1965 AMC 12 Problem 39

Question

A foreman noticed an inspector checking a

3

-inch hole with a

2

-inch plug and a

1

-inch plug and suggested that two more gauges be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter

d

of each, to the nearest hundredth of an inch, is:

0.87

0.86

0.83

0.75

0.71

Use radii

\frac{3}{2},

1,

and

\frac{1}{2},

and let a new gauge have radius

r

If its center is

(u,v)

, subtract its three tangency-distance equations

Place the hole’s center at the origin. The

2

-inch and

1

-inch plug centers are

(0,-\frac{1}{2})

and

(0,1).

Let a new gauge have radius

r

and center

(u,v).

Tangency to the two plugs gives

\begin{gathered} u^2+(v+\frac{1}{2})^2=(1+r)^2,\\ u^2+(v-1)^2=(\frac{1}{2}+r)^2, \end{gathered}

while internal tangency to the hole gives

u^2+v^2=(\frac{3}{2}-r)^2.

Subtracting pairs of equations yields

v=\frac12+\frac r3

and

v=\frac32-2r.

Thus

r=\frac{3}{7},

so

d=2r=\frac{6}{7}\approx0.86.

Therefore, the correct answer is B.

Problem 39 in other years