Difference between revisions of "2005 AMC 12B Problems/Problem 6"
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+ | == Solution 3 (Stewart's Theorem) == | ||
+ | Let <math>BD=k</math>. Then, by Stewart's Theorem, | ||
+ | |||
+ | <math>2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2 | ||
+ | \implies k^2-2k-15=0 | ||
+ | \implies k=\boxed{3}</math> | ||
+ | |||
+ | ~apsid | ||
+ | |||
== See also == | == See also == |
Revision as of 12:16, 4 September 2021
- The following problem is from both the 2005 AMC 12B #6 and 2005 AMC 10B #10, so both problems redirect to this page.
Contents
Problem
In , we have and . Suppose that is a point on line such that lies between and and . What is ?
Solutions
Solution 1
Draw height (Perpendicular line from point C to line AD). We have that . From the Pythagorean Theorem, . Since , , and , so .
Solution 2 (Trig)
After drawing out a diagram, let . By the Law of Cosines, . In , we have , and using the identity and Law of Cosines one more time: . The only positive value for is , which gives the length of . Thus the answer is .
~Bowser498
Solution 3 (Stewart's Theorem)
Let . Then, by Stewart's Theorem,
~apsid
See also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.