Angle-chasing is too hard, ft. EGMO 2025/3

(Glen here.)

Last week, I had the privilege of coordinating for the European Girls' Mathematical Olympiad, which meant that I was involved in creating the marking scheme and grading scripts for one of the problems. But before doing any of that, I first had to test-solve the problem I was assigned. This post will be about my thought process during this test-solve. Diagrams are scanned in from my rough work, so I'm sorry if they look horrible.

First, the problem:

(EGMO 2025/3) Let $ABC$ be an acute triangle. Points $B, D, E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Suppose triangle $ADE$ is acute, and let $H$ be its orthocentre. Points $P$ and $Q$ lie on lines $BM$ and $CN$, respectively, such that $D, H, M,$ and $P$ are concyclic and pairwise different, and $E, H, N,$ and $Q$ are concyclic and pairwise different. Prove that $P, Q, N,$ and $M$ are concyclic.

Initial thoughts

• Here's a diagram:

• Having $H$ outside of the triangle is kind of annoying, so here's another diagram. It seems that it would make more sense to view $\triangle ADE$ as the "main" triangle.

• This looks like something I could bash. Sure, there are circles, but there are lots of ways to get around that. (But also I am surrounded by people who actually like geometry and I don't want to get judged, so I don't pursue this.)
• Let's extend $BM$ and $CN$ to meet at $T$. Then, it suffices to show that $TM\cdot TP = TN \cdot TC$.
• I can hear Ker Yang in my head screaming Reim!!! - right. One could equivalently prove that $B,P,Q,C$ are concyclic.
• It seems that $P,Q$ are weird points, so we might try to get rid of them...looking at the alternate formulation with $T$, it suffices to show that $T$ lies on the radical axis of the two circles, which lets us get rid of $P,Q$, which seems nice.
• This reminds me of IMO 2019/2, in the sense that both problems have symmetrically constructed circles and made me think of Ker Yang screaming Reim!!!, but I can't actually remember what the statement of IMO 2019/2 was, so whatever.

Investigating $T$

• Surely $T$ should be a nice point. Maybe it lies on on $AH$? No, that doesn't look right.
• But $B,M$ and $C,N$ are defined with respect to the main triangle via length ratios in a symmetric way, so maybe $T$ lies on the median.
• Indeed, letting $BM$ intersect the median $AL$ at $T'$, we can get $AT':T'L$ by Menelaus and the ratio should be the same should be the same on the other side, so indeed $T$ lies on the median.
• Looks like the ratio is $1:2$, so $T$ is the midpoint of $AG$ (where $G$ is the centroid), which seems nice?
• I guess that means $BG||MT$, which is nice, but I don't see how to continue.

Bashing

• Oh well, desperate times call for desperate measures. This should really be bashable: we can get the centre of $(HMD)$ by intersecting the perpendicular bisectors of $HD$ and $MD$, then we just need to check that $TH$ is perpendicular to the line through the centres $O_1$ of $(HMD)$ and $O_2$ of $(HNE)$ (which we get for free via symmetry). The former gradient is nice since $T$ and $H$ are nice, so it can't be that bad.
• I have faith that I can get this out before someone notices, so I decide to give it a shot. Set up in the usual way, so $A(0,1)$, $D(a,0)$, $E(b,0)$, so $H(0,-ab)$, $M\left(\frac{a}2,\frac12\right)$. (Note: if you want to practise your bashing, try following the outline above and see if you get the same expressions that I do below, which are reproduced verbatim from my working.)
• The perpendicular bisector of $MD$ is $y = a\left(x - \frac{3a}4\right) + \frac14 = ax + \frac{1-3a^2}4$.
• The perpendicular bisector of $MH$ is $y = -\frac1b\left(x - \frac{a}2\right) - \frac{ab}2 = -\frac1b x + \frac{a-ab^2}{2b}$.
• Intersecting: $$x = \frac{\frac{a-ab^2}{2b} - \frac{1-3a^2}4}{a+\frac1b} = \frac{\frac1{4b}(2a-2ab^2-b+3a^2b)}{\frac1b(ab+1)} = \frac{3a^2b+2a-b-2ab^2}{4(ab+1)}$$ and putting this into the first equation $$y = \frac{3a^3b+2a^2-ab-2a^2b^2+1-3a^2+ab-3a^3b}{4(ab+1)} = \frac{1-a^2-2a^2b^2}{4(ab+1)}.$$
• We are so incredibly lucky that the shared denominator of $x,y$ is symmetric, so when we swap $a,b$, they're the same.
• To calculate the gradients, we just need to look at the numerators. Moreover, the symmetric terms in the $y$ numerator cancel out, and swapping $a,b$ for $x$ gives the same terms, but with different coefficients. One may immediately see that the gradient of $O_1O_2$ is $$\frac{b^2-a^2}{5(a^2b-ab^2) + 3(a-b)} = -\frac{a+b}{5ab+3}.$$
• On the other hand, gradient $TH$ is $\frac{\frac23 + ab}{\frac{2(a+b)}3} = \frac{2+3ab}{2(a+b)}$...what?!

