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1969 IMO Problems/Problem 4

Revision as of 17:26, 7 August 2024 by Pf02 (talk | contribs)

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: [1]

Remarks, added by pf02, August 2024

It is worth repeating here the note hal9v4ik makes on https://aops.com/community/p376814. This problem is a consequence of a particular case of Thebault's problem III (the Sawayama-Thebault-Streefkerk theorem; see https://en.wikipedia.org/wiki/Th%C3%A9bault%27s_theorem). However, this should not be viewed as a solution to the problem. I believe the point of the problem is proving the Sawayama-Thebault-Streefkerk theorem in a particular case. Also, the theorem is not well known, so it should not be used as a reference.

Below I will give another solution, based on analytic (coordinate) geometry. It is not elegant, but it is very straightforward and simple (except for the computations which can be tedious at times.)

Solution 2

TO BE CONTINUED. SAVING FOR NOW, SO THAT I DON'T LOSE WORK DONE SO FAR

See Also

1969 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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