1982 AMC 12 Problem 14

Attempt Problem 14 of the 1982 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 1982 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

In the adjoining figure, points

B

and

C

lie on line segment

AD,

and

AB,

BC,

and

CD

are diameters of circles

O,

N,

and

P,

respectively. Circles

O,

N,

and

P

all have radius

15,

and the line

AG

is tangent to circle

P

G.

AG

intersects circle

N

at points

E

and

F,

then chord

EF

has length

20

15\sqrt2

24

25

none of these

Use right triangle

APG

to find the distance from

N

to line

AG

A chord at distance

d

from a circle’s center has length

2\sqrt{r^2-d^2}

Here

AP=75,

PG=15,

and

AG\perp PG.

The distance from

N

AG

scales with

\frac{AN}{AP}=\frac{45}{75},

so it is

15(\frac{45}{75})=9.

Therefore the chord in the radius-

15

circle has length

2\sqrt{15^2-9^2}=2\sqrt{144}=24.

Therefore, the correct answer is C.

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