Let $ABC$ be an acute-angled triangle, and let $AA_1, BB_1, CC_1$ be its altitudes. Points $A', B', C'$ are chosen on the segments $AA_1, BB_1, CC_1$, respectively, so that $\angle BA'C = \angle AC'B = \angle CB'A = 90^{o}$. Let segments $AC'$ and $CA'$ intersect at $B"$; points $A", C"$ are defined similarly. Prove that hexagon $A'B"C'A"B'C"$ is circumscribed.
Problem
Source: Caucasus MO 2023, senior, day 2, P8
Tags: geometry
kvedula2004
04.08.2023 17:58
Note that $AB'^2=AB_1\cdot AC=AC_1\cdot AB=AC'^2$, so we must have $AB'=AC'$. This means that $AB'A''C'$ is a kite, since $\angle AB'A''=\angle AC'A''=90^{\circ}$, and it must have an incircle, $\omega_1$. Define $\omega_2$ and $\omega_3$ similarly. If $\omega_1$, $\omega_2$, and $\omega_3$ were pairwise different circles, then using Monge's Theorem implies that $A'$, $B'$, and $C'$ are collinear, contradiction. Therefore, at least two of these circles are the same, and this circle will be tangent to all sides of the hexagon in question by definition.
ReaperGod
04.08.2023 18:10
you need to prove that the points are cyclic
kvedula2004
04.08.2023 18:26
ReaperGod wrote: you need to prove that the points are cyclic The definition of circumscribed is that it has an incircle: see here. For further measure: Claim: $A'B"C'A"B'C"$ cannot be inscribed in a circle. Proof: Supposed that all of the vertices lie on a circle. Note that $\angle C''A'B''=90^{\circ}$, so $\overline{B''C''}$ would be a diameter. However, by similar logic, we must have $\overline{A''B''}$ and $\overline{A''C''}$ be diameters as well, contradiction.