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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(...")
 
(→‎Problem)
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MP("H",(0,6),NE);MP("L",(-8,2),S);MP("M",(8,2),S);
 
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:
  

Revision as of 16:17, 27 December 2014

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

$\fbox{D}$

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

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