Converting rectangle to hexagon
True, False, or Conditionally True?
We can cut any rectangle into three pieces using straight lines, and rearrange these pieces into a convex hexagon whose sides are all of the same length.
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3 solutions
Moderator note:
Notice that the final equation is equivalent to X 2 = 4 S 2 + ( L − X ) 2 , or that X = 8 L 4 L 2 + S 2 .
But because we also need X < L (implied constraint in the diagram), hence this gives us S 2 < 4 L 2 , or that S < 2 L like in Marta's analysis.
If the short and long sides of the rectangle are s and l, and the length of the black lines is x, it is not too hard to show that 4 s 2 + ( l − x ) 2 = x 2 which is easily solved for x.
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I am not really following. Can you please explain?
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He is claiming that, given the sides of the rectangle s and l, you can always find a value x that corresponds to where in the picture the 3 lines meet. (Though he does not prove that x is always less than l)
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@Alex Li – Yes, you can always find a solution. If the condition failed, just replace L by S and S by L or rotate the rectangle 90 degrees.
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@Ossama Ismail – I see that you added a figure to explain this. That is great!
How can you be sure that you can actually find such black lines?
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I agree, the solution writer only shown that it's true for 2 configurations, but he didn't show that it's true for all possible configurations.
To construct the solution, we proceed as shown below.
Make E is a midpoint of the shorter side (or midpoint of any side if A B C D is a square). Find midpoint F of A E . Draw a line perpendicular to A E through F . G is an intersection of this line with A B . This construction guarantees that A G = G E . Cut along G E , and symmetrically below. Move pieces as shown above.
The construction will fail, that is the line through F will not intersect the segment A B , only if B C > 2 A B or it is equal to it. Since we picked the smaller of the two sides (or one of the sides of a square), this cannot happen.
First of all, I thought that I have to come up with a regular Hexagon. So I hit false. Because otherwise I would have known is true. The hypothesis didn't state that the second form is just a guide.
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I do not think it is necessary to state that the second form is just a guide. Pictures are always guides. Somebody said, "Geometry is the art of making correct statements by looking at wrong figures."
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True, but I agree that making that image a bit less regular would have been probably fair. It's okay to not give away the solution, but actually misleading is less recommended.
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@Marta Reece – I don't understand. Doesn't the problem ask for regular hexagon? Why do we want to make it less regular?
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@Christopher Boo – The problem asks for a hexagon with equal sides. The solution has equal sides but not equal angles. So it is not a regular hexagon. What we could do is make the image less regular in a similar way.
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@Marta Reece – Oh right, what was I thinking. Thanks for the clarification!
No, remember that all sides of the hexagon must have the same length.
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I have modified the image to satisfy that requirement and added section explaining how the requirement is satisfied.
The line through F has to be perpendicular to AE, not AB
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I made the correction. Sorry about not catching it before and thanks for pointing it out.
I have understood that sides of the exagon had to be pieces of the border of the rectangle, the figure could be full of color to be more clear.
Answer: Yes ..
Note:: all black lines have the same length.
Assume the length of the black line = X and the rectangle dimensions are S and L .
To find X , solve X = 4 S 2 + ( L − X ) 2