1990 USAMO Problems/Problem 5: Difference between revisions
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== Problem == | == Problem == | ||
An acute-angled triangle <math>ABC</math> is given in the plane. The circle with diameter <math>\, AB \,</math> intersects altitude <math>\, CC' \,</math> and its extension at points <math>\, M \,</math> and <math>\, N \,</math>, and the circle with diameter <math>\, AC \,</math> intersects altitude <math>\, BB' \,</math> and its extensions at <math>\, P \,</math> and <math>\, Q \,</math>. Prove that the points <math>\, M, N, P, Q \,</math> lie on a common circle. | |||
== Solution == | == Solution == | ||
{{solution}} | {{solution}} | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 11:15, 10 February 2008
Problem
An acute-angled triangle 
is given in the plane. The circle with diameter 
intersects altitude 
and its extension at points 
and 
, and the circle with diameter 
intersects altitude 
and its extensions at 
and 
. Prove that the points 
lie on a common circle.
Solution
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Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
