2004 USAMO Problems/Problem 1
Problem
(Titu Andreescu) Let be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
When does equality hold?
Solution
By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence Now we factor the desired expression into
. Temporarily discarding the case where
and
, we can divide through by the
to get the simplified expression
.
Now, draw diagonal . By the law of cosines,
. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that
. Cosine is monotonically decreasing on this interval, so by setting
at the extreme values, we see that
. Applying the law of cosines analogously to
and
, we see that
; we hence have
and
.
We wrap up first by considering the second inequality. Because ,
. This latter expression is of course greater than or equal to
because the inequality can be rearranged to
, which is always true. Multiply the first inequality by
and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.
Equality occurs when and
, or when
is a kite.
Resources
2004 USAMO (Problems • Resources) | ||
Preceded by First problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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