Difference between revisions of "2020 AMC 8 Problems/Problem 24"
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Each red square has side length <math>(1+d)</math>, so by solving <math>\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}</math>, we obtain <math>d = \frac{1}{4}</math>. The actual fraction of the total area covered by the gray tiles will be slightly less than <math>\frac{1}{(1+d)^2}</math>, which implies <math>\frac{1}{(1+d)^2} > \frac{64}{100} \iff \frac{1}{1+d} > \frac{4}{5} \iff d < \frac{1}{4}</math>. Hence <math>d</math> (and thus <math>\frac{d}{s}</math>, since we are assuming <math>s=1</math>) is less than <math>\frac{1}{4}</math>, and the only choice that satisfies this is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>. | Each red square has side length <math>(1+d)</math>, so by solving <math>\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}</math>, we obtain <math>d = \frac{1}{4}</math>. The actual fraction of the total area covered by the gray tiles will be slightly less than <math>\frac{1}{(1+d)^2}</math>, which implies <math>\frac{1}{(1+d)^2} > \frac{64}{100} \iff \frac{1}{1+d} > \frac{4}{5} \iff d < \frac{1}{4}</math>. Hence <math>d</math> (and thus <math>\frac{d}{s}</math>, since we are assuming <math>s=1</math>) is less than <math>\frac{1}{4}</math>, and the only choice that satisfies this is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>. | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/t8MVmKEyUhw | ||
+ | |||
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== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Revision as of 22:11, 19 January 2023
Contents
[hide]Problem 24
A large square region is paved with gray square tiles, each measuring
inches on a side. A border
inches wide surrounds each tile. The figure below shows the case for
. When
, the
gray tiles cover
of the area of the large square region. What is the ratio
for this larger value of
Solution 1
The area of the shaded region is . To find the area of the large square, we note that there is a
-inch border between each of the
pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of
times the length of the border, i.e.
. Adding this to the total length of the consecutive squares, which is
, the side length of the large square is
, yielding the equation
. Taking the square root of both sides (and using the fact that lengths are non-negative) gives
, and cross-multiplying now gives
.
Note: Once we obtain to ease computation, we may take the reciprocal of both sides to yield
so
Multiplying both sides by
yields the same answer as before. ~peace09
Solution 2
Without loss of generality, we may let (since
will be determined by the scale of
, and we are only interested in the ratio
). Then, as the total area of the
gray tiles is simply
, the large square has area
, making the side of the large square
. As in Solution 1, the side length of the large square consists of the total length of the gray tiles and
lots of the border, so the length of the border is
. Since
if
, the answer is
.
Solution 3 (using answer choices)
As in Solution 2, we let without loss of generality. For sufficiently large
, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown:
Each red square has side length
, so by solving
, we obtain
. The actual fraction of the total area covered by the gray tiles will be slightly less than
, which implies
. Hence
(and thus
, since we are assuming
) is less than
, and the only choice that satisfies this is
.
Video Solution
Please like and subscribe!
Video Solution by OmegaLearn
https://youtu.be/UpCURw5Moig?t=31
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
[edit: false link]
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1515
~Interstigation
Video Solution by STEMbreezy
https://youtu.be/wq8EUCe5oQU?t=353
~STEMbreezy
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.