Difference between revisions of "1996 USAMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | Let the triangle be ABC. Assume A is the largest angle. Let AD be the altitude. Assume AB ≤ AC, so that BD ≤ BC/2. If BD > BC/ | + | Let the triangle be <math>ABC</math>. Assume <math>A</math> is the largest angle. Let <math>AD</math> be the altitude. Assume <math>AB ≤ AC</math>, so that <math>BD ≤ BC/2</math>. If <math>BD > \frac{BC}{3}</math>, then reflect in <math>AD</math>. If <math>B'</math> is the reflection of <math>B'</math>, then <math>B'D = BD</math> and the intersection of the two triangles is just <math>ABB'</math>. But <math>BB' = 2BD > 2/3 BC</math>, so <math>ABB'</math> has more than <math>\frac{2}{3}</math> the area of <math>ABC</math>. |
− | If BD < BC/3, then reflect in the angle bisector of C. The reflection of A' is a point on the segment BD and not D. (It lies on the line BC because we are reflecting in the angle bisector. A'C > DC because ∠CAD < ∠CDA = | + | If <math>BD < BC/3</math>, then reflect in the angle bisector of <math>C</math>. The reflection of <math>A'</math> is a point on the segment <math>BD</math> and not <math>D</math>. (It lies on the line <math>BC</math> because we are reflecting in the angle bisector. <math>A'C > DC</math> because <math>∠CAD < ∠CDA = 90^{\circ}</math>. Finally, <math>A'C ≤ BC</math> because we assumed <math>∠B</math> does not exceed <math>∠A</math>). The intersection is just <math>AA'C</math>. But <math>\frac{[AA'C]}{[ABC]} = \frac{CA'}{CB} > CD/CB ≥ 2/3</math>. |
== See Also == | == See Also == |
Revision as of 18:33, 29 October 2019
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.