75th Romanian Mathematical Olympiad

Let $H_A$ be the reflection of $H$ across line $BC$. Then $H_A$ lies on the circumcircle of triangle $ABC$. Since $A_{1}H=A_1{H}_A$, we get $A_{1}O+A_1{H}_A=R=O{H}_A$. Hence, $A_1$ lies on segment $O{H}_A\cap BC$, and is uniquely defined by the given condition. We now prove that $\angle BH{A}_1=\angle B$ and $\angle CH{A}_1=\angle C$.

Suppose without loss of generality that $\angle B\le \angle C$. Let $A_O$ be the point diametrically opposite to $A$ on the circle. Then:

$\angle BH{A}_1=\angle BH{H}_A-\angle A_{1}H{H}_A=\angle C-\angle O{H}_{A}A=\angle C-\angle A_{O}A{H}_A=\angle C-(\angle A-\angle BA{A}_O-\angle H_{A}AC)=\angle C-(\angle A-(90^{\circ }-\angle C)-\angle BC{A}_O)=\angle C-(\angle A-(90^{\circ }-\angle C)-(90^{\circ }-\angle C))=\angle C-(\angle A-180^{\circ }+2\angle C)=\angle B$.

Similarly, we find that $\angle CH{A}_1=\angle C$.

$$\frac{B{A}_1}{C{A}_1}=\frac{A_{BH{A}_1}}{A_{CH{A}_1}}=\frac{BH\cdot H{A}_1\cdot \sin (\angle BH{A}_1)}{CH\cdot H{A}_1\cdot \sin (\angle CH{A}_1)}=\frac{2R\cos B\sin B}{2R\cos C\sin C}=\frac{\sin (2B)}{\sin (2C)}.$$

Applying the same process for $B_1$ and $C_1$, we obtain: $$\frac{C{B}_1}{A{B}_1}=\frac{\sin (2C)}{\sin (2A)},\frac{A{C}_1}{B{C}_1}=\frac{\sin (2A)}{\sin (2B)}.$$

Multiplying the three ratios gives: $$\frac{B{A}_1}{C{A}_1}\cdot \frac{C{B}_1}{A{B}_1}\cdot \frac{A{C}_1}{B{C}_1}=\frac{\sin (2B)}{\sin (2C)}\cdot \frac{\sin (2C)}{\sin (2A)}\cdot \frac{\sin (2A)}{\sin (2B)}=1.$$ Thus, the cevians $A{A}_1,B{B}_1$, and $C{C}_1$ are concurrent. Given that: $\overrightarrow{A{A}_1}+\overrightarrow{B{B}_1}+\overrightarrow{C{C}_1}=\overrightarrow{0}$, the concurrency point must be the centroid, implying these are medians. Therefore, $A_1,B_1,C_1$ are midpoints of $BC,CA,AB$ respectively, and hence: $$\frac{\sin (2A)}{\sin (2B)}=1\Rightarrow \sin (2A)=\sin (2B),$$ and, similarly, $\sin (2A)=\sin (2C)$.

Since triangle $ABC$ is acute and the sine function is injective in $(0^{\circ },180^{\circ })$, so $2A=2B=2C\Rightarrow A=B=C$. Therefore, triangle $ABC$ is equilateral.