Is there a way to cut a 10" round cake into 3 pieces that can fit into a square box smaller than 10"?

Question

1) It's an imprenetrable fruitcake, requiring a bandsaw to cut it
2) The box is exactly as high as the fruitcake, and the bandsaw is only able to make cuts vertical to the fruitcake's bottom.

Duke is off to a good start with a box size of 9.99. 10 points goes to the answer with the smallest box.

Duke, you're right, I should have been more clear about this problem, and asked for how small of a square box can anyone get away with. As for the impentrable fruitcake, it isn't meant to be eaten, but to be given away as a holiday gift---in a smaller box, of course.

Answer 1

Yes, cut 2 very small circular segments with perpendicular chords, which can be placed in a corner, then the 2 of the sides of the original box (parallel to the chords) can be moved inside to touch the chords.
Example: circle x² + y² = 25 inscribed into a square with
sides x = ±5 and y = ± 5. Make 2 cuts along the lines
x = 4.99 and y = 4.99, rearrange the pieces into a square with sides x = 4.99, x = -5, y = 4.99, y = -5, side length 9.99 instead of original 10.

EDIT (after having read the additional details) Well, going on with the above approach I tried to determine the length L of a chord that can fit in the corner being put along the side of the circumscribed regular octagon /it is 2R(√2-1) long, so L<2R(√2-1)/. If we cut a piece with a chord at x, then
L = 2√(R² - x²), then 2 similar triangles lead to the proportion:
L/(2R(√2-1)) = (x-R(√2-1))/(R(2-√2)),
which in turn leads to the equation:
√2(x - R(√2-1)) = 2√(R² - x²),
the only positive root being
x = (R/3)(√2 - 1 + 2√(√2)) ≈ 0.930876*R
For R = 5 /see the example above/ this yields x ≈ 4.65, so the box size becomes little more than 5 + 4.65 = 9.65. It's easy to see that a circular segment with this chord length has an angle between the chord and the tangent in an endpoint less than 45 degrees, what means that the curvature of the arc will not cause parts of it to lean out of the box if the angle between the chord and the box sides is 45 degrees.

I don't claim that 9.65 is the minimum, it is only a possible solution.

Final note to the author: You should state the additional details more precisely - it's all clear about the 10 points, but I am much more interested who will eat the cake?

Answer 2

Why would you need to cut such a cake?
Unless you have a set of bandsaws instead of teeth.

Edit: A holiday gift? You must really dislike someone to even think of giving them such a cake.

Answer 3

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Answer 4

since your question does not indicate how tall the box is. you can cut the cake into pieces slightly larger than that of a triangle (one side would be rounded of course) and then just stack them up.

Answer 5

it depends, wat type of cake is it?