2008 AMC 12B Problem 9

Below is the professionally curated solution for Problem 9 of the 2008 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12B solutions, or check the answer key.

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Concepts:chord perpendicular bisector Pythagorean Theorem

Difficulty rating: 1500

Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6.$ Point $C$ is the midpoint of the minor arc $AB.$ What is the length of the line segment $AC?$

$\sqrt{10}$

$\dfrac{7}{2}$

$\sqrt{14}$

$\sqrt{15}$

$4$

Solution:

Let $O$ be the center and $D$ the point where $\overline{OC}$ meets $\overline{AB}.$ Since $C$ is the midpoint of arc $AB,$ $\overline{OC}$ is the perpendicular bisector of the chord, so $AD = 3.$

In right triangle $ADO,$ $OD = \sqrt{5^2 - 3^2} = 4,$ so $DC = OC - OD = 5 - 4 = 1.$

Then in right triangle $ADC,$ $AC = \sqrt{AD^2 + DC^2} = \sqrt{3^2 + 1^2} = \sqrt{10}.$

Thus, the correct answer is A.

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