2008 AMC 10B Problems/Problem 10

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Problem

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 $AC$ ?

$\mathrm{(A)}\ \sqrt{10}\qquad\mathrm{(B)}\ \frac{7}{2}\qquad\mathrm{(C)}\ \sqrt{14}\qquad\mathrm{(D)}\ \sqrt{15}\qquad\mathrm...$

Solution

Let the center of the circle be $O$ , and let $D$ be the intersection of $\overline{AB}$ and $\overline{OC}$ (then $D$ is the midpoint of $\overline{AB}$ ). $OA=OB=5$ , since they are both radii.

By the Pythagorean Theorem, $OD = \sqrt{OA^2 - DA^2} = 4$ , and by subtraction, $CD=OC - OD = 1$ .

Using the Pythagorean Theorem again, $AC= \sqrt{AD^2 + CD^2} = \sqrt{3^2+1^2}=\sqrt{10} \Longrightarrow \textbf{(A)}$ .

pen d = linewidth(0.7); pathpen = d; pointpen = black; pen f = fontsize(9);path p = CR((0,0),5);pair O = (0,0), A=(5,0), B = ...

2008 AMC 10B (Problems • Resources)
Preceded by Problem 9	Followed by Problem 11
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2008 AMC 10B Problems/Problem 10

From AoPSWiki

Problem

Solution

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