Let $ d$ be a line and points $ M,N$ on the $ d$. Circles $ \alpha,\beta,\gamma,\delta$ with centers $ A,B,C,D$ are tangent to $ d$, circles $ \alpha,\beta$ are externally tangent at $ M$, and circles $ \gamma,\delta$ are externally tangent at $ N$. Points $ A,C$ are situated in the same half-plane, determined by $ d$. Prove that if exists an circle, which is tangent to the circles $ \alpha,\beta,\gamma,\delta$ and contains them in its interior, then lines $ AC,BD,MN$ are concurrent or parallel.
Problem
Source: Romania Junior TST Day 3 Problem 4
Tags: geometry, geometric transformation, homothety, geometry proposed
¬[ƒ(Gabriel)³²¹º]¼
13.06.2008 21:36
nice and simple problem: it's equivalent to prove that $ \frac{\overline{AM}}{\overline{BM}}=\frac{\overline{CN}}{\overline{DN}}$ and it's equivalent to this: Let a cicle $ \Gamma$ with center O and radius R, a chord EF of $ \Gamma$ with distance $ d$ from O and a point P on EF. Let the 2 circle $ \Gamma_a$ and $ \Gamma_b$ tangent to EF on P and tangent respectively to the minor and bigger arch EF and with radius respectively $ R_a$ and $ R_b$. Then $ \frac{R_b}{R_a}$ not dipend from choice of P. Call A and B the center of $ \Gamma_a$ and $ \Gamma_b$. Call K the projection of O on AB. Then we have: $ \overline{OA}^2 - \overline{AK}^2 = \overline{OB}^2 - \overline{BK}^2$ $ (R-R_a)^2 - (R_a + d)^2 = (R-R_b)^2 - (R_b - d)^2$ $ \frac{R_b}{R_a} = \frac{R+d}{R-d}$
The QuattoMaster 6000
10.07.2008 06:01
Let $ \omega$ be the circle tangent to $ \alpha$, $ \beta$, $ \gamma$, and $ \delta$ and let its center be $ O$. Also, let the tangency points between $ \alpha$, $ \beta$, $ \gamma$, and $ \delta$ with $ \omega$ be $ W$, $ X$, $ Z$, and $ Y$ respectively. Let $ XM$ and $ YN$ meet at $ T$ and $ WM$ and $ ZN$ meet at $ U$ and let $ MN$ meet $ \omega$ at $ P$ and $ Q$ so that $ MP < NP$ and $ NQ > MQ$. Consider the homothety around $ X$ that maps $ \beta$ to $ \omega$. Let it map $ d$ to $ d'$. Notice that since $ d$ is tangent to $ \beta$, we have that $ d'$ is tangent to $ \omega$. Let the tangency point between $ d'$ and $ \omega$ be $ T'$. Also, consider two points $ K$ and $ L$ on $ d'$ so that $ L$ is on the same half-plane as $ A$ and $ B$ and $ K$ is on the same half-plane as $ D$ and $ C$. We have that $ \angle T'QP = \angle LT'P$ because of the tangency between $ \omega$ and $ d'$. We also have that $ \angle T'PQ = \angle LT'P$ since $ d'\parallel PQ$. Thus, $ \angle T'PQ = \angle T'QP$, so $ T'$ is the midpoint of arc $ PWQ$. Therefore, $ XT'$ passes through the midpoint of arc $ PWQ$, so $ XM$ passes through the midpoint of this arc because of the homothety. Similarly does $ YN$, so $ T = T'$, so $ T$ is the midpoint of arc $ PWQ$. Similarly, $ U$ is the midpoint of arc $ PXQ$. Thus, we have that $ \angle TQP = \angle TPQ = \angle TYQ$ and $ \angle NTQ = \angle YTQ$, so $ \triangle NTQ\sim \triangle QTY$. Thus, $ \frac {NT}{QT} = \frac {QT}{TY}\implies QT^2 = TY\cdot NT$. Similarly, we can find that $ TM\cdot TX = TP^2 = QT^2 = TY\cdot NT$ and $ UM\cdot UW = UP^2 = UQ^2 = UN\cdot UZ$. Now, the homotheties imply that $ \frac {MX}{XT} = \frac {MB}{OT}$ and $ \frac {WM}{WU} = \frac {AM}{OU} = \frac {AM}{OT}$. Thus, $ \frac {AM}{MB} = \frac {WM}{MX}\cdot \frac {XT}{WU}$. Similarly, we can find that $ \frac {CN}{DN} = \frac {ZN}{NY}\cdot \frac {TY}{ZU}$. Notice that since $ WTUZ$ and $ TZYU$ are cyclic, we have that $ \frac {WM}{MX} = \frac {MT}{MU}$ and $ \frac {ZN}{NY} = \frac {NT}{NU}$. This means that $ \frac {AM}{MB} = \frac {MT}{MU}\cdot \frac {XT}{WU} = \frac {TP^2}{UP^2}$ and
\[ \frac {CN}{DN} = \frac {NT}{NU}\cdot \frac {TY}{ZU} = \frac {QT^2}{UQ^2} = \frac {TP^2}{UP^2} = \frac {AM}{MB}
\]
Now, $ AM\perp MN$ since and $ CD\perp MN$ since $ \alpha$ and $ \beta$ are tangent to $ MN$ at $ M$ and $ \gamma$ and $ \delta$ are tangent to $ MN$ at $ N$. Thus, $ AB\parallel CD$. Also, we have just shown that $ \frac {AM}{MB} = \frac {CN}{DN}$. From this, it is well known that $ AC$, $ MN$, and $ BD$ must concur.