Difference between revisions of "2002 AMC 12B Problems/Problem 23"

Revision as of 17:01, 25 February 2008

Problem

In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$ ?

$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$

Solution

Let $D$ be the foot of the median from $A$ to $\overline{BC}$ , and we let $AD = BC = 2a$ . Then by the Law of Cosines on $\triangle ABD, \triangle ACD$ , we have $\begin{align*} 1^2 &= a^2 + (2a)^2 - 2(a)(2a)\cos ADB \\ 2^2 &= a^2 + (2a)^2 - 2(a)(2a)\cos ADC \end{align*}$

Since $\cos ADC = \cos (180 - ADB) = -\cos ADB$ , we can add these two equations and get

$5 = 10a^2$

Hence $a = \frac{1}{\sqrt{2}}$ and $BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}$ .

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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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 == Problem ==
-In <math>\triangle ABC</math>, we have <math>AB = 1</math> and <math>AC = 2</math>. Side <math>\overline{BC}</math> and the [[median]] from <math>A</math> to <math>\overline{BC}</math> have the same length. What is <math>BC</math>?
+In <math>\triangle ABC</math>, we have <math>AB = 1</math> and <math>AC = 2</math>. Side <math>\overline{BC}</math> and the [[Median of a triangle|median]] from <math>A</math> to <math>\overline{BC}</math> have the same length. What is <math>BC</math>?
 <math>\mathrm{(A)}\ \frac{1+\sqrt{2}}{2}

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Difference between revisions of "2002 AMC 12B Problems/Problem 23"

Revision as of 17:01, 25 February 2008

Problem

Solution

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