AMC 10 Geometry Math Olympiad USA Math Olympiad

Rectangle Pattern | AMC-10A, 2016 | Problem 10

Try this beautiful problem from Geometry based on Rectangle Pattern from AMC-10A, 2016, Problem 10. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on Rectangle Pattern from AMC 10A, 2016, Problem 10.

Rectangle Pattern- AMC-10A, 2016- Problem 10

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?

Rectangle Pattern Problem
  • \(1\)
  • \(2\)
  • \(4\)
  • \(6\)
  • \(8\)

Key Concepts




Check the Answer

But try the problem first…

Answer: \(2\)

Suggested Reading

AMC-10A (2016) Problem 10

Pre College Mathematics

Try with Hints

First hint

Rectangle Pattern Problem figure

Given that length of the inner rectangle be $x$. Therefore the area of that rectangle is $x \cdot 1=x$
The second largest rectangle has dimensions of $x+2$ and 3 , Therefore area $3 x+6$. Now area of the second shaded rectangle= $3 x+6-x=2 x+6$

can you finish the problem……..

Second Hint

Rectangle Pattern Problem figure

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

can you finish the problem……..

Final Step

Now The problem states that $x, 2 x+6,2 x+14$ is an arithmetic progression,i.e the common difference will be same . So we can say $(2 x+6)-(x)=(2 x+14)-(2 x+6) \Longrightarrow x+6=8 \Longrightarrow x=2$

Therefore the side length =\(2\)

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