Difference between revisions of "1989 USAMO Problems/Problem 4"
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+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 19:11, 18 July 2016
Problem
Let be an acute-angled triangle whose side lengths satisfy the inequalities
. If point
is the center of the inscribed circle of triangle
and point
is the center of the circumscribed circle, prove that line
intersects segments
and
.
Solution
Consider the lines that pass through the circumcenter . Extend
,
,
to
,
,
on
,
,
, respectively.
We notice that passes through sides
and
if and only if
belongs to either regions
or
.
Since , we let
,
,
.
We have
Since divides angle
into two equal parts, it must be in the region marked by the
of angle
, so
is in
.
Similarly, is in
and
. Thus,
is in their intersection,
. From above, we have
passes through
and
.
See Also
1989 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.