Girls European Mathematical Olympiad

Let $\omega$ be the circumcircle of $△ABC$. Reflecting $\omega$ in line $BC$, we obtain circle ${\omega }^{\prime }$ which, obviously, contains points $H$ and $P^{\prime }$. Let $M$ be the midpoint of $BC$. As triangle $△ABC$ is acute-angled, then $H$ and $O$ lie inside this triangle.

Let us assume that $\angle CAB=60^{\circ }$. Since $$\angle COB=2\angle CAB=120^{\circ }=180^{\circ }-60^{\circ }=180^{\circ }-\angle CAB=\angle CHB,$$ hence $O$ lies on ${\omega }^{\prime }$. Reflecting $O$ in line $BC$, we obtain point $O^{\prime }$ which lies on $\omega$ and this point is the center of ${\omega }^{\prime }$. Then $O{O}^{\prime }=2OM=2R\cos \angle CAB=AH$, so $AH=O{O}^{\prime }=H{O}^{\prime }=AO=R$, where $R$ is the radius of $\omega$ and, naturally, of ${\omega }^{\prime }$. Then quadrilateral $AH{O}^{\prime }O$ is a rhombus, so $A$ and $O^{\prime }$ are symmetric to each other with respect to $HO$. As $H,G$ and $O$ are collinear (Euler line), then $\angle GAH=\angle H{O}^{\prime }G$. Diagonals of quadrilateral $GOP{O}^{\prime }$ intersect at $M$. Since $\angle BOM=60^{\circ }$, so $$OM=M{O}^{\prime }=\mathrm{ctg}60^{\circ }\cdot MB=\frac{MB}{\sqrt{3}}.$$ As $3\cdot MO\cdot M{O}^{\prime }=M{B}^2=MB\cdot MC=MP\cdot MA=3MG\cdot MP$, then $GOP{O}^{\prime }$ is cyclic. Since $BC$ is a perpendicular bisector of $O{O}^{\prime }$, so the circumcircle of quadrilateral $GOP{O}^{\prime }$ is symmetrical with respect to $BC$. Thus $P^{\prime }$ also belongs to the circumcircle of $GOP{O}^{\prime }$, hence $\angle G{O}^{\prime }{P}^{\prime }=\angle GP{P}^{\prime }$. Note that $\angle GP{P}^{\prime }=\angle GAH$ since $AH\parallel P{P}^{\prime }$. And as it was proved $\angle GAH=\angle H{O}^{\prime }G$, then $\angle H{O}^{\prime }G=\angle G{O}^{\prime }{P}^{\prime }$. Thus triangles $△H{O}^{\prime }G$ and $△G{O}^{\prime }{P}^{\prime }$ are equal and hence $HG=G{P}^{\prime }$.

