The point $K{}$ is the middle of the arc $BAC$ of the circumcircle of the triangle $ABC$. The point $I{}$ is the center of its inscribed circle $\omega$. The line $KI$ intersects the circumcircle of the triangle $ABC$ at $T{}$ for the second time. Prove that the circle passing through the midpoints of the segments $BC, BT$ and $CT$ is tangent to the circle which is symmetric to $\omega$ with respect to $BC$.
Problem
Source: Russian TST 2018, Day 9 P2 (Group NG), P4 (Groups A & B)
Tags: geometry, tangency
03.04.2023 09:04
Bump for this
11.02.2024 17:07
Bump....
08.05.2024 22:01
JafoNoksy wrote: In file you can see my solution. Very nice solution and problem!
18.08.2024 17:34
We claim this circle is ninepoint circle reflected over $MI$ which finishes by feurbach tangency, and so it suffices to show that $MI$ passes through the midpoint of $AT'$ call it $W$.It suffices to show $WMI$ are collinear and $W$ is on NPC, the latter follows by homoethety of $2$ at $A$ and the former, let nagel cevian meet farther incircle at $L$ then well known $AL \parallel MI$, so suffices to show reflection of $T'$ over $I$ lies on nagellian but follows as $MT'$ is refl of nagellian in $I$.
15.09.2024 15:57
Perhaps a more reasonable solution. Let $T$' be the reflection of $T$ over $BC$. Now take homothety at $T$' with factor $2$. This leads to proving that this homothetized circle is tangent to the circumcircle. Work with classical incentre setup. $t = \frac{k-p}{k\overline{p}-1} = -\frac{2abc+a^2b+a^2c}{2a+b+c}$. Now we compute $t' = b^2+c^2+b^2c^2\frac{2a+b+c}{2abc+a^2b+a^2c}$ and hence $-i' = 2(ab+bc+ca)+b^2+c^2+b^2c^2\frac{2a+b+c}{2abc+a^2b+a^2c} = \frac{(2a+b+c)(ab+bc+ca)^2}{a(2bc+ab+ac)}$. Hence we conclude that the power of $I$' with respect to the circumcircle is $i'\overline{i'} = \frac{(a+b+c)^2(ab+bc+ca)^2}{a^2b^2c^2} = (\overline{i}i)^2$ and thus we are done as this clearly implies that $Pow(I',(ABC)) = (2Rr)^2$.