Team Selection Test for IMO 2024

Let us define $B_a$, $B_c$, $C_a$, $C_b$ similarly. Let the circle passing through $B$, $C$, $A_b$, $A_c$ be ${\omega }_a$ and let us define ${\omega }_b$, ${\omega }_c$ analogously. We will start with a lemma that we will use repeatedly.

Lemma: Let $ABC$ be a triangle, $O$ be its circumcenter and $AO\cap BC=D$. Let $E$, $F$ be points on the lines $AB$, $AC$ respectively such that $DB=DE$ and $DC=DF$. Then we have $BC\parallel EF$.

Proof: Let the midpoints of the segments $BE$ and $CF$ be $K$ and $L$, respectively. Since $\angle AKL=\angle ADL=\angle B$. Let $B{A}_c\cap C{A}_b=X$.

Claim 1: $X$ lies on the radical axis of the circles ${\omega }_b$ and ${\omega }_c$.

Proof: Let ${\omega }_c\cap C{A}_b=C_1$ and ${\omega }_b\cap B{A}_c=B_1$. We will prove that $B$, $C$, $B_1$, $C_1$ are concyclic which implies the claim. Since $\angle C_a{C}_{1}C=\angle C_a{C}_{b}C=\angle A$ we have that $A$, $B$, $C_a$, $C_1$ are concyclic. Using the lemma, we have $A_b{C}_b\parallel B_a{B}_c$ hence $A_b$, $C_b$, $A$, $C$ are concyclic and $B$ lies on the radical axis of the circles ${\omega }_c$ and $(A{A}_b{C}_1{C}_a)$ therefore $B$, $C_1$, $C_a$ are collinear. Hence we have $\angle B{C}_{1}C=180^{\circ }-\angle A$ and $\angle B{B}_{1}C=180^{\circ }-\angle A$ can be obtained similarly. $\square$

Let $M_a$ be the midpoint of the segment $O_b{O}_c$ and let $O$ be the circumcenter of the triangle $ABC$.

Claim 2: $H{M}_a\parallel O{O}_a$.

Proof: $A$ lies on the radical axis of the circles ${\omega }_b$, ${\omega }_c$ and from Claim 1 we have $AX\perp O_b{O}_c$. Let $BE$, $CF$ be the altitudes of the triangle and $A_b{A}_c\cap BC=T$. Let the second intersection of the line $H{M}_a$ with the circle $(AFHE)$ be $R$ and the foot of the perpendicular from $H$ to the line $AX$ be $S$. Since the points $O_a$, $O_b$, $O_c$ lie on the altitudes of the triangle $ABC$, we have $(O_b,O_c;M_a,\infty )=-1$ and projecting this to the circle $(AFHE)$ from the point $H$ we obtain $(E,F;R,S)=-1$. Then projecting this from $A$ to the line $BC$ we have $(C,B;AR\cap BC,AX\cap BC)=-1$ hence we must have $AR\cap BC=T$. Since $R$ lies on the circle with diameter $AH$ we obtain $H{M}_a\perp AT$. Finally, since the line $AT$ is the radical axis of the circles ${\omega }_a$ and $(ABC)$ we have $O{O}_a\perp AT$ which implies

$$H{M}_a\parallel O{O}_a.$$ $\square$

Now, let $G_1$ be the centroid of the triangle $O_a{O}_b{O}_c$, we will prove that $G_1$ is the midpoint of $HG$. Triangles $H{M}_a{M}_c$ and $O{O}_a{O}_c$ have sides that are parallel to each other, hence they are similar and the similarity ratio is equal to $\frac{O_b{O}_c}{M_b{M}_c}=2$. Therefore the intersection of the lines $O_a{M}_a$ and $HG$ divides both segments with the ratio $2:1$ and this point is in fact $G_1$ since $O_a{M}_a$ is a median, and it is also the midpoint of $HG$ hence we are done.