Difference between revisions of "2010 USAMO Problems/Problem 4"

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==Problem==
 
==Problem==
Let <math>ABC</math> be a triangle with <math>\angle A = 90^{\degree}</math>. Points <math>D</math>
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Let <math>ABC</math> be a triangle with <math>\angle A = 90^{\circ}</math>. Points <math>D</math>
 
and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle
 
and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle
 
ABD = \angle DBC</math> and <math>\angle ACE = \angle ECB</math>. Segments <math>BD</math> and
 
ABD = \angle DBC</math> and <math>\angle ACE = \angle ECB</math>. Segments <math>BD</math> and

Revision as of 11:48, 6 May 2010

Problem

Let $ABC$ be a triangle with $\angle A = 90^{\circ}$. Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and $CE$ meet at $I$. Determine whether or not it is possible for segments $AB, AC, BI, ID, CI, IE$ to all have integer lengths.

Solution

We know that angle $BIC = \frac{3\pi}{4}$, as the other two angles in triangle $BIC$ add to $\frac{\pi}{4}$. Assume that only $AB, BC, BI$, and $CI$ are integers. Using the Law of Cosines on triangle BIC,

$BC^2 = BI^2 + CI^2 - 2BI*CI*cos \frac{3\pi}{4}$. Observing that $BC^2 = AB^2 + AC^2$ and that $cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$, we have

$AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}$

$\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}$

Since the right side of the equation is a rational number, the left side (i.e. $\sqrt{2}$) must also be rational. Obviously since $\sqrt{2}$ is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for $AB, BC, BI$, and $CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.

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