Oops

• I stare at my bash trying to figure out what went wrong, but I can't find anything wrong with it.
• Eventually I decide to just check that my $T$ was correct: I check the gradient between $\left(\frac{a+b}6,\frac23\right)$ and $M$ and compare it to the gradient of $MB$ and they are...different. Oh.
• I look at the Menelaus diagram again: oops, we have $\frac{AT}{TL} = \frac{AM}{MD}\frac{BD}{BL} = \frac23$ aaaaaaaaaaahhhhhhhhhhh.
• Ok, so I should have had $T\left(\frac{a+b}5,\frac35\right)$, so gradient $TH$ is $$\frac{\frac35+ab}{\frac{a+b}5} = \frac{3+5ab}{a+b}.$$ Phew.

So that's a solution for this problem. My bash fit into a tiny corner of my page: the following picture is (approximately) a 7cm by 12cm rectangle.

It later turned out that this was efficient enough to get added as an official solution, which is nice, I guess? Anyway, back to the problem.

Back to synthetic

• I realise that I have ignored the possible Reim approach. But even after applying Reim, we still have to deal with $P,Q$ which are not nice, and I don't know how to deal with that. Sorry Ker Yang.
• I return to trying to show that $T$ is on the radical axis, which lets us get rid of $P,Q$.
• The length ratio on $T$ is kind of cursed: I don't really know what to do with $2:3$.
• It might be possible that we don't actually need $M,N$ to be the midpoints, but maybe all we need is that $MN||BC$? It's not immediately clear to me whether this might be true and I don't have a ruler or a compass, so I check it on geogebra. (In a real contest, I would have probably just modified my coordinate bash to see what would happen. Or maybe I'd have used the ruler and compass that I totally would have brought to draw a better diagram.) It turns out to be false, which means that the $2:3$ ratio actually matters?! Ew.
• The circle $(DMH)$ is kind of weird; I don't really know what to do with it.
• I realise that this is exactly the sort of problem that I would normally solve by bashing, which means I really don't know how to approach it synthetically.
• Maybe I could inverflect to swap $D\leftrightarrow N$ and $E\leftrightarrow M$? But $H$ doesn't go to anywhere nice.
• I could try to do some length-chasing to compute the power of point at $T$, but that doesn't seem any less bashy than the coordinates approach.
• Adding $G$ back to the median, $T,G$ divide it into a $6:4:5$ ratio, which is...sad.

