(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A_c$, $ B_c$; $ C_1$ is the common point of $ AA_c$ and $ BB_c$. Points $ A_1$, $ B_1$ are defined similarly. Prove that circle $ A_1B_1C_1$ passes through the circumcenter of triangle $ ABC$.
Problem
Source: Sharygin contest. The final raund. 2008. Grade 9. First day. Problem 4
Tags: geometry, circumcircle, angle bisector, geometry unsolved
plane geometry
01.09.2008 10:33
see http://www.mathlinks.ro/Forum/viewtopic.php?t=223753 sin∠ABB1/ sin∠CBB1 * sin∠CAA1/ sin∠BAA1 * sin∠BCC1/ sin∠ACC1= sin∠B0BC/ sin∠B0BA * sin∠BAA0/ sin∠CAA0 * sin∠ACC0/sin∠BCC0 = B0A/B0C * BA0/CA0 * BC0/AC0 =1 We have AA1,BB1,CC1 are concurrent,denote the intersection is D We can acquire O,D,C1,B1 are concyclic and OP is diameter O,A1,B1,D are concyclic and OP is diameter So A1,B1,C1,D,O are concyclic with OP its diameter
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sunken rock
20.03.2009 14:35
As it can be seen from above proof, CC1 is symmedian in triangle ABC and angle bisector of angle AC1B. Knowing that the symmedian passes through the common point of the tangents to (ABC) at A and B, we get OC1 and CC1 perpendicular, hence OC1 is angle bisector as well. Inverting about O so that the circle O transforms to it self, the circle (OAB) is transformed into AB, hence the image of C1 is AB∩OC1=C2, similarly getting A2 and B2. Taking into account the above statements, we get C_2A/C_2B = (b/a)^2 and other 2 similar relations which, multiplied, show that A2, B2 and C2 are collinear, the conclusion follows (a,b and c being the side lengths of triangle ABC). Best regards, sunken rock
Scorpius
06.07.2009 19:43
Can someone solve this problem using complex numbers ? I tried to, but I failed. At least give a hint or something.
anantmudgal09
14.04.2016 08:53
By USAMO 2008/2, we know that $C_1$ is the midpoint of the chord intercepted by the $C$-symmedian in the circumcircle of triangle $ABC$. Now, inverting about $O$ maps $C_1$ to the intersection point of the tangent to the circle at $C$ and line $AB$. Thus, inversion maps the circumcircle of $A_1B_1C_1$ to the trilinear polar of the symmedian point $K$ of triangle $ABC$ and so, points $A_1,B_1,C_1,O$ are con cyclic.