2002 AMC 12B Problem 23

Problem 23 of 25HarderGeometry

In

\triangle ABC,

we have

AB=1

and

AC=2.

Side

BC

and the median from

A

to

BC

have the same length. What is

BC?

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Solution

Let

M

be the midpoint of

BC,

set

AM=2a,

and let

\theta=\angle AMB,

so

\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,

so

a=\dfrac{\sqrt2}{2}

and

BC=2a=\sqrt2.

Thus, the correct answer is C.

More practice

Concepts: median (geometry) · law of cosines

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.