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:

$0$

$1$

$\sqrt2$

$\dfrac1{\sqrt2}+\dfrac12$

$\dfrac1{\sqrt2}+1$

Answer: D

Concepts:circle chord trapezoid calculus

Difficulty rating: 2650

Small Hint:

Let $JH=HG=x,$ so $OG=a-2x$ and $OH=a-x$

Big Hint:

Express the two half-chords with the Pythagorean theorem, then let $x$ approach $0$

Solution:

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.

1968 AMC 12 Problem 35

Answer: D

Difficulty rating: 2650

Small Hint:

Big Hint:

Solution:

Problem 35 in Other Years