THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

We first show that $Y$ lies on the circle $ABC$. To this end, let $Y^{\prime }$ be the point on the ray $MD$, emanating from $M$, such that $M{Y}^{\prime }\cdot MD=M{A}^2$. The triangles $MA{Y}^{\prime }$ and $MDA$ are therefore similar and have opposite orientations. Since $M{Y}^{\prime }\cdot MD=M{A}^2=M{C}^2$, so are the triangles $MC{Y}^{\prime }$ and $MDC$. Hence, using directed angles, $\angle A{Y}^{\prime }C=\angle A{Y}^{\prime }M+\angle M{Y}^{\prime }C=\angle MAD+\angle DCM=\angle CDA=\angle ABC$, so $Y^{\prime }$ lies on the circle $ABC$.

Let the lines $AD$ and $BC$ meet at $Z$. The equality of directed angles $\angle PDZ=\angle PBC=\angle PBZ$ shows that $Z$ lies on the circle $BDP$; and the equality of directed angles $\angle Y^{\prime }BZ=\angle Y^{\prime }BC=\angle Y^{\prime }AC=\angle Y^{\prime }AM=\angle Y^{\prime }DZ$ shows that so does $Y^{\prime }$. Since the angle $ADC$ is acute, $MA\ne MD$, so $M{Y}^{\prime }\ne MD$, and $Y^{\prime }$ is therefore the point where $MD$ crosses the circle $BDP$ again. Consequently, $Y^{\prime }=Y$, so the latter lies on the circle $ABC$, by the preceding paragraph.



Next, we show that $FY$ is the internal bisectrix of the angle $AYC$. To this end, simply refer to the internal angle bisectrix theorem and the similar triangles above, to write $$\frac{FA}{FC}=\frac{AD}{CD}=\frac{AD}{AM}\cdot \frac{CM}{CD}=\frac{YA}{YM}\cdot \frac{YM}{YC}=\frac{YA}{YC}.$$ Let now $BE$, the internal bisectrix of the angle $ABC$, cross the circle $ABC$ again at $B^{\prime }$, so the latter is the midpoint of the arc $AC$ not containing $B$. The line $B^{\prime }Y$ is therefore the external bisectrix of the angle $AYC$, so it is perpendicular to $FY$, by the preceding paragraph.

Let the line $\ell$ through $P$ and parallel to $AC$ meet the line $B^{\prime }Y$ at $S$. To prove the required perpendicularity, we show $PQ$ and $PS$ perpendicular.

Use directed angles to write $\angle PSY=\angle (AC,B^{\prime }Y)=\angle ACY+\angle CY{B}^{\prime }=\angle ACY+\angle CA{B}^{\prime }=\angle ACY+\angle B^{\prime }CA=\angle B^{\prime }CY=\angle B^{\prime }BY=\angle PBY$, and infer that $S$ lies on the circle $BDP$. Restated, the line through $Y$ and perpendicular to $FY$ passes through the point where $\ell$ crosses the circle $BDP$ again, which is $S$. Similarly, the line through $X$ and perpendicular to $EX$ passes through the point where $\ell$ crosses the circle $BDP$ again, which is $S$. Consequently, $QX$ and $QY$ are perpendicular to $SX$ and $SY$, respectively, so $Q$ lies on the circle $BDP$ of which $QS$ is a diameter. The conclusion follows.

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Alternative solution.

Begin by noticing that, since $\angle ABC=\angle ADC$, the circles $ABC$ and $ACD$ are reflections of one another across $M$.

We first show that $X$ lies on the circle $ACD$. To this end, let the ray $MB$, emanating from $M$, cross the circle $ACD$ at $X_1$, and let $X^{\prime }$ be the reflection of $X_1$ across $M$. By the preceding, $X^{\prime }$ lies on the circle $ABC$ and $A{X}_{1}C{X}^{\prime }$ is a parallelogram, so, using directed angles, $\angle D{X}_{1}B=\angle D{X}_{1}A+\angle A{X}_{1}B=\angle DCA+\angle A{X}_1{X}^{\prime }=\angle DCA+\angle C{X}^{\prime }{X}_1=\angle DCA+\angle C{X}^{\prime }B=\angle DCA+\angle CAB=\angle (CD,AB)$. The equality of directed angles $\angle DPB=\angle PDC+\angle (CD,AB)+\angle ABP=\angle (CD,AB)$ therefore implies that $\angle D{X}_{1}B=\angle DPB$, so $X_1$ lies on the circle $BDP$. It follows that $X_1$ and $X$ coincide, so $X$ lies on the circle $ACD$. Similarly, $Y$ lies on the circle $ABC$.

