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> | + | 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> | 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> |
Revision as of 15:58, 5 June 2020
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.