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.

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## 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).Log in to reply

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 like Timmy but the equation gets really messy. So I gave up.

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I admit that

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.Log 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].

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Jon H.'s comment contained the numerical answer only. He didn't post a fully worked solution.

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Silly rule…

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Silly or not, I want to respect the rules and do everything by the book.

Now would anyone like to help me? An approach? A cryptic hint maybe?

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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.

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Thanks a lot!

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!

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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.

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I try to use cosine rule to solve it, but the equation goes complicated...

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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.

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How do you know where I live? Dhaka is a big place. And you can't see my address by viewing my profile (that is only for the staff to see).

Have you been spying on me?:)

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I'm not spying on you. I first came to know about you and the fact that your college is Notre Dame from Brilliant. One day Tushar(you know him) was talking something about a boy named Mursalin who studies in Notre Dame at our English coaching. So I thought it might have been the same Mursalin from Brilliant and so I asked some things about you out of

'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.Log in to reply

Why are you being so apologetic? You did nothing wrong. Cheer up!

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ifI manage to solve it).Log in to reply

Silly person...

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