Difference between revisions of "1968 AHSME Problems/Problem 35"

Revision as of 12:10, 21 February 2017

Problem

draw(circle((0,0),10, 0, 180),black+linewidth(.75));
draw((-10,0)--(10,0),black+linewidth(.75));
draw((-sqrt(96),2)--(sqrt(96),2),black+linewidth(.75));
draw((-8,6)--(8,6),black+linewidth(.75));
draw((0,0)--(0,10),black+linewidth(.75));
draw((-8,6)--(-8,2),black+linewidth(.75));
draw((8,6)--(8,2),black+linewidth(.75));
dot((0,0));
MP("O",(0,0),S);
MP("a",(5,0),S);
MP("J",(0,10),N);
MP("D",(sqrt(96),2),E);
MP("C",(-sqrt(96),2),W);
MP("F",(8,6),E);
MP("E",(-8,6),W);
MP("G",(0,2),NE);
MP("H",(0,6),NE);
MP("L",(-8,2),S);
MP("M",(8,2),S);
 (Error making remote request. Unknown error_msg)

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 $K$ (sq. in.) represent the area of trapezoid $CDFE$ and let $R$ (sq. in.) 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:

$\text{(A)} 0\quad\text{(B)} 1\quad\text{(C)} \sqrt{2}\quad\text{(D)} \frac{1}{\sqrt{2}}+\frac{1}{2}\quad\text{(E)} \frac{1}{\sqrt{2}}+1$

Solution

Let $OG = a - 2h$ , where $h = JH = HG$ . Since the areas of rectangle $EHGL$ and trapezoid $EHGC$ are both half of rectangle $CDFE$ and trapezoid $EFDC$ , respectively, the ratios between their areas will remain the same, so let us consider rectangle $EHGL$ and trapezoid $EHGC$ . Draw radii $OC$ and $OE$ , both of which obviously have length $a$ . By the Pythagorean theorem, the length of $EH$ is $\sqrt{a^2 - (OG + h)^2}$ , and the length of $CG$ is $\sqrt{a^2 - OG^2}$ . It follows that the area of rectangle $EHGL$ is $EH * HG = h\sqrt{a^2 - (OG + h)^2}$ while the area of trapezoid $EHGC$ is $\frac{HG}{2}(EH + CG)$ $= \frac{h}{2}(\sqrt{a^2 - (OG + h)^2} + \sqrt{a^2 - OG^2})$ . Now, we want to find the limit, as $OG$ approaches $a$ , of $\frac{K}{R}$ . Note that this is equivalent to finding the same limit as $h$ approaches $0$ . Substituting $a - 2h$ into $OG$ yields that trapezoid $EHGC$ has area $\frac{h}{2}(\sqrt{a^2 - (a - 2h + h)^2} + \sqrt{a^2 - (a - 2h)^2}) =$ $\frac{h}{2}(\sqrt{2ah - h^2} + \sqrt{4ah - 4h^2})$ and that rectangle $EHGL$ has area $h\sqrt{a^2 - (a - 2h + h)^2} = h(\sqrt{2ah - h^2})$ . Our answer thus becomes $\lim_{h\rightarrow 0}\frac{\frac{h}{2}(\sqrt{2ah - h^2} + \sqrt{(4ah - 4h^2})}{h(\sqrt{2ah - h^2})} = \lim_{h\rightarrow 0}\left[\frac{1}{2} * \frac{\sqrt{h}(\sqrt{2a - h} + 2\sqrt{a - h})}{\sqrt{h}(\sqrt{2a - h})}\right]$ $\implies \frac{1}{2} * \frac{\sqrt{2a} + 2\sqrt{a}}{\sqrt{2a}} = \frac{1}{2}\left(1 + \frac{2}{\sqrt{2}}\right) = \frac{1}{2}+\frac{1}{\sqrt{2}} \textbf{ (D)}.$

1968 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 34	Followed by Problem 35
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Difference between revisions of "1968 AHSME Problems/Problem 35"

Revision as of 12:10, 21 February 2017

Problem

Solution

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