1996 USAMO Problems/Problem 3
Problem
Let be a triangle. Prove that there is a line
(in the plane of triangle
) such that the intersection of the interior of triangle
and the interior of its reflection
in
has area more than
the area of triangle
.
Solution
Let the triangle be . Assume
is the largest angle. Let
be the altitude. Assume $AB ≤ AC$ (Error compiling LaTeX. Unknown error_msg), so that $BD ≤ BC/2$ (Error compiling LaTeX. Unknown error_msg). If
, then reflect in
. If
is the reflection of
, then
and the intersection of the two triangles is just
. But
, so
has more than
the area of
.
If , then reflect in the angle bisector of
. The reflection of
is a point on the segment
and not
. (It lies on the line
because we are reflecting in the angle bisector.
because $∠CAD < ∠CDA = 90^{\circ}$ (Error compiling LaTeX. Unknown error_msg). Finally, $A'C ≤ BC$ (Error compiling LaTeX. Unknown error_msg) because we assumed $∠B$ (Error compiling LaTeX. Unknown error_msg) does not exceed $∠A$ (Error compiling LaTeX. Unknown error_msg)). The intersection is just
. But $\frac{[AA'C]}{[ABC]} = \frac{CA'}{CB} > CD/CB ≥ 2/3$ (Error compiling LaTeX. Unknown error_msg).
See Also
1996 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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