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

(Solution)
(Solution)
Line 3: Line 3:
  
 
==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/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 > 2/3 BC, so ABB' has more than 2/3 the area of ABC.
+
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 = 90o. Finally, A'C ≤ BC because we assumed ∠B does not exceed ∠A). The intersection is just AA'C. But area AA'C/area ABC = CA'/CB > CD/CB ≥ 2/3.
+
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 17:33, 29 October 2019

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 $AB ≤ AC$ (Error compiling LaTeX. Unknown error_msg), so that $BD ≤ BC/2$ (Error compiling LaTeX. Unknown error_msg). 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 $BB' = 2BD > 2/3 BC$, 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 $∠CAD < ∠CDA = 90^{\circ}$ (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 $AA'C$. But $\frac{[AA'C]}{[ABC]} = \frac{CA'}{CB} > CD/CB ≥ 2/3$ (Error compiling LaTeX. Unknown error_msg).

See Also

1996 USAMO (ProblemsResources)
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. AMC logo.png