hence the only value of RHS modulo 8 in equation1 is either 3 or 5 none of which is possible. (Using the fact that square of any odd number leaves remainder 1 on dividing by 8).
I forgot the proof that I had. I will finish it later.
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?
We assume that $$$s$$$ is not even, i.e, $$$s$$$ is odd, and arrive at contradiction.
Have you remembered it back yet?