2008 AMC 10B Problem 10

Question

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

The line from the center through

C

perpendicularly bisects chord

AB

Find the center-to-chord distance, subtract from

5

to reach

C,

then use the right triangle with leg

3

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.

Problem 10 in other years