by using the I-cut-and-you-choose protocol

Divide the pizza in 8 even slices and tell them to deal with their own problems on their own instead of bugging me about them

Well that's pretty obvious. Pie are square, so she just needs to eat the core root half of the pizza. EZPZ.

>>4171
My attempt: If the square pizza has a side length of 10, then draw a circle with radius 5 in the center. Inscribe a square in said circle. That's Nina's portion.
What if Nina wants her portion to be circular? I think that involves squaring the circle, so you can't construct it using compass and straightedge. You'd have to measure it out with a ruler or something.

ninaaaaaaaaaaaaa

>>4172
It's going to be the same area either way so whether it's square or circle getting the exact autism side lengths or radius is easy.

For a circle it's just multiply the radius by sqrt(1/2) and measure that out before using whatever to cut out a perfect circle which probably won't work since that's hard and you're probably not that good with your hands. So the square would probably work best. So just convert the circle area to square area and there you go, side lengths are (sqrt(pi/2)*r). Get out a calculator, get your measuring tape, and cut that cold pizza!

I realized that you could just cut off the crust first and then split the pizza in half. Maybe Chris doesn't like his crust to be cut off.

>>4176
But the Chris would have a lot more of the 'zza, the crust + his half.

This was the first idea that came to mind.

Assuming the pizza is a circle, cut a concentric circle inside the pizza with a radius square-root 2 over 2 of the pizza's radius. This will assure they each get an equal portion of the pizza.

(1) Starting with the area of a circle equation
(2) The area of Nina's pizza (A_N) and the area of Chris' pizza (A_C) must equal the area of the whole pizza (A_P)
(3) From the problem statement we know that the A_N must equal A_C
(4) Plug the value for A_C into eq. (2)
(5) Substitute eq (1) for the areas where P is the radius of the whole pizza and N is the radius of Nina's pizza.
(6) Algebra
(7) Square both sides and discard the negative roots since a pizza can't have negative radius
(8) Divide by root 2 and rationalize the denominator, to get the radius of Nina's pizza in terms of the whole pizza.

>>4179
Anonymous.... Everything else is right... but... it's πr^2 for area....

Pizzas are pies huh

>>4179
>>4180
Actually, looking over it again, I realized I solved it completely differently.

>>4179
>>4184
Decided to take a more interesting approach and see if I can come up with a solution that doesn't involve a concentric circle.

The first shape that comes to mind is, a rectangle inscribed inside the circle but that doesn't seem to be possible. As the rectangle dimensions needed to create equal area portions result in one of the sides being longer than the circles diameter. I used pythagorean theorem to calculate chord length in terms of apothem and radius. The apothem can be rewriten as the rectangle width/2. So I can use this to write the area of the rectangle in terms of chord length and radius. Then by applying the 1/2 circle area constraint from >>4184 on the rectangle area I could do a bunch of algebra and solve for chord length in terms of radius. I may have messed up my algebra and lost track of roots but I ended up getting length ~= 1.82 * radius. Taking this value and plugging it back into the chord formula to solve for the other side of the rectangle I get width ~= 2.706 * radius. This width value is obviously larger than circle's diameter of 2 * radius.

Seeing as a rectangle doesn't work, I doubt any regular, inscribed, polygon fits the conditions (a triangle probably can).
I wonder if there is any inscribed polygon that can satisfy the OP constraints and do these polygons have to be convex or have to be concave or can they be either (excluding triangles)?

>>4185
My immediate thought would be an octogon since 8 pizza slices and whatnot, but I'm too tired to solve it now, should work though, I think.

measure the area of the pie
now impose a circle of exactly half that area into the center of the pie, and cut that circle out
if you want an actual mathy answer I don't have it, but if it's area/2 in the center then it's exactly half and chris gets all the crust which satisfies the prompt

>>4168
are those lagrange points