Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
Line 21: Line 21:
 
== Solution ==
 
== Solution ==
 
Let <math>OG = a - 2h</math>, where <math>h = JH = HG</math>. Since the areas of rectangle <math>EHGL</math> and trapezoid <math>EHGC</math> are both half of rectangle <math>CDFE</math> and trapezoid <math>EFDC</math>, respectively, the ratios between their areas will remain the same, so let us consider rectangle <math>EHGL</math> and trapezoid <math>EHGC</math>. Draw radii <math>OC</math> and <math>OE</math>, both of which obviously have length <math>a</math>. By the Pythagorean theorem, the length of <math>EH</math> is <math>\sqrt{a^2 - (OG + h)^2}</math>, and the length of <math>CG</math> is <math>\sqrt{a^2 - OG^2}</math>. It follows that the area of rectangle <math>EHGL</math> is <math>EH * HG = h\sqrt{a^2 - (OG + h)^2}</math> while the area of trapezoid <math>EHGC</math> is  <math>\frac{HG}{2}(EH + CG)</math> <math>= \frac{h}{2}(\sqrt{a^2 - (OG + h)^2} + \sqrt{a^2 - OG^2})</math>. Now, we want to find the limit, as <math>OG</math> approaches <math>a</math>, of <math>\frac{K}{R}</math>. Note that this is equivalent to finding the same limit as <math>h</math> approaches <math>0</math>. Substituting <math>a - 2h</math> into <math>OG</math> yields that trapezoid <math>EHGC</math> has area <math>\frac{h}{2}(\sqrt{a^2 - (a - 2h + h)^2} + \sqrt{a^2 - (a - 2h)^2}) =</math> <math>\frac{h}{2}(\sqrt{2ah - h^2} + \sqrt{4ah - 4h^2})</math> and that rectangle <math>EHGL</math> has area <math>h\sqrt{a^2 - (a - 2h + h)^2} = h(\sqrt{2ah - h^2})</math>. Our answer thus becomes
 
Let <math>OG = a - 2h</math>, where <math>h = JH = HG</math>. Since the areas of rectangle <math>EHGL</math> and trapezoid <math>EHGC</math> are both half of rectangle <math>CDFE</math> and trapezoid <math>EFDC</math>, respectively, the ratios between their areas will remain the same, so let us consider rectangle <math>EHGL</math> and trapezoid <math>EHGC</math>. Draw radii <math>OC</math> and <math>OE</math>, both of which obviously have length <math>a</math>. By the Pythagorean theorem, the length of <math>EH</math> is <math>\sqrt{a^2 - (OG + h)^2}</math>, and the length of <math>CG</math> is <math>\sqrt{a^2 - OG^2}</math>. It follows that the area of rectangle <math>EHGL</math> is <math>EH * HG = h\sqrt{a^2 - (OG + h)^2}</math> while the area of trapezoid <math>EHGC</math> is  <math>\frac{HG}{2}(EH + CG)</math> <math>= \frac{h}{2}(\sqrt{a^2 - (OG + h)^2} + \sqrt{a^2 - OG^2})</math>. Now, we want to find the limit, as <math>OG</math> approaches <math>a</math>, of <math>\frac{K}{R}</math>. Note that this is equivalent to finding the same limit as <math>h</math> approaches <math>0</math>. Substituting <math>a - 2h</math> into <math>OG</math> yields that trapezoid <math>EHGC</math> has area <math>\frac{h}{2}(\sqrt{a^2 - (a - 2h + h)^2} + \sqrt{a^2 - (a - 2h)^2}) =</math> <math>\frac{h}{2}(\sqrt{2ah - h^2} + \sqrt{4ah - 4h^2})</math> and that rectangle <math>EHGL</math> has area <math>h\sqrt{a^2 - (a - 2h + h)^2} = h(\sqrt{2ah - h^2})</math>. Our answer thus becomes
−
<cmath>\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}[\frac{1}{2} * \frac{\sqrt{h}(\sqrt{2a - h} + 2\sqrt{a - h})}{\sqrt{h}(\sqrt{2a - h})}]</cmath>
+
<cmath>\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]</cmath>
−
<cmath>\implies \frac{1}{2} * \frac{\sqrt{2a} + 2\sqrt{a}}{\sqrt{2a}} = \frac{1}{2}(1 + \frac{2}{\sqrt{2}}) = \frac{1}{2}+\frac{1}{\sqrt{2}} \textbf{  (D)}.</cmath>
+
<cmath>\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)}.</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 07:27, 11 January 2015

Problem

[asy] draw(circle((0,0),10),black+linewidth(.75)); fill((-11,0)--(11,0)--(11,-11)--(-11,-11)--cycle,white); 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 $EH * HG = h\sqrt{a^2 - (OG + h)^2}$ 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 \[\lim_{h\rightarrow 0}\frac{\frac{h}{2}(\sqrt{2ah - h^2} + \sqrt{(4ah - 4h^2})}{h(\sqrt{2ah - h^2})} = \lim_{h\rightarrow 0}\left[\frac{1}{2} * \frac{\sqrt{h}(\sqrt{2a - h} + 2\sqrt{a - h})}{\sqrt{h}(\sqrt{2a - h})}\right]\] \[\implies \frac{1}{2} * \frac{\sqrt{2a} + 2\sqrt{a}}{\sqrt{2a}} = \frac{1}{2}\left(1 + \frac{2}{\sqrt{2}}\right) = \frac{1}{2}+\frac{1}{\sqrt{2}} \textbf{ (D)}.\]

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1968_AHSME_Problems/Problem_35&oldid=66828"