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.
(Similarly to the proposition at the beginning of the proof, are collinear if
.)
Writing down the equality above for , we get an equation for
which is easy to solve, and after a few computations we get
.
Thus
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.
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 |