Let O be a circumcenter of an acute-angled triangle ABC. Consider two circles ω and Ω inscribed in the angle BAC in such way that ω is tangent from the outside to the arc BOC of a circle circumscribed about the triangle BOC; and the circle Ω is tangent internally to a circumcircle of triangle ABC. Prove that the radius of Ω is twice the radius ω.
Problem
Source: SRMC 2015
Tags: geometry
chomikchomik
03.04.2015 23:53
solution:
I used Feuerbach's theorem for 9-point circle.
Denote midpoints of AC and AB by M and N respectively. Let circle BOC intersect lines AB and AC in points L and K respectively different from B and C. Homothety 2:1 with centre A transforms circle ACB into circle AMN. It suffices to show that circles AMN and ω are internally tangent.
Inversion with center in A and constant circle BOC transforms circle AMN into line XY where K and L are midpoints of segments AX and AY respectively. Circle KLBC is a 9-point circle of triangle AXY, because angles ACY and KNA are similar and right (inversion and easy lemma). From Feuerbach's theorem the excircle of traingle AXY is externally tangent to circle BCKL. Circle ω is transformed under inversion into such circle, because it's externally tangent to circle BCKL and lines AC and AB. There are only two such circles. In this case it's ω and it's image, which is further from A.
It means that the image of ω is tangent to line XY. It ends the proof, because XY is image of circle AMN.
TelvCohl
04.04.2015 16:55
My solution: Let $ \Omega $ touch $ AC, AB $ at $ B_1, C_1 $, respectively . Let $ \omega $ touch $ AC, AB $ at $ B_2, C_2 $, respectively . From Mannheim theorem $ \Longrightarrow $ the midpoint of $ B_1C_1 $ is the Incenter $ I $ of $ \triangle ABC $ . From the problem Prove that KL bisects BI $ \Longrightarrow B_2C_2 $ is the perpendicular bisector of $ AI $ , so $ \Omega $ is the image of $ \omega $ under homothety $ \mathbf{H}(A,2) \Longrightarrow $ the radius of $ \Omega $ is twice the radius of $ \omega $ . Q.E.D
nguyenhaan2209
20.07.2019 18:25
Let I-incenter, BI,CI-(O)=D,E, DE-AC,AB=F,G, N=(DFIC)-(EGIB) then BNC=BEI+CDI=BOC so N on (BOC). Notice GNF=360-GNI-FNI=GBI+FCI=GLF/2 so N on (L). Finally, EGN+NBC=EGN+NGI+IBC=EGI+IBC=EBI+IBC=FNC so tangency hence q.e.d
parmenides51
10.12.2022 11:33
Let $O$ be a circumcenter of an acute-angled triangle $ABC$. Consider two circles $\omega$ and $\Omega$ inscribed in the angle $\angle BAC$ in such way that $\omega$ is tangent from the outside to the arc $BOC$ of a circle circumscribed about the triangle $BOC$, and the circle $\Omega$ is tangent internally to a circumcircle of triangle $ABC$. Prove that the radius of $\Omega$ is twice the radius $\omega$.