Difference between revisions of "2008 AIME I Problems/Problem 10"
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clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); | clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); | ||
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− | Applying the [[triangle inequality]] to <math>ADE</math>, we see that <math>AD | + | Applying the [[triangle inequality]] to <math>ADE</math>, we see that <math>AD > 20\sqrt {7}</math>. However, if <math>AD</math> is strictly greater than <math>20\sqrt {7}</math>, then the circle with radius <math>10\sqrt {21}</math> and center <math>A</math> does not touch <math>DC</math>, which implies that <math>AC > 10\sqrt {21}</math>, a contradiction. Therefore, <math>AD = 20\sqrt {7}</math>. |
It follows that <math>A</math>, <math>D</math>, and <math>E</math> are collinear, and also that <math>ADC</math> and <math>ACF</math> are <math>30-60-90</math> triangles. Hence <math>AF = 15\sqrt {7}</math>, and | It follows that <math>A</math>, <math>D</math>, and <math>E</math> are collinear, and also that <math>ADC</math> and <math>ACF</math> are <math>30-60-90</math> triangles. Hence <math>AF = 15\sqrt {7}</math>, and |
Revision as of 17:37, 1 March 2009
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)
Applying the triangle inequality to , we see that
. However, if
is strictly greater than
, then the circle with radius
and center
does not touch
, which implies that
, a contradiction. Therefore,
.
It follows that ,
, and
are collinear, and also that
and
are
triangles. Hence
, and

Finally, the answer is .
Solution 2
No restrictions are set on the lengths of the bases, so for calculational simplicity let . Since
is a
triangle,
.

The answer is . Note that while this is not rigorous, the above solution shows that
is indeed the only possibility.
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 |