Difference between revisions of "1995 IMO Problems/Problem 1"
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==Problem== | ==Problem== | ||
Let <math>A,B,C,D</math> be four distinct points on a line, in that order. The circles with diameters <math>AC</math> and <math>BD</math> intersect at <math>X</math> and <math>Y</math>. The line <math>XY</math> meets <math>BC</math> at <math>Z</math>. Let <math>P</math> be a point on the line <math>XY</math> other than <math>Z</math>. The line <math>CP</math> intersects the circle with diameter <math>AC</math> at <math>C</math> and <math>M</math>, and the line <math>BP</math> intersects the circle with diameter <math>BD</math> at <math>B</math> and <math>N</math>. Prove that the lines <math>AM,DN,XY</math> are concurrent. | Let <math>A,B,C,D</math> be four distinct points on a line, in that order. The circles with diameters <math>AC</math> and <math>BD</math> intersect at <math>X</math> and <math>Y</math>. The line <math>XY</math> meets <math>BC</math> at <math>Z</math>. Let <math>P</math> be a point on the line <math>XY</math> other than <math>Z</math>. The line <math>CP</math> intersects the circle with diameter <math>AC</math> at <math>C</math> and <math>M</math>, and the line <math>BP</math> intersects the circle with diameter <math>BD</math> at <math>B</math> and <math>N</math>. Prove that the lines <math>AM,DN,XY</math> are concurrent. | ||
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+ | == Hint == | ||
+ | Radical axis and radical center! Think RADICAL-ly! | ||
+ | |||
Revision as of 22:45, 27 August 2014
Contents
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
.
We know that via standard formulae, so quadrilaterals
and
are cyclic. 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.