Next, we show that $Q$ lies on the circle $BDP$. To this end, let the perpendicular bisectrix of the segment $AC$ cross the circle $ABC$ at $B^{\prime }$ and $M_1$, and the circle $ACD$ at $D^{\prime }$ and $M_2$, so that $B,D^{\prime }$ and $M_1$ all lie in the same half-plane relative to the line $AC$. Notice that $B^{\prime }$ and $D^{\prime }$ lie on the bisectrices $BP$ and $DP$, respectively, of the angles $ABC$ and $ADC$, respectively. Further, notice that $$\frac{BA\cdot X^{\prime }A}{BC\cdot X^{\prime }C}=\frac{\text{area }BA{X}^{\prime }}{\text{area }BC{X}^{\prime }}=\frac{MA}{MC}=1,$$ and refer to the fact that $BE$ bisects the angle $ABC$, to write $$\frac{EA}{EC}=\frac{BA}{BC}=\frac{X^{\prime }C}{X^{\prime }A}=\frac{XA}{XC},$$ and infer that $XE$ bisects the angle $AXC$, so $M_2$ lies on the line through $E,Q,X$.

Similarly, $M_1$ lies on the line through $F,Q,Y$. Use directed angles to write $\angle XQY=\angle M_{2}Q{M}_1=\angle Q{M}_2{M}_1+\angle M_2{M}_{1}Q=\angle X{M}_2{D}^{\prime }+\angle B^{\prime }{M}_{1}Y=\angle XD{D}^{\prime }+\angle B^{\prime }BY=\angle XDP+\angle PBY=\angle XBP+\angle PBY=\angle XBY$, and infer that $Q$ lies on the circle $BDP$.

Finally, since $M_1$ and $M_2$ are reflections of one another across $M$, the quadrangle $X{M}_1{X}^{\prime }{M}_2$ is a parallelogram, so, using directed angles, $\angle XQP=\angle XBP=\angle X^{\prime }B{B}^{\prime }=\angle X^{\prime }{M}_1{B}^{\prime }=\angle X{M}_2{M}_1$. This shows that $PQ$ and $M_1{M}_2$ are parallel; since the latter is perpendicular to $AC$, so is the former.

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Alternative solution.

If one of the vertices $B$, $D$ lies on the perpendicular bisectrix of the diagonal $AC$, so do both $P$ and $Q$, and the conclusion follows.

Assuming now that neither $B$ nor $D$ lie on the perpendicular bisectrix of the diagonal $AC$, we show that $P$ and $Q$ project orthogonally to the same point on the line $AC$. To this end, project $P$ and $Q$ on the line $AC$ to $P^{\prime }$ and $Q^{\prime }$, respectively, and notice that the two coincide if and only if, in terms of directed segments, $E{P}^{\prime }/F{P}^{\prime }=E{Q}^{\prime }/F{Q}^{\prime }$. Notice that $$\frac{E{P}^{\prime }}{F{P}^{\prime }}=\frac{\tan \angle EFP}{\tan \angle FEP}=\frac{\tan \angle AFD}{\tan \angle AEB}\text{and}\frac{E{Q}^{\prime }}{F{Q}^{\prime }}=\frac{\tan \angle EFQ}{\tan \angle FEQ},$$ where angles are all directed, so it is sufficient to show that $$(\tan \angle AEB)(\tan \angle EFQ)=(\tan \angle AFD)(\tan \angle FEQ).$$ To this end, let the line through $B$ and perpendicular to $BE$ meet the line $AC$ at $T$. We will prove that $B,E,T,X$ are concyclic, so, in terms of directed angles, $\angle FEQ=\angle TEX=\pi /2+\angle EBM$. Similarly, $\angle EFQ=\pi /2+\angle FDM$, so it is sufficient to show that $$(\tan \angle AEB)(\tan \angle MBE)=(\tan \angle AFD)(\tan \angle MDF).$$ With reference to the sine law, straightforward calculations show that the left-hand member is $(\tan \angle EBA{)}^2$, the right-hand member is $(\tan \angle FDC{)}^2$, and the conclusion follows by equality of the two angles.

We still have to prove that $B,E,T,X$ are concyclic. Since $BT$ and $BX$ are the two bisectrices of the angle $ABC$, the cross-ratio $(T,E;A,C)$ is harmonic, so $ME\cdot MT=M{A}^2$.

It is therefore sufficient to show that $M{A}^2=MB\cdot MX$. To this end, let the lines $AB$ and $CD$ meet at $W$, and let the lines $AD$ and $BC$ meet at $Z$. Use directed angles to write $\angle PDZ=\angle PBC=\angle PBZ$ and infer that $Z$ lies on the circle $BDP$; similarly, so does $W$, and the quadrangle $BDWZ$ is therefore cyclic.

Reflect $C$ in the circle $BDWZ$, centered at $O$, to obtain $C^{\prime }$. Since $A$ lies on the polar of $C$ with respect to this circle, the lines $A{C}^{\prime }$ and $OC$ are perpendicular, so $C^{\prime }$ lies on the circle $\gamma$ on diameter $AC$. It follows that reflection in the circle $BDWZ$ sends $\gamma$ to itself, so the two circles are orthogonal. Consequently, the power of $M$ with respect to the circle $BDWZ$ is the square of the radius of $\gamma$; that is, $MB\cdot MX=M{A}^2$. This completes the proof.