2012 AMC 12A Problem 16

Below is the professionally curated solution for Problem 16 of the 2012 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12A solutions, or check the answer key.

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Concepts:law of cosines chord circle

Difficulty rating: 1870

16.

Circle $C_1$ has its center $O$ lying on circle $C_2.$ The two circles meet at $X$ and $Y.$ Point $Z$ in the exterior of $C_1$ lies on circle $C_2$ and $XZ = 13,$ $OZ = 11,$ and $YZ = 7.$ What is the radius of circle $C_1?$

$5$

$\sqrt{26}$

$3\sqrt{3}$

$2\sqrt{7}$

$\sqrt{30}$

Solution:

Let $r$ be the radius of $C_1,$ so $OX = OY = r.$ These are equal chords of $C_2,$ so they subtend equal angles at $Z:$ $\angle XZO = \angle OZY.$

Applying the Law of Cosines to triangles $XZO$ and $YZO,$ $\frac{13^2 + 11^2 - r^2}{2 \cdot 13 \cdot 11} = \frac{7^2 + 11^2 - r^2}{2 \cdot 7 \cdot 11}.$

Clearing denominators and solving gives $r^2 = 30,$ so $r = \sqrt{30}.$

Thus, the correct answer is E.

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