# 1969 IMO Problems/Problem 4

## Problem

A semicircular arc $\gamma$ is drawn with $AB$ as diameter. $C$ is a point on $\gamma$ other than $A$ and $B$, and $D$ is the foot of the perpendicular from $C$ to $AB$. We consider three circles, $\gamma_1, \gamma_2, \gamma_3$, all tangent to the line $AB$. Of these, $\gamma_1$ is inscribed in $\triangle ABC$, while $\gamma_2$ and $\gamma_3$ are both tangent to $CD$ and $\gamma$, one on each side of $CD$. Prove that $\gamma_1, \gamma_2$, and $\gamma_3$ have a second tangent in common.

## Solution

Denote the triangle sides $a = BC, b = CA, c = AB$. Let $\omega$ be the circumcircle of the right angle triangle $\triangle ABC$ centered at the midpoint $O$ of its hypotenuse $c = AB$. Let $R, S, T$ be the tangency points of the circles $K_1, K_2, K_3$ with the line AB. In an inversion with the center $A$ and positive power $r_A^2 = AC^2 = b^2$ ( $r_A$ being the inversion circle radius), the line AB is carried into itself, the circle $\omega$ is carried into the altitude line $CD$ and the altitude line $CD$ into the circle $\omega$. This implies that the circle $K_3$ intersecting the inversion circle $A$ is carried into itself, but this is possible only if the circle $K_3$ is perpendicular to the inversion circle $A$. It follows that the tangency point $T$ of the circle $K_3$ is the intersection of the inversion circle $(A, r_A = b)$ with the line $AB$. Similarly, in an inversion with the center B and positive power $r_B^2 = BC^2 = a^2$ ( $r_B$ being the inversion circle radius), the line AB is carried into itself, the circle $\omega$ is carried into the altitude line $CD$ and the altitude line $CD$ into the circle $\omega$. This implies that the circle $K_2$ intersecting the inversion circle $B$ is carried into itself, but this is possible only if the circle $K_2$ is perpendicular to the inversion circle $B$. It follows that the tangency point S of the circle $K_2$ is the intersection of the inversion circle $(B, r_B = a)$ with the line $AB$.

The distance between the tangency points S, T is the equal to ST = AT - AS = AT - (AB - BS) = AC - (AB - BC) = a + b - c. The radius r of the incircle $K_1$ of the right angle triangle $\triangle ABC$ is equal to $r = \frac{|\triangle ABC|}{s} = \frac{ab}{a + b + c} = \frac{a + b - c}{2} = s - c$

where $|\triangle ABC|$ and s are the area and semiperimeter of the triangle $\triangle ABC$, for example, because of an obvious identity $(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab$

or just because the angle $\angle C = 90^\circ$ is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then $AR' = AT - \frac{ST}{2} = b - r = b - \frac{a + b - c}{2} = \frac{c + b - a}{2} = s - a = AR$

Therefore, the points $R' \equiv R$ are identical and the midpoint of the segment ST is the tangency point R of the incircle $K_1$ with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles $K_2, K_3$ are tangent to the incircle $K_1$. Radii $r_2, r_3$ of the circles $K_2, K_3$ are now easily calculated: $r_2 = SD = BS - BD = a - \frac{a^2}{c}$ $r_3 = TD = AT - AD = b - \frac{b^2}{c}$

Denote $I, I_2, I_3$ the centers of the circles $K_1, K_2, K_3$. The line $I_2I_3$ cuts the midline RI of the trapezoid $STI_3I_2$ at the distance from the point R equal to $\frac{SI_2 + TI_3}{2} = \frac{r_2 + r_3}{2} = \frac{a + b}{2} - \frac{a^2 + b^2}{2c} = \frac{a + b - c}{2} = r = RI$

As a result, the centers $I_2, I, I_3$ are collinear (in fact, I is the midpoint of the segment $I_2I_3$). The common center line $I_2I_3$ and the common external tangent AB of the circles $K_1, K_2, K_3$ meet at their common external homothety center $H \equiv I_2I_3 \cap AB$ and the other common external tangent of the circles $K_2, K_3$ from the common homothety center H is a tangent to the circle $K_1$ as well.

The above solution was posted and copyrighted by yetti. The original thread can be found here: