Tuesday 29 October 2013

euclidean geometry - How big is my pizza, if I know its slices' sizes?




I bought a box of frozen pizza: eight slices, baked and then frozen, stacked in a box. The packaging assured me that I was holding an 18-inch[-diameter] pizza. That got me thinking: how do I know they're not lying?



Assume (though this may not be true of the pizza) that we have a circle[1] $S=\partial D$ centered at a point $O$ and four chords (closed line segments properly embedded in $D$) that intersect in a single point $C$. (Possibly $C=O$.) You may assume also (because this seems reasonable for pizza) that the angles made between adjacent chords (in the cyclic order around $C$) is between $30^\circ$ and $60^\circ$ and that the distance between $C$ and $O$ is less than the distance between $C$ and $S$.



Is it possible to use the lengths of the segments from $C$ to $S$ and the angles between adjacent segments to find the diameter of $D$? (If not, then would it be possible if my "you may assume also" assumptions were tightened a little?) If so, how?



(Of course, it's possible to find the diameter by measuring the arclengths of the curved parts of the slices of pizza, adding them up, and dividing by $\pi$. But I'm wondering if there's a way to do it from the sidelengths and tip-of-the-slice angles of the pizza slices.)







[1] A geometric circle, meaning the locus of points a certain distance from $O$, not just a topological circle.


Answer



Let a pizza triangle be the convex envelope of the vertices of a pizza slice. The area of any pizza triangle is a bit smaller then the area of its pizza slice, but not so much (especially given the condition on the angles). It is easy to compute the area of a pizza triangle through the sine theorem, so a simple criterion is given by computing the sum of the areas of the pizza triangles and compare it with the area of a regular octagon inscribed in a circle with diameter $18$ inches.



Anyway, given two opposite pizza slices it is not difficult to compute the radius of the disk from which they have been cut, since two opposite pizza slices give a cyclic quadrilateral for which it is not difficult to compute the side lengths and the area given $a,b,c,d,\theta$:



enter image description here



hence the circumradius is provided by Parameshvara's formula:




$$ R = \frac{1}{4\Delta}\sqrt{(l_1 l_3+l_2 l_4)(l_1 l_2 + l_3 l_4)(l_1 l_4+l_2 l_3)} $$
where $l_1,l_2,l_3,l_4$ are the side lengths of the cyclic quadrilateral depicted above: they can be computed through the cosine theorem. Also notice that Ptolemy's theorem gives: $$l_1 l_3+l_2 l_4=(a+c)(b+d).$$



Another possible approach is the following. We have:
$$\text{pow}_\Gamma(C) = ac = bd = R^2-OC^2, $$
so we just need to find $OC^2$. If we take $M$ and $N$ as the midpoints of the chords in the picture above, it is trivial that $OC$ is the diameter of the circumcircle of $CMN$, and we may compute the circumradius of $CMN$ through the sine theorem:
$$\frac{OC}{2}=\frac{MN}{2\sin\theta}$$
then the length of $MN$ through the cosine theorem, so that:
$$ OC^2 = \frac{1}{\sin^2\theta}\left(\left(\frac{a-c}{2}\right)^2+\left(\frac{b-d}{2}\right)^2-\frac{|a-c||b-d|}{2}\cos\theta\right) $$
and:





$$ R^2 = ac+\frac{1}{4\sin^2\theta}\left[(a-c)^2+(b-d)^2-2|a-c||b-d|\cos\theta\right].$$




If you do not know which couples of slices are "antipodal", well, they are not difficult to recognize: antipodal slices must have the same angle $\theta$ and fulfill $ac=bd$ (the intersecting chord theorem).


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