In △ ABC, we have AB=1 and AC=2. Side BC and the median from A to BC have the same length. What is BC?

2002 AMC 12B Problem 23

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23.

\triangle ABC,

we have

AB=1

and

AC=2.

Side

BC

and the median from

A

BC

have the same length. What is

BC?

\dfrac{1+\sqrt{2}}{2}

\dfrac{1+\sqrt{3}}{2}

\sqrt{2}

\dfrac{3}{2}

\sqrt{3}

Let

M

be the midpoint of

BC,

AM=2a,

and

\theta=\angle AMB

Apply the Law of Cosines to

\triangle ABM

and

\triangle AMC

(angles

\theta

and

180^\circ-\theta

) and add

Let

M

be the midpoint of

BC,

set

AM=2a,

and let

\theta=\angle AMB,

\angle AMC=180^\circ-\theta.

With

BM=CM=a,

so that

BC=2a,

the Law of Cosines in

\triangle ABM

and

\triangle AMC

gives

a^2+4a^2-4a^2\cos\theta=1,

a^2+4a^2+4a^2\cos\theta=4.

Adding,

10a^2=5,

a=\dfrac{\sqrt2}{2}

and

BC=2a=\sqrt2.

Thus, the correct answer is C.