Look, I didn't bash a geometry problem!

This is Glen again.

There's been a lack of regular olympiad content recently, and since no one posted this week I thought I'd attempt a problem and write about my thought process. There was some discussion recently on the retirees' Discord server about APMO, so I figured I'd look at the most recent iteration, and since I wanted to be able to solve the problem in a reasonable amount of time, I attempted Q1.

Here's the problem:

(APMO 2024/1) Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $BQX$ and $CPX$ intersect at a point $Y \neq X$. Prove that the points $A, X$, and $Y$ are collinear.

Preliminary thoughts

First and foremost, here's a diagram:

Sorry about the quality - I'm just scanning my working and copy-pasting the diagrams. I think it makes it more authentic, anyway.

My first thought was that it's probably bashable. Here's a reasonable-looking approach:

• Set up $A,B,C$ in the usual way with $BC$ the $x$-axis, then get $D,E$ by intersecting some $y=c$ with $AB,AC$.
• Set $X$ to be some random point and solve for $P,Q$.
• Solve for circumcentre of $BQX$ and the $CPX$ one is similar. Call these $O_1,O_2$.
• $XY$ is the line through $X$ perpendicular to $O_1O_2$. Actually, it just suffices to show that $AX\perp O_1O_2$.

 This shouldn't be that bad, since ultimately gradient $AX$ is nice, so stuff should cancel. But also, there are like 5 variables and also this is a Q1 so really I should have enough faith in myself to be able to find a synethetic solution in a shorter time than it takes to carry out the bash.

Other initial ideas:

• This construction of intersecting circles to get $Y$ reminds me a little of the construction of a spiral centre.
• If we let $(BXQ)$ intersect $AB$ again at $K$ and $(CXP)$ intersect $AC$ again at $AL$, then it suffices to show that $AB\cdot AK = AC \cdot AL$, which is equivalent to $DKEL$ cyclic (by Reim).
• Returning to the spiral centre thing, $Y$ is the spiral centre of $KQ$ and $PL$, but that's not very useful, since in the second bullet point we have pretty much gotten rid of $Y$.
• $X$ looks like a good point to invert about as a bunch of things pass through it and the parallel condition goes to something similar. There could also be some sort of shenanigans where we swap $D,Q$ and $E,P$ or something. But this seems overkill for a Q1.

Reframing the problem

Let's instead try to draw the points in a different order. We can instead draw the trapezoid $DEPQ$ first. Then, $B,C$ are points on the line $PQ$, and $A = BD \cap CE$. We're left with showing that $BD,CE$ and the radical axis of $(BXQ),(CXP)$ are concurrent.

What does this look like algebraically? The circumcentre $O_1$ of $(BXQ)$ is on the perpendicular bisector of $XQ$, which is a fixed line. As $B$ moves linearly (at constant velocity) along $PQ$, $O_1$ moves linearly along this line. If we fix $C$ and move $B$, this means $BD\cap CE$ and the radical axis of the $(BXQ),(CXP)$ move "cross-ratio-ly", so it suffices to check 3 special cases of $B$. So we have to check $3\times 3 = 9$ special cases, which is a lot, and also, overkill. Oops. This paragraph probably didn't make any sense if you haven't heard of moving points; I'll probably write a short appendix at the end of this.

Desperate inversion attempt

I was sort of out of ideas, so I tried to invert the diagram because why not.

The triangle is $XQ'P'$, and the circle is what $BC$ gets sent to. The line on top was me failing to convert the parallel condition. What I should have drawn is $(XD'E')$ tangent to $(XP'Q')$, which is just equivalent to $D'E'||Q'P'$. In any case, the lines $BD$ and $CE$ become circles through $X$, and that seemed like too much effort so I gave up.

Reframing the problem again

Let's try to draw the circles and $BC$ first, this time. I tried and failed a couple of times; here are some failed attempts:

Here's a successful attempt:

Here's what I want to show now:

Let $X$ be an intersection point of two circles $\Gamma_1$ and $\Gamma_2$. Let a line intersect $\Gamma_1$ at $B,Q$ and $\Gamma_2$ at $P,C$, with $B,Q,P,C$ in that order. Let $A$ be a (variable) point on the radical axis of $\Gamma_1,\Gamma_2$. Let $PX$ intersect $AB$ at $D$ and $QX$ intersect $AC$ at $E$. Prove that $DE||BC$.

This isn't quite the same as the original problem, but we can probably get between the two by some sort of phantom point argument.

I tried to extend $PX,QX$ to intersect the respective circles again, but I didn't see how any spiral stuff would help.

I also constructed the intersection of the radical axis and $BC$ - this has to satisfy some length condition that depends only on $B,Q,P,C$ (and not $X$). And then, the key realisation came: $E$ (and similarly, $D$) is only defined by half the points in the diagram!

