Pizza puzzle, find maximum area of this piece?

With an awful lot of work, (though it was exciting), I get -

(-27) + (18)√2 + (-18)√3 + (27/2)√6

With circle centre on (0, 0), begin at (0, 6) with the

clockwise rotation order of 30, 120, 90, 45, 60, 15.

EDIT:

I guess you're not kidding, Scythian. I tested 6 hopeful diagrams, of which this one

contained the only right-angled triangle, so I thought this must be it. Something must

have gone astray with my initial programming to cut down the number of possibilities.

EDIT:

After correcting a few silly mistakes, I now get -

Area = (-351) + (459/2)√2 + (-459/2)√3 + (351/2)√6 ≈ 5.9418

I hope that's it because I'm getting dizzy and Wolfram's about to blow a gasket.

That's with the order: 45, 15, 60, 30, 90, 120, clockwise from (0, 6).

EDIT: Hold your horses for a while, because, even though I'm getting some answers,

I think I'm getting wrong areas for the particular orders. Something is definitely fishy.

EDIT: Ok, I think I have it now, once I fixed up a couple of minuses that should have

been pluses and vice versa.

The circular order of 15, 90, 60, 45, 30, 120 seems to give the largest area,

so I'll get the answer to you asap.

EDIT: Area = (18) + (-45/