Difference between revisions of "1996 USAMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | 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 \le AC</math>, so that <math>BD \le 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 | + | 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 \le AC</math>, so that <math>BD \le 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 > \frac{2}{3} BC</math>, so <math>ABB'</math> has more than <math>\frac{2}{3}</math> the area of <math>ABC</math>. |
− | 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>\angle{CAD} < \angle{CDA} = 90^{\circ}</math>. Finally, <math>A'C \le BC</math> because we assumed <math>\angle B</math> does not exceed <math>\angle A</math>). The intersection is | + | 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>\angle{CAD} < \angle{CDA} = 90^{\circ}</math>. Finally, <math>A'C \le BC</math> because we assumed <math>\angle B</math> does not exceed <math>\angle A</math>). The intersection is at least <math>AA'C</math>. But <math>\frac{[AA'C]}{[ABC]} = \frac{CA'}{CB} > CD/CB \ge 2/3</math>. |
== See Also == | == See Also == |
Latest revision as of 09:29, 29 August 2023
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
, so that
. 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
. Finally,
because we assumed
does not exceed
). The intersection is at least
. But
.
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.