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

## 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 $BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}$.

## Alternate Solution

From Stewart's Theorem, we have $(2)(1/2a)(2) + (1)(1/2a)(1) = (a)(a)(a) + (1/2a)a(1/2a).$ Simplifying, we get $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.$