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The diameter of a closed, topologically bounded region of the plane is the greatest distance between two points in the region. Example: the diameter of a rectangle is the length of its diagonal. Of all the regions whose diameter equals 1, one of them, call it Rmax,  encloses the largest area.

Can you prove, or disprove, that Rmax also encloses all other regions of diameter 1? That is, that all other regions of diameter 1 can be made to fit inside Rmax?

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Spoiler

Assume Rmax can contain every region of diameter one.

Fit a circle of diameter one into Rmax.  Pick any point in Rmax which is not in the circle of diameter one.  Now look at the line through this new point and the center of the circle.  Notice the distance between the point and the far intersection between the circle and line is greater than one (the diameter of the circle plus the distance to the point).  Since Rmax cannot have diameter greater than one, there is no point outside the circle and inside Rmax.  Therefore Rmax must be a circle of diameter one.

Now take an equilateral triangle of diameter one.  The lengths between each of the three vertices of the triangle is one.  The only line segments of length one contained inside the circle are ones that intersect the center of the circle at their midpoint.  Then you can fully contain at most one of the sides of the triangle inside the circle.  Therefore a circle of diameter one cannot contain an equilateral triangle of diameter one.

Since we've reached a contradiction, Rmax cannot contain every region of diameter one.

 

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Here goes...

Rmax must be a circle of diameter one. To prove this, assume it isn’t. Then all other possible diameters of Rmax must be one or shorter than one. If this is true, then Rmax can fit inside a circle of diameter one, which has greater area than Rmax. Therefore Rmax doesn’t enclose the maximum area. By contradiction then, Rmax must be a circle of diameter one.

Not my cleanest proof, but I’m thrilled to be the first response.

Edited by CynPyn
Had hoped to hide my answer... don’t know how

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10 hours ago, CynPyn said:

Here goes...

Rmax must be a circle of diameter one. To prove this, assume it isn’t. Then all other possible diameters of Rmax must be one or shorter than one. If this is true, then Rmax can fit inside a circle of diameter one, which has greater area than Rmax. Therefore Rmax doesn’t enclose the maximum area. By contradiction then, Rmax must be a circle of diameter one.

Not my cleanest proof, but I’m thrilled to be the first response.

Hi @CynPyn and welcome to the Den.

Let's accept from this that  Rmax is the unit diameter circle.

Now imagine a rectangle with unit diameter (diagonal has length 1.)
That can be made to fit into
Rmax .

The question is does every unit-diameter region into Rmax ?

 

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This is a great example of a puzzle “where intuition might suggest wrong answer.” Thanks Bonanova, and shout out to Event Horizon for showing us the right answer!

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