1979 IMO Problems/Problem 3
Problem
Two circles in a plane intersect. is one of the points of intersection. Starting simultaneously from two points move with constant speed, each travelling along its own circle in the same sense. The two points return to simultaneously after one revolution. Prove that there is a fixed point in the plane such that the two points are always equidistant from
Solution 1
Let be the centers of the 2 circles and let the points B, C move on the circles , recpectively. Let A' be the other intersection of the 2 circles different from the point A. Since the angles are equal, the isosceles triangles are similar. Consequently, the angles are equal and the angles
are also equal. The angles are equal and the angles are also equal. As a result, the point A' is on the side BC of the triangle , which is always similar to the fixed triangle .
Lemma (Yaglom's Geometric Transformations II, pp 68-69). Assume that a figure F' (the triangle ) moves in such a way that it remains at all times similar to a fixed figure F (the triangle ) and in addition, some point of the figure F' does not move at all (the point A). If a point B of the figure F' describes a curve (the circle ), then any other point M of the figure F' describes a curve similar to the curve .
Consequently, the midpoint M of the segment BC decribes a circle (Q). When the points B, C, become identical with the point A, the midpoint M of BC also becomes identical with A. Hence, the circle (Q) passes through the point A. The line BC always passes through A'. Just after the start of motion counter-clockwise, the points B, C, A' follow on the line BC in this order. The points B, C both pass through the point A' one after another and just before the end of one period of motion, the points A', B, C follow on the line BC In this order. From the continuity principle, the midpoint M of BC also passes through the point A' and consequently, the circle (Q), the locus of the midpoints M, also passes through the point A'. Thus the circle (Q) belongs to the same pencil with the circles and it is centered on their center line . Let p be the perpendicular bisector of the segment BC intersectong the circle (Q) at a point P different from the point M. The point A' and the circle (Q) passing through this point are fixed and the angle is right. Therefore, A'P is a fixed diameter of the circle (Q), which means that the point P is also fixed. Q.E.D.
When the points B, C are both diametrally opposite to the point A on their respective circles, the segment is the midline of the triangle , i.e., , . The cyclic quadrilateral AA'MP inscribed in the circle (Q) is then a rectangle. Its circumcenter Q is the intersection of its diagonals AM, A'P intersecting at their midpoint and the midpoint of AM is identical with the midpoint of . Thus the fixed point P is the reflection of the intersection A' of the circles in the midpoint Q of the segment .
This solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]
Solution 2
Let and be the antipodes of on the two circles and let be the foot of the altitude from , which is the other intersection point of the two circles. Also, let be the midpoint of , and construct rectangle . Our claim is that is the fixed point. We let and be the two points; by the condition the angles are the same. So we have a spiral similarity Now let be the midpoint of . By spiral similarity, since maps to , it follows , , , and are cyclic too. So actually lies on the circumcircle of rectangle , meaning , hence as needed.
Remark: The special point can be identified by selecting the special case , and , .
This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]
Solution 3
Let the points on the circles be and . Then, let the the other intersection of the two circles, and and be the centers of the circles containing and , respectively. WLOG, let us name and such that . Let be the midpoint of , and the antipode of with respect to the circle centered at with radius be . Let be where this such circle hits . We first note thatand alsomeaning that since , , and , then SAS similarity gets us that . This means we have so and because , then and similarly, we also have . This means thatmeaning that is the midpoint of . Now, since is the antipode of , then so for all and for some fixed and we are done.
This solution was posted and copyrighted by kevinmathz. The original thread for this problem can be found here: [3]
See Also
1979 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |