The incircle $\omega$ of triangle $ABC$ touches $BC, CA, AB$ at points $A_1, B_1$ and $C_1$ respectively, $P$ is an arbitrary point on $\omega$. The line $AP$ meets the circumcircle of triangle $AB_1C_1$ for the second time at point $A_2$. Points $B_2$ and $C_2$ are defined similarly. Prove that the circumcircle of triangle $A_2B_2C_2$ touches $\omega$.
Problem
Source: Sharygin Correspondence Round 2024 P4
Tags: geometry
InterLoop
06.03.2024 14:11
xonk state things for rigor
We denote by $I$ the incenter of $\triangle ABC$.
Claim 1. Given a circle $\Omega$ with radius $x$, a circle with radius $x/2$ passing through the center of $\Omega$, call it $\Gamma$, is tangent to $\Omega$.
$\textsf{\textbf{Proof of Claim.}}$ For the sake of contradiction, assume that $\Omega$ intersects $\Gamma$ at two distinct points, $P$ and $Q$. Then we know that $OP = OQ = x$, however we know that the diameter of $\Gamma$ also has length $x$. This implies that $OP$ and $OQ$ are both diameters of $\Gamma$, which implies $P = Q$, a contradiction.
$\blacksquare$
Claim 2. The points $A_2$, $B_2$, $C_2$, $P$, $I$ are concyclic, and the diameter of this circle is $IP$.
$\textsf{\textbf{Proof of Claim.}}$ Using angles directed modulo $180^{\circ}$, note that
$$\angle PA_2I = \angle AA_2I = \angle AB_1I = 90^{\circ}$$$$\angle PB_2I =\angle BB_2I = \angle BC_1I = 90^{\circ}$$$$\angle PC_2I = \angle CC_2I = \angle CA_1I = 90^{\circ}$$and thus $\angle PA_2I = \angle PB_2I =\angle PC_2I$ which implies $A_2B_2C_2PI$ is concyclic. Clearly $IP$ is the diameter of this circle since all the aforementioned angles are $90^{\circ}$.
$\blacksquare$
{Now let the radius of $\omega$ be $r$. Since $P$ lies on $\omega$, the length $IP = r$. However now note that since $IP$ is the diameter of the circumcircle of $A_2B_2C_2$, the radius of $A_2B_2C_2$ is $r/2$. Now we use $\textsf{\textbf{Claim 1}}$ which proves the problem
Om245
06.03.2024 17:31
Note $AI$ is diameter in $(AB_1C_1)$ hence $\angle IA_2A = 90=\angle PA_2I$. similarly we get $\angle PB_2I=\angle PC_2I=90$. Hence we get $A_2,B_2,C_2$ lie on circle with diameter $PI$. And as center of this circle will lie on $PI$ we get $(A_2B_2C_2)$ is tangent to $\omega$.
sami1618
09.06.2024 01:08
We claim that $A_2$, $B_2$, and $C_2$ all lie on $(PI)$, which is clearly sufficient. We prove this for $A_2$ with the other proofs being identical. Notice $\angle IB_1A=\angle IC_1A=90^{\circ}$, so $IA$ is the diameter of $(AB_1C_1)$, thus $\angle PA_2I=\angle AA_2I=90^{\circ}$, as desired.
ehuseyinyigit
06.08.2024 07:11
Just observed $I$ is the Miquel-Point and then thought about Miquel-Point Properties for one dumpty hour . Here is the solution: $I$ is the Miquel Point, which means $(AB_1C_1)\cap (B_1CA_1)\cap (A_1BC_1)=I$. Then observe that $\angle IC_1A=90^{\circ}$ implies $AI$ being diameter of $(AB_1C_1)$. Similarly $BI$ and $CI$ is diameter of $(A_1BC_1)$ and $(A_1CB_1)$, respectively. By Thales Thorem, one can easily get $\angle AA_2I=\angle BB_2I=\angle CC_2I=90^{\circ}$. Thus, $PB_2A_2IC_2$ is concyclic, which implies $(A_2B_2C_2)$ is tangent to $w$ at $P$.
khanhnx
06.08.2024 12:00
We have $\angle{PA_2I} = \angle{PB_2I} = \angle{PC_2I} = 90^{\circ}$ then $P, A_2, B_2, C_2, I$ lie on circle with diameter $IP$. From this, it's easy to see that $(A_2B_2C_2)$ tangents $\omega$ at $P$