# 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 $ABC$.