1956 AMC 12 Problem 14

Attempt Problem 14 of the 1956 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 1956 AMC 12 solutions, or check the answer key.

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

14.

The points

A,

B,

and

C

are on a circle

O.

The tangent line at

A

and the secant

BC

intersect at

P,

B

lying between

C

and

P.

BC=20

and

PA=10\sqrt3,

then

PB

equals:

5

10

10\sqrt3

20

30

Use

PA^2=PB\cdot PC

PB=t,

then

PC=t+20

Let

PB=t.

Since

B

lies between

P

and

C,

we have

PC=t+20.

The tangent-secant theorem gives

(10\sqrt3)^2=t(t+20).

Thus

t^2+20t-300=0,

(t-10)(t+30)=0.

A length is positive, so

t=10.

Therefore, the correct answer is B.

Problem 14 in other years