Let $ABC$ be a triangle and $P$ be an interior point. Let $\omega_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ internally and tangent to the circumcircle of $ABC$ at $A_1$ internally and let $\Gamma_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ externally and tangent to the circumcircle of $ABC$ at $A_2$ internally. Define $B_1$, $B_2$, $C_1$, $C_2$ analogously. Let $O$ be the circumcentre of $ABC$. Prove that the lines $A_1A_2$, $B_1B_2$, $C_1C_2$ and $OP$ are concurrent.
Problem
Source: 2023 Turkey NMO 2nd Round P2
Tags: geometry, circumcirle, tangent, tangent circles, external tangent, concurrent lines
electrovector
22.12.2023 03:30
I am not sure of this solution rn but I think this has previously been examined before. Draw tangents at $A_1$ and $A_2$ and let them intersect at $A_3$. Considering the power axis of degenerate circle $P$ and $(ABC)$ gives the solution using La Hire etc.
As mentioned above draw the tangents at $A_1$ and $A_2$ wrt $(ABC)$ and let them intersect at $A_3$. Since $A_3A_1 = A_3A_2$, the power of $A_3$ wrt to the circles tangent to $(ABC)$ and $(BPC)$ is the same, therefore it lies on the power axis, namely $BC$.
Since $A_3B \cdot A_3C = A_3D^2$, its powers wrt to $(ABC)$ and $D$ are the same. Defining $B_3$ and $C_3$ similarly, we have $A_3$, $B_3$ and $C_3$ are collinear and $OP$ is perpendicular to this line. Let the pole of this line wrt to $(ABC)$ be $X$. Using La Hire a couple of times gives us $X$ lies on $A_1A_2$, $B_1B_2$ and $C_1C_2$. Also $OX$ is perpendicular to this line which gives us $X \in OP$ and we are done.
Note: this is a cute problem but the case $P=I$ was already established in some competition (don’t remember which one though)
IAmTheHazard
22.12.2023 19:54
Sad problem. Upon inversion at $P$ fixing $(ABC)$, $\omega_A$ and $\Gamma_A$ get sent to lines tangent to $(ABC)$ and parallel to $\overline{B'C'}$, so they're the intersections of the perpendicular bisector of $\overline{B'C'}$ with $(ABC)$. The concurrency condition is equivalent to $(PA_1'A_2')$ etc. being intersecting again, with the circumcenter $O$ lying on their common radical axis, but this is trivial as $OA_1'\cdot OA_2'=OB_1'\cdot OB_2'=OC_1'\cdot OC_2'$ with the common value as the squared radius of $(ABC)$. (to prove coaxiality of three circles that are known to pass through the same point, it suffices to demonstrate another point with equal power to all three)
VicKmath7
22.12.2023 20:15
Alternatively, Pascal also works.
Let $X_iP \cap (ABC)=X_{i+2}$ for $X \in \{B, C\}$ and $i=1, 2$. Observe that $B_3, B_4$ are midpoints of the two arcs of $(ABC)$, formed by the tangent at $P$ to $(APC)$, so $B_3B_4$ passes through $O$ and similarly $C_3C_4$ passes through $O$. If $B_1B_2 \cap C_1C_2=X$ and $B_1C_3 \cap C_2B_4=Y$, Pascal's theorem for $B_1B_3B_4C_2C_4C_3$ yields $P, Y, O$ collinear, and Pascal's theorem for $B_2B_1C_3C_1C_2B_4$ yields $X, Y, P$ collinear. Thus, $X \in OP$, which easily finishes by applying this twice more.
xeroxia
23.12.2023 03:58
My long solution as follows. There may be some shortcut statements, if any don't hesitate to inform me. --- Let $\Pi = (ABC)$, $\Pi_A = (BPC)$, and the radius of $\Pi$ be $R$. The radical axis of each pair of the three circles intersect at a single point. That is, the radical axis of $\omega_A$ with $\Gamma_A$, the radical axis of $\omega_A$ with $\Pi$, and the radical axis of $\Gamma_A$ with $\Pi$ intersect at a single point. Let this point be $O_A$. $O_AP = O_AA_1 = O_AA_2$. Let $(O_A, |O_AP|) = \pi_A$. Since the tangent of $\Pi_A$ at $O_AP$ and the tangent of $\Pi$ at $O_AA_2$ are equal, $O_A$ is on the radical axis of these two circles (That is, $O_A \in BC$). In this case, $$OO_A^2 - R^2 = O_AP^2 \qquad (1)$$. Now, let us define $O_B$ and $\pi_B$ for $\Pi_B = (APC)$ in a similar way. $$OO_B^2 - R^2 = O_BP^2 \qquad (2)$$The radical axis of $\pi_A$ and $\pi_B$ passes through $P$ and is perpendicular to $O_AO_B$. From $(1)-(2)$, $$OO_A^2 - OO_B^2 = O_AP^2 - O_BP^2 \qquad (3)$$, hence $OP \perp O_AO_B$. Therefore, the radical axis of $\pi_A$ and $\pi_B$ is $OP$. Let the intersection of $A_1A_2$ and $B_1B_2$ be $K$. Since the point $K$ is on the radical axis of both $\Pi$ with $\pi_A$ and $\Pi$ with $\pi_B$, it will also be on the radical axis of $\pi_A$ with $\pi_B$. Hence, $K$ will be on $OP$. In other words, if the intersection point of $OP$ and $A_1A_2$ is $K$, then $B_1B_2$ will also pass through this point $K$. Similarly, $C_1C_2$ will also pass through this point $K$.
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