Subfield: Geometry


Question: Let $H$ be the orthocenter and $G$ be the centroid of acute-angled triangle $\triangle A B C$ with $A B \neq A C$. The line $A G$ intersects the circumcircle of $\triangle A B C$ at $A$ and $P$. Let $P^{\prime}$ be the reflection of $P$ in the line $B C$. Prove that $\angle C A B=60^{\circ}$ if and only if $H G=G P^{\prime}$.


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

Let us assume that $\angle C A B=60^{\circ}$. Since

$$
\angle C O B=2 \angle C A B=120^{\circ}=180^{\circ}-60^{\circ}=180^{\circ}-\angle C A B=\angle C H B,
$$

hence $O$ lies on $\omega^{\prime}$. Reflecting $O$ in line $B C$, we obtain point $O^{\prime}$ which lies on $\omega$ and this point is the center of $\omega^{\prime}$. Then $O O^{\prime}=2 O M=2 R \cos \angle C A B=A H$, so $A H=O O^{\prime}=H O^{\prime}=A O=R$, where $R$ is the radius of $\omega$ and, naturally, of $\omega^{\prime}$. Then quadrilateral $A H O^{\prime} O$ is a rhombus, so $A$ and $O^{\prime}$ are symmetric to each other with respect to $H O$. As $H, G$ and $O$ are collinear (Euler line), then $\angle G A H=\angle H O^{\prime} G$. Diagonals of quadrilateral $G O P O^{\prime}$ intersects at $M$. Since $\angle B O M=60^{\circ}$, so

$$
O M=M O^{\prime}=\operatorname{ctg} 60^{\circ} \cdot M B=\frac{M B}{\sqrt{3}}
$$

As $3 M O \cdot M O^{\prime}=M B^{2}=M B \cdot M C=M P \cdot M A=3 M G \cdot M P$, then $G O P O^{\prime}$ is a cyclic. Since $B C$ is a perpendicular bisector of $O O^{\prime}$, so the circumcircle of quadrilateral $G O P O^{\prime}$ is symmetrical with respect to $B C$. Thus $P^{\prime}$ also belongs to the circumcircle of $G O P O^{\prime}$, hence $\angle G O^{\prime} P^{\prime}=\angle G P P^{\prime}$. Note that $\angle G P P^{\prime}=\angle G A H$ since $A H \| P P^{\prime}$. And as it was proved $\angle G A H=\angle H O^{\prime} G$, then $\angle H O^{\prime} G=\angle G O^{\prime} P^{\prime}$. Thus triangles $\triangle H O^{\prime} G$ and $\triangle G O^{\prime} P^{\prime}$ are equal and hence $H G=G P^{\prime}$.

Now we will prove that if $H G=G P^{\prime}$ then $\angle C A B=60^{\circ}$. 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 $H C \perp C A^{\prime}$ since $A B \| C A^{\prime}$ 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 $H G=G P^{\prime}$ it follows that $\triangle H G O^{\prime}$ is equal to $\triangle P^{\prime} G O^{\prime}$. Therefore $H$ and $P^{\prime}$ are symmetric with respect to $G O^{\prime}$. Hence $G O^{\prime} \perp H P^{\prime}$ and $G O^{\prime} \| A^{\prime} P^{\prime}$. Let $H G$ intersect $A^{\prime} P^{\prime}$ at $K$ and $K \not \equiv O$ since $A B \neq A C$. We conclude that $H G=G K$, because line $G O^{\prime}$ is midline of the triangle $\triangle H K A^{\prime}$. Note that $2 G O=H G$. since $H O$ is Euler line of triangle $A B C$. So $O$ is midpoint of segment $G K$. Because of $\angle C M P=\angle C M P^{\prime}$, then $\angle G M O=\angle O M P^{\prime}$. Line $O M$, that passes through $O^{\prime}$, is an external angle bisector of $\angle P^{\prime} M A^{\prime}$. Also we know that $P^{\prime} O^{\prime}=O^{\prime} A^{\prime}$, then $O^{\prime}$ is the midpoint of arc $P^{\prime} M A^{\prime}$ of the circumcircle of triangle $\triangle P^{\prime} M A^{\prime}$. It



