2001 AIME II Problems/Problem 13
Problem
In quadrilateral ,
and
,
,
, and
. The length
may be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution
Extend and
to meet at
. Then, since
and
, we know that
. Hence
, and
is isosceles. Then
.
![[asy] /* We arbitrarily set AD = x */ real x = 60^.5, anglesize = 28; pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7); pair A=(0,0), D=(x,0), B=IP(CR(A,8),CR(D,10)), E=(-3x/5,0), C=IP(CR(E,16),CR(D,64/5)); D(MP("A",A)--MP("B",B,NW)--MP("C",C,NW)--MP("D",D)--cycle); D(B--D); D(A--MP("E",E)--B,d); D(anglemark(D,A,B,anglesize));D(anglemark(C,D,A,anglesize));D(anglemark(A,B,D,anglesize));D(anglemark(E,C,D,anglesize));D(anglemark(A,B,D,5/4*anglesize));D(anglemark(E,C,D,5/4*anglesize)); MP("10",(B+D)/2,SW);MP("8",(A+B)/2,W);MP("6",(B+C)/2,NW); [/asy]](http://latex.artofproblemsolving.com/4/1/7/417316517e7273e0d7f31dbbae8e0b80e30a8d54.png)
Using the similarity, we have:
The answer is .
Extension: To Find , use Law of Cosines on
to get
Then since
use Law of Cosines on
to find
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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