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"

(Created page with "== 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(...")
 
m (→‎See also)
 
(14 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
<asy>
 
<asy>
−
draw(circle((0,0),10),black+linewidth(.75));
+
draw(arc((0,0),10, 0, 180),black+linewidth(.75));
−
fill((-11,0)--(11,0)--(11,-11)--(-11,-11)--cycle,white);
 
 
draw((-10,0)--(10,0),black+linewidth(.75));
 
draw((-10,0)--(10,0),black+linewidth(.75));
 
draw((-sqrt(96),2)--(sqrt(96),2),black+linewidth(.75));
 
draw((-sqrt(96),2)--(sqrt(96),2),black+linewidth(.75));
Line 10: Line 9:
 
draw((8,6)--(8,2),black+linewidth(.75));
 
draw((8,6)--(8,2),black+linewidth(.75));
 
dot((0,0));
 
dot((0,0));
−
MP("O",(0,0),S);MP("a",(5,0),S);
+
MP("O",(0,0),S);
−
MP("J",(0,10),N);MP("D",(sqrt(96),2),E);MP("C",(-sqrt(96),2),W);
+
MP("a",(5,0),S);
−
MP("F",(8,6),E);MP("E",(-8,6),W);MP("G",(0,2),NE);
+
MP("J",(0,10),N);
−
MP("H",(0,6),NE);MP("L",(-8,2),S);MP("M",(8,2),S);
+
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>
 
</asy>
−
 
 
In this diagram the center of the circle is <math>O</math>, the radius is <math>a</math> inches, chord <math>EF</math> is parallel to chord <math>CD</math>. <math>O</math>,<math>G</math>,<math>H</math>,<math>J</math> are collinear, and <math>G</math> is the midpoint of <math>CD</math>. Let <math>K</math> (sq. in.) represent the area of trapezoid <math>CDFE</math> and let <math>R</math> (sq. in.) represent the area of rectangle <math>ELMF.</math> Then, as <math>CD</math> and <math>EF</math> are translated upward so that <math>OG</math> increases toward the value <math>a</math>, while <math>JH</math> always equals <math>HG</math>, the ratio <math>K:R</math> becomes arbitrarily close to:
 
In this diagram the center of the circle is <math>O</math>, the radius is <math>a</math> inches, chord <math>EF</math> is parallel to chord <math>CD</math>. <math>O</math>,<math>G</math>,<math>H</math>,<math>J</math> are collinear, and <math>G</math> is the midpoint of <math>CD</math>. Let <math>K</math> (sq. in.) represent the area of trapezoid <math>CDFE</math> and let <math>R</math> (sq. in.) represent the area of rectangle <math>ELMF.</math> Then, as <math>CD</math> and <math>EF</math> are translated upward so that <math>OG</math> increases toward the value <math>a</math>, while <math>JH</math> always equals <math>HG</math>, the ratio <math>K:R</math> becomes arbitrarily close to:
  
Line 21: Line 26:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{D}</math>
+
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>LMFE</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 <cmath>EH\cdot HG = h\sqrt{a^2 - (OG + h)^2}</cmath> while the area of trapezoid <math>EHGC</math> is  <cmath>\frac{HG}{2}(EH + CG)=\frac{h}{2}\left(\sqrt{a^2 - (OG + h)^2} + \sqrt{a^2 - OG^2}\right).</cmath>
 +
 
 +
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 <cmath>\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)</cmath> and that rectangle <math>EHGL</math> has area <cmath>h\sqrt{a^2 - (a - 2h + h)^2} = h\left(\sqrt{2ah - h^2}\right).</cmath> Our answer thus becomes
 +
 
 +
<cmath>
 +
\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*}
 +
</cmath>
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1968|num-b=34|num-a=35}}   
+
{{AHSME 35p box|year=1968|num-b=34|after=Last Problem}}   
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:52, 16 August 2023

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 $LMFE$ 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 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Last Problem
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 • 31 • 32 • 33 • 34 • 35
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=197032"