Difference between revisions of "2016 AMC 10A Problems/Problem 10"

(Video Solution)
m (Video Solutions)
 
(3 intermediate revisions by the same user not shown)
Line 51: Line 51:
 
x =2\implies \boxed{\textbf B}</math>
 
x =2\implies \boxed{\textbf B}</math>
  
==Video Solution (CREATIVE THINKING)==
+
==Video Solutions==
https://youtu.be/tyRN1WyasOI
+
i) https://youtu.be/tyRN1WyasOI (Creative Thinking)
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
  
 
+
ii) https://youtu.be/XXX4_oBHuGk?t=791
 
 
 
 
== Video Solution ==
 
https://youtu.be/XXX4_oBHuGk?t=791
 
  
 
~IceMatrix
 
~IceMatrix
  
== Another Video Solution ==
+
iii) https://youtu.be/6lozP3dgr_0
 
 
https://youtu.be/6lozP3dgr_0
 
  
 
~savannahsolver
 
~savannahsolver

Latest revision as of 04:52, 18 December 2024

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 $1$ 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]

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8$

Solution

Let the length of the inner rectangle be $x$.

Then the area of that rectangle is $x\cdot1 = x$.

The second largest rectangle has dimensions of $x+2$ and $3$, making its area $3x+6$. The area of the second shaded area, therefore, is $3x+6-x = 2x+6$.

The largest rectangle has dimensions of $x+4$ and $5$, making its area $5x + 20$. The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is $(5x+20) - (3x+6) = 2x + 14$.

The problem states that $x, 2x+6, 2x+14$ is an arithmetic progression, meaning that the terms in the sequence increase by the same amount each term.

Therefore, $(2x+6) - (x) = (2x+14) - (2x+6)\implies x+6 = 8\implies x =2\implies \boxed{\textbf B}$

Video Solutions

i) https://youtu.be/tyRN1WyasOI (Creative Thinking)

~Education, the Study of Everything

ii) https://youtu.be/XXX4_oBHuGk?t=791

~IceMatrix

iii) https://youtu.be/6lozP3dgr_0

~savannahsolver

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png