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.


(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 \(ABC\).


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 \(ABC\).