Now we will prove that if $HG=G{P}^{\prime }$ then $\angle CAB=60^{\circ }$. Reflecting $A$ with respect to $M$, we get $A^{\prime }$. Then, as it was said in the first part of solution, points $B,C,H$ and $P^{\prime }$ belong to ${\omega }^{\prime }$. Also it is clear that $A^{\prime }$ belongs to ${\omega }^{\prime }$. Note that $HC\perp C{A}^{\prime }$ since $AB\parallel C{A}^{\prime }$ and hence $H{A}^{\prime }$ is a diameter of ${\omega }^{\prime }$. Obviously, the center $O^{\prime }$ of circle ${\omega }^{\prime }$ is midpoint of $H{A}^{\prime }$. From $HG=G{P}^{\prime }$ it follows that $△HG{O}^{\prime }$ is equal to $△P^{\prime }G{O}^{\prime }$. Therefore $H$ and $P^{\prime }$ are symmetric with respect to $G{O}^{\prime }$. Hence $G{O}^{\prime }\perp H{P}^{\prime }$ and $G{O}^{\prime }\parallel A^{\prime }{P}^{\prime }$. Let $HG$ intersect $A^{\prime }{P}^{\prime }$ at $K$ and $K\notin O$ since $AB\ne AC$. We conclude that $HG=GK$, because line $G{O}^{\prime }$ is midline of the triangle $△HK{A}^{\prime }$. Note that $2GO=HG$, since $HO$ is Euler line of triangle $ABC$. So $O$ is midpoint of segment $GK$. Because of $\angle CMP=\angle CM{P}^{\prime }$, then $\angle GMO=\angle OM{P}^{\prime }$. Line $OM$, that passes through $O^{\prime }$, is an external angle bisector of $\angle P^{\prime }M{A}^{\prime }$. Also we know that, then $O^{\prime }$ is the midpoint of arc $P^{\prime }M{A}^{\prime }$. It follows that quadrilateral $P^{\prime }M{O}^{\prime }{A}^{\prime }$ is cyclic, then $\angle O^{\prime }M{A}^{\prime }=\angle O^{\prime }{P}^{\prime }{A}^{\prime }=\angle O^{\prime }{A}^{\prime }{P}^{\prime }$. Let $OM$ and $P^{\prime }{A}^{\prime }$ intersect at $T$. Triangles $△T{O}^{\prime }{A}^{\prime }$ and $△A^{\prime }{O}^{\prime }M$ are similar, hence $\frac{O^{\prime }{A}^{\prime }}{O^{\prime }M}=\frac{O^{\prime }T}{O^{\prime }{A}^{\prime }}$. In other words, $O^{\prime }M\cdot O^{\prime }T=O^{\prime }{A}^{\prime 2}$. Using Menelaus' theorem for triangle $△HK{A}^{\prime }$ and line $T{O}^{\prime }$, we obtain that $$\frac{A^{\prime }{O}^{\prime }}{O^{\prime }H}\cdot \frac{HO}{OK}\cdot \frac{KT}{T{A}^{\prime }}=3\cdot \frac{KT}{T{A}^{\prime }}=1.$$ It follows that $\frac{KT}{T{A}^{\prime }}=\frac{1}{3}$ and $K{A}^{\prime }=2KT$. Using Menelaus' theorem for triangle $T{O}^{\prime }{A}^{\prime }$ and line $HK$ we get $$1=\frac{O^{\prime }H}{H{A}^{\prime }}\cdot \frac{A^{\prime }K}{KT}\cdot \frac{TO}{O{O}^{\prime }}=\frac{1}{2}\cdot 2\cdot \frac{TO}{O{O}^{\prime }}=\frac{TO}{O{O}^{\prime }}.$$ It means that $TO=O{O}^{\prime }$, so $O^{\prime }{A}^{\prime 2}=O^{\prime }M\cdot O^{\prime }T=O{O}^{\prime 2}$. Hence $O^{\prime }{A}^{\prime }=O{O}^{\prime }$ and consequently, $O\in {\omega }^{\prime }$. Finally we conclude that $2\angle CAB=\angle BOC=180^{\circ }-\angle CAB$, so $\angle CAB=60^{\circ }$.

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Alternative solution.

Let $O^{\prime }$ and $G^{\prime }$ denote the reflection of $O$ and $G$ respectively, with respect to the line $BC$. We then need to show $\angle CAB=60^{\circ }$ iff $G^{\prime }{H}^{\prime }=G^{\prime }P$. Note that $△H^{\prime }OP$ is isosceles and hence $G^{\prime }{H}^{\prime }=G^{\prime }P$ is equivalent to $G^{\prime }$ lying on the bisector $\angle H^{\prime }OP$. Let $\angle H^{\prime }AP=\epsilon$. By the assumption $AB\ne AC$, we have $\epsilon \ne 0$. Then $\angle H^{\prime }OP=2\angle H^{\prime }AP=2\epsilon$, hence $G^{\prime }{H}^{\prime }=G^{\prime }P$ iff $\angle G^{\prime }O{H}^{\prime }=\epsilon$. But $\angle G{O}^{\prime }H=\angle G^{\prime }O{H}^{\prime }$. Let $D$ be the midpoint of $O{O}^{\prime }$. It is known that $\angle GDO=\angle GAH=\epsilon$. Let $F$ be the midpoint of $HG$. Then $HG=FO$ (Euler line). Let $\angle G{O}^{\prime }H=\delta$. We then have to show $\delta =\epsilon$ iff $\angle CAB=60^{\circ }$. But by similarity ($△GDO\sim △F{O}^{\prime }O$) we have $\angle F{O}^{\prime }O=\epsilon$. Consider the circumcircles of the triangles $F{O}^{\prime }O$ and $G{O}^{\prime }H$. By the sine law and since the segments $HG$ and $FO$ are of equal length we deduce that the circumcircles of the triangles $F{O}^{\prime }O$ and $G{O}^{\prime }H$ are symmetric with respect to the perpendicular bisector of the segment $FG$ iff $\delta =\epsilon$. Obviously, $O^{\prime }$ is the common point of these two circles. Hence $O^{\prime }$ must be fixed after the symmetry about the perpendicular bisector of the segment $FG$.

