2008 AMC 12B Problem 9

Attempt Problem 9 of the 2008 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12B solutions, or check the answer key.

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9.

Points AA and BB are on a circle of radius 55 and AB=6.AB = 6. Point CC is the midpoint of the minor arc AB.AB. What is the length of the line segment AC?AC?

10\sqrt{10}

72\dfrac{7}{2}

14\sqrt{14}

15\sqrt{15}

44

Answer: A
Concepts:chordperpendicular bisectorPythagorean Theorem
Difficulty rating: 1500
Solution:

Let OO be the center and DD the point where OC\overline{OC} meets AB.\overline{AB}. Since CC is the midpoint of arc AB,AB, OC\overline{OC} is the perpendicular bisector of the chord, so AD=3.AD = 3.

In right triangle ADO,ADO, OD=5232=4,OD = \sqrt{5^2 - 3^2} = 4, so DC=OCOD=54=1.DC = OC - OD = 5 - 4 = 1.

Then in right triangle ADC,ADC, AC=AD2+DC2AC = \sqrt{AD^2 + DC^2} =32+12= \sqrt{3^2 + 1^2} =10.= \sqrt{10}.

Thus, the correct answer is A.

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