International Mathematical Olympiad Shortlisted Problems

Denote by $\omega$ the circumcircle of the triangle $ABC$, and let $\angle ACB=\gamma$. Note that the condition $\gamma <\angle CBA$ implies $\gamma <90^{\circ }$. Since $\angle PBA=\gamma$, the line $PB$ is tangent to $\omega$, so $PA\cdot PC=P{B}^2=P{D}^2$. By $\frac{PA}{PD}=\frac{PD}{PC}$ the triangles $PAD$ and $PDC$ are similar, and $\angle ADP=\angle DCP$.

Next, since $\angle ABQ=\angle ACB$, the triangles $ABC$ and $AQB$ are also similar. Then $\angle AQB=\angle ABC=\angle ARC$, which means that the points $D,R,C$, and $Q$ are concyclic. Therefore $\angle DRQ=\angle DCQ=\angle ADP$.

Figure 1

Now from $\angle ARB=\angle ACB=\gamma$ and $\angle PDB=\angle PBD=2\gamma$ we get $$\angle QBR=\angle ADB-\angle ARB=\angle ADP+\angle PDB-\angle ARB=\angle DRQ+\gamma =\angle QRB,$$ so the triangle $QRB$ is isosceles, which yields $QB=QR$.

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Alternative solution.

Again, denote by $\omega$ the circumcircle of the triangle $ABC$. Denote $\angle ACB=\gamma$. Since $\angle PBA=\gamma$, the line $PB$ is tangent to $\omega$.

Let $E$ be the second intersection point of $BQ$ with $\omega$. If $V^{\prime }$ is any point on the ray $CE$ beyond $E$, then $\angle BE{V}^{\prime }=180^{\circ }-\angle BEC=180^{\circ }-\angle BAC=\angle PAB$; together with $\angle ABQ=\angle PBA$ this shows firstly, that the rays $BA$ and $CE$ intersect at some point $V$, and secondly that the triangle $VEB$ is similar to the triangle $PAB$. Thus we have $\angle BVE=\angle BPA$. Next, $\angle AEV=\angle BEV-\gamma =\angle PAB-\angle ABQ=\angle AQB$; so the triangles $PBQ$ and $VAE$ are also similar.

Let $PH$ be an altitude in the isosceles triangle $PBD$; then $BH=HD$. Let $G$ be the intersection point of $PH$ and $AB$. By the symmetry with respect to $PH$, we have $\angle BDG=\angle DBG=\gamma =\angle BEA$; thus $DG\parallel AE$ and hence $\frac{BG}{GA}=\frac{BD}{DE}$. Thus the points $G$ and $D$ correspond to each other in the similar triangles $PAB$ and $VEB$, so $\angle DVB=\angle GPB=90^{\circ }-\angle PBQ=90^{\circ }-\angle VAE$. Thus $VD\perp AE$.

Let $T$ be the common point of $VD$ and $AE$, and let $DS$ be an altitude in the triangle $BDR$. The points $S$ and $T$ are the feet of corresponding altitudes in the similar triangles $ADE$ and $BDR$, so $\frac{BS}{SR}=\frac{AT}{TE}$. On the other hand, the points $T$ and $H$ are feet of corresponding altitudes in the similar triangles $VAE$ and $PBQ$, so $\frac{AT}{TE}=\frac{BH}{HQ}$. Thus $\frac{BS}{SR}=\frac{AT}{TE}=\frac{BH}{HQ}$, and the triangles $BHS$ and $BQR$ are similar.

Finally, $SH$ is a median in the right-angled triangle $SBD$; so $BH=HS$, and hence $BQ=QR$.

Figure 2

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Alternative solution.

Denote by $\omega$ and $O$ the circumcircle of the triangle $ABC$ and its center, respectively. From the condition $\angle PBA=\angle BCA$ we know that $BP$ is tangent to $\omega$.

Let $E$ be the second point of intersection of $\omega$ and $BD$. Due to the isosceles triangle $BDP$, the tangent of $\omega$ at $E$ is parallel to $DP$ and consequently it intersects $BP$ at some point $L$. Of course, $PD\parallel LE$. Let $M$ be the midpoint of $BE$, and let $H$ be the midpoint of $BR$. Notice that $\angle AEB=\angle ACB=\angle ABQ=\angle ABE$, so $A$ lies on the perpendicular bisector of $BE$; thus the points $L,A,M$, and $O$ are collinear. Let ${\omega }_1$ be the circle with diameter $BO$. Let $Q^{\prime }=HO\cap BE$; since $HO$ is the perpendicular bisector of $BR$, the statement of the problem is equivalent to $Q^{\prime }=Q$.

Consider the following sequence of projections (see Fig. 3). 1. Project the line $BE$ to the line $LB$ through the center $A$. (This maps $Q$ to $P$.) 2. Project the line $LB$ to $BE$ in parallel direction with $LE$. ($P\mapsto D$.) 3. Project the line $BE$ to the circle $\omega$ through its point $A$. ($D\mapsto R$.) 4. Scale $\omega$ by the ratio $\frac{1}{2}$ from the point $B$ to the circle ${\omega }_1$. ($R\mapsto H$.) 5. Project ${\omega }_1$ to the line $BE$ through its point $O$. ($H\mapsto Q^{\prime }$.)

We prove that the composition of these transforms, which maps the line $BE$ to itself, is the identity. To achieve this, it suffices to show three fixed points. An obvious fixed point is $B$ which is fixed by all the transformations above. Another fixed point is $M$, its path being $M\mapsto L\mapsto E\mapsto E\mapsto M\mapsto M$.

Figure 3 Figure 4

In order to show a third fixed point, draw a line parallel with $LE$ through $A$; let that line intersect $BE,LB$ and $\omega$ at $X,Y$ and $Z\ne A$, respectively (see Fig. 4). We show that $X$ is a fixed point. The images of $X$ at the first three transformations are $X\mapsto Y\mapsto X\mapsto Z$. From $\angle XBZ=\angle EAZ=\angle AEL=\angle LBA=\angle BZX$ we can see that the triangle $XBZ$ is isosceles. Let $U$ be the midpoint of $BZ$; then the last two transformations do $Z\mapsto U\mapsto X$, and the point $X$ is fixed.