The 35th Japanese Mathematical Olympiad

For three distinct points $X_1,X_2,X_3$, when the line $X_1{X}_2$ is rotated counterclockwise by an angle $\theta$ about $X_1$ to coincide with the line $X_1{X}_3$, this $\theta$ is denoted by $\angle X_2{X}_1{X}_3$. Note that differences of $180^{\circ }$ are disregarded. First, we show the following lemma concerning the Circle of Apollonius.

Lemma 1. Let $STU$ be a triangle satisfying $ST\ne SU$, and let $\omega$ be a circle passing through $S$. When an inversion with center $S$ is performed, mapping $T$ and $U$ to $T^{\prime }$ and $U^{\prime }$ respectively, the circle $\omega$ is the Apollonius circle for the segment $TU$ passing through $S$ if and only if $\omega$ is mapped by this inversion to the perpendicular bisector of the segment $T^{\prime }{U}^{\prime }$.

Proof of Lemma 1. Let $V$ and $W$ be the intersection points of the internal and external angle bisectors of $\angle TSU$ with the line $TU$ respectively, and let $V^{\prime }$ and $W^{\prime }$ be their respective images under the inversion. From $ST:SU=VT:VU=WT:WU$, the circumcircle of the triangle $SVW$ is the Apollonius circle for the segment $TU$ passing through $S$, so it suffices to show that the circumcircle of the triangle $SVW$ is mapped to the perpendicular bisector of the segment $T^{\prime }{U}^{\prime }$ by the inversion. Since $V^{\prime }$ and $W^{\prime }$ are the intersection points of the internal and external angle bisectors of $\angle T^{\prime }S{U}^{\prime }$ with the circumcircle of the triangle $S{T}^{\prime }{U}^{\prime }$ respectively, they are the midpoints of the arc $T^{\prime }{U}^{\prime }$ not containing $S$ and the arc $T^{\prime }{U}^{\prime }$ containing $S$ on the circumcircle of the triangle $S{T}^{\prime }{U}^{\prime }$ respectively. Therefore, the image of the circumcircle of triangle $SVW$ under the inversion, which is the line $V^{\prime }{W}^{\prime }$, is the perpendicular bisector of the segment $T^{\prime }{U}^{\prime }$. Thus, the lemma is proved. ■

Let ${\Omega }_A$ be the Apollonius circle for the side $BC$ passing through $A$, and similarly define ${\Omega }_B$ and ${\Omega }_C$. Note that $B_1$ and $B_2$ lie on ${\Omega }_B$, and $C_1$ and $C_2$ lie on ${\Omega }_C$.

Lemma 2. $A_1,A_2$ lie on ${\Omega }_A$.

Proof of Lemma 2. Since ${\Omega }_A$ is symmetric with respect to the line $BC$, $A_1$ is on ${\Omega }_A$ if and only if $A_2$ is on ${\Omega }_A$. We will show that $A_2$ lies on ${\Omega }_A$. Let $P$ and $Q$ be the two intersection points of ${\Omega }_B$ and ${\Omega }_C$. Then, $$BP:CP=\frac{AP\cdot BC}{AC}:\frac{AP\cdot BC}{AB}=AB:AC$$ Therefore, $P$ lies on ${\Omega }_A$, and similarly for $Q$. Let $\sigma$ be the inversion with respect to a circle centered at $P$ with radius 1. Then, for any points $Y,Z$ different from $P$, the triangle $PYZ$ is similar to the triangle $P\sigma (Z)\sigma (Y)$, so we obtain $$\sigma (A)\sigma (B)=\frac{AB}{AP\cdot BP}=\frac{BC}{BP\cdot CP}=\sigma (B)\sigma (C).$$ Similarly, $\sigma (A)\sigma (C)=\sigma (B)\sigma (C)$ also holds, so the triangle $\sigma (A)\sigma (B)\sigma (C)$ is equilateral. Also, by Lemma 1, $\sigma ({\Omega }_A)$ is the perpendicular bisector of the segment $\sigma (B)\sigma (C)$, and similarly for $\sigma ({\Omega }_B)$ and $\sigma ({\Omega }_C)$. Therefore, $\sigma (Q)$ is the center of the equilateral triangle $\sigma (A)\sigma (B)\sigma (C)$, and the line $\sigma (B)\sigma (B_2)$ and the line $\sigma (C)\sigma (C_2)$ are symmetric with respect to the line $\sigma (A)\sigma (Q)$. Also, the four points $\sigma (B),\sigma (B_2),\sigma (C),\sigma (C_2)$ are concyclic, and since $\sigma (B)$ and $\sigma (C)$ are symmetric with respect to the line $\sigma (A)\sigma (Q)$, $\sigma (B_2)$ and $\sigma (C_2)$ are also symmetric with respect to the line $\sigma (A)\sigma (Q)$. Here, from the condition that $B_2,C_2$ do not lie on the circumcircle of the triangle $ABC$, note that the circumcircle of the triangle $\sigma (A)\sigma (B)\sigma (C_2)$ and the circumcircle of the triangle $\sigma (A)\sigma (B_2)\sigma (C)$ are distinct. Since $\sigma (A_2)$ is the intersection point of these two circles other than $\sigma (A)$, it lies on the line $\sigma (A)\sigma (Q)$, and this line is $\sigma ({\Omega }_A)$, so $A_2$ lies on ${\Omega }_A$.