Let $Z$ be the intersection of the radical axis and $BC$. Then, we can determine $\frac{AE}{EC}$ by Menelaus: $$\frac{AE}{EC} = \frac{AX}{XZ}\frac{QZ}{QC}.$$

Similarly, $$\frac{AD}{DB} = \frac{AX}{XZ}\frac{PZ}{PB}.$$

But $$\frac{QC}{QZ} = 1 + \frac{ZC}{QZ} = 1 + \frac{ZB}{PZ} = \frac{PB}{PZ},$$ so we are done.

Recovering the original formulation

Here's a phantom point argument, as promised: let $BD$ intersect $XY$ at $A'$, and then let $A'C$ intersect $QX$ at $E'$. Then what we've shown above shows that $DE'||BC||DE$, so $E=E'$ and hence $A=CE\cap BD$ and $A' = CE' \cap BD$ coincide.

Alternatively, we could reuse the proof of the alternate statement to prove the original version directly: let $AX$ intersect $BC$ at $Z$. Then, the two Menelaus computations above show that $Z$ is on the radical axis, so $A$ is as well.

Looking at AoPS

The first solution is a short angle-chase that shows the second bullet point in my initial thoughts: $DKEL$ cyclic. Oh well, I am allergic to angle-chasing. I suppose that if I had drawn an accurate diagram I might have noticed that $X$ was also on the circle.

There are a bunch of solutions with DDIT, which honestly makes me feel a bit better about my overkill ideas.

Someone also mentioned a 4-page coordinate bash, so I'm glad I didn't try that. I am, however, also confident that the bash is much shorter than that.

Revisiting the inversion approach

If you actually do the inversion properly (exercise!) with the knowledge that $DKXEL$ cyclic, the diagram ends up looking suspiciously similar to the original one. But instead, $D,E$ swap roles with $Q,P$, $B,C$ swap roles with $K,L$, and $A$ swaps roles with $Y$. One might then hope for something a bit less nebulous than "swapping roles", e.g. a proof along the lines "There's a Möbius transformation/inversion that swaps $D\leftrightarrow Q,E\leftrightarrow P,B\leftrightarrow K,C\leftrightarrow L,X\leftrightarrow\infty$, so $A\rightarrow Y$, hence done." Unfortunately, I think this question has too many degrees of freedom for that to work. There is indeed a Möbius transformation that swaps $D\leftrightarrow Q,E\leftrightarrow P,X\leftrightarrow\infty$, but this doesn't necessarily swap $B$ and $K$. Rather, it sends $K$ to some $B'$ for which the corresponding $K'$ (second intersection between $(B'QX)$ and $B'D$) is where $B$ gets sent. So that's somewhat disappointing.

It's possible that the DDIT approach is the way to fix this, but I'm not sure.

Bonus/appendix: big preserve cross-ratio

There's some general theory about moving points, but I'm using a baby version that only needs knowledge of cross-ratios. Ker Yang calls this big preserve cross-ratio because that's what it's called when you directly translate it from Chinese.

Let's fix $D,E,X,Q,P,C$, and vary $B$ linearly along $PQ$. We can define $A = BD\cap CE$, and also the radical axis to be the line through $X$ perpendicular to $O_1O_2$. Suppose that for three fixed points $B^1,B^2,B^3$ (I'm using superscripts so I don't clash notation with the $O_i$), $A$ indeed lies on the radical axis. Call these points $A^1,A^2,A^3$.

Now, for general $B$, projection through $D$ gives $(A,A^1;A^2,A^3) = (B,B^1;B^2,B^3)$.

On the other hand, since $O_1$ moves linearly with $B$, we also have $(O_1,O_1^1;O_1^2,O_1^3) = (B,B^1;B^2,B^3)$. Alternatively, this is projection through the point at infinity in the direction perpendicular to $BC$.

But now, $$X(Y,Y^1;Y^2,Y^3) = O_2(O_1,O_1^1;O_1^2,O_1^3) = (B,B^1;B^2,B^3) = X(A,A^1;A^2,A^3).$$

Since $XA^i$ coincides with $XY^i$ for $i=1,2,3$, then $XY$ coincides with $XA$, i.e. $A,X,Y$ collinear.

It remains to find $B^1,B^2,B^3$. The usual trick is to guess some degenerate/special cases:

• $B = P$: the radical axis is $XP$, so $A$ lies on it since it lies on $BD$ (which is now $PD$).
• $B = C$: the radical axis is $XC$, and $A = C$.
• $B = \infty_{PQ}$ (point at infinity on $PQ$): $BD$ is just $DE$, so $A=E$. On the other hand, $(BQX)$ is just $QX$, so the "radical axis" is $XE$.

So it turns out that this could be done without varying $C$ as well, which is nice. (I was expecting something even more horrible along the lines of: pick $3$ nice points for $B$, for each of those pick $3$ nice points for $C$, then check $9$ cases.)

Again, I should reiterate that this is massively overkill for this problem. This method does work for harder problems as well, however, so I expect at some point someone who is more familiar with this sort of stuff than I am will write a post about it.