Matching known configurations

• I recall that $HM$ is a nice line: it intersects $(ADE)$ at the antipode $E'$ of $E$ and the second intersection of $(EH)$ (circle with diameter $EH$) and $(ADE)$. Can't believe I forgot that.
• I draw in a bunch of antipodes and reflections of $H$ across the sides of the triangle and try to angle-chase (sigh) but don't get anywhere (double sigh).
• It really looks like there should be some solution along these lines because there are so many parallel lines and midpoints, but I can't see anything.
• There's some super messy-looking diagram on my working that I clearly spent a lot of time on, but I can't really tell what I was doing:

• Let's go back to the $6:4:5$ ratio. $6:4$ is just $3:2$, which is still bad, but maybe we can work with that. It would be nice if we could get this involved with a Menelaus with a $1:2$ ratio...oh, wait. Don't $O,G,H$ satisfy that?
• I can't remember whether it's $OG:GH$ or $HG:GO$ that's $1:2$, so I quickly recall what they are in coordinates and get that $HG:GO = 1:2$. Hence, applying Menelaus tells us that $HT$ intersects $AO$ at midpoint $AO$. Progress at last!

• Let $O'$ be the midpoint of $AO$. Then, it suffices to show that $O'$ is on the radical axis of the two circles. How do we compute this?
• Hold on: $O'M = \frac12 OB = \frac12 OC = O'N$. This looks like it might go somewhere.
• Actually, by Thales, $O'$ is the circumcentre of $AMON$.
• If we to extend $O'M$ to meet $(HMD)$ again at $X$, and similarly construct $Y$ on the other side, so we want $O'M\cdot O'X = O'N\cdot O'Y$, and then the equal lengths cancel out and we just need $O'X = O'Y$.

Moving lengths around

• I notice that $(HMD)$ goes to a possibly nicer circle when reflected across $M$: it's now $(AE'M)$, where $E'$ is the antipode of $E$ on $(ADE)$.
• Let $O_1'$ be the reflection of $O'$ across $M$, and $X'$ be the second intersection of $(AC'M)$ and line $BM$. Then, it suffices to show that $O_1'X'$ is equal to the corresponding length on the other side.
• But $O_1'X' = O_1'M + MX' = O_1'M + MO' + O'X'$ (directed lengths), so it suffices to show that $O'X' = O'Y'$ (in a directed sense, i.e. either $X',Y'$ both lie on rays $O'M, O'N$ or both don't).
• This just means that $O'$ lies on the radical axis of the $(AE'M)$ and $(AD'N)$, i.e. that these the radical axis of these two circles is just $AO$.

• We can now restate the problem, relabelling $\triangle ADE$ as $\triangle ABC$:

Given a triangle $ABC$, let $B',C'$ be the antipodes of $B,C$ on $(ABC)$. Let $M,N$ be the midpoints of $AB,AC$ and $O$ be the circumcentre of $\triangle ABC$. Prove that the radical axis of $(AC'M)$ and $(AB'N)$ is $AO$.

• This looks a lot better! All the points are now nice points, so surely this has to work. 

Overkill

• I notice that all the three circles in this new formulation pass through $A$, so inversion would probably be helpful. Indeed, this becomes a problem with only lines!

Given a triangle $ABC$, let $B',C'$ be on the line $BC$ such that $AB'\perp AB$, $AC'\perp AC$. Let $M,N$ be the reflections of $A$ over $B,C$ respectively. Prove that $MC'$ and $NB'$ intersect on the altitude of $A$.

• This is trivial with coordinate geometry. But the whole point of this endeavour was to find a non-bash solution.
• This is almost certainly a known problem, but I can't place it and also I don't know how to solve it synthetically, because I have always solved anything that remotely resembles this by bashing.
• Even so, there are only lines so it can't be hard, but I can't figure out how to prove it.
• After failing for a bit, I try to label some lengths but this ends up being dangerously close to a trigo bash, which is 1) a bash 2) worse than the coordinates approach, anyway.
• Maybe some harmonics thing might work? There are right angles, after all.
• But right angles need to be paired with angle bisectors to create harmonics, which is aren't present here.
• We could try the next best thing, which is cross-ratios. Let $MC'$ and $NB'$ intersect the line through $A$ parallel to $BC$ at $P,Q$. Let $MC'$ and $NB'$ intersect the altitude through $A$ at $Z,Z'$. Then, $$(P,M;C',Z) = A(P,M;C',Z) = A(Z',B';N,Q) = (Z',B';N,Q)$$ where the middle equality comes from rotating the pencil by $90^\circ$.

