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2002 AMC 12B Problems/Problem 23

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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 $BC$ ?

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

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

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

See also

2002 AMC 12B (Problems • 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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2002_AMC_12B_Problems/Problem_23"

Category: Introductory Geometry Problems

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

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