2002 AMC 12B Problem 23

Below is the professionally curated solution for Problem 23 of the 2002 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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Concepts:median (geometry)law of cosines

Difficulty rating: 1820

23.

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 $BC?$

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

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

$\sqrt{2}$

$\dfrac{3}{2}$

$\sqrt{3}$

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

Adding, $10a^2=5,$ so $a=\dfrac{\sqrt2}{2}$ and $BC=2a=\sqrt2.$

Thus, the correct answer is C.

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