Difference between revisions of "2001 AIME II Problems/Problem 13"
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== Solution 2 == | == Solution 2 == | ||
− | 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 | + | 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>. |
== See also == | == See also == |
Revision as of 16:16, 7 January 2017
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 .
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.