Difference between revisions of "1990 USAMO Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | + | Let <math>A'</math> be the intersection of the two circles. <math>AA'</math> is perpendicular to both <math>BA', CA'</math> implying <math>B, C, A'</math> are collinear. Since <math>A'</math> is the foot of the altitude from <math>A</math>, <math>A, H, A'</math> are concurrent, where <math>H</math> is the orthocentre. | |
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| + | Now, <math>H</math> is also the intersection of <math>BB', CC'</math> which means that <math>AA', MN, PQ</math> are concurrent. Since <math>A, M, N, A'</math> and <math>A, P, Q, A'</math> are cyclic, <math>M, N, P, Q</math> are cyclic by the radical axis theorem. | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 03:02, 9 August 2009
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
Let 
be the intersection of the two circles. 
is perpendicular to both 
implying 
are collinear. Since is the foot of the altitude from 
, 
are concurrent, where 
is the orthocentre.
Now, is also the intersection of 
which means that 
are concurrent. Since 
and 
are cyclic, 
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 | ||
