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 June 10th, 2011, 02:31 PM #11 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Largest square inside a cube From the upper left corner of that cube to point A, let that length be a. The side length will be s. You will get two equations relating a to s. Once you solve that system you'll discover the only solution to that picture is when a=1/4 and s=3/4*sqrt(2) as proglote showed. With a little work you can show that point A must be on that edge it is shown to be on in order for that shape to be a square, therefore this problem is solved.
 June 10th, 2011, 04:55 PM #12 Senior Member     Joined: Feb 2010 Posts: 658 Thanks: 115 Re: Largest square inside a cube Well I certainly agree that the four sides of ABCD are each $\dfrac{3\sqrt{2}}{4}$ But that doesn't make it a square. It doesn't even make it a rhombus. There is still more to show.
 June 10th, 2011, 06:49 PM #13 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 My guess is 1x1 square.
June 10th, 2011, 07:26 PM   #14
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Re: Largest square inside a cube

Quote:
 Originally Posted by mrtwhs Well I certainly agree that the four sides of ABCD are each $\dfrac{3\sqrt{2}}{4}$ But that doesn't make it a square. It doesn't even make it a rhombus. There is still more to show.
yes....I agree with you.....

Can anybody explain it's a square?

 June 10th, 2011, 08:19 PM #15 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Let A(0, 1/4, 1), B(1/4, 0, 0), C(1, 3/4, 0) and D(3/4, 1, 1). Use the distance formula to determine AB = CD = 3?(2)/4. AB = <1/4, -1/4, -1> and CD = <-1/4, 1/4, 1>, since AB = -CD, they're parallel. AD = <3/4, 3/4, 0> and BC = <3/4, 3/4, 0>, since AD = BC, they're parallel. AB · AD = 3/16 + -3/16 + -0 = 0, AD · CD = -3/16 + 3/16 + 0 = 0, AB · BC = 3/16 + -3/16 + -0 = 0 and BC · CD = -3/16 + 3/16 + 0 = 0. Hence ? = 90° for all four internal angles. Therefore ABCD is a square with the maximum sides of 3?(2)/4.
June 11th, 2011, 06:18 AM   #16
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Re: Largest square inside a cube

Quote:
 Originally Posted by The_Fool From the upper left corner of that cube to point A, let that length be a. The side length will be s. You will get two equations relating a to s. Once you solve that system you'll discover the only solution to that picture is when a=1/4 and s=3/4*sqrt(2) as proglote showed. With a little work you can show that point A must be on that edge it is shown to be on in order for that shape to be a square, therefore this problem is solved.
I still don't know how to prove that there is no "bigger" square satisfying the conditions of the problem..

 June 11th, 2011, 03:50 PM #17 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Largest square inside a cube For simplicity, the center of the square and cube are placed at the origin. Wherever the square is placed inside the cube it will have the same side length. Obviously, to maximize side length, the square must be in the center of the cube and all corners of the square much touch the cube. If they don't, then there is a slightly larger square at a slightly different position that will defeat it. Ultimately, the square will be in the center. To describe this square you need four variables. The side length, the tilt from the z axis, the angle of one side of the square from the x or y axis, and the rotation of the square about the origin with the axis of rotation perpendicular to the square. You have a constraint that for all 4 points of the square it must touch the cube's wall. Using those 4 variables you can show where all 4 corners of the square is. Since this is a constraint problem you may have to use the Lagrange multiplier.

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