1983 AMC 12 Problem 10

Attempt Problem 10 of the 1983 AMC 12 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 1983 AMC 12 solutions, or check the answer key.

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10.

Segment

AB

is both a diameter of a circle of radius

1

and a side of an equilateral triangle

ABC.

The circle also intersects

AC

and

BC

at points

D

and

E,

respectively. The length of

AE

\frac32

\frac53

\frac{\sqrt3}{2}

\sqrt3

\frac{2+\sqrt3}{2}

Because

AB

is a diameter,

\angle AEB

is a right angle

Triangle

ABE

is a

30^\circ

60^\circ

90^\circ

triangle with hypotenuse

AB=2

Since

E

is on the circle with diameter

AB,

\angle AEB=90^\circ.

Also

\angle ABE=60^\circ

because

ABC

is equilateral. Thus

ABE

is a

30^\circ

60^\circ

90^\circ

triangle with hypotenuse

AB=2,

AE=\sqrt3.

Therefore, the correct answer is D.

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