Let $R$ be the intersection point, other than $A$, of the circumcircle of the triangle $AB{C}_1$ and the circumcircle of the triangle $A{B}_{1}C$. If the two circles are tangent, then let $R=A$, and the line $AR$ is taken to be the common tangent line to the two circles at $A$. Also, since $B_1,C_1$ are points in the interior of the triangle $ABC$, note that $R$ is different from $B$ and $C$. Then, $$\begin{array}{rl}\angle BRC& =\angle BRA+\angle ARC\\ & =\angle B{C}_{1}A+\angle A{B}_{1}C\\ & =\angle A{C}_{2}B+\angle C{B}_{2}A\\ & =\angle A{A}_{2}B+\angle C{A}_{2}A\\ & =\angle C{A}_{2}B\\ & =\angle B{A}_{1}C.\end{array}$$ Therefore, the circumcircle of the triangle $A_{1}BC$ also passes through $R$. Here, since $A_1$ is a point in the interior of the triangle $ABC$, $R$ is also different from $A$.

Proof of Lemma 3. Let ${\Gamma }_A$ be the arc $BC$ of the circumcircle of the triangle $A_{1}BC$ that contains $A_1$. Let ${\Gamma }_B$ be the arc $CA$ of the circumcircle of the triangle $A{B}_{1}C$ that contains $B_1$. Let ${\Gamma }_C$ be the arc $AB$ of the circumcircle of the triangle $AB{C}_1$ that contains $C_1$. Then, since $A_1$ is in the interior of the triangle $ABC$, ${\Gamma }_A$ is inside the circumcircle of the triangle $ABC$, and the part of the circumcircle of the triangle $A_{1}BC$ not containing ${\Gamma }_A$ is outside the circumcircle of the triangle $ABC$. Since the same holds for ${\Gamma }_B$ and ${\Gamma }_C$, either all of ${\Gamma }_A,{\Gamma }_B,{\Gamma }_C$ contain $R$, or none of them contain $R$. Also, if none of them contain $R$, then $R$ must be on the opposite side of the line $BC$ from $A_1$, on the opposite side of the line $CA$ from $B_1$, and on the opposite side of the line $AB$ from $C_1$. But since $A_1,B_1,C_1$ are all in the interior of the triangle $ABC$, this is a contradiction. Thus ${\Gamma }_A,{\Gamma }_B,{\Gamma }_C$ all pass through $R$. Here, if we assume $P=R$ (where $P$ is from Lemma 2), then $P=A_1=B_1=C_1$, which contradicts the assumption that $A_1,B_1,C_1$ are distinct. So note that $P\ne R$. Let $\tau$ be the operation of applying $\sigma$ defined in Lemma 2 followed by an inversion centered at $\sigma (R)$. By Lemma 1, $\tau ({\Omega }_A)$ is the Apollonius circle for the segment $\tau (B)\tau (C)$ passing through $\tau (A)$. Furthermore, since the four points $A_1,B,C,R$ are concyclic, $\tau (A_1)$ lies on the line $\tau (B)\tau (C)$. Therefore, $\tau (A_1)$ is the intersection of the internal or external bisector of $\angle \tau (B)\tau (A)\tau (C)$ with the line $\tau (B)\tau (C)$, and similarly for $\tau (B_1)$ and $\tau (C_1)$. Here, since ${\Gamma }_A,{\Gamma }_B,{\Gamma }_C$ all contain $R$, $\tau ({\Gamma }_A),\tau ({\Gamma }_B),\tau ({\Gamma }_C)$ all contain the point at infinity $\tau (R)$. Therefore, $\tau (A_1),\tau (B_1),\tau (C_1)$ are all intersection points of external angle bisectors with the opposite sides. Also, $$\frac{\tau (B)\tau (A_1)}{\tau (A_1)\tau (C)}\cdot \frac{\tau (C)\tau (B_1)}{\tau (B_1)\tau (A)}\cdot \frac{\tau (A)\tau (C_1)}{\tau (C_1)\tau (B)}=\frac{\tau (A)\tau (B)}{\tau (C)\tau (A)}\cdot \frac{\tau (B)\tau (C)}{\tau (A)\tau (B)}\cdot \frac{\tau (C)\tau (A)}{\tau (B)\tau (C)}=1.$$ Therefore, by the converse of Menelaus's Theorem, the three points $\tau (A_1),\tau (B_1),\tau (C_1)$ are collinear. Thus the lemma is proved. ■

