# 2005 AMC 12B Problems/Problem 6

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)