Difference between revisions of "2010 USAMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | 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 | ||
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<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 , as the other two angles in triangle
add to
. Assume that only
, and
are integers. Using the Law of Cosines on triangle BIC,
. Observing that
and that
, we have
Since the right side of the equation is a rational number, the left side (i.e. ) must also be rational. Obviously since
is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for
, and
to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.