# Difference between revisions of "1995 IMO Problems/Problem 1"

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== 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. | 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. | ||

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+ | ==Solution 2== | ||

+ | Let <math>AM</math> and <math>PT</math> intersect at <math>X</math>. Now, assume that <math>X, N, P</math> are not collinear. In that case, let <math>XD</math> intersect the circle with diameter <math>BD</math> at <math>N'</math>. | ||

+ | |||

+ | We know that <math><AMC = <BND = <ATP = 90^\circ</math> via standard formulae, so quadrilaterals <math>AMPT</math> and <math>DNPT</math> are cyclic. Hence, by Power of a Point, <cmath>XM * XA = XP * XT = XN * XD.</cmath> However, because <math>X</math> lies on radical axis <math>TP</math> of the two circles, we have <cmath>XM * XA = XN' * XD.</cmath> Hence, <math>XD = XD'</math>, a contradiction since <math>D</math> and <math>D'</math> are distinct. We therefore conclude that <math>X, N, D</math> are collinear, which gives the concurrency of <math>AM, PT</math>, and <math>DN</math>. This completes the problem. | ||

==See also== | ==See also== | ||

[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |

## Revision as of 22:37, 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.

## 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.

## 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.