2000 AIME II Problems/Problem 8
Contents
Problem
In trapezoid , leg
is perpendicular to bases
and
, and diagonals
and
are perpendicular. Given that
and
, find
.
Solution
Solution 1
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)
Solution 2
Let . Dropping the altitude from
and using the Pythagorean Theorem tells us that
. Therefore, we know that vector
and vector
. Now we know that these vectors are perpendicular, so their dot product is 0.
As above, we can solve this quadratic to get the positve solution
.
Solution 3
Let and
. From Pythagoras with
, we obtain
. Since
and
are perpendicular diagonals of a quadrilateral, then
, so we have
Substituting
and simplifying yields
and the quadratic formula gives
. Then from
, we plug in
to find
.
Solution 4
Let be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.