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: [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

We start by stating a simple proposition. Let $L$ be a line and three circles with centers $C_1, C_2, C_3$ tangent to the line $L$ , all on one side. Denote $A_1, A_2, A_3$ the points on the line where the circles touch it. Then there is a second line tangent to the three circles if and only if $C_1, C_2, C_3$ are collinear, and this is the case if and only if $(\overline{C_1A_1} - \overline{C_2A_2})/(\overline{C_2A_2} - \overline{C_3A_3}) = (\overline{A_1A_2})/(\overline{A_2A_3})$

I will not give the full proof of this proposition, and in fact, we need to know only that if the equality of the two fractions holds, then a tangent to two circles is tangent to the third as well. The idea is that if the equality holds then triangles $\triangle{C_1A_1B}, \triangle{C_2A_2B}, \triangle{C_3A_3B}$ are similar, which implies that $C_1, C_2, C_3$ are collinear, which implies that a tangent to two of the circles is tangent to the third as well.

Now let us prove the problem. Look at the following figure:

We use the notation given in the problem, and we denote $O$ the center of the circle $\gamma$ , $C_1, C_2, C_3$ the centers of the circles $\gamma_1, \gamma_2,\gamma_3$ , and $A_1, A_2, A_3$ the points where the circles touch the line $AB$ . Let $AB$ be the $x$ -axis, $CD$ be the $y$ -axis, $D$ be the origin, and the orientations of axes as shown in the picture. Let $a = \overline{OD}$ . We think of $a$ as positive, $a > 0$ . There is no loss of generality in assuming that the circle is of radius $1$ , and that $D$ is to the right of $O$ . This implies the following coordinates:

$D = (0, 0)$

$A = (-1 - a, 0)$

$B = (1 - a, 0)$

$O = (-a, 0)$

The circle has the equation $(x + a)^2 + y^2 = 1$

Then $C = (0, \sqrt{1 - a^2})$

We will also make use of the point $O_4 = (-a, -1)$ on the circle, vertically down from $O$ .

The plan is to calculate the coordinates of $C_1, C_2, C_3$ (the centers of the circles $\gamma_1, \gamma_2,\gamma_3$ ) as expressions in $a$ . Then we will be able to verify that $(\overline{C_1A_1} - \overline{C_2A_2})/(\overline{C_2A_2} - \overline{C_3A_3}) = (\overline{A_1A_2})/(\overline{A_2A_3})$ , which will have solved the problem.

Calculating the coordinates of $C_2 = (x_2, y_2)$ is easy. Take in account that the distance from $C_2$ to line $AB$ has to be the same as the distance to the circle $\gamma$ . In other words, $y_2 = ($ the radius of $\gamma) - \overline{C_2O} = 1 - \sqrt{(x_2 + a)^2 + y_2^2}$ . Also, take in account that $C_2$ is on the bisector of $\angle{ADC}$ , which implies $x_2 + y_2 = 0$ .

From this system of two equations with two unknowns we get

$-x_2 = 1 - \sqrt{(x_2 + a)^2 + x_2^2}$

Solve this, and get $x_2 = 1 - a \pm \sqrt{2 - 2a}$ . Since $-1 < x_2 < 0$ , we see that only $x_2 = 1 - a - \sqrt{2 - 2a}$ is acceptable. Therefore

$C_2 = (1 - a - \sqrt{2 - 2a}, -(1 - a) + \sqrt{2 - 2a}$

By a similar computation, we get

$C_3 = (-1 - a + \sqrt{2 + 2a}, -1 - a + \sqrt{2 + 2a})$

Now we want to compute the coordinates of $C_1 = (x_1, y_1)$ . We could look for the point on $CO_4$ such that the distance to $AB$ , which is $y_1$ , equals the distance to $AC$ . But there is an easier way because we know classical formulas for computing the radius of the incircle of a triangle. Thus, if we denote by $A_r$ and $s$ the area and the semiperimeter of the triangle $\triangle{ABC}$ , we know that

$y_1 =$ the radius of the incircle $= A_r/s = \frac{\overline{AB} \ \overline{DC}}{2} \ \frac{2}{\overline{AC} + \overline{BC} + \overline{AB}} = ... = \frac{\sqrt{2 + 2a} + \sqrt{2 - 2a} - 2}{2}$

(I skipped some obvious computations at the last equality).

An easy way to compute $x_1$ is to impose that $C, C_1, O_4$ are collinear.

(Similarly to the proposition at the beginning of the proof, $M = (m_1, m_2), N = (n_1, n_2), P = (p_1, p_2)$ are collinear if $(n_1 - m_1)/(n_2 - m_2) = (p_1 - m_1)/(p_2 - m_2)$ .)

Writing down the equality above for $C, C_1, O_4$ , we get an equation for $x_1$ which is easy to solve, and after a few computations we get $x_1$ . Thus

$C_1 = \left( \frac{\sqrt{2 + 2a} - \sqrt{2 - 2a} - 2a}{2}, \frac{\sqrt{2 + 2a} + \sqrt{2 - 2a} - 2}{2} \right)$

Now we can verify easily that $(\overline{C_1A_1} - \overline{C_2A_2})/(\overline{C_2A_2} - \overline{C_3A_3}) = (\overline{A_1A_2})/(\overline{A_2A_3})$ . In fact, we can see that $A_1$ is the midpoint between $A_2$ and $A_3$ , and the radius of $\gamma_1$ is the arithmetic mean of the radii of $\gamma_2$ and $\gamma_3$ . This concludes the proof.

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

Page

Toolbox

Search

1969 IMO Problems/Problem 4

Contents

Problem

Solution

Remarks, added by pf02, August 2024

Solution 2

See Also