2008 AMC 10B Problem 10

Attempt Problem 10 of the 2008 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10B solutions, or check the answer key.

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10.

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 $AC?$

$\sqrt{10}$

$\dfrac{7}{2}$

$\sqrt{14}$

$\sqrt{15}$

$4$

Answer: A

Concepts:chord Pythagorean Theorem circle

Difficulty rating: 1170

Solution:

Let $O$ be the center and $D$ the midpoint of $AB.$ Then $OD\perp AB$ with $AD=3,$ so $OD=\sqrt{5^2-3^2}=4.$

Since $C$ is the midpoint of the minor arc, $O,$ $D,$ $C$ are collinear and $DC=OC-OD=5-4=1.$

Then $AC=\sqrt{AD^2+DC^2}$ $=\sqrt{3^2+1^2}$ $=\sqrt{10}.$

Thus, the correct answer is A.

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