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^{\ | + | 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 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
. 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.