1962 AMC 12 Problem 25

Attempt Problem 25 of the 1962 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 1962 AMC 12 solutions, or check the answer key.

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

25.

Given square

ABCD

with side

8

feet. A circle is drawn through vertices

A

and

D

and tangent to side

BC.

The radius of the circle, in feet, is:

4

4\sqrt2

5

5\sqrt2

6

Place

A=(0,0),

D=(0,8),

and

BC

on the line

x=8

The center lies on the perpendicular bisector of

AD

, and its distance to

BC

equals the radius

Place the center at

(h,4).

Since the circle passes through

A,

r^2=h^2+16.

Tangency to

x=8

gives

r=8-h.

Therefore

h^2+16=(8-h)^2,

h=3

and

r=5.

Thus, the correct answer is C.

Problem 25 in other years