2011 AIME I Problems/Problem 2
Problem
In rectangle ,
and
. Points
and
lie inside rectangle
so that
,
,
,
, and line
intersects segment
. The length
can be expressed in the form
, where
,
, and
are positive integers and
is not divisible by the square of any prime. Find
.
Solution
Let us call the point where intersects
point
, and the point where
intersects
point
. Since angles
and
are both right angles, and angles
and
are congruent due to parallelism, right triangles
and
are similar. This implies that
. Since
,
. (
is the same as
because they are opposite sides of a rectangle.) Now, we have a system:
Solving this system (easiest by substitution), we get that:
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:
and
Notice that adding these two sides would give us twelve plus the overlap . This means that:
Since isn't divisible by any perfect square, our answer is: