The bisectors of angles $A$, $B$, and $C$ of triangle $ABC$ meet for the second time its circumcircle at points $A_1$, $B_1$, $C_1$ respectively. Let $A_2$, $B_2$, $C_2$ be the midpoints of segments $AA_1$, $BB_1$, $CC_1$ respectively. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are similar.
Problem
Source: Sharygin Finals 2023 8.2
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023, similar triangles
Minkowsi47
02.08.2023 13:24
Straightforward with complex bash
euclides05
02.08.2023 13:43
Let the circumircle of ABC be the unit circle and the complex coordinates of $A, B, C$ be $a^2, b^2, c^2$ respectively. Then $a_1= -bc, b_1= -ca, c_1= -ab$, and
$a_2= \frac{a^2-bc}{2}, b_2= \frac{b^2-ca}{2}, c_2= \frac{c^2-ab}{2}$.
We compute $\frac{b_1-a_1}{c_1-a_1}= \frac{\overline{b_2}-\overline{a_2}}{\overline{c_2}-\overline{a_2}}= \frac{c(b-a)}{b(c-a)}$ therefore triangles $A_1B_1C_1$ and $A_2B_2C_2$ are similar having opposite orientation.
Quidditch
02.08.2023 15:06
Here’s a simple synthetic one. Let $O,I$ be the circumcenter, incenter, respectively. Note that $OA_2\perp AA_1$, so $\angle OA_2I=90^{\circ}$. Thus, $A_2$ lies on the circle with diameter $OI$. Similarly, $B_2,C_2$ also lie on that circle. Hence, $A_2,B_2,C_2,I$ are concyclic. We finish by angle chasing: $$\angle B_2A_2C_2=\angle B_2IC_2=180^{\circ}-\angle BIC=90^{\circ}-\frac{1}{2}BAC=\angle ACB+\angle ABC=\angle C_1A_1B_1$$as desired.
sami1618
17.03.2024 21:26
Just use mass-points