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

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<math>
 
<math>
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\mathrm{(A)}\ 3      \qquad
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\textbf{(A) }\ 3      \qquad
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\mathrm{(B)}\ 2\sqrt{3}      \qquad
+
\textbf{(B) }\ 2\sqrt{3}      \qquad
−
\mathrm{(C)}\ 4      \qquad
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\textbf{(C) }\ 4      \qquad
−
\mathrm{(D)}\ 5      \qquad
+
\textbf{(D) }\ 5      \qquad
−
\mathrm{(E)}\ 4\sqrt{2}
+
\textbf{(E) }\ 4\sqrt{2}
 
</math>
 
</math>
  
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== Solution ==
+
== Solutions ==
−
Draw height <math>CH</math>. 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>.
 
  
−
<math>[asy]
+
=== Solution 1 ===
−
draw((0,0)--(1.5,4)--(3,0)--cycle);
+
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>.
−
draw((3,0)--(5,0)--(1.5,4)--cycle);
+
 
−
[/asy]</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.

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)

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

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