AoPS - Problem

Source: 46th International Tournament of Towns, Senior A-Level P5, Fall 2024

Tags: geometry

gnoka

15.11.2024 14:22

Given a circle ${\omega }_{1}$ , and a circle ${\omega }_{2}$ inside it. An arbitrary circle ${\omega }_{3}$ is chosen which is tangent to the two latter circles and both tangencies are internal. The tangency points are linked by a segment. A tangent line to ${\omega }_{2}$ is drawn through the meet point of this segment and the circle ${\omega }_{2}$ . Thus a chord of the circle ${\omega }_{3}$ is obtained. Prove that the ends of all such chords (obtained by all possible choices of ${\omega }_{3}$ ) belong to a fixed circle. Pavel Kozhevnikov

Attachments:

View attachments. Note $CD=CF$ and $BG=BF$, so we have that $BG$ is parallel to $AD$. Therefore, the tangent at $G$ is perpendicular to $AD$. Hence, $AH=AI$. Thus, we note this length must be $\sqrt{d(A, HI)^2-d(C, HI)^2+CD^2}$. Let $\angle CAB=\theta$. We know from $CA+CB=R_1-R_3+R_3-R_2=R_1-R_2$ that $C$ lies on an ellipse with foci at $A$ and $B$. We have $CD=R_1 - AC$. Also, $d(A, HI)=AB\cos\theta + R_2$. Finally, $d(C, HI)=AB\cos\theta + R_2 - AC$. There is $AC = \frac{p\epsilon}{1-\epsilon\cos\theta}$. Here, $\epsilon = \frac{AB}{R_1-R_2}$, and $p\epsilon = \frac{(R_1-R_2)(1-\epsilon^2)}{2}$. Thus, the square of the length $AH=AI$ is equal to $(AB\cos\theta + R_2)^2 - (AB\cos\theta + R_2 - \frac{p\epsilon}{1-\epsilon\cos\theta})^2 + (R_1-\frac{p\epsilon}{1-\epsilon\cos\theta})^2 = \frac{p\epsilon}{1-\epsilon\cos\theta} \cdot (2AB\cos\theta+2R_2-2R_1) +R_1^2$. The $\frac{1}{1-\epsilon\cos\theta}$ has coefficient $(R_1-R_2)(1-\epsilon^2)(R_2-R_1)+R_1^2 = AB^2+2R_1R_2-R_2^2$. The $\frac{\cos\theta}{1-\epsilon\cos\theta}$ has coefficient $(R_1-R_2)(1-\epsilon^2)AB-R_1^2\epsilon = AB(R_1-R_2)-\frac{AB}{R_1-R_2}(R_1^2+AB^2)$. Hence, we see that indeed this is of the form $\frac{a-a\epsilon\cos\theta}{1-\epsilon\cos\theta}$, making it a constant.