Difference between revisions of "2001 AIME II Problems/Problem 13"
Mathgeek2006 (talk | contribs) (Undo revision 61762 by XXQw3rtyXx (talk)) |
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− | '''Extension''': Find <math>AD</math> | + | '''Extension''': To Find <math>AD</math>, use Law of Cosines on <math>\triangle BCD</math> to get <math>\cos(\angle BCD)=\frac{13}{20}</math> |
+ | Then since <math>\angle BCD=\angle ABD</math> use Law of Cosines on <math>\triangle ABD</math> to find <math>AD=2\sqrt{15}</math> | ||
== See also == | == See also == |
Revision as of 16:05, 19 January 2016
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.