Team Selection Test for IMO, 2024

Let $M$ be the Miquel point of the quadrilateral defined by the lines $AB$, $AC$, $BC$, $IO$. We will prove that $M$ lies on all three circles. Since the statement is symmetric, we will only show that $M$ lies on the circle with diameter $AA_2$.

Define $S = A_1 \cap BC$, or equivalently, as the point satisfying $(B, C; D, S) = -1$. Let $A_3$ be the foot of the perpendicular from $S$ to $IO$. We will eventually show that $A_2 = A_3$.

Claim 1: $\angle BA_3S = \angle SA_3C = \angle BAC$.
Proof: Since $(B, C, D, S) = -1$, $SA_3$ is the polar of $D$ with respect to the circle $(ABC)$. Therefore we have $OA_3 \cdot OD = R^2$, where $R$ is the circumradius of $ABC$. Then we obtain $DA_3 \cdot DO = DO^2 - OA_3 \cdot DO = DO^2 - R^2 = DB \cdot DC$ hence $B$, $C$, $O$, $A_3$ are concyclic and $\angle BA_3C = 2 \angle BAC$. The conclusion then follows as we have $DA_3 \perp A_3S$ and $(B, C, D, S) = -1$. $\square$

Claim 2: $\angle AMA_3 = 90^\circ$.
Proof: Let $Q$, $R$ be the projections from $A_3$ to $AB$, $AC$ respectively. Then, we have
$$\frac{BF}{BQ} = \frac{BF}{BA_3} \cdot \frac{BA_3}{BQ} = \frac{\cos BAC}{\sin AFE} \cdot \frac{1}{\sin AEF} = \frac{CE}{CA_3} \cdot \frac{CR}{CR} = \frac{CE}{CB}$$

Let $A'$ be the reflection of $A$ with respect to the line $IO$.

Claim 3: $A_3$, $A_1$, $A'$ are collinear.
Proof: Since $AA'$, $A_3S$ are both perpendicular to $IO$, it suffices to show that $\frac{AA'}{A_3S} = \frac{AA_1}{A_1S}$. Let $AS \cap IO = K$. We have $\frac{AA'}{A_3S} = 2\frac{AA'/2}{A_3S} = 2\frac{AK}{KS}$. Hence we need to prove that $2 \cdot AK \cdot A_1S = KS \cdot AA_1$, which is well known since these points satisfy $(A, A_1; K, S) = -1$. $\square$

Let $T$ be the circumcircle of $ABC$ and let $T$ be the second intersection of $AD$ and $R$.

Claim 4: $T$ lies on the line $\overline{A'A_1A_3}$.
Proof: From Claim 1, we know that $Pow(D, T) = DT \cdot DA = DA_3 \cdot DO$ hence $T$, $A$, $A_3$, $O$ are concyclic. Then we find $\angle A'TA_3 = 180^\circ - \angle AOA'/2 = \angle A'TA'$ hence $T \in A_3A'$. $\square$

Claim 5: $T$, $A'$, $X$ are collinear.
Proof: Take the reflection of the triangle $ABC$ with respect to line $IO$. Then, $A'B'C$ is also a triangle with the same incircle and circumcircle as $ABC$. Let $T_0$ be the second intersection of $A'X$ and $R$. Then, from the Dual of Desargues' Involution Theorem in the degenerate complete quadrilateral $ABXC$, there is an involution swapping pairs $(A'B, A'C)$, $(A'B', A'C')$ and $(A'A, A'X)$, and involutions on circles are inversions hence $BC$, $B'C'$, $AT_0$ must concur and $T = T_0$. $\square$

Therefore, $A_3 = A_2$ and $M$ lies on the circle with diameter $AA_2$ and we are done.