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1968 AHSME Problems/Problem 35

Revision as of 12:19, 21 February 2017 by Hiabc (talk | contribs) (made sol more readable)

Problem

[asy] draw(arc((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); [/asy] 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 \[EH\cdot HG = h\sqrt{a^2 - (OG + h)^2}\] while the area of trapezoid $EHGC$ is \[\frac{HG}{2}(EH + CG)=\frac{h}{2}\left(\sqrt{a^2 - (OG + h)^2} + \sqrt{a^2 - OG^2}\right).\]

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 \[\frac{h}{2}\left(\sqrt{a^2 - (a - 2h + h)^2} + \sqrt{a^2 - (a - 2h)^2}\right) =\frac{h}{2}\left(\sqrt{2ah - h^2} + \sqrt{4ah - 4h^2}\right)\] and that rectangle $EHGL$ has area \[h\sqrt{a^2 - (a - 2h + h)^2} = h\left(\sqrt{2ah - h^2}\right).\] Our answer thus becomes

\begin{align*} \lim_{h\rightarrow 0}\dfrac{\frac{h}{2}\bigl(\sqrt{2ah - h^2} + \sqrt{4ah - 4h^2}\bigr)}{h\bigl(\sqrt{2ah - h^2}\bigr)} &= \lim_{h\rightarrow 0}\left[\dfrac{1}{2}\cdot\dfrac{\sqrt{h}\bigl(\sqrt{2a - h} + 2\sqrt{a - h}\bigr)}{\sqrt{h}\bigl(\sqrt{2a - h}\bigr)}\right] \\ \implies \frac{1}{2}\cdot\frac{\sqrt{2a} + 2\sqrt{a}}{\sqrt{2a}} &= \frac{1}{2}\left(1 + \frac{2}{\sqrt{2}}\right) = \boxed{\textbf{(D) }\frac{1}{2}+\frac{1}{\sqrt{2}}.} \end{align*}

See also

1968 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 35
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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