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 $A{A}_2$.

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: $\angle B{A}_{3}S=\angle S{A}_{3}C=\angle BAC$. Proof: Since $(B,C,D,S)=-1$, $S{A}_3$ is the polar of $D$ with respect to the circle $(ABC)$. Therefore we have $O{A}_3\cdot OD=R^2$, where $R$ is the circumradius of $ABC$. Then we obtain $D{A}_3\cdot DO=D{O}^2-O{A}_3\cdot DO=D{O}^2-R^2=DB\cdot DC$ hence $B$, $C$, $O$, $A_3$ are concyclic and $\angle B{A}_{3}C=2\angle BAC$. The conclusion then follows as we have $D{A}_3\perp A_{3}S$ and $(B,C,D,S)=-1$. $\square$

Claim 2: $\angle AM{A}_3=90^{\circ }$. Proof: Let $Q$, $R$ be the projections from $A_3$ to $AB$, $AC$ respectively. Then, we have $$\frac{BF}{BQ}=\frac{BF}{B{A}_3}\cdot \frac{B{A}_3}{BQ}=\frac{\cos BAC}{\sin AFE}\cdot \frac{1}{\sin AEF}=\frac{CE}{C{A}_3}\cdot \frac{CR}{CR}=\frac{CE}{CB}$$

Let $A^{\prime }$ be the reflection of $A$ with respect to the line $IO$.

Claim 3: $A_3$, $A_1$, $A^{\prime }$ are collinear. Proof: Since $A{A}^{\prime }$, $A_{3}S$ are both perpendicular to $IO$, it suffices to show that $\frac{A{A}^{\prime }}{A_{3}S}=\frac{A{A}_1}{A_{1}S}$. Let $AS\cap IO=K$. We have $\frac{A{A}^{\prime }}{A_{3}S}=2\frac{A{A}^{\prime }/2}{A_{3}S}=2\frac{AK}{KS}$. Hence we need to prove that $2\cdot AK\cdot A_{1}S=KS\cdot A{A}_1$, 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 $\overline{A^{\prime }{A}_1{A}_3}$. Proof: From Claim 1, we know that $Pow(D,T)=DT\cdot DA=D{A}_3\cdot DO$ hence $T$, $A$, $A_3$, $O$ are concyclic. Then we find $\angle A^{\prime }T{A}_3=180^{\circ }-\angle AO{A}^{\prime }/2=\angle A^{\prime }T{A}^{\prime }$ hence $T\in A_3{A}^{\prime }$. $\square$

Claim 5: $T$, $A^{\prime }$, $X$ are collinear. Proof: Take the reflection of the triangle $ABC$ with respect to line $IO$. Then, $A^{\prime }{B}^{\prime }C$ is also a triangle with the same incircle and circumcircle as $ABC$. Let $T_0$ be the second intersection of $A^{\prime }X$ and $R$. Then, from the Dual of Desargues' Involution Theorem in the degenerate complete quadrilateral $ABXC$, there is an involution swapping pairs $(A^{\prime }B,A^{\prime }C)$, $(A^{\prime }{B}^{\prime },A^{\prime }{C}^{\prime })$ and $(A^{\prime }A,A^{\prime }X)$, and involutions on circles are inversions hence $BC$, $B^{\prime }{C}^{\prime }$, $A{T}_0$ must concur and $T=T_0$. $\square$

Therefore, $A_3=A_2$ and $M$ lies on the circle with diameter $A{A}_2$ and we are done.