Let $\gamma,\gamma_0,\gamma_1,\gamma_2$ be four circles in plane,such that $\gamma_i$ is interiorly tangent to $\gamma$ in point $A_i$,and $\gamma_i$ and $\gamma_{i+1}$ are exteriorly tangent in point $B_{i+2}$,$i=0,1,2$(the indexes are reduced modulo $3$).The tangent in $B_i$,common for circles $\gamma_{i-1}$ and $\gamma_{i+1}$,intersects circle $\gamma$ in point $C_i$,situated in the opposite semiplane of $A_i$ with respect to line $A_{i-1}A_{i+1}$.Prove that the three lines $A_iC_i$ are concurrent.
Problem
Source: Stars of Mathematics 2015 Senior Level #2
Tags: geometry, abstract algebra
Dukejukem
10.01.2016 21:22
Let $R$ be the circumradius of $ABCDEF.$ Note that
\begin{align*}
\frac{\sin\angle CAD}{\sin\angle EAD} \cdot \frac{\sin\angle ECF}{\sin\angle ACF} \cdot \frac{\sin\angle AEB}{\sin\angle CEB} &= \frac{2R\sin\angle CAD}{2R\sin\angle EAD} \cdot \frac{2R\sin\angle ECF}{2R\sin\angle ACF} \cdot \frac{2R\sin\angle AEB}{2R\sin\angle CEB} \\ \\
&= \frac{CD}{DE} \cdot \frac{EF}{FA} \cdot \frac{AB}{BC},
\end{align*}where the last step follows by the Law of Sines. The desired result now follows from Trigonometric Ceva in $\triangle ACE.$
that in cyclic hexagon $ABCDEF$, the diagonals $AD, BE, CF$ are concurrent iff $AB \cdot CD \cdot EF = BC \cdot DE \cdot FA.$ In this case, it is sufficient to show that $A_1C_2 \cdot A_3C_1 \cdot A_2C_3 = C_2A_3 \cdot C_1A_2 \cdot C_3A_1. \quad (\star)$ ______________________________________________________________________________________________________________________________________________________ Lemma: Let $\Gamma$ and $\omega$ be two circles tangent at a point $A.$ Let $B, C$ be points on $\Gamma$ and draw tangents $BY, CZ$ to $\omega.$ Then $\tfrac{AB}{AC} = \tfrac{BY}{CZ}.$ Proof of Lemma: Let the common tangent to $\Gamma$ and $\omega$ meet $BC$ at $X$ and set $\lambda = -\tfrac{AB^2}{AC^2}.$ It is well-known that $\tfrac{BX}{XC} = \lambda$, implying that $X = \tfrac{B + \lambda C}{1 + \lambda}.$ Now, define the function $f : \mathbb{R}^2 \mapsto \mathbb{R}$ by $f(P) = \text{pow}(P, \omega) - \text{pow}(P, \Gamma).$ It is well-known that $f$ is linear. Then since $X$ lies on the radical axis of $\Gamma$ and $\omega$, linearity yields \[0 = f(X) = \frac{1}{1 + \lambda}f(B) + \frac{\lambda}{1 + \lambda}f(C) = \frac{BY^2}{1 + \lambda} + \frac{\lambda \cdot CZ^2}{1 + \lambda}.\]Thus, multiplication by $1 + \lambda$ and rearrangement yields the desired result. $\blacksquare$ ______________________________________________________________________________________________________________________________________________________ Now, applying the lemma for $\gamma_0$ and $\gamma_1$, we deduce that $\tfrac{A_1C_2}{A_1C_3} = \tfrac{B_2C_2}{B_3C_3}.$ Analagously, one may obtain $\tfrac{A_3C_1}{A_3C_2} = \tfrac{B_1C_1}{B_2C_2}$ and $\tfrac{A_2C_3}{A_2C_1} = \tfrac{B_3C_3}{B_1C_1}.$ Then multiplication yields $\tfrac{A_1C_2}{A_1C_3} \cdot \tfrac{A_3C_1}{A_3C_2} \cdot \tfrac{A_2C_3}{A_2C_1} = 1$, which implies $(\star).$ $\square$