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
![[asy] size(400); 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("<math>A</math>",A,S); label("<math>B</math>",B,NW); label("<math>C</math>",C,NE); label("<math>D</math>",D,SE); label("<math>E</math>",E,W); label("<math>F</math>",F,S); clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); [/asy]](http://latex.artofproblemsolving.com/f/7/2/f7201be26289c39a083ac81d13f6b797afb48c03.png)
Applying the triangle inequality to , we see that
On the other hand, if
is strictly greater than
, then the circle with radius
and center
does not touch
, which implies that
, a contradiction. Hence
It follows that
are collinear, and also that
and
are 30-60-90 triangles. Hence
, and
Hence the answer to this problem is 25+7, or 32.
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 | ||
All AIME Problems and Solutions |