Difference between revisions of "2010 USAMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | We know that angle <math>BIC = | + | 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, |
− | <math>BC^2 = BI^2 + CI^2 - 2BI*CI*cos | + | </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 |
− | <math>AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}< | + | </math>AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}<math> |
− | <math>\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}< | + | </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}< | + | 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 22:58, 9 May 2010
Problem
Let be a triangle with
. Points
and
lie on sides
and
, respectively, such that
and
. Segments
and
meet at
. Determine whether or not it is possible for
segments
to all have integer lengths.
Solution
We know that angle , as the other two angles in triangle
add to 45^{\circ}
AB, BC, BI
CI
BC^2 = BI^2 + CI^2 - 2BI*CI*cos 135^{\circ}
BC^2 = AB^2 + AC^2
cos 135^{\circ} = -\frac{\sqrt{2}}{2}
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}
\sqrt{2}
\sqrt{2}
AB, BC, BI
CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.