1995 IMO Problems/Problem 1

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

Think about Radical Axis, Power of a Point and Radical Center.

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 $PN\cdot PB=PM\cdot PC$ 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 name of the circle $AMND$ be $O$ . 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 $PT$ (a subsegment of $XY$) intersect at $Z$. Now, assume that $Z, N, P$ are not collinear. In that case, let $ZD$ intersect the circle with diameter $BD$ at $N'$ and the circle through $D, P, T$ at $N''$.

We know that $\angle AMC = \angle BND = \angle ATP = 90^\circ$ via standard formulae, so quadrilaterals $AMPT$ and $DNPT$ are cyclic. Thus, $N'$ and $N''$ are distinct, as none of them is $N$. 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, $ZN'' = ZN'$, a contradiction since $D$ and $D'$ are distinct. We therefore conclude that $Z, N, D$ are collinear, which gives the concurrency of $AM, XY$, and $DN$. This completes the problem.

Solution 3

Let $AM$ and $XY$ intersect at $Z$. Because $\angle AMC = \angle BND = \angle APT = 90^\circ$, we have quadrilaterals $AMPT$ and $DNPT$ cyclic. Therefore, $Z$ lies on the radical axis of the two circumcircles of these quadrilaterals. But $Z$ also lies on radical axis $XY$ of the original two circles, so the power of $Z$ with respect to each of the four circles is all equal to $ZM * ZA$. Hence, $Z$ lies on the radical axis $DN$ of the two circles passing through $D$ and $N$, as desired.

Discussion

Lemma: The radical axis of two pairs of circles $O_1$, $O_2$ and $O_3$, $O_4$ are the same line $l$. Furthermore, $O_1$ and $O_2$ intersect at $A$ and $B$, and $O_3$ and $O_4$ intersect at $C$ and $D$. Then $A, B, C,$ and $D$ are concyclic.

The proof of this lemma is trivial using the argument in Solution 3 and applying the converse of Power of a Point.

Note that this Problem 1 is a corollary of this lemma. This lemma is an effective way to relate four circles, just as the radical center can relate three circles.

Solution 1 also gives a trivial lemma that can also be useful:

Lemma 2: Chords $AB$ of $\omega_1$ and $CD$ of $\omega_2$ intersect on the segment $XY$ formed from the intersections of the two circles. Then $A, B, C, D$ are concyclic.

Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL!

See also

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