2011 USAMO Problems/Problem 5
Contents
Problem
Let be a given point inside quadrilateral
. Points
and
are located within
such that
,
,
,
. Prove that
if and only if
.
Solution
First note that if and only if the altitudes from
and
to
are the same, or
. Similarly
iff
.
If we define , then we are done if we can show that S=1.
By the law of sines, and
.
So,
By the terms of the problem, . (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.)
Rearranging yields .
Applying the law of sines to the triangles with vertices at P yields .
Solution 2
Lemma. If and
are not parallel, then
are concurrent.
Proof. Let and
meet at
. Notice that with respect to triangle
,
and
are isogonal conjugates. With respect to triangle
,
and
are isogonal conjugates. Therefore,
and
lie on the reflection of
in the angle bisector of
, so
are collinear. Hence,
are concurrent at
.
Now suppose but
is not parallel to
. Then
and
are not parallel and thus intersect at a point
. But then
also passes through
, contradicting
. A similar contradiction occurs if
but
is not parallel to
, so we can conclude that
if and only if
.
See also
2011 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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