1969 IMO Problems/Problem 4
Problem
A semicircular arc is drawn with as diameter. is a point on other than and , and is the foot of the perpendicular from to . We consider three circles, , all tangent to the line . Of these, is inscribed in , while and are both tangent to and , one on each side of . Prove that , and have a second tangent in common.
Solution
Denote the triangle sides . Let be the circumcircle of the right angle triangle centered at the midpoint of its hypotenuse . Let be the tangency points of the circles with the line AB. In an inversion with the center and positive power ( being the inversion circle radius), the line AB is carried into itself, the circle is carried into the altitude line and the altitude line into the circle . This implies that the circle intersecting the inversion circle is carried into itself, but this is possible only if the circle is perpendicular to the inversion circle . It follows that the tangency point of the circle is the intersection of the inversion circle with the line . Similarly, in an inversion with the center B and positive power ( being the inversion circle radius), the line AB is carried into itself, the circle is carried into the altitude line and the altitude line into the circle . This implies that the circle intersecting the inversion circle is carried into itself, but this is possible only if the circle is perpendicular to the inversion circle . It follows that the tangency point S of the circle is the intersection of the inversion circle with the line .
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 of the right angle triangle is equal to
where and s are the area and semiperimeter of the triangle , for example, because of an obvious identity
or just because the angle is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then
Therefore, the points are identical and the midpoint of the segment ST is the tangency point R of the incircle with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles are tangent to the incircle . Radii of the circles are now easily calculated:
Denote the centers of the circles . The line cuts the midline RI of the trapezoid at the distance from the point R equal to
As a result, the centers are collinear (in fact, I is the midpoint of the segment ). The common center line and the common external tangent AB of the circles meet at their common external homothety center and the other common external tangent of the circles from the common homothety center H is a tangent to the circle 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 be a line and three circles with centers tangent to the line , all on one side. Denote 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 are collinear, and this is the case if and only if
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 are similar, which implies that 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 the center of the circle , the centers of the circles , and the points where the circles touch the line . Let be the -axis, be the -axis, be the origin, and the orientations of axes as shown in the picture. Let . We think of as positive, . There is no loss of generality in assuming that the circle is of radius , and that is to the right of . This implies the following coordinates:
The circle has the equation
Then
We will also make use of the point on the circle, vertically down from .
The plan is to calculate the coordinates of (the centers of the circles ) as expressions in . Then we will be able to verify that , which will have solved the problem.
Calculating the coordinates of is easy. Take in account that the distance from to line has to be the same as the distance to the circle . In other words, the radius of . Also, take in account that is on the bisector of , which implies .
From this system of two equations with two unknowns we get
Solve this, and get . Since , we see that only is acceptable. Therefore
By a similar computation, we get
Now we want to compute the coordinates of . We could look for the point on such that the distance to , which is , equals the distance to . 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 and the area and the semiperimeter of the triangle , we know that
the radius of the incircle
(I skipped some obvious computations at the last equality).
An easy way to compute is to impose that are collinear.
Use that are collinear if and only if .
Writing down the equality above for , we get
Solve for and simplify, after which we get
Now we can verify easily that . In fact, we can see that is the midpoint between and , and the radius of is the arithmetic mean of the radii of and . This concludes the proof.
[Solution by pf02, August 2024]
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 |