Subfield: Geometry


Question: Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $A B C$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A, B, C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $A O P, B O P$, and $C O P$ by $O_{A}, O_{B}$, and $O_{C}$, respectively. The lines $\ell_{A}, \ell_{B}$, and $\ell_{C}$ perpendicular to $B C, C A$, and $A B$ pass through $O_{A}, O_{B}$, and $O_{C}$, respectively. Prove that the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$ is tangent to the line $O P$.

Solution:
As usual, we denote the directed angle between the lines $a$ and $b$ by $\sphericalangle(a, b)$. We frequently use the fact that $a_{1} \perp a_{2}$ and $b_{1} \perp b_{2}$ yield $\sphericalangle\left(a_{1}, b_{1}\right)=\sphericalangle\left(a_{2}, b_{2}\right)$.

Let the lines $\ell_{B}$ and $\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B} L_{C}$; moreover, $O_{A}, O_{B}$, $O_{C}$ all lie on the perpendicular bisector of $O P$.

<img_3848>

Claim 1. The points $L_{B}, P, O_{A}$, and $O_{C}$ are concyclic.

Proof. Since $O$ is symmetric to $P$ in $O_{A} O_{C}$, we have

$$
\sphericalangle\left(O_{A} P, O_{C} P\right)=\sphericalangle\left(O_{C} O, O_{A} O\right)=\sphericalangle(C P, A P)=\sphericalangle(C B, A B)=\sphericalangle\left(O_{A} L_{B}, O_{C} L_{B}\right) .
$$

Denote the circle through $L_{B}, P, O_{A}$, and $O_{C}$ by $\omega_{B}$. Define the circles $\omega_{A}$ and $\omega_{C}$ similarly. Claim 2. The circumcircle of the triangle $L_{A} L_{B} L_{C}$ passes through $P$.

Proof. From cyclic quadruples of points in the circles $\omega_{B}$ and $\omega_{C}$, we have

$$
\begin{aligned}
\sphericalangle\left(L_{C} L_{A}, L_{C} P\right) & =\sphericalangle\left(L_{C} O_{B}, L_{C} P\right)=\sphericalangle\left(O_{A} O_{B}, O_{A} P\right) \\
& =\sphericalangle\left(O_{A} O_{C}, O_{A} P\right)=\sphericalangle\left(L_{B} O_{C}, L_{B} P\right)=\sphericalangle\left(L_{B} L_{A}, L_{B} P\right) .
\end{aligned}
$$

Claim 3. The points $P, L_{C}$, and $C$ are collinear.

Proof. We have $\sphericalangle\left(P L_{C}, L_{C} L_{A}\right)=\sphericalangle\left(P L_{C}, L_{C} O_{B}\right)=\sphericalangle\left(P O_{A}, O_{A} O_{B}\right)$. Further, since $O_{A}$ is the centre of the circle $A O P, \sphericalangle\left(P O_{A}, O_{A} O_{B}\right)=\sphericalangle(P A, A O)$. As $O$ is the circumcentre of the triangle $P C A, \sphericalangle(P A, A O)=\pi / 2-\sphericalangle(C A, C P)=\sphericalangle\left(C P, L_{C} L_{A}\right)$. We obtain $\sphericalangle\left(P L_{C}, L_{C} L_{A}\right)=$ $\sphericalangle\left(C P, L_{C} L_{A}\right)$, which shows that $P \in C L_{C}$.



Similarly, the points $P, L_{A}, A$ are collinear, and the points $P, L_{B}, B$ are also collinear. Finally, the computation above also shows that

$$
\sphericalangle\left(O P, P L_{A}\right)=\sphericalangle(P A, A O)=\sphericalangle\left(P L_{C}, L_{C} L_{A}\right),
$$

which means that $O P$ is tangent to the circle $P L_{A} L_{B} L_{C}$.