2008 AMC 12B Problems/Problem 9

Revision as of 00:21, 15 August 2011 by Minsoens (talk | contribs)

Problem 9

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 $AC = \sqrt{5^2+5^2-2*5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$, which is answer choice $\boxed{\text{A}}$.

Solution 2

[asy] defaultpen(fontsize(8)); pair A=(-3,4), B=(3,4), C=(0,5), D=(0,4), O=(0,0); D(Circle(O,5)); D(O--B--A--O--C);D(A--C--B); label("$A$",A,(-1,1));label("$O$",O,(0,-1));label("$B$",B,(1,1));label("$C$",C,(0,1));label("$D$",D,(-1,-1)); [/asy]

Enlarge.png
Figure 1

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 $AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow \boxed{\text{A}}$.

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
Invalid username
Login to AoPS