$\delta =\epsilon$ we have $\epsilon =\delta$ iff $△HO{O}^{\prime }$ is isosceles. But $H{O}^{\prime }=H^{\prime }O=R$, and so $$\epsilon =\delta \iff O{O}^{\prime }=R\iff OD=\frac{R}{2}\iff \cos \angle CAB=\frac{1}{2}\iff \angle CAB=60^{\circ }.$$

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Alternative solution.

Let $H^{\prime }$ and $G^{\prime }$ denote the reflection of points $H$ and $G$ with respect to the line $BC$. It is known that $H^{\prime }$ belongs to the circumcircle of $△ABC$. The equality $HG=G{P}^{\prime }$ is equivalent to $H^{\prime }{G}^{\prime }=G^{\prime }P$. As in the solution 2, it is equivalent to the statement that point $G^{\prime }$ belongs to the perpendicular bisector of $H^{\prime }P$, which is equivalent to $O{G}^{\prime }\perp H^{\prime }P$, where $O$ is the circumcenter of $△ABC$.

Let points $A(a)$, $B(b)$, and $C(c=-\bar{b})$ belong to the unit circle in the complex plane. Point $G$ has coordinate $g=(a+b-\bar{b})/3$. Since $BC$ is parallel to the real axis, point $H^{\prime }$ has coordinate $h^{\prime }=\bar{a}=\frac{1}{a}$.

Point $P(p)$ belongs to the unit circle, so $\bar{p}=\frac{1}{p}$. Since $a,p,g$ are collinear we have $$\frac{p-a}{g-a}=\frac{\overline{p-a}}{\overline{g-a}}.$$ After computation we get $p=\frac{g-a}{1-ga}$. Since $G^{\prime }(g)$ is the reflection of $G$ with respect to the chord $BC$, we have $g^{\prime }=b+(-\bar{b})-b(-\bar{b})\bar{g}=b-\bar{b}+\bar{g}$. Let $b-\bar{b}=d$. We have $\bar{d}=-d$. So $$g=\frac{a+d}{3},\bar{g}=\frac{\bar{a}-d}{3},g^{\prime }=d+\bar{g}=\frac{\bar{a}+2d}{3},{\bar{g}}^{\prime }=\frac{a-2d}{3},\text{and}p=\frac{g-a}{1-\bar{g}a}=\frac{d-2a}{2+ad}.(1)$$ It is easy to see that $\overrightarrow{O{G}^{\prime }}\perp \overrightarrow{H^{\prime }{P}^{\prime }}$ is equivalent to $$\frac{g^{\prime }}{h^{\prime }-p}=-\left(\frac{g^{\prime }}{h^{\prime }-p}\right)=-\frac{g^{\prime }}{\frac{1}{h^{\prime }}-\frac{1}{p}}=\frac{g^{\prime }{h}^{\prime }p}{h^{\prime }-p}$$ since $h^{\prime }$ and $p$ belong to the unit circle (note that $H^{\prime }\ne P$ because $AB\ne AC$). This is equivalent to $g^{\prime }=\overline{g^{\prime }{h}^{\prime }}p$ and from (1), after easy computations, this is equivalent to $$a^2{g}^2+a^2+d^2+1=(a^2+1)(d^2+1)=0.$$ We cannot have $a^2+1=0$, because then $a=\pm i$, but $AB\ne AC$. Hence $d=b-\bar{b}=\pm i$, and the pair $\{b,c=-\bar{b}\}$ is either $\{-\frac{\sqrt{3}}{2}+\frac{i}{2},\frac{\sqrt{3}}{2}+\frac{i}{2}\}$ or $\{-\frac{\sqrt{3}}{2}-\frac{i}{2},\frac{\sqrt{3}}{2}-\frac{i}{2}\}$. Both cases are equivalent to $\angle BAC=60^{\circ }$ which completes the proof.