Difference between revisions of "1990 USAMO Problems/Problem 5"
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== See Also == | == See Also == | ||
− | {{USAMO box|year=1990|num-b=4|after= | + | {{USAMO box|year=1990|num-b=4|after=Last 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] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 19:17, 18 July 2016
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.
See Also
1990 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last 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.