MarioNintendo Geometry/Algebra problem about pizza

[MarioNintendo]

Hey guys, I need help from you majors in math and stuff. While I was eating pizza today (with fork and knife, otherwise this problem wouldn't exist) I was wondering at what height we'd have to cut a slice in order to end up with two pieces of the same area. Kinda silly, I know, but I'm beginning University tomorrow after 6 months of not doing any kind of maths (I was more physics oriented) and I wanted to refresh my memory.

Here's how far I got. I got a bit lazy in the end, but I was more curious at how you guys thought this problem could be solved. Seems fairly easy and definitely possible, just need the right tools to work it.




(I pretended the sides of the pizza were all the same size so I could have 60° angles)

Thanks! Hope I didn't screw up at some part... D:

[dag12]

Use the r:r^2:r^3 theorem.

For two similar 2D shapes, if the lengths are in the proportion of r:1, then the areas are in the proportion of r^2:1

So that just means just cut it so that one of the two pieces is an equilateral triangle of side length x/sqrt(2), where x is the length of one of the sides of the original pizza.

Edit: So that means from the top, using a 30-60-90 triangle relation, you would need to cut it at the point x*sqrt(3)/2sqrt(2) (from the top of the pizza).

[MarioNintendo]

Originally Posted by dag12

Use the r:r^2:r^3 theorem.

For two similar 2D shapes, if the lengths are in the proportion of r:1, then the areas are in the proportion of r^2:1

So that just means just cut it so that one of the two pieces is an equilateral triangle of side length x/sqrt(2), where x is the length of one of the sides of the original pizza.

Edit: So that means from the top, using a 30-60-90 triangle relation, you would need to cut it at the point x*sqrt(3)/2sqrt(2) (from the top of the pizza).

Whoa, I don't recall that theorem, but somehow it makes so much sense, no wonder I couldn't think of it...!

I can't figure out how you came to those answers, though, I'd need to re-do it on paper XD

[LongGone]

Are you eating hexagonal pizzas?

[MarioNintendo]

Originally Posted by LongGone

Are you eating hexagonal pizzas?

:s you found my weakness

[Reincarnate]

I would just do

Area of slice: BH/2
Thus
Area of minitriangle: b*(H-h)/2
Area of trapezoid: h*(b+B)/2

Set equal and solve for h

b*(H-h)/2 = h*(b+B)/2
bH-bh = hb+hB
bH = hb+hB+bh
bH = h(b+B+b)
h = bH/(2b+B)

But we don't want to care about what b is. We just want to write this as a function of h.

Note that, due to similar triangles:
(H-h)/(b/2) = H/(B/2)
(B/2)*(H-h) = H*(b/2)
b = 2*(B/2)*(H-h)/H

So:

h = bH/(2b+B) from earlier, substitute in the b expression
h = (2*(B/2)*(H-h)/H)*H/(2*(2*(B/2)*(H-h)/H)+B)
h in this case reduces to

h = H*(2-sqrt(2))/2
or
h = 0.2928932188134H

In other words, it's a tad under 30% of the way up the pizza slice (from the bottom)

[Reincarnate]

Actually let me double check that with dag's results

He says x*sqrt(3)/(2*sqrt(2)) from the tip of the pizza works
In this case 2*H/sqrt(3) = x

So
(2*H/sqrt(3))*sqrt(3)/(2*sqrt(2)) + H*(2-sqrt(2))/2 should equal H
And indeed it does

Because
(2*H/sqrt(3))*sqrt(3)/(2*sqrt(2)) reduces to H/sqrt(2)
and
H/sqrt(2) + H*(2-sqrt(2))/2 = H(1/sqrt(2) + (2-sqrt(2))/2) = H(sqrt(2)/2 + (2-sqrt(2))/2) = H(2/2) = H

[qqwref]

I'm bored, so... what happens if the pizza is circular, with radius x, and you cut a 60 degree slice?

The area of the section is πx²/6, and we want the equilateral triangle on top to have an area of half of that or πx²/12. If it has side length w, its area is w²√(3/4), and then we have
πx²/12 = w²√(3/4) => w² = πx²/6√3 => w = x√(π/6√3).

Now what's h? It's the radius of the circle (or x) minus the height of the equilateral triangle (or w√(3/4)), so:
h = x - x√(π/8√3)