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
(a subsegment of
) 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.