Difference between revisions of "1968 AHSME Problems/Problem 35"
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== Problem == | == Problem == | ||
<asy> | <asy> | ||
− | draw( | + | draw(arc((0,0),10, 0, 180),black+linewidth(.75)); |
− | |||
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 20: | Line 26: | ||
== 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> | + | 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>. |
− | <cmath>\lim_{h\rightarrow 0}\ | + | |
− | + | 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| | + | {{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 01:52, 16 August 2023
Problem
In this diagram the center of the circle is
, the radius is
inches, chord
is parallel to chord
.
,
,
,
are collinear, and
is the midpoint of
. Let
(sq. in.) represent the area of trapezoid
and let
(sq. in.) represent the area of rectangle
Then, as
and
are translated upward so that
increases toward the value
, while
always equals
, the ratio
becomes arbitrarily close to:
Solution
Let , where
. Since the areas of rectangle
and trapezoid
are both half of rectangle
and trapezoid
, respectively, the ratios between their areas will remain the same, so let us consider rectangle
and trapezoid
.
Draw radii and
, both of which obviously have length
. By the Pythagorean Theorem, the length of
is
, and the length of
is
. It follows that the area of rectangle
is
while the area of trapezoid
is
Now, we want to find the limit, as approaches
, of
. Note that this is equivalent to finding the same limit as
approaches
. Substituting
into
yields that trapezoid
has area
and that rectangle
has area
Our answer thus becomes
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 |
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