Difference between revisions of "1968 AHSME Problems/Problem 35"
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− | 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 <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> | + | 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 <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})} = \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})} = \frac{1}{2} * \frac{\sqrt{h}(\sqrt{2a - h} + 2\sqrt{a - h})}{\sqrt{h}(\sqrt{2a - h})}</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}(1 + \frac{2}{\sqrt{2}}) = \frac{1}{2}+\frac{1}{\sqrt{2}} \textbf{ (D)}.</cmath> |
Revision as of 18:05, 27 December 2014
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
, 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 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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