In $\triangle ABC$, we have $AB = 1$ and $AC = 2$. Side $\overline{BC}$ and the [median](https://artofproblemsolving.com/wiki/index.php/Median_of_a_triangle) 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 1

Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$ extended past $B$. Let $AD = x$ and $BD = y$.
Using the Pythagorean Theorem, we obtain the equations
\begin{align*}  x^2 + y^2 = 1 \hspace{0.5cm}(1)\\ x^2 + y^2 + 2ya + a^2 = 4a^2 \hspace{0.5cm}(2)\\ x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3) \end{align*}
Subtracting $(1)$ equation from $(2)$ and $(3)$, we get
\begin{align*} 2ya + a^2 = 4a^2 - 1 \hspace{0.5cm}(4)\\ 4ya + 4a^2 = 3 \hspace{0.5cm}(5) \end{align*}
Then, subtracting $2 \times (4)$ from $(5)$ and rearranging, we get $10a^2 = 5$, so $BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}$
~greenturtle 11/28/2017

Solution 2
[2002 12B AMC-23.png](https://artofproblemsolving.com/wiki/index.php/File:2002_12B_AMC-23.png)
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](https://artofproblemsolving.com/wiki/index.php/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 $BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}$.

Solution 3
From [Stewart's Theorem](https://artofproblemsolving.com/wiki/index.php/Stewart%27s_Theorem), we have $(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.$ Simplifying, we get $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.$
- awu2014

Solution 4 [Pappus's Median Theorem]
There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have $\triangle{ABC}$, and you draw a median from point $A$ to side $BC$ (label this as $M$), then: $(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}$. Note that $b$ is the length of side $\overline{AC}$, $c$ is the length of side $\overline{AB}$, and $a$ is length of side $\overline{BC}$. Let $MB = MC = x$. Then $AM = 2x$. Now, we can plug into the formula given above: $AM = 2x$, $b = 2$, $c = 1$, and $a = 2x$. After some simple algebra, we find $x = \dfrac{\sqrt{2}}{2}$. Then, $BC = \boxed{\sqrt{2}} \implies \boxed{C}$. 
-Flames
Note: Pappus's Median Theorem is just a special case of Stewart's Theorem, with $m = n$. ~Puck_0

Video Solution by TheBeautyofMath
[https://youtu.be/jEVMgWKQIW8](https://artofproblemsolving.comhttps://youtu.be/jEVMgWKQIW8)
~IceMatrix