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 04:02, 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 |