Difference between revisions of "1995 IMO Problems/Problem 1"
(New page: 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>. Th...) |
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+ | ==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. | ||
== Solution == | == Solution == | ||
+ | Since <math>M</math> is on the circle with diameter <math>AC</math>, we have <math>\angle AMC=90</math> and so <math>\angle MCA=90-A</math>. We simlarly find that <math>\angle BND=90</math>. Also, notice that the line <math>XY</math> is the radical axis of the two circles with diameters <math>AC</math> and <math>BD</math>. Thus, since <math>P</math> is on <math>XY</math>, we have <math>PN\cdotPB=PM\cdot PC</math> and so by the converse of Power of a Point, the quadrilateral <math>MNBC</math> is cyclic. Thus, <math>90-A=\angle MCA=\angle BNM</math>. Thus, <math>\angle MND=180-A</math> and so quadrilateral <math>AMND</math> is cyclic. Let the circle which contains the points <math>AMND</math> be cirle <math>O</math>. Then, the radical axis of <math>O</math> and the circle with diameter <math>AC</math> is line <math>AM</math>. Also, the radical axis of <math>O</math> and the circle with diameter <math>BD</math> is line <math>DN</math>. Since the pairwise radical axes of 3 circles are concurrent, we have <math>AM,DN,XY</math> are concurrent as desired. | ||
− | + | ==See also== | |
+ | |||
+ | [[Category:Olympiad Geometry Problems]] |
Revision as of 21:48, 1 September 2008
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. 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.