Difference between revisions of "1990 USAMO Problems/Problem 5"
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{{USAMO box|year=1990|num-b=4|after=Final Question}} | {{USAMO box|year=1990|num-b=4|after=Final Question}} | ||
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356630#356630 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356630#356630 Discussion on AoPS/MathLinks] | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
Revision as of 19:49, 3 July 2013
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 (other than 
). 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
