# Spiral similarity solves an IMO problem

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While recovering from a knee surgery, I entertained myself by solving a geometry problem from the last International Mathematical Olympiad. My solution, shown below, is an example of using plane transformations (spiral similarity, in this case) to prove geometric statements. ## Problem

(IMO-2014, P4)

Points $$P$$ and $$Q$$ lie on side $$BC$$ of acute-angled triangle $$ABC$$ such that $$\angle PAB=\angle BCA$$ and $$\angle CAQ = \angle ABC$$. Points $$M$$ and $$N$$ lie on lines $$AP$$ and $$AQ$$, respectively, such that $$P$$ is the midpoint of $$AM$$ and $$Q$$ is the midpoint of $$AN$$. Prove that the lines $$BM$$ and $$CN$$ intersect on the circumcircle of triangle $$ABC$$.

## Solution

Let $$\angle BAC = \alpha$$.

$\angle APB = \pi - \angle PAB - \angle PBA = \pi - \angle ACB - \angle CBA = \alpha$ Let $$B_1$$ and $$C_1$$ be such points that $$B$$ and $$C$$ are midpoints of $$AB_1$$ and $$AC_1$$, respectively.

Consider a spiral similarity $$h$$ such that $$h(B_1)=A$$ and $$h(B)=C$$ (it necessarily exists).

Now we shall prove that $$h(M)=N$$, i.e. that $$h$$ transforms the green $$\triangle B_1BM$$ into the magenta $$\triangle ACN$$ .

Being a spiral similarity, $$h$$ rotates all lines by the same angle. It maps $$B_1B$$ to $$AC$$, therefore that angle equals $$\angle(B_1B, AC)=\pi-\alpha$$. (We need to be careful to measure all rotations in the same direction; on my drawing it is clockwise.)

$$h(A)=C_1$$, since $$h$$ preserves length ratios. So $$h(AM)$$ (where $$AM$$ denotes the line, not the segment) is a line that passes through $$h(A)=C_1$$. It also needs to be parallel to $$BC$$, because $$\angle (AM,BC)=\pi-\alpha$$ is the rotation angle of $$h$$. $$C_1B_1$$ is the unique such line ($$C_1B_1 \parallel BC$$ by the midline theorem).

Since $$h(AM)=C_1B_1$$ and $$h(MB_1)=NA$$, $h(M)=h(AM\cap MB_1)=h(AM)\cap h(MB_1)=C_1B_1\cap NA=N.$

Now that we know that $$h(BM)=CN$$, we can deduce that $$\angle BZC=\angle(BM,CN)=\pi-\alpha$$ (the rotation angle). And because $$\angle BAC+\angle BZC=\pi$$, $$Z$$ lies on the circumcircle of $$ABC$$.