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

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==Solution==
 
==Solution==
  
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We know that angle <math>BIC = \frac{3\pi}{4}</math>, as the other two angles in triangle <math>BIC</math> add to <math>\frac{\pi}{4}</math>. Assume that only AB, BC, BI, and CI are integers. Using the [[Law of Cosines]] on triangle BIC,
+
We know that angle <math>BIC = \frac{3\pi}{4}</math>, as the other two angles in triangle <math>BIC</math> add to <math>\frac{\pi}{4}</math>. Assume that only <math>AB, BC, BI</math>, and <math>CI</math> are integers. Using the [[Law of Cosines]] on triangle BIC,
  
 
<math>BC^2 = BI^2 + CI^2 - 2BI*CI*cos \frac{3\pi}{4}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> and that <math>cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}</math>, we have
 
<math>BC^2 = BI^2 + CI^2 - 2BI*CI*cos \frac{3\pi}{4}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> and that <math>cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}</math>, we have
Line 9: Line 9:
 
<math>\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}</math>
 
<math>\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}</math>
  
−
Since the right side of the equation is a rational number, the left side (i.e. <math>\sqrt{2}</math>) must also be rational. Obviously since <math>\sqrt{2}</math> is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for <math>AB, BC, BI, and CI</math> to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.
+
Since the right side of the equation is a rational number, the left side (i.e. <math>\sqrt{2}</math>) must also be rational. Obviously since <math>\sqrt{2}</math> is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for <math>AB, BC, BI</math>, and <math>CI</math> to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.

Revision as of 23:52, 5 May 2010

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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