Jer

I remember figuring this rule out on my own back in high school, where to solve a problem, I needed to know which numbers have an odd number of factors.

Thanks for sharing. I was bracing for a challenge, but everything is right there.

So sweet. We had a kitty named Huckleberry who looked a lot this yours many years ago. I'll try and find a photo.

I like math now.

I'd like to say I did that, but I doubled the 27 three times. When I got 216 I realized that would have been easier.

Some "let them eat cake" energy going on there

It's funny because it's true.
(Or the least untrue things he's posted.)

I don't think your argument is sound. There's no guarantee that for any P the triangular cross-section can be adjusted by epsilon and intersect all six sides. On a cube-like shape it does, but for my shape it sometimes can't. Does every F=6, V=8 shape have some P that does work?

Desmos | 3D Graphing Calculator

Looks like you can get a hexagon. At least for this simple case. www.desmos.com/3d/brrufqrvj4

No. A tetrahedron with the tips of two corners sliced off fits the description. I'm having trouble picturing whether it has a hexagonal cross-section.

You guessed correctly. I don't have that one, just the ones that skip by constant amounts.

I think I've cracked it. For any even number of people, there are none (I think). For 7, there are 5. For 9, just 3. (I don't know if I'm missing some sporadic cases, though.)

Another arrangement is acebd, which is its own dual, matching in the same order: ACEBD.

Call your arrangement aedbc, and it matches ADBEC after each rotation pictures. The alternate arrangement adbec matches AEDCB. Swap lower-case letters with upper.

Desmos | Graphing Calculator

www.desmos.com/calculator/r...

It seems like the area is not unique. If the triangle is zero the area is 90.25. If it takes up half the square, the isoceles right triangle has area 361(3-2sqrt2) = 61.94. I think there is a missing constraint based on the Pythagoras tag.

The paper doesn't say every shape checked has the property, it only implies no counterexamples had been found. From the paper:
As of today, most of these polyhedra are known to have Rupert’s property, but some remain open.

My damned backwards brain that still mixes up left and right: ∀ is an upside-down "A" bc it's the indefinite article that means there is one, or it exists. ∃ is a backwards E bc it means "for Every."

Does $200 worth of groceries fill a whole bag, still?

Desmos | Graphing Calculator

Spoiler in the graph. www.desmos.com/calculator/y...

Paddy 7/31/25

Yes this one is easier. When a=sqrt(3)/3, Area=sqrt(3)/2 - 2/3 www.desmos.com/calculator/a...

Paddy 7/30/25

Area when a=3 is 9/4 - sqrt(3) Not too bad. Work shown in Desmos. www.desmos.com/calculator/6...

Those are the extremes, but A can be folded to anywhere along BC. The boundary is a smooth curve.

Desmos | Graphing Calculator

Nice one. Here's a Desmos I whipped up. www.desmos.com/calculator/2...

folded rectangle solutions

For completeness: the solutions. The first (black) has the nice rational crease of length 7.5, the second (orange) is a root of a degree 5 polynomial as found by WolframAlpha to be about 13.5907 I wonder if any perimeters have two rational solutions?
www.desmos.com/calculator/d...

ways to fold a rectangle perimeter formula

Perimeter as a function of x, color coded as in the first graph. Time to start solving your question. y=39/11 hits both the black and orange portions.
www.desmos.com/calculator/5...

Ways to fold a rectangle

My perimeter is 4 (can scale by 11 later) with sides AB=a and BC=2-a. You can see the crease will pass through different sides based on where a lies with respect to the interval [2/3, 4sqrt(3)-6] www.desmos.com/calculator/i...

There are two solutions. If the rectangle is 6 by 16, the crease is 7.5. If the rectangle is 10.233 by 11.776 the crease is about 13.591. I will follow up with some Desmos graphs. There are three ways this crease can go.

This was the Pioneer Valley All Day Sing. The Western Mass Convention is two days in early March. (Also check out the Berkshire Foothills All Day in November.) Everything else checks out for sure.

Trace of midpoints

I kept n=2, but that's just a scale factor, and I used k instead of b. Coordinates of foci on lines 29-30 vertices on 31-34.
www.desmos.com/calculator/d...