2011 USAMO Problems/Problem 3
In hexagon , which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy
,
, and
. Furthermore
,
, and
. Prove that diagonals
,
, and
are concurrent.
Contents
[hide]Solutions
Solution 1
Let ,
, and
,
,
,
,
intersect
at
,
intersect
at
, and
intersect
at
. Define the vectors:
Clearly,
.
Note that . By sliding the vectors
and
to the vectors
and
respectively, then
. As
is isosceles with
, the base angles are both
. Thus,
. Similarly,
and
.
Next we will find the angles between ,
, and
. As
, the angle between the vectors
and
is
. Similarly, the angle between
and
is
, and the angle between
and
is
. Thus, the angle between
and
is
, or just
in the other direction if we take it modulo
. Similarly, the angle between
and
is
, and the angle between
and
is
.
And since , we can arrange the three vectors to form a triangle, so the triangle with sides of lengths
,
, and
has opposite angles of
,
, and
, respectively. So by the law of sines:
and the triangle with sides of length
,
, and
has corrosponding angles of
,
, and
. But then triangles
,
, and
. So
,
, and
, and
,
, and
are the reflections of the vertices of triangle
about the sides. So
,
, and
concur at the orthocenter of triangle
.
Solution 2
We work in the complex plane, where lowercase letters denote point affixes. Let denote hexagon
. Since
, the condition
is equivalent to
.
Construct a "phantom hexagon" as follows: let
be a triangle with
,
, and
(this is possible since
by the angle conditions), and reflect
over its sides to get points
, respectively. By rotation and reflection if necessary, we assume
and
have the same orientation (clockwise or counterclockwise), i.e.
. It's easy to verify that
for
and opposite sides of
have equal lengths. As the corresponding sides of
and
must then be parallel, there exist positive reals
such that
,
, and
. But then
, etc., so
If
, then
must be similar to
and the conclusion is obvious. Otherwise, since
and
, we must have
and
. Construct parallelograms
and
; if
is the reflection of
over
and
is the reflection of
over
, then by simple angle chasing we can show that
and
. But
means
and
must be linearly dependent (note that
and
), so we must have
. But then C'D'\parallel A'F'$, which is impossible, so we're done.
Alternatively, use the projection formula with unit circle$ (Error compiling LaTeX. Unknown error_msg)(A'C'E')$.
See Also
2011 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |