1968 AMC 12 Problem 35

Question

In this diagram the center of the circle is

O,

the radius is

a

inches, chord

EF

is parallel to chord

CD,

O,

G,

H,

J

are collinear, and

G

is the midpoint of

CD.

Let

K

(square inches) represent the area of trapezoid

CDFE

and let

R

(square inches) represent the area of rectangle

ELMF.

Then, as

CD

and

EF

are translated upward so that

OG

increases toward the value

a,

while

JH

always equals

HG,

the ratio

K:R

becomes arbitrarily close to:

0

1

\sqrt2

\dfrac1{\sqrt2}+\dfrac12

\dfrac1{\sqrt2}+1

Let

JH=HG=x,

so

OG=a-2x

and

OH=a-x

Express the two half-chords with the Pythagorean theorem, then let

x

approach

0

Let

JH=HG=x.

Then

OG=a-2x

and

OH=a-x.

The half-chords are

\begin{aligned} GD&=2\sqrt{x(a-x)},\\ HF&=\sqrt{x(2a-x)}. \end{aligned}

Since both figures have height

x,

\begin{aligned} \frac KR&=\frac{x(GD+HF)}{2x(HF)}\\ &=\frac12+\frac{GD}{2HF}. \end{aligned}

As

x

approaches

0,

the latter fraction approaches

\frac{2\sqrt{ax}}{2\sqrt{2ax}}=\frac1{\sqrt2}.

Thus

K:R

approaches

\frac{1}{\sqrt2}+\frac{1}{2}.

Therefore, the correct answer is D.

Problem 35 in other years