AoPS - Problem

Source: 2015 Saudi Arabia BMO TST II p3

Tags: geometry, circumcircle, collinear, incenter

parmenides51

24.07.2020 15:49

Let $ABC$ be a triangle, $H_a, H_b$ and $H_c$ the feet of its altitudes from $A, B$ and $C$, respectively, $T_a, T_b, T_c$ its touchpoints of the incircle with the sides $BC, CA$ and $AB$, respectively. The circumcircles of triangles $AH_bH_c$ and $AT_bT_c$ intersect again at $A'$. The circumcircles of triangles $BH_cH_a$ and $BT_cT_a$ intersect again at $B'$. The circumcircles of triangles $CH_aH_b$ and $CT_aT_b$ intersect again at $C'$. Prove that the points $A',B',C'$ are collinear. Malik Talbi

Let $H$ and $I$ be the orthocenter and incenter of $ABC$ respectively. Because of right angles, we know that $H$ is on $(AH_bH_c)$ and $I$ is on $(AT_bT_c)$. Now we will show that $A'$ lies on line $HI$. Since $HH_b$ and $IT_b$ are both perpendicular to $CA$, $HH_b \parallel IT_b$ and it suffices to show $\measuredangle A'IT_b = \measuredangle A'HH_b$. Indeed, $\measuredangle A'IT_b = \measuredangle A'AT_b = \measuredangle A'AH_b = \measuredangle A'HH_b$. Similarly, $B'$ and $C'$ lie on line $HI$ too, thus proving that $A', B', C'$ are collinear.