Area of a set of squares

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Godot

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(sorry, im not sure if this goes in geometry or intermediate geometry)
WVbbuLF_SQG_pn-cGk73BQ.jpg
Question:If all the shaded quadrilaterals are squares, and their combined area is S, how do you prove that x^2/2=<S=<X^2

i started out by marking one of the sides of the smaller shaded squares as a parameter(b), then finding the equation for the area(f(x)=x^2-4xb-8b^2) then finding the derivative (f'(x)=2x-4b) but after that i got stuck, can anyone tell me what to do next?
 
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Well, one side of the equality is trivial. You automatically know that S <= X2, because X2 is the entire area of the square. Proving that the area of the shaded squares is more than half the area of the whole square is a bit more difficult. Just looking at the picture, it seems like the four smaller squares are the same size, but we cannot assume that. Are you given any information to indicate whether this is or isn't the case?
 
ksdhart2 said:
Just looking at the picture, it seems like the four smaller squares are the same size, but we cannot assume that. Are you given any information to indicate whether this is or isn't the case?
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ah sorry, yes the 4 smaller squares are equal. I would've posted the full question but it isn't in English so it wouldn't have helped much
 
mmm4444bot said:
Hmm? Are you saying that you comprehend the other language, but cannot translate it to English?
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No I would've posted a picture of the question, I just forgot to add that we know that the smaller squares were equal. Everything else was there
 
Godot said:
… the equation for the [shaded] area f(x)=x^2-4xb-8b^2 then finding the derivative f'(x)=2x-4b …
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Is that substraction a typo?

The derivative gives the rate at which the shaded area is changing with respect to a change in x. What's your strategy for using a rate to prove the given inequality?

So, x is changing and b is fixed. I had been thinking it was the other way around.
 
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Actually, I don't think you even need the derivative. There's a simpler way to go about it. Let b be the side length of one of the smaller shaded squares, and c be the side length of the larger shaded square. We then know that S = 4b2 + c2 and X = 2b + c. Thus, X2 = (2b + c)2 = 4b2 + 4bc + c2. We need to prove that X2/2 <= S, or equivalently, S >= X2/2. So, let's try a proof by contradiction and assume the opposite. Assume that S < X2/2. Working through a few steps:

  • S < X2/2
  • 4b2 + c2 < (4b2 + 4bc + c2)/2
  • 4b2 + c2 < 2b2 + 2bc + c2/2

I trust you can finish up from here. Eventually you'll reach a statement that is impossible, and your proof by contradiction is complete.
 
Godot said:
Thats what was wrong, [so I have] made x fixed and b the variable and now it works perfectly fine
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Glad to hear it.

I would like to see your proof. I'm curious to see the application of f'(b). :cool:
 
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