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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 <math>AB, BC, BI</math>, and <math>CI</math> are integers. Using the [[Law of Cosines]] on triangle BIC,
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We know that angle <math>BIC = 135^{\circ}</math>, as the other two angles in triangle <math>BIC</math> add to 45^{\circ}<math>. Assume that only </math>AB, BC, BI<math>, and </math>CI<math> are integers. Using the [[Law of Cosines]] on triangle BIC,
  
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<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
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</math>BC^2 = BI^2 + CI^2 - 2BI*CI*cos 135^{\circ}<math>. Observing that </math>BC^2 = AB^2 + AC^2<math> and that </math>cos 135^{\circ} = -\frac{\sqrt{2}}{2}<math>, we have
  
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<math>AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}</math>
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</math>AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}<math>
  
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<math>\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}</math>
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</math>\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}<math>
  
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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.
+
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$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.

Revision as of 21:58, 9 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 $BIC = 135^{\circ}$, as the other two angles in triangle $BIC$ add to 45^{\circ}$. Assume that only$AB, BC, BI$, and$CI$are integers. Using the [[Law of Cosines]] on triangle BIC,$BC^2 = BI^2 + CI^2 - 2BI*CI*cos 135^{\circ}$. Observing that$BC^2 = AB^2 + AC^2$and that$cos 135^{\circ} = -\frac{\sqrt{2}}{2}$, we have$AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}$$ (Error compiling LaTeX. Unknown error_msg)\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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