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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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 +
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== Hint ==
 +
Radical axis and radical center! Think RADICAL-ly!
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Revision as of 21:45, 27 August 2014

Problem

Let $A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. The line $XY$ meets $BC$ at $Z$. Let $P$ be a point on the line $XY$ other than $Z$. The line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$, and the line $BP$ intersects the circle with diameter $BD$ at $B$ and $N$. Prove that the lines $AM,DN,XY$ are concurrent.


Hint

Radical axis and radical center! Think RADICAL-ly!


Solution

Since $M$ is on the circle with diameter $AC$, we have $\angle AMC=90$ and so $\angle MCA=90-A$. We simlarly find that $\angle BND=90$. Also, notice that the line $XY$ is the radical axis of the two circles with diameters $AC$ and $BD$. Thus, since $P$ is on $XY$, 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 $MNBC$ is cyclic. Thus, $90-A=\angle MCA=\angle BNM$. Thus, $\angle MND=180-A$ and so quadrilateral $AMND$ is cyclic. Let the circle which contains the points $AMND$ be cirle $O$. Then, the radical axis of $O$ and the circle with diameter $AC$ is line $AM$. Also, the radical axis of $O$ and the circle with diameter $BD$ is line $DN$. Since the pairwise radical axes of 3 circles are concurrent, we have $AM,DN,XY$ are concurrent as desired.

Solution 2

Let $AM$ and $PT$ intersect at $Z$. Now, assume that $Z, N, P$ are not collinear. In that case, let $ZD$ intersect the circle with diameter $BD$ at $N'$.

We know that $<AMC = <BND = <ATP = 90^\circ$ via standard formulae, so quadrilaterals $AMPT$ and $DNPT$ are cyclic. Hence, by Power of a Point, \[ZM * ZA = ZP * ZT = ZN * ZD.\] However, because $Z$ lies on radical axis $TP$ of the two circles, we have \[ZM * ZA = ZN' * ZD.\] Hence, $ZD = ZD'$, a contradiction since $D$ and $D'$ are distinct. We therefore conclude that $Z, N, D$ are collinear, which gives the concurrency of $AM, PT$, and $DN$. This completes the problem.

See also

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