Let $D$ be the intersection of the internal angle bisector of $\angle BAC$ with the line $BC$. This point $D$ lies on ${\Omega }_A$. Also, let $O_A$ be the center of ${\Omega }_A$. Then $O_A$ lies on the line $BC$. Therefore, $$\angle O_{A}AB=\angle O_{A}AD+\angle DAB=\angle AD{O}_A+\angle CAD=\angle ACD$$ from which it follows that the line $A{O}_A$ is tangent to the circumcircle of triangle $ABC$ at $A$, so $O_A{A}^2=O_{A}B\cdot O_{A}C$ holds. From this and $O_{A}A=O_A{A}_2$, $O_A{A}_2^2=O_{A}B\cdot O_{A}C$ holds. Therefore, the line $A_2{O}_A$ is tangent to the circumcircle of the triangle $A_{2}BC$ at $A_2$. Also, under the inversion with respect to ${\Omega }_A$, $B$ and $C$ are interchanged, and since $P,Q$ defined in Lemma 2 are invariant, ${\Omega }_B$ and ${\Omega }_C$ are interchanged by this inversion. Therefore, if $B_2^{\prime }$ is the image of $C_2$ under this inversion, then $B_2^{\prime }$ lies on ${\Omega }_B$. Here, the four points $B,B_2^{\prime },C,C_2$ are concyclic, and this circle is invariant under the inversion with respect to ${\Omega }_A$, so $B_2^{\prime }=B_2$. Therefore, $O_A$ lies on the line $B_2{C}_2$, and since $O_A{A}_2^2=O_A{B}_2\cdot O_A{C}_2$, the circumcircle of the triangle $A_2{B}_2{C}_2$ is tangent to the line $A_2{O}_A$. Therefore, $$\begin{array}{rl}\angle A_1{B}_1{C}_1& =\angle A_{1}R{C}_1\\ & =\angle A_{1}RB+\angle BR{C}_1\\ & =\angle A_{1}CB+\angle BA{C}_1\\ & =\angle BC{A}_2+\angle C_{2}AB\\ & =\angle BC{A}_2+\angle C_2{A}_{2}B\\ & =\angle B{A}_2{O}_A+\angle C_2{A}_2{O}_A+\angle O_A{A}_{2}B\\ & =\angle C_2{A}_2{O}_A\\ & =\angle C_2{B}_2{A}_2\end{array}$$

and, similarly $\angle A_1{C}_1{B}_1=\angle B_2{C}_2{A}_2$, so the result is shown.