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February 16th, 2015, 10:29 AM  #1 
Newbie Joined: Feb 2015 From: United States Posts: 1 Thanks: 0  Can anyone help me? (inscribed squares)
Hi there! I am struggling with this problem, I am just not sure what to do! I had an idea for what to do for letter a, but I'm not sure about it. And if you could help with b too, that would be great! 
February 16th, 2015, 10:57 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond 
Each square (other than the first) has an area of 1/2 of the preceding square. See if you can figure out why.

February 16th, 2015, 01:36 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra 
If you can't think of an algebraic way to show that, consider what happens when you fold a square along the edges of the next one. Count the thicknesses of paper that make up each square.

February 16th, 2015, 06:38 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond 
Let $\displaystyle M$ be the midpoints of the sides of the square with sides $\displaystyle S_n$. Then the vertices of the square with sides $\displaystyle S_{n1}$ are at $\displaystyle M$. By the Pythagorean theorem, $\displaystyle \left(\frac12 S_n\right)^2+\left(\frac12 S_n\right)^2=S^2_{n1}$ $\displaystyle \frac14 S^2_n+\frac14 S^2_n=S^2_{n1}$ $\displaystyle \frac12 S^2_n=S^2_{n1}$ Hence the general term is $\displaystyle \left(\frac12\right)^{n1}$ and the first five terms are $\displaystyle 1,\frac12,\frac14,\frac18,\frac{1}{16}$. Last edited by greg1313; February 16th, 2015 at 07:40 PM. 

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