2008 AMC 12B Problem 9

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 radius through the arc midpoint

C

perpendicularly bisects chord

AB,

meeting it at

D

with

AD = 3

Find the center-to-chord distance

OD,

then

DC = 5 - OD,

and apply the Pythagorean theorem to

\triangle ADC

Let

O

be the center and

D

the point where

\overline{OC}

meets

\overline{AB}.

Since

C

is the midpoint of arc

AB,

\overline{OC}

is the perpendicular bisector of the chord, so

AD = 3.

In right triangle

ADO,

OD = \sqrt{5^2 - 3^2} = 4,

so

DC = OC - OD = 5 - 4 = 1.

Then in right triangle

ADC,

AC = \sqrt{AD^2 + DC^2}

= \sqrt{3^2 + 1^2}

= \sqrt{10}.

Thus, the correct answer is A.

Problem 9 in other years