Let $ ABC$ be a triangle, and $ D$ be a point where incircle touches side $ BC$. $ M$ is midpoint of $ BC$, and $ K$ is a point on $ BC$ such that $ AK\perp BC$. Let $ D'$ be a point on $ BC$ such that $ \frac{D'M}{D'K}=\frac{DM}{DK}$. Define $ \omega_{a}$ to be circle with diameter $ DD'$. We define $ \omega_{B},\omega_{C}$ similarly. Prove that every two of these circles are tangent.
Problem
Source: Iranian National Olympiad (3rd Round) 2007
Tags: geometry, geometric transformation, rotation, radical axis, geometry proposed
prowler
27.08.2007 23:12
This problem is interesting. Just an observation: Denote by $ \Omega_{9}$ the center of the nine-point circle and by $ L$ the projection of $ \Omega_{9}$ onto $ BC$. Clearly $ (KMDD')$ form a harmonic division. Let $ T_{1}$ be the intersection of the circles $ d_{KM}, d_{DD'}$ - with diameters $ KM$ and $ DD'$. Let $ L'$ be the projection of $ T_{1}$ onto $ BC$. Then pairs $ (KM), (DD'),(L',\infty)$ are in the involution on $ (BC)$ that is determined by the rotation of the angle $ 90^{\circ}$ around the vertex $ T_{1}$. Thus we have $ (KMDL') = (KMD'\infty)$, which gives us $ KL' = ML'$, therefore $ L\equiv L'$. Hence the radical axes of $ d_{KM}, d_{DD'}$ passes through $ \Omega_{9}$. In other words $ L$ is the radical center of $ C(\Omega_{9},\frac{R}{2}),d_{KM}, d_{DD'}$. Hope I calculated everything correct or again I did not understood involution right.
lomos_lupin
04.09.2007 01:26
Hard and technical problem . Did anyone solve this in the exam time, Omid?
Lemma1. Let $ C$ be a circle with center $ O$ and radius $ R$ and $ A,B$ two point collinear with $ O$. If $ AO*BO = R^{2}$ then any circle passing through $ A, B$ is perpendicular to $ C$.
Lemma2. If $ C_{1}, C_{2}$ are two tangent circles and $ C_{3}$ is circle perpendicular to both, then $ O_{3}$ center of $ C_{3}$ is on the common tangent of $ C_{1},C_{2}$.
Lemmas proof. Exercise !
Let $ D''$ be the midpoint of $ DD'$. Since $ (D',K,D,M) =-1$ then $ D''K*D''M = (\frac{DD'}{2})^{2}$. So by lemma1, $ \omega_{a}$ is perpendicular to Nine-point circle. Also note that $ \omega_{a}$ is perpendicular to incircle. Hence by lemma2, its center, D'', is on the common tangent of incircle and nine-point circle.
Let the common point of incircle and nine-point be $ T$.
In a similar way, the centers of $ \omega_{b},\omega_{c}$ are also on the tangent line at $ T$. Now we have three circles with their centers all on a line and all 3 of them passing through a point on this line. Since they are not identical, they are all tangent to each other.
Any mistakes?
Amir.S
04.09.2007 14:22
not hard at all. let F be fuerbach point of $ \triangle ABC$ ,let a tanget at F to the incircle intersect BC at N, we have $ NK.NM=NF^{2}=ND^{2}$ hence N is midpoint of DD' ,hence a circle with diameter DD' pass through Fuerbach point. the same is for $ \omega_{b},\omega_{c}$ and as proved their centers lie on a tangent line to incircle at F
BaBaK Ghalebi
04.09.2007 17:24
as far as I know yeah almost everyone...