I had recently opened a discussion thread about a geometry problem from an online-math-camp-test after the test was over. However, it turns out I broke some rule. There was a 24-hour rule ("no problems are to be discussed until 24 hours have passed") and I somehow missed that rule. Now that 24 hours *have* passed, it's completely okay to discuss about that problem.

I've deleted the previous discussion thread and I'm really sorry for the inconvenience.

Here's the problem I was talking about:

In \(△XYZ\),

\(2∠X=3∠Y\) and \(YZ=x\), \(XZ=y\), \(XY=z\). Find a triangle that has positive integer side lengths and satisfies the equation:

\((x^2−y^2)(x^2+zx−y^2)=y^2z^2\)

Good luck and thanks in advance!

Problem source: RGS Math Olympiad-June 2013.

This problem was also recycled in BdMO online math camp exam-1.

## Comments

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TopNewestWell, I worked on this problem a bit.

Let \( \angle X = 3\theta\) & \(\angle Y = 2\theta\). Then \(\angle Z = \pi - 5\theta\). Applying sine rule, we get:

\( \frac{x}{\sin{3\theta}} = \frac{y}{\sin{2\theta}} = \frac{z}{\sin{5\theta}} = k \) (Say)

Putting in the given equation, it is observed that the equation turns out to be an identity for such a \(k\) & \(\theta\).

As such you want to find a real

\(k\)such that \(k \sin{2\theta}, k \sin{3\theta}\) & \(k \sin{5\theta}\) are all integers, which apparently doesn't seem possible (by noting cases when sine function has 'good looking' rational & irrational values). – Paramjit Singh · 3 years, 11 months agoLog in to reply

Timmy but the equation gets really messy. So I gave up. – Mursalin Habib · 3 years, 11 months ago

Thanks for the help. Notice that the angles don't have to be integers. So, trying to find cases when sine function has 'good looking' rational & irrational values obviously won't work. I tried using the cosine law likeLog in to reply

anglesdon't need to be integers,but a conclusion can be made:If the sines have ugly values,obviously no \(k\) will work for all

3of them (by inspection), so we limit ourselves to rationals & those irrationals (for which we can easily define a conjugate, in the prospect of multiplying it to \(k\) to receive an integral value). Trial shows no such rational values exist (sine function has many less rational values, had it been tan function,it would have been less difficult!) & trying to find such irrational values which have similar conjugates for such \(\theta\) so that all angles are less than \(180^{\circ}\) also doesn't come up. – Paramjit Singh · 3 years, 11 months agoLog in to reply

Could you explain a bit more? What kind of trial?

I thought the sine function could take any rational value in the range \([-1, 1]\) [in this case, \((0, 1]\)].

Am I saying something wrong?

Do you mean to say this problem has no solution? I know for a fact that it does. \((x,y,z)=(144,135,31)\) is the solution [credit goes to John H. See the previous deleted discussion]. – Mursalin Habib · 3 years, 11 months ago

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– Paramjit Singh · 3 years, 11 months ago

Oh, I guess \(\theta = 6.106...\) suffices.(as per the assumption,I haven't checked yet). Clearly a very ugly rational(or maybe irrational) sine value :). Please link me to his solution.Log in to reply

Jon H.'s comment contained the numerical answer only. He didn't post a fully worked solution. – Mursalin Habib · 3 years, 11 months ago

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Here's a starter. Currently I have not been able to complete the solution, but I will (probably) post it soon. I am not sure if this is the most elegant way to do this though: it is certainly very tedious.

Let \(X= 3 \theta \) and \(Y= 2 \theta \). From the cosine rule, \(x^2= y^2 + z^2 - 2yz \cos3\theta \). Then, \(x^2-y^2= z(z - 2y \cos3\theta) \). Also note that \(x^2-y^2+zx= z(z+x-2y \cos3\theta) \). We can now substitute these values in the given equation:

\[ z^2 (z - 2y \cos3\theta)(z+x-2y \cos3\theta)= y^2z^2 \] \[\implies (z - 2y \cos 3\theta)(z+x-2y \cos3\theta)= y^2 \] \[\implies (z - 2y \cos 3\theta)(z+x-2y \cos 3\theta)= z^2 + x^2 - 2xz\cos 2\theta \] If you expand the L.H.S, the \(z^2\) term cancels out. The equation, still, will be a gigantic one. – Sreejato Bhattacharya · 3 years, 11 months ago

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This is the way I approached the problem. But I gave up when I saw the ugly-looking equation. It looks like you're onto something. So post your solution solution when you're done. This problem has been nagging me for 3 months. It's time to put it to rest.

Once again, thank you so much! – Mursalin Habib · 3 years, 11 months ago

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– Sreejato Bhattacharya · 3 years, 11 months ago

The only reason I think it might work is that it helps for the cancellation of \(y^2\). I expanded the L.H.S, and as expected, the \(z^2\) term also cancels out. But what remains is a mammoth equation, which is very difficult to interpret. So I'd say let's not expand the brackets, and make a few more trigonometric substitutions to see what we get. What I have got so far is not even close to a concrete solution.Log in to reply

Silly rule… – Tim Vermeulen · 3 years, 11 months ago

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Now would anyone like to help me? An approach? A cryptic hint maybe? – Mursalin Habib · 3 years, 11 months ago

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I'm not sure if this would work: first solve the equation as a diophantine equation. there may be infinity number of solutions or only one may be. Narrow down the possible solutions by triangle inequality. Then check out for which of those solutions, 2X=3Y is true. [ i really am not good at math, so sorry if it is another silly method ]. By the way, I live in NAM garden, near to you. – Aiman Rafeed · 3 years, 11 months ago

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Have you been spying on me?:) – Mursalin Habib · 3 years, 11 months ago

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'just'curiosity (not for spying) and came to know.... btw, was my suggestion able to help you out? Even a bit? Really sorry to waste your time if it didn't. – Aiman Rafeed · 3 years, 11 months agoLog in to reply

Why are you being so apologetic? You did nothing wrong. Cheer up! – Mursalin Habib · 3 years, 11 months ago

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– Aiman Rafeed · 3 years, 11 months ago

I seriously didn't notice the smiley...ha ha. Anyway, let me know as soon as you solve it. And if you can, post the solution here in brief.Log in to reply

ifI manage to solve it). – Mursalin Habib · 3 years, 11 months agoLog in to reply

I try to use cosine rule to solve it, but the equation goes complicated... – Timmy Ben · 3 years, 11 months ago

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Silly person... – Justin Wong · 3 years, 11 months ago

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