1954 AMC 12 Problem 9

Attempt Problem 9 of the 1954 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 1954 AMC 12 solutions, or check the answer key.

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

A point

P

is outside a circle and is

13

inches from the center. A secant from

P

cuts the circle at

Q

and

R

so that the external segment of the secant

\overline{PQ}

9

inches and

\overline{QR}

7

inches. The radius of the circle is:

3

in.

4

in.

5

in.

6

in.

7

in.

The secant power is the external length times the whole secant length

Equate

9(9+7)

with

13^2-r^2

The power of

P

PQ\cdot PR=9(16)=144.

It is also

13^2-r^2,

169-r^2=144.

Hence

r^2=25

and

r=5

inches.

Thus, the correct answer is C.

Problem 9 in other years