Difference between revisions of "2005 AMC 12B Problems/Problem 6"

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.

Problem

In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$ ?

$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$

Solutions

Solution 1

Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . By the Pythagorean Theorem, $CH=\sqrt{48}$ . Since $CD=8$ , $HD=\sqrt{8^2-48}=\sqrt{16}=4$ , and $BD=HD-1$ , so $BD=\boxed{\textbf{(A) }3}$ .

Solution 2 (Trig)

After drawing out a diagram, let $\angle{ABC}=\theta$ . By the Law of Cosines, $7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}$ . In $\triangle CBD$ , we have $\angle{CBD}=(180-\theta)$ , and using the identity $\cos(180-\theta)=-\cos{\theta}$ and Law of Cosines one more time: $8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0$ . The only positive value for $x$ is $3$ , which gives the length of $\overline{BD}$ . Thus the answer is $\boxed{\textbf{(A) }3}$ .

~Bowser498

Solution 3 (Stewart's Theorem)

Let $BD=k$ . Then, by Stewart's Theorem,

$2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2 \implies k^2+2k-15=0 \implies k=\boxed{\textbf{(A) }3}$

~apsid

~edited by always90degrees (tiny fix, sign error)

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.

@@ Line 23: / Line 23: @@
 == Solution 3 (Stewart's Theorem) ==
-Let <math>BD=k</math>. Then, by 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^2+2k-15=0
-\implies k=\boxed{3}</math>
+\implies k=\boxed{\textbf{(A) }3}</math>
 ~apsid
+~edited by always90degrees (tiny fix, sign error)
 == See also ==

Page

Toolbox

Search