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

Revision as of 13:35, 15 December 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 $\triangle ABC$ , 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,

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

~apsid

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 18: / Line 18: @@
 === 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{\textbf{(A) }3}</math>.
 ~Bowser498

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