Difference between revisions of "2020 AMC 8 Problems/Problem 24"
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+ | ==Problem 24== | ||
A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math> | A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math> | ||
+ | <asy> | ||
+ | draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); | ||
+ | filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); | ||
+ | filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); | ||
+ | filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); | ||
+ | filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); | ||
+ | filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); | ||
+ | filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); | ||
+ | filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); | ||
+ | filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); | ||
+ | filldraw((12,12)--(12,9)--(9,9)--(9,12)--cycle, mediumgray); | ||
+ | </asy> | ||
+ | |||
+ | <math>\textbf{(A) }\frac{6}{25} \qquad \textbf{(B) }\frac{1}{4} \qquad \textbf{(C) }\frac{9}{25} \qquad \textbf{(D) }\frac{7}{16} \qquad \textbf{(E) }\frac{9}{16}</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | The area of the shaded region is <math>(24s)^2</math>. To find the area of the large square, we note that there is a <math>d</math>-inch border between each of the <math>23</math> pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of <math>23+2 = 25</math> times the length of the border, i.e. <math>25d</math>. Adding this to the total length of the consecutive squares, which is <math>24s</math>, the side length of the large square is <math>(24s+25d)</math>, yielding the equation <math>\frac{(24s)^2}{(24s+25d)^2}=\frac{64}{100}</math>. Taking the square root of both sides (and using the fact that lengths are non-negative) gives <math>\frac{24s}{24s+25d}=\frac{8}{10} = \frac{4}{5}</math>, and cross-multiplying now gives <math>120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \frac{d}{s} = \frac{24}{100} = \boxed{\textbf{(A) }\frac{6}{25}}</math>. | ||
+ | |||
+ | |||
+ | Note: Once we obtain <math>\tfrac{24s}{24s+25d} = \tfrac{4}{5},</math> to ease computation, we may take the reciprocal of both sides to yield <math>\tfrac{24s+25d}{24s} = 1 + \tfrac{25d}{24s} = \tfrac{5}{4},</math> so <math>\tfrac{25d}{24s} = \tfrac{1}{4}.</math> Multiplying both sides by <math>\tfrac{24}{25}</math> yields the same answer as before. ~peace09 | ||
+ | |||
+ | ==Solution 2== | ||
+ | Without loss of generality, we may let <math>s=1</math> (since <math>d</math> will be determined by the scale of <math>s</math>, and we are only interested in the ratio <math>\frac{d}{s}</math>). Then, as the total area of the <math>576</math> gray tiles is simply <math>576</math>, the large square has area <math>\frac{576}{0.64} = 900</math>, making the side of the large square <math>\sqrt{900}=30</math>. As in Solution 1, the side length of the large square consists of the total length of the gray tiles and <math>25</math> lots of the border, so the length of the border is <math>d = \frac{30-24}{25} = \frac{6}{25}</math>. Since <math>\frac{d}{s}=d</math> if <math>s=1</math>, the answer is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>. | ||
+ | Nice. | ||
+ | |||
+ | ==Solution 3 (using answer choices)== | ||
+ | As in Solution 2, we let <math>s = 1</math> without loss of generality. For sufficiently large <math>n</math>, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: | ||
<asy> | <asy> | ||
draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); | draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); | ||
Line 12: | Line 40: | ||
filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); | filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); | ||
filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); | filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); | ||
+ | |||
+ | for(int i = 1; i <= 13; i += 4){ | ||
+ | draw((1,i)--(13,i), red); | ||
+ | draw((i,1)--(i,13), red); | ||
+ | } | ||
</asy> | </asy> | ||
+ | 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 by Math-X (First understand the problem!!!)== | |
+ | https://youtu.be/UnVo6jZ3Wnk?si=Qs8zS0YEfg1iP-Y2&t=5584 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution(🚀 Just 2 m)== | ||
+ | |||
+ | https://www.youtube.com/watch?v=Vnk73Kd8t4o&list=PL73YVYWi-yG8Exr884k6y3eq8VBFMZRIF&index=24 | ||
+ | |||
+ | <i>~Education, the Study of Everything </i> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/t8MVmKEyUhw | ||
+ | |||
+ | Please like and subscribe! | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/UpCURw5Moig?t=31 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/NHYB0VI3dcY | ||
+ | |||
+ | ~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== | ||
+ | {{AMC8 box|year=2020|num-b=23|num-a=25}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:38, 21 January 2024
Contents
- 1 Problem 24
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (using answer choices)
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution(🚀 Just 2 m)
- 7 Video Solution
- 8 Video Solution by OmegaLearn
- 9 Video Solution by WhyMath
- 10 Video Solution by Interstigation
- 11 Video Solution by STEMbreezy
- 12 See also
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
.
Nice.
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 by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=Qs8zS0YEfg1iP-Y2&t=5584
~Math-X
Video Solution(🚀 Just 2 m)
https://www.youtube.com/watch?v=Vnk73Kd8t4o&list=PL73YVYWi-yG8Exr884k6y3eq8VBFMZRIF&index=24
~Education, the Study of Everything
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.