Difference between revisions of "2005 AMC 12B Problems/Problem 6"
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<math> | <math> | ||
| − | \ | + | \textbf{(A) }\ 3 \qquad |
| − | \ | + | \textbf{(B) }\ 2\sqrt{3} \qquad |
| − | \ | + | \textbf{(C) }\ 4 \qquad |
| − | \ | + | \textbf{(D) }\ 5 \qquad |
| − | \ | + | \textbf{(E) }\ 4\sqrt{2} |
</math> | </math> | ||
| − | == Solution == | + | == Solutions == |
| − | Draw height <math>CH</math>. We have that <math>BH=1</math>. | + | |
| + | === Solution 1 === | ||
| + | Draw height <math>CH</math> (Perpendicular line from point C to line AD). We have that <math>BH=1</math>. By 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{\textbf{(A) }3}</math>. | ||
| + | |||
| + | === 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{\textbf{(A) }3}</math>. | ||
| + | |||
| + | ~Bowser498 | ||
| + | |||
| + | == 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{\textbf{(A) }3}</math> | ||
| + | |||
| + | ~apsid | ||
| + | |||
| + | ~edited by always90degrees (tiny fix, sign error) | ||
== See also == | == See also == | ||
Latest revision as of 14:29, 5 July 2022
- 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 
. By 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
~edited by always90degrees (tiny fix, sign error)
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. 
