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Difference between revisions of "1990 USAMO Problems/Problem 5"

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== Solution ==
 
== Solution ==
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Let <math>A'</math> be the intersection of the two circles. <math>AA'</math> is perpendicular to both <math>BA'</math>, <math>CA'</math> implying <math>B</math>, <math>C</math>, <math>A'</math> are collinear. Since <math>A'</math> is the foot of the altitude from <math>A</math>: <math>A</math>, <math>H</math>, <math>A'</math> are concurrent, where <math>H</math> is the orthocentre.
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Let <math>A'</math> be the intersection of the two circles (other than <math>A</math>). <math>AA'</math> is perpendicular to both <math>BA'</math>, <math>CA'</math> implying <math>B</math>, <math>C</math>, <math>A'</math> are collinear. Since <math>A'</math> is the foot of the altitude from <math>A</math>: <math>A</math>, <math>H</math>, <math>A'</math> are concurrent, where <math>H</math> is the orthocentre.
  
 
Now, <math>H</math> is also the intersection of <math>BB'</math>, <math>CC'</math> which means that <math>AA'</math>, <math>MN</math>, <math>PQ</math> are concurrent. Since <math>A</math>, <math>M</math>, <math>N</math>, <math>A'</math> and <math>A</math>, <math>P</math>, <math>Q</math>, <math>A'</math> are cyclic, <math>M</math>, <math>N</math>, <math>P</math>, <math>Q</math> are cyclic by the radical axis theorem.
 
Now, <math>H</math> is also the intersection of <math>BB'</math>, <math>CC'</math> which means that <math>AA'</math>, <math>MN</math>, <math>PQ</math> are concurrent. Since <math>A</math>, <math>M</math>, <math>N</math>, <math>A'</math> and <math>A</math>, <math>P</math>, <math>Q</math>, <math>A'</math> are cyclic, <math>M</math>, <math>N</math>, <math>P</math>, <math>Q</math> are cyclic by the radical axis theorem.

Revision as of 20:29, 18 August 2012

Problem

An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $\, AB \,$ intersects altitude $\, CC' \,$ and its extension at points $\, M \,$ and $\, N \,$, and the circle with diameter $\, AC \,$ intersects altitude $\, BB' \,$ and its extensions at $\, P \,$ and $\, Q \,$. Prove that the points $\, M, N, P, Q \,$ lie on a common circle.

Solution

Let $A'$ be the intersection of the two circles (other than $A$). $AA'$ is perpendicular to both $BA'$, $CA'$ implying $B$, $C$, $A'$ are collinear. Since $A'$ is the foot of the altitude from $A$: $A$, $H$, $A'$ are concurrent, where $H$ is the orthocentre.

Now, $H$ is also the intersection of $BB'$, $CC'$ which means that $AA'$, $MN$, $PQ$ are concurrent. Since $A$, $M$, $N$, $A'$ and $A$, $P$, $Q$, $A'$ are cyclic, $M$, $N$, $P$, $Q$ are cyclic by the radical axis theorem.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

1990 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Final Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
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