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]
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 |