Difference between revisions of "2000 AIME II Problems/Problem 8"
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
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Revision as of 19:30, 4 July 2013
Problem
In trapezoid , leg
is perpendicular to bases
and
, and diagonals
and
are perpendicular. Given that
and
, find
.
Solution
Let be the height of the trapezoid, and let
. Since
, it follows that
, so
.
Let be the foot of the altitude from
to
. Then
, and
is a right triangle. By the Pythagorean Theorem,
The positive solution to this quadratic equation is .
![[asy] size(200); pathpen = linewidth(0.7); pair C=(0,0),B=(0,110^.5),A=(11^.5,B.y),D=(10*11^.5,0),E=foot(A,C,D); D(MP("A",A,(2,.5))--MP("B",B,W)--MP("C",C)--MP("D",D)--cycle); D(A--C);D(B--D);D(A--E,linetype("4 4") + linewidth(0.7)); MP("\sqrt{11}",(A+B)/2,N);MP("\sqrt{1001}",(A+D)/2,NE);MP("\sqrt{1001}",(A+D)/2,NE);MP("x",(B+C)/2,W);MP("y",(D+C)/2);D(rightanglemark(B,IP(A--C,B--D),C,20)); [/asy]](http://latex.artofproblemsolving.com/3/3/c/33cba5bae648539c2b01af0bfdabeeb2f45e237b.png)
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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