IMO 2019 Shortlisted Problems

Let $N$ and $M$ be the midpoints of the $\mathrm{arcs}\text{overparen}BC$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\angle FAE=\angle BAC=\angle BNC$, the right-angled kites $AFIE$ and $NBMC$ are similar. Consider the spiral similarity $\phi$ (dilation in case of $AB=AC$) that moves $AFIE$ to $NBMC$. The directed angle in which $\phi$ changes directions is $\angle (AF,NB)$, same as $\angle (AP,NP)$ and $\angle (AQ,NQ)$; so lines $AP$ and $AQ$ are mapped to lines $NP$ and $NQ$, respectively. Line $EF$ is mapped to $BC$; we can see that the intersection points $P=EF\cap AP$ and $Q=EF\cap AQ$ are mapped to points $BC\cap NP$ and $BC\cap NQ$, respectively. Denote these points by $P^{\prime }$ and $Q^{\prime }$, respectively.



Let $L$ be the midpoint of $BC$. We claim that points $P,Q,D,L$ are concyclic (if $D=L$ then line $BC$ is tangent to circle $PQD$). Let $PQ$ and $BC$ meet at $Z$. By applying Menelaus' theorem to triangle $ABC$ and line $EFZ$, we have $$\frac{BD}{DC}=\frac{BF}{FA}\cdot \frac{AE}{EC}=-\frac{BZ}{ZC},$$ so the pairs $B,C$ and $D,Z$ are harmonic. It is well-known that this implies $ZB\cdot ZC=ZD\cdot ZL$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $BC$, because the cross-ratio is preserved.) Hence, $ZD\cdot ZL=ZB\cdot ZC=ZP\cdot ZQ$ by the power of $Z$ with respect to the circumcircle; this proves our claim.

By $\angle MP{P}^{\prime }=\angle MQ{Q}^{\prime }=\angle ML{P}^{\prime }=\angle ML{Q}^{\prime }=90^{\circ }$, the quadrilaterals $MLP{P}^{\prime }$ and $MLQ{Q}^{\prime }$ are cyclic. Then the problem statement follows by $$\begin{array}{rl}\angle DPA+\angle AQD& =360^{\circ }-\angle PAQ-\angle QDP=360^{\circ }-\angle PNQ-\angle QLP\\ & =\angle LPN+\angle NQL=\angle P^{\prime }ML+\angle LM{Q}^{\prime }=\angle P^{\prime }M{Q}^{\prime }=\angle PIQ\end{array}.$$

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Alternative solution.

Define the point $M$ and the same spiral similarity $\phi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F,E$ to $B,C$ is the Miquel point of the lines $FE,BC,BF$ and $CE$; that is, the second intersection of circles $ABC$ and $AEF$. Denote that point by $S$.

By $\phi (F)=B$ and $\phi (E)=C$ the triangles $SBF$ and $SCE$ are similar, so we have $$\frac{SB}{SC}=\frac{BF}{CE}=\frac{BD}{CD}.$$ By the converse of the angle bisector theorem, that indicates that line $SD$ bisects $\angle BSC$ and hence passes through $M$.

Let $K$ be the intersection point of lines $EF$ and $SI$. Notice that $\phi$ sends points $S,F,E,I$ to $S,B,C,M$, so $\phi (K)=\phi (FE\cap SI)=BC\cap SM=D$. By $\phi (I)=M$, we have $KD\parallel IM$.



We claim that triangles $SPI$ and $SDQ$ are similar, and so are triangles $SPD$ and $SIQ$. Let ray $SI$ meet the circumcircle again at $L$. Note that the segment $EF$ is perpendicular to the angle bisector $AM$. Then by $\angle AML=\angle ASL=\angle ASI=90^{\circ }$, we have $ML\parallel PQ$. Hence, $\widehat{PL}=\widehat{MQ}$ and therefore $\angle PSL=\angle MSQ=\angle DSQ$. By $\angle QPS=\angle QMS$, the triangles $SPK$ and $SMQ$ are similar. Finally, $$\frac{SP}{SI}=\frac{SP}{SK}\cdot \frac{SK}{SI}=\frac{SM}{SQ}\cdot \frac{SD}{SM}=\frac{SD}{SQ}$$ shows that triangles $SPI$ and $SDQ$ are similar. The second part of the claim can be proved analogously.

Now the problem statement can be proved by $$\angle DPA+\angle AQD=\angle DPS+\angle SQD=\angle QIS+\angle SIP=\angle QIP.$$

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Alternative solution.

Denote the circumcircle of triangle $ABC$ by $\Gamma$, and let rays $PD$ and $QD$ meet $\Gamma$ again at $V$ and $U$, respectively. We will show that $AU\perp IP$ and $AV\perp IQ$. Then the problem statement will follow as $$\angle DPA+\angle AQD=\angle VUA+\angle AVU=180^{\circ }-\angle UAV=\angle QIP.$$

Let $M$ be the midpoint of $\mathrm{arc}\widehat{BUVC}$ and let $N$ be the midpoint of $\mathrm{arc}\widehat{CAB}$; the lines $AIM$ and $AN$ being the internal and external bisectors of angle $BAC$, respectively, are perpendicular. Let the tangents drawn to $\Gamma$ at $B$ and $C$ meet at $R$; let line $PQ$ meet $AU,AI,AV$ and $BC$ at $X,T,Y$ and $Z$, respectively.

As in Solution 1, we observe that the pairs $B,C$ and $D,Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B,C$ and $U,P$ are also harmonic. Analogously, the pair $V,Q$ is harmonic with $B,C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\Gamma$ by its harmonic pair with respect to $B,C$. In particular, the inversion maps points $B,C,N,U,V$ to points $B,C,M,P,Q$, respectively.

Combine the inversion with projecting $\Gamma$ from $A$ to line $PQ$; the points $B,C,M,P,Q$ are projected to $F,E,T,P,Q$, respectively.



The combination of these two transformations is projective map from the lines $AB,AC$, $AN,AU,AV$ to $IF,IE,IT,IP,IQ$, respectively. On the other hand, we have $AB\perp IF$, $AC\perp IE$ and $AN\perp AT$, so the corresponding lines in these two pencils are perpendicular. This proves $AU\perp IP$ and $AV\perp IQ$, and hence completes the solution.