Difference between revisions of "1968 AHSME Problems/Problem 35"
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== 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 | + | 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>. |
− | <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 == |
Revision as of 12:19, 21 February 2017
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 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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