Difference between revisions of "1968 AHSME Problems/Problem 35"
(→Solution) |
m (→See also) |
||
Line 40: | Line 40: | ||
== 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 00:52, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.