• So we are done, right? No...we're only done if those points are harmonic. What we do want is $(P,M;C',Z) = (Q,N;B',Z')$, which is not the same thing. That's really weird.
• I am confused by this for quite a while before I realise that I still haven't used one condition: that $B,C$ are the midpoints of $AM,AN$. This means that $PC'=C'M$ and $QB'=B'N$, which means that $$\frac{MZ}{ZP} = (P,M;C',Z) = (Z',B';N,Q) = \frac{NZ'}{Z'Q}.$$
• Now, letting $D$ be the foot of perpendicular of $A$ on $MN$, the LHS is $\frac{DZ}{ZA}$ while the RHS is $\frac{DZ'}{Z'A}$, which tells us that $Z=Z'$ and at last, we are done.

Both solutions combined took me about 2.5-3 hours in total. I'm guessing that the bash took me about 10-15 minutes, including fixing the mistake.

Post-mortem

Comparing to the official solutions

We went through the official solutions, and it turned out that there were much quicker synthetic solutions. The intended solution is the first one on AoPS (post #2), which uses Reim (as I guessed), except it's used on two new points that you have to draw into the diagram (which I didn't) instead of $BC$. After you construct these two points, every step is just some angle-chasing.

The other main method is to show that $O'$ lies on the radical axis. There's a much faster way than what I did via angle-chasing: see posts #3 and #8 or AoPS, for example.

I realise the solution I came up with doesn't talk about angles at all, which reminds me of Yan Sheng's comment on my post on APMO 2024/1:

So, yeah, that seems like a recurring theme. Maybe I should have learnt how to do geometry properly when I was younger.

Remarks on the bash

My coordinate geometry solution used a couple of tiny synthetic observations, but the bash is still very much feasible without them:

• Instead of the Menelaus argument to get $AT:TL$, one could just intersect $BM$ and $CN$ computationally. (In fact, this would probably have been faster for me because then I wouldn't have made the stupid mistake.)
• Even if one doesn't reframe the argument in terms of $T$ being on the radical axis, it's not hard to straight-up check that the four points are cyclic: one could instead check that the perpendicular bisectors of $MP$, $MN$ and $CQ$ are concurrent. This is nicer than it sounds and is basically just one more step from what I've already done: the perpendicular bisector of $MP$ is the line through $O_1$ (which we already know) perpendicular to $BM$. One just needs to plug in $x=\frac{a+b}4$ (the equation for the perpendicular bisector of $MN$) and check that the $y$ expression is symmetric.

This is, I think, a good example of abusing symmetry in a question to simplify a bash - if you start with a setup that allows for that, you are putting yourself in a very good position.

Simplifying the overkill

I can't believe I failed this Junior Team-level geometry problem. There's a much simpler way to finish this which I only realised while typing up this blog post. The trick to a (reasonable) synthetic is to think in coordinates but write down synthetic things. Let $\ell$ be the altitude of $A$. Then, it suffices to show that $\frac{d(M,\ell)}{d(C',\ell)} = \frac{d(N,\ell)}{d(B',\ell)}$. Letting $D$ be the foot of the perpendicular from $A$, this is equivalent to showing that $DB \times DB' = DC \times DC'$, which is true since both sides are equal to $DA^2$ (or $-DA^2$ if you're directing lengths).

A faster finish

Here's a quicker way to show $O'X=O'Y$, courtesy of the one perfect scorer of this year's EGMO: We just need to show $XH||BC$, so $YH||BC$ by symmetry and so $XY||BC$. Indeed, $\angle XHD = \angle XMD = \angle O'MA = 90^\circ - \angle DEA = \angle EDH$, so we're done. None of the official solutions solved the problem by showing $O'X=O'Y$ (and my approach was rejected because it was too unhinged), so that's pretty nice.

My diagram was not anywhere close to being accurate enough for me to have guessed this, but it's cool that $XY$ actually passes through a useful point (i.e. $H$). Kudos to her for finding this.

Anyway, that's it for now. Stay tuned for next week's post on the rest of the EGMO problems.