Difference between revisions of "2001 AIME II Problems/Problem 13"
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Draw a line from <math>B</math>, parallel to <math>\overline{AD}</math>, and let it meet <math>\overline{CD}</math> at <math>M</math>. Note that <math>\triangle{DAB}</math> is similar to <math>\triangle{BMC}</math> by AA similarity, since <math>\angle{ABD}=\angle{MCB}</math> and since <math>BM</math> is parallel to <math>CD</math> then <math>\angle{BMC}=\angle{ADM}=\angle{DAB}</math>. Now since <math>ADMB</math> is an isosceles trapezoid, <math>MD=8</math>. By the similarity, we have <math>MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}</math>, hence <math>CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}</math>. | Draw a line from <math>B</math>, parallel to <math>\overline{AD}</math>, and let it meet <math>\overline{CD}</math> at <math>M</math>. Note that <math>\triangle{DAB}</math> is similar to <math>\triangle{BMC}</math> by AA similarity, since <math>\angle{ABD}=\angle{MCB}</math> and since <math>BM</math> is parallel to <math>CD</math> then <math>\angle{BMC}=\angle{ADM}=\angle{DAB}</math>. Now since <math>ADMB</math> is an isosceles trapezoid, <math>MD=8</math>. By the similarity, we have <math>MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}</math>, hence <math>CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/NsQbhYfGh1Q?t=75 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == |
Revision as of 03:27, 2 February 2021
Contents
[hide]Problem
In quadrilateral ,
and
,
,
, and
. The length
may be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
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
Solution 2
Draw a line from , parallel to
, and let it meet
at
. Note that
is similar to
by AA similarity, since
and since
is parallel to
then
. Now since
is an isosceles trapezoid,
. By the similarity, we have
, hence
.
Video Solution
https://youtu.be/NsQbhYfGh1Q?t=75
~ pi_is_3.14
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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