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### Geometry of a Pizza

```
Date: 5/31/96 at 2:41:58
From: Anonymous
Subject: Geometry with circles

I have a pizza. The radius is 10 inches long. The pizza was cut into
16 equal slices. When 1 slice was left, my brother and I both wanted
it, so we agreed to cut it in half, but I like the crust more than he
does, so we decided to cut it in a way perpendicular to the "usual
way."

In other words, the two pieces would not be symmetrical. One piece
would contain all topping, and the other would contain some topping
and some crust.

Finally, the question: How far up the radius from the center of the
circle will I need to cut so we will both have an equal area of pizza?
Note: the area of the crust and topping make no difference. Just think
of it as a regular circle.

My first strategy was to simply find where to cut a triangle in the
same way above, but I had no luck. At this point I can think of no
other plan. Could you help me out? Thanks!
```

```
Date: 5/31/96 at 3:30:28
From: Doctor Pete
Subject: Re: Geometry with circles

This is a pretty ugly problem (the answer is a little messy).  I'm not
sure what level of math you are familiar with, but the solution I've
got requires trigonometry.

So, take a piece (1/16th) of the pizza. Then its area is Pi*r^2/16 =
100*Pi/16 = 25*Pi/4.  Half of this, 25*Pi/8, is the area of the slice
you want. So if you were to cut the slice perpendicularly at its axis
of symmetry ("the usual way") at some point, x inches, away from the
center, you'd get a triangle (the topping part), and a funny-shaped
trapezoid-like thingy (the crust part).  So the area of the triangle
must be 25*Pi/8.
Here's a picture (it's a bit wider than it actually is):
.
/|\
/ |A\
/  |  \
/   |   \
/    |x   \
/     |     \
/      |_     \
/_______|_|_____\
b

where angle A = Pi/16 (this is easy to see). But tan(A) = b/x, so
b = x tan(A). It follows that the area of this triangle is b*x = x^2*
tan(A) = x^2*tan(Pi/16).  But this is equal to 25*Pi/8, so solving for
x:
x = sqrt(25*Pi/(8*tan(Pi/16)))
= 5/2 * sqrt(Pi/(2*tan(Pi/16)))

Most people would be content to leave the solution at that, but using
some half-angle identities, we have:

tan(Pi/16) = sqrt((1-cos(Pi/8))/(1+cos(Pi/8)))
= sqrt((1-sqrt((1+cos(Pi/4))/2))/(1+sqrt((1+cos(Pi/
4))/2)))

sqrt(1-sqrt((1+sqrt(2)/2)/2))
= -----------------------------
sqrt(1+sqrt((1+sqrt(2)/2)/2))

= sqrt(7+4*sqrt(2)-(4+2*sqrt(2))*sqrt(2+sqrt(2))).

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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