2008 AMC 12B Problems/Problem 9

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 $AC$ ?

$\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4$

Solutions

Solution 1

Let $\alpha$ be the angle that subtends the arc $AB$ . By the law of cosines, $6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$ implies $\cos(\alpha) = 7/25$ .

The half-angle formula says that $\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ . The law of cosines tells us $AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$ , which is answer choice $\boxed{\text{A}}$ .

Solution 2

Define $D$ as the midpoint of line segment $\overline{AB}$ , and $O$ the center of the circle. Then $O$ , $C$ , and $D$ are collinear, and since $D$ is the midpoint of $AB$ , $m\angle ODA=90\deg$ and so $OD=\sqrt{5^2-3^2}=4$ . Since $OD=4$ , $CD=5-4=1$ , and so $AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow \boxed{\text{A}}$ .

$[asy] pen d = linewidth(0.7); pathpen = d; pointpen = black; pen f = fontsize(9); path p = CR((0,0),5); pair O = (0,0), A=(5,0), B = IP(p,CR(A,6)), C = IP(p,CR(A,3)), D=IP(A--B,O--C); D(p); D(MP("A",A,E)--D(MP("O",O))--MP("B",B,NE)--cycle); D(A--MP("C",C,ENE),dashed+d); D(O--C,dashed+d); D(rightanglemark(O,D(MP("D",D,W)),A)); MP("5",(A+O)/2); MP("3",(A+D)/2,SW); [/asy]$

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 12 Problems and Solutions

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2008 AMC 12B Problems/Problem 9

Contents

Problem

Solutions

Solution 1

Solution 2

See Also