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

(New page: 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>. Th...)
 
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==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.  
  
  
 
== Solution ==
 
== Solution ==
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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.
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==See also==
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[[Category:Olympiad Geometry Problems]]

Revision as of 20:48, 1 September 2008

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.

See also

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