1995 IMO Problems/Problem 1
Problem
Let be four distinct points on a line, in that order. The circles with diameters and intersect at and . The line meets at . Let be a point on the line other than . The line intersects the circle with diameter at and , and the line intersects the circle with diameter at and . Prove that the lines are concurrent.
Hint
Radical axis and radical center! Think RADICAL-ly!
Solution
Since is on the circle with diameter , we have and so . We simlarly find that . Also, notice that the line is the radical axis of the two circles with diameters and . Thus, since is on , we have $PN\cdotPB=PM\cdot PC$ (Error compiling LaTeX. Unknown error_msg) and so by the converse of Power of a Point, the quadrilateral is cyclic. Thus, . Thus, and so quadrilateral is cyclic. Let the circle which contains the points be cirle . Then, the radical axis of and the circle with diameter is line . Also, the radical axis of and the circle with diameter is line . Since the pairwise radical axes of 3 circles are concurrent, we have are concurrent as desired.
Solution 2
Let and intersect at . Now, assume that are not collinear. In that case, let intersect the circle with diameter at and the circle through at .
We know that via standard formulae, so quadrilaterals and are cyclic. Thus, and are distinct, as none of them is . Hence, by Power of a Point, However, because lies on radical axis of the two circles, we have Hence, , a contradiction since and are distinct. We therefore conclude that are collinear, which gives the concurrency of , and . This completes the problem.
Solution 3
Let and intersect at . Because , we have quadrilaterals and cyclic. Therefore, lies on the radical axis of the two circumcircles of these quadrilaterals, so . But as a result lies on the radical axis of the two original circles (with diameters and ), so lies on as well.