# 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 20:48, 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. ! 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.