Difference between revisions of "2016 AMC 10A Problems/Problem 10"
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== Problem == | == Problem == | ||
| + | A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle? | ||
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<asy> | <asy> | ||
size(6cm); | size(6cm); | ||
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== Solution == | == Solution == | ||
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Let the length of the inner rectangle be <math>x</math>. | Let the length of the inner rectangle be <math>x</math>. | ||
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x =2\implies \boxed{\textbf B}</math> | x =2\implies \boxed{\textbf B}</math> | ||
| − | ==Video Solution== | + | ==Video Solution (CREATIVE THINKING)== |
| + | https://youtu.be/tyRN1WyasOI | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | == Video Solution == | ||
https://youtu.be/XXX4_oBHuGk?t=791 | https://youtu.be/XXX4_oBHuGk?t=791 | ||
~IceMatrix | ~IceMatrix | ||
| + | |||
| + | == Another Video Solution == | ||
https://youtu.be/6lozP3dgr_0 | https://youtu.be/6lozP3dgr_0 | ||
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~savannahsolver | ~savannahsolver | ||
| − | ==See Also== | + | == See Also == |
{{AMC10 box|year=2016|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2016|ab=A|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:56, 25 June 2023
Contents
Problem
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is 
foot wide on all four sides. What is the length in feet of the inner rectangle?
![[asy] size(6cm); defaultpen(fontsize(9pt)); path rectangle(pair X, pair Y){ return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle; } filldraw(rectangle((0,0),(7,5)),gray(0.5)); filldraw(rectangle((1,1),(6,4)),gray(0.75)); filldraw(rectangle((2,2),(5,3)),white); label("$1$",(0.5,2.5)); draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead)); draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead)); label("$1$",(1.5,2.5)); draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead)); draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead)); label("$1$",(4.5,2.5)); draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead)); draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead)); label("$1$",(4.1,1.5)); draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead)); draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead)); label("$1$",(3.7,0.5)); draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead)); draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead)); [/asy]](http://latex.artofproblemsolving.com/0/e/8/0e82b393f0623e820199f3756d84f20a00a00d4d.png)
Solution
Let the length of the inner rectangle be 
.
Then the area of that rectangle is 
.
The second largest rectangle has dimensions of 
and 
, making its area 
. The area of the second shaded area, therefore, is 
.
The largest rectangle has dimensions of 
and 
, making its area 
. The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is 
.
The problem states that 
is an arithmetic progression, meaning that the terms in the sequence increase by the same amount each term.
Therefore,
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://youtu.be/XXX4_oBHuGk?t=791
~IceMatrix
Another Video Solution
~savannahsolver
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