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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 $O M$ and $P^{\prime} A^{\prime}$ intersect at $T$. Triangles $\triangle T O^{\prime} A^{\prime}$ and $\triangle A^{\prime} O^{\prime} M$ are similar, hence $O^{\prime} A^{\prime} / O^{\prime} M=O^{\prime} T / O^{\prime} A^{\prime}$. In the other words, $O^{\prime} M \cdot O^{\prime} T=O^{\prime} A^{\prime 2}$. Using Menelaus' theorem for triangle $\triangle H K A^{\prime}$ and line $T O^{\prime}$, we obtain that

$$
\frac{A^{\prime} O^{\prime}}{O^{\prime} H} \cdot \frac{H O}{O K} \cdot \frac{K T}{T A^{\prime}}=3 \cdot \frac{K T}{T A^{\prime}}=1
$$

It follows that $K T / T A^{\prime}=1 / 3$ and $K A^{\prime}=2 K T$. Using Menelaus' theorem for triangle $T O^{\prime} A^{\prime}$ and line $H K$ we get that

$$
1=\frac{O^{\prime} H}{H A^{\prime}} \cdot \frac{A^{\prime} K}{K T} \cdot \frac{T O}{O O^{\prime}}=\frac{1}{2} \cdot 2 \cdot \frac{T O}{O O^{\prime}}=\frac{T O}{O O^{\prime}}
$$

It means that $T O=O O^{\prime}$, 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 $2 \angle C A B=\angle B O C=180^{\circ}-\angle C A B$, so $\angle C A B=60^{\circ}$.
<img_4029>
Let $O^{\prime}$ and $G^{\prime}$ denote the reflection of $O$ and $G$, respectively, with respect to the line $B C$. We then need to show $\angle C A B=60^{\circ}$ iff $G^{\prime} H^{\prime}=G^{\prime} P$. Note that $\triangle H^{\prime} O P$ is isosceles and hence



$G^{\prime} H^{\prime}=G^{\prime} P$ is equivalent to $G^{\prime}$ lying on the bisector $\angle H^{\prime} O P$. Let $\angle H^{\prime} A P=\varepsilon$. By the assumption $A B \neq A C$, we have $\varepsilon \neq 0$. Then $\angle H^{\prime} O P=2 \angle H^{\prime} A P=2 \varepsilon$, hence $G^{\prime} H^{\prime}=G^{\prime} P$ iff $\angle G^{\prime} O H^{\prime}=\varepsilon$. 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 $\angle G D O=\angle G A H=\varepsilon$. Let $F$ be the midpoint of $H G$. Then $H G=F O$ (Euler line). Let $\angle G O^{\prime} H=\delta$. We then have to show $\delta=\varepsilon$ iff $\angle C A B=60^{\circ}$. But by similarity $\left(\triangle G D O \sim \triangle F O^{\prime} O\right.$ ) we have $\angle F O^{\prime} O=\varepsilon$. Consider the circumcircles of the triangles $F O^{\prime} O$ and $G O^{\prime} H$. By the sine law and since the segments $H G$ and $F O$ 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 $F G$ iff $\delta=\varepsilon$. 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 $F G$ iff $\delta=\varepsilon$ so we have $\varepsilon=\delta$ iff $\triangle H O O^{\prime}$ is isosceles. But $H O^{\prime}=H^{\prime} O=R$, and so

$$
\varepsilon=\delta \Longleftrightarrow O O^{\prime}=R \Longleftrightarrow O D=\frac{R}{2} \Longleftrightarrow \cos \angle C A B=\frac{1}{2} \Longleftrightarrow \angle C A B=60^{\circ} \text {. }
$$