Dtjytyk

Can anyone help with Q25, above? I have tried applying the sine and cosine rules, arguments about similar triangles, and general diagram-chasing to no avail.

I am somewhat confused by the condition that there be only a finite number $m^2 + 2m -1$ of possible integer perimeters $p$. Could one not extend $TR$ arbitrarily far to pick up infinitely many such values for $p$, or does that break the angle-bisector condition?

Many thanks in advance.

So, we know $m$, $n$, and the angle bisector condition. $frac{RS}{RQ} = frac{n}{m}$ by the angle bisector, $RS+RQ$ is an integer since the perimeter is, and we have the triangle inequality conditions
$$|RS-RQ|< m+n < RS+RQ$$

With that, we leave the geometry behind. We should also note the other conditions in the problem: $n>m$ and $m+n$ is an integer multiple of $n-m$.

Let $m+n=a(n-m)$ and $RS=bn$. Then $RQ=bm$, and the triangle inequality conditions become
$$b(n-m) < a(n-m),quad m+n < b(n+m)$$
or equivalently $1 < b < a$. We know $a$ is a natural number, but $b$ could be anything. To narrow down the possibilities for $b$, we note that $b(n+m)$ is an integer. It's strictly between $m+n$ and $a(m+n)$, which leads to $(a-1)(m+n)-1$ possibilities. Now, we set this equal to $m^2+2m-1$ and do some algebra:
begin{align*}(a-1)(m+n)-1 &= m^2 + 2m -1\
frac{(m+n)^2}{n-m} - (m+n) &= m^2+2m\
(m+n)^2 - (n^2-m^2) &= m(m+2)(n-m)\
2m^2+2mn &= m^2n -m^3 + 2mn - 2m^2\
4m^2 &= m^2(n-m)end{align*}
And there, the answer $n-m=4$ falls out.

Why are there only finitely many possible integer values for the perimeter? It's the triangle inequality coupled with the angle bisector condition. At the lower extreme for the perimeter, the triangle degenerates to the segment $QS$ traced twice, $R=T$ and the perimeter is $2(m+n)$. At the upper extreme, the triangle degenerates to the segment $RS$ traced twice, $RS=frac{m+n}{n-m}cdot TS$, and the perimeter is $frac{2n(m+n)}{n-m}$.

That $n-m$ in the denominator is also why we have that condition that $n-m$ divides $m+n$, by the way. If we don't force the maximum perimeter to be an integer, the count gets messier.

Sorry, I don't have any reputation so I can't leave this as a comment, I just wanted to name the source of this question.

This is from the 2019 Fermat Competition, a national competition for grade 11 students in Canada distributed by the University of Waterloo. This contest was held about 3 weeks ago, this is why it probably isn't on AOPS or other sites yet. Full results for the competition are already in though so there is no fear of these comments spoiling anything.

I am a high school teacher in Canada and some of my students wrote this contest. None of them had any idea how to do this question, and I had no idea where to even start! I am glad that this was answered so that I can share the solution with some of them!

This isn't a full solution yet, but here are some observations re-framing the problem that might be useful. It was too long to put in the comments and additionally makes it clear why there cannot be infinitely many possibilities.

Firstly, the condition that $RT$ is an angle bisector means that the ratio of $RQ$ and $RS$ is fixed at $frac{m}{n}$. This means that the vertex $R$ must lie on the circle of Apollonius defined by $QTS$ (i.e. the locus of points $R$ such that $frac{RQ}{RS} = frac{m}{n}$ is a circle). The circle passes through $R$ and $T$ and its center lies on the extension of $QS$. You can work out the radius and center (I get a radius of $frac{mn}{n-m}$ and the center's distance from $Q$ of $frac{m^2}{n-m}$).

Secondly, the condition that the perimeter is $p$ means that we want the sum of the distances from $R$ to $Q$ and $S$ to be $p - (n + m)$. That means that $R$ must lie on the ellipse with foci $Q$ and $S$, and major axis parameter $p - (n + m)$.

So we have a circle and an ellipse, with the circle passing through a point on the axis, which means they must cross unless one is entirely contained inside the other. Viewing $m$ and $n$ as fixed (hence fixing the circle), we see that the ellipse grows with $p$, and for large enough $p$ must stop intersecting the circle.

Now, the number of possible perimeter values is just the number of choices for $p$ that have the circle and ellipse cross. Presumably there's a slick way to do that count, but I haven't found it yet. You could probably brute force the Cartesian equations for the two, but it's late and I'll leave that for someone else. :D