Difference between revisions of "2005 AMC 12B Problems/Problem 6"
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− | == Solution == | + | == Solutions == |
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+ | === Solution 1 === | ||
Draw height <math>CH</math> (Perpendicular line from point C to line AD). We have that <math>BH=1</math>. From the [[Pythagorean Theorem]], <math>CH=\sqrt{48}</math>. Since <math>CD=8</math>, <math>HD=\sqrt{8^2-48}=\sqrt{16}=4</math>, and <math>BD=HD-1</math>, so <math>BD=\boxed{\text{(A)}3}</math>. | Draw height <math>CH</math> (Perpendicular line from point C to line AD). We have that <math>BH=1</math>. From the [[Pythagorean Theorem]], <math>CH=\sqrt{48}</math>. Since <math>CD=8</math>, <math>HD=\sqrt{8^2-48}=\sqrt{16}=4</math>, and <math>BD=HD-1</math>, so <math>BD=\boxed{\text{(A)}3}</math>. | ||
− | == Solution | + | === Solution 2 (Trig) === |
After drawing out a diagram, let <math>\angle{ABC}=\theta</math>. By the Law of Cosines, <math>7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}</math>. In <math>\triangle CBD</math>, we have <math>\angle{CBD}=(180-\theta)</math>, and using the identity <math>\cos(180-\theta)=-\cos{\theta}</math> and Law of Cosines one more time: <math>8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0</math>. The only positive value for <math>x</math> is <math>3</math>, which gives the length of <math>\overline{BD}</math>. Thus the answer is <math>\boxed{\text{A}}</math>. | After drawing out a diagram, let <math>\angle{ABC}=\theta</math>. By the Law of Cosines, <math>7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}</math>. In <math>\triangle CBD</math>, we have <math>\angle{CBD}=(180-\theta)</math>, and using the identity <math>\cos(180-\theta)=-\cos{\theta}</math> and Law of Cosines one more time: <math>8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0</math>. The only positive value for <math>x</math> is <math>3</math>, which gives the length of <math>\overline{BD}</math>. Thus the answer is <math>\boxed{\text{A}}</math>. |
Revision as of 04:45, 12 January 2021
- The following problem is from both the 2005 AMC 12B #6 and 2005 AMC 10B #10, so both problems redirect to this page.
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
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.