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Difference between revisions of "1996 USAMO Problems/Problem 3"

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Let <math>ABC</math> be a triangle. Prove that there is a line <math>l</math> (in the plane of triangle <math>ABC</math>) such that the intersection of the interior of triangle <math>ABC</math> and the interior of its reflection <math>A'B'C'</math> in <math>l</math> has area more than <math>\frac{2}{3}</math> the area of triangle <math>ABC</math>.
 
Let <math>ABC</math> be a triangle. Prove that there is a line <math>l</math> (in the plane of triangle <math>ABC</math>) such that the intersection of the interior of triangle <math>ABC</math> and the interior of its reflection <math>A'B'C'</math> in <math>l</math> has area more than <math>\frac{2}{3}</math> the area of triangle <math>ABC</math>.
  
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==Hint==
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==Solution==
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Without loss of generality, set A(0,0), B(a,1), and C(b,1). The line of reflection should be y = k for some constant k.
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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>.
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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>.
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== See Also ==
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{{USAMO newbox|year=1996|num-b=2|num-a=4}}
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{{MAA Notice}}
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[[Category:Olympiad Geometry Problems]]

Latest revision as of 08:29, 29 August 2023

Problem

Let $ABC$ be a triangle. Prove that there is a line $l$ (in the plane of triangle $ABC$) such that the intersection of the interior of triangle $ABC$ and the interior of its reflection $A'B'C'$ in $l$ has area more than $\frac{2}{3}$ the area of triangle $ABC$.

Solution

Let the triangle be $ABC$. Assume $A$ is the largest angle. Let $AD$ be the altitude. Assume $AB \le AC$, so that $BD \le BC/2$. If $BD > \frac{BC}{3}$, then reflect in $AD$. If $B'$ is the reflection of $B$, then $B'D = BD$ and the intersection of the two triangles is just $ABB'$. But $BB' = 2BD > \frac{2}{3} BC$, so $ABB'$ has more than $\frac{2}{3}$ the area of $ABC$.

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 $\angle{CAD} < \angle{CDA} = 90^{\circ}$. Finally, $A'C \le BC$ because we assumed $\angle B$ does not exceed $\angle A$). The intersection is at least $AA'C$. But $\frac{[AA'C]}{[ABC]} = \frac{CA'}{CB} > CD/CB \ge 2/3$.

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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