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

Revision as of 18:32, 28 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 $PT$ (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 $<AMC = <BND = <ATP = 90^\circ$ 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 $<AMC = <BND = <APT = 90^\circ$ , 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!

@@ Line 17: / Line 17: @@
 == Solution 3==
-Let <math>AM</math> and <math>DN</math> intersect at <math>Z</math>. Because <math><AMC = <BND = <APT = 90^\circ</math>, we have quadrilaterals <math>AMPT</math> and <math>DNPT</math> cyclic. Therefore, <math>Z</math> lies on the radical axis of the two circumcircles of these quadrilaterals, so <math>ZM * ZA = ZN * ZD</math>. But as a result <math>Z</math> lies on the radical axis of the two original circles (with diameters <math>AC</math> and <math>BD</math>), so <math>Z</math> lies on <math>TP</math> as well.
+Let <math>AM</math> and <math>XY</math> intersect at <math>Z</math>. Because <math><AMC = <BND = <APT = 90^\circ</math>, we have quadrilaterals <math>AMPT</math> and <math>DNPT</math> cyclic. Therefore, <math>Z</math> lies on the radical axis of the two circumcircles of these quadrilaterals. But <math>Z</math> also lies on radical axis <math>XY</math> of the original two circles, so the power of <math>Z</math> with respect to each of the four circles is all equal to <math>ZM * ZA</math>. Hence, <math>Z</math> lies on the radical axis <math>DN</math> of the two circles passing through <math>D</math> and <math>N</math>, as desired.
+==Discussion==
+''Lemma:'' The radical axis of two pairs of circles <math>O_1</math>, <math>O_2</math> and <math>O_3</math>, <math>O_4</math> are the same line <math>l</math>. Furthermore, <math>O_1</math> and <math>O_2</math> intersect at <math>A</math> and <math>B</math>, and <math>O_3</math> and <math>O_4</math> intersect at <math>C</math> and <math>D</math>. Then <math>A, B, C,</math> and <math>D</math> 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 <math>AB</math> of <math>\omega_1</math> and <math>CD</math> of <math>\omega_2</math> intersect on the segment <math>XY</math> formed from the intersections of the two circles. Then <math>A, B, C, D</math> are concyclic.
+Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL!
 ==See also==
 [[Category:Olympiad Geometry Problems]]

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