2008 AIME I Problems/Problem 10
Problem
Let be an isosceles trapezoid with
whose angle at the longer base
is
. The diagonals have length
, and point
is at distances
and
from vertices
and
, respectively. Let
be the foot of the altitude from
to
. The distance
can be expressed in the form
, where
and
are positive integers and
is not divisible by the square of any prime. Find
.
Solution
Solution 1
![[asy] size(300); defaultpen(1); pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0); draw(F--C--B--A); draw(E--A--D--C); draw(A--C,dashed); draw(circle(A,abs(C-A)),dotted); label("\(A\)",A,S); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,SE); label("\(E\)",E,N); label("\(F\)",F,S); clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); [/asy]](http://latex.artofproblemsolving.com/9/0/9/9091a509366e2723a7fb556a8255537bb23c13b1.png)
Assuming that is a triangle and applying the triangle inequality, we see that
. However, if
is strictly greater than
, then the circle with radius
and center
does not touch
, which implies that
, a contradiction. As a result, A, D, and E are collinear. Therefore,
.
Thus, and
are
triangles. Hence
, and
![$EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}$](http://latex.artofproblemsolving.com/d/2/3/d23856decb7cf64df12003938a5fb775faab773e.png)
Finally, the answer is .
Solution 2
Extend through
, to meet
(extended through
) at
.
is an equilateral triangle because of the angle conditions on the base.
If then
, because
and therefore
.
By simple angle chasing, is a 30-60-90 triangle and thus
,
and
Similarly is a 30-60-90 triangle and thus
.
Equating and solving for ,
and thus
.
and
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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