1995 IMO Problems/Problem 1

Revision as of 22:37, 27 August 2014 by Suli (talk | contribs) (Solution)

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.


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. ! Undefined control sequence.) 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 $X$. Now, assume that $X, N, P$ are not collinear. In that case, let $XD$ 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, \[XM * XA = XP * XT = XN * XD.\] However, because $X$ lies on radical axis $TP$ of the two circles, we have \[XM * XA = XN' * XD.\] Hence, $XD = XD'$, a contradiction since $D$ and $D'$ are distinct. We therefore conclude that $X, N, D$ are collinear, which gives the concurrency of $AM, PT$, and $DN$. This completes the problem.

See also

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