IMO 2022 · Problem 4

Let \(ABCDE\) be a convex pentagon such that \(BC = DE\). Assume that there is a point \(T\) inside \(ABCDE\) with \(TB = TD\), \(TC = TE\) and \(\angle ABT = \angle TEA\). Let line \(AB\) intersect lines \(CD\) and \(CT\) at points \(P\) and \(Q\), respectively. Assume that the points \(P, B, A, Q\) occur on their line in that order. Let line \(AE\) intersect lines \(CD\) and \(DT\) at points \(R\) and \(S\), respectively. Assume that the points \(R, E, A, S\) occur on their line in that order. Prove that the points \(P, S, Q, R\) lie on a circle.

Official text of the problem. Official problem paper (PDF)

Solution

By convexity and the prescribed orders, the supporting lines \(AB,AE,CD\) bound a triangle \(APR\) containing the pentagon. The points on its side \(PR\) occur in the order \(P,C,D,R\), and \(T\) lies strictly inside this triangle.

Moreover, \(Q,T,C\) occur in that order: \(C,T\) lie on the same side of \(AB\), while \(Q,T\) lie on the same side of \(CD\). Similarly, \(S,T,D\) occur in that order.

The three length equalities give \[ \triangle TBC\cong\triangle TDE \] by SSS, so \[ \angle BTC=\angle DTE. \] Since \(BQ\) and \(BA\) are the same ray, and \(ES\) and \(EA\) are the same ray, the given angle equality gives \[ \angle TBQ=\angle TES. \] Also, \(TQ\) is opposite to \(TC\), and \(TS\) is opposite to \(TD\). Consequently, \[ \angle BTQ =180^\circ-\angle BTC =180^\circ-\angle DTE =\angle ETS. \] Thus \[ \triangle TBQ\sim\triangle TES. \] In particular, \[ \frac{TQ}{TS}=\frac{TB}{TE}=\frac{TD}{TC}, \qquad \angle BQT=\angle EST. \] Using the point orders, set \[ \alpha=\angle PQC=\angle TSR. \]

The length ratio above can be rewritten as \[ \frac{TC}{TS}=\frac{TD}{TQ}. \] Furthermore, \(\angle CTD=\angle STQ\), because their respective sides are opposite rays. Hence, by SAS, \[ \triangle TCD\sim\triangle TSQ. \] Set \[ \beta=\angle TCD=\angle TSQ. \]

In triangle \(PCQ\), the rays \(CP\) and \(CD\) are opposite, while \(CQ\) and \(CT\) coincide. Therefore \[ \angle PCQ=180^\circ-\beta, \] and the angle sum gives \[ \angle RPQ=\angle CPQ =180^\circ-(180^\circ-\beta)-\alpha =\beta-\alpha. \]

At \(S\), ray \(SR\) lies inside \(\angle TSQ\). Indeed, \(A\) lies between \(P\) and \(Q\), so ray \(SA\), which is ray \(SR\), lies between \(SP\) and \(SQ\). Also, \(T\) lies inside triangle \(SPR\), so ray \(ST\) lies between \(SP\) and \(SR\). It follows that \[ \angle RSQ =\angle TSQ-\angle TSR =\beta-\alpha =\angle RPQ. \]

Finally, \(P\) and \(S\) lie on the same side of \(RQ\), since \(A\) lies inside both segments \(PQ\) and \(RS\). The equality \(\angle RSQ=\angle RPQ\) therefore proves that \(P,S,Q,R\) lie on a circle.

The solution was checked against the official IMO shortlist solutions.