Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 IMO Problems/Problem 1"
(→‎Solution)
Line 5: Line 5:
 
== 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.
 +
 +
==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 21:37, 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.


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 $X$. 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_IMO_Problems/Problem_1&oldid=63233"