# Mathematics puzzle - biggest and smallest possible half (yes, 'half' - read the damn question)

This is a spinoff from here in the thread about when the moon is visible to the smallest set of people on Earth…

The (simplified) puzzle scenario is as follows:
You have a perfectly spherical orange and an unlimited supply of pins

You stick pins in any arrangement you like, all over the surface of the orange, then you cut it perfectly in half (vertical slice only from stem to blossom end) so that one of the halves has the largest possible count of pins.

Is it always the case that the other half of the orange has the lowest possible count of pins, or could you have cut it somewhere else to achieve that?

For the sake of this exercise, assume it is impossible to cut through the exact point where a pin is placed, so you probably need to think of this in terms of clustering the pins at stations or segments (like the numbers on a clock) with the cut line passing in between - but the stations need not be evenly distributed, as long as you can cut between them and end up with two perfect halves of orange.

BTW, the reason I didn’t think the answer to this question is just “Duh, of course it is!”, is because of things like nontransitive dice. (Appreciate the one thing isn’t like the other - I only mean that I wonder if the properties of ‘biggest half/smallest half’ might be nontransitive, somehow).

Suppose you fix the amount of pins in the whole orange at P. Then the amount of pins in one half determines the amount of pins in the other half; specifically, the one is P minus the other. Thus, maximizing one half’s pin count is as good as minimizing the other half’s pin count. That reasoning is perfectly correct.

Concur. If you subtract the possible maximum from a finite number, you are left with the possible minimum. How that maximum is determined is irrelevant.

Yeah… stated that way, it’s a lot less puzzling a puzzle - thanks.

I concur with the exception of pointing out that there can be duplicate maximums and minimums. E.g., if you lay out four groups around the orange like:

1 2
2 3

The maximum would be 5 and the minimum 3, but you can cut vertically or horizontally.

OP said unlimited so we need to allow for an infinite number of pins.

Wouldn’t this make the concepts of minimum and maximum meaningless? We can’t talk about different magnitudes of infinity. Pin counts are strictly integers.

If one half has the maximum number of pins , the other side has the minimum.
If one half has the minimum number of pins, the other side has the maximum

There MAY be another place to cut it to get halves with maximum and minimum pins …being the same number of pins.

Originally, they were going to be pins with numbers on them; I managed to work out that would be unnecessary all on my own.

If we’re cutting the orange into quarters, I think we can make it so that the theoretical quarter with the maximum pin count overlaps with the quarter with the minimum pin count.

Ohhhhh.
What if we use Banach-Tarski?

My first instinct was that you can’t, but you’re right. If you have a 12-slice orange, put the following number of pins in each slice:

1, 0, 5; 0, 2, 0; 2, 2, 2; 2, 2, 2

The maximum quarter would contain the pins of the 3rd, 4th, and 5th slices: 7 pins. The minimum would consist of the 4th, 5th, and 6th slices: 2 pins. ∎

I think he wants a planar slice.

Doesn’t this hinge on the definition of the stem as the middle and definition of “cutting in a different place?”

Let’s assume we have a hundred pins. Put 50 in the upper left side of the orange (with the stem the middle), and 40 on the upper right. Now put ten on the lower left. If you cut it in the middle, you get 60 max and 40 min on the other side. But if you cut it in half perpendicular to the stem, you get 90 and 10, a lower minimum.

So was the intent of “cutting in a different place” with the same middle point?

A cut doesn’t have a “middle point”, as you’re describing it. Any two cuts will intersect in two opposite points, and you can put the point of intersection anywhere you like.