# 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.

## 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. ! Undefined control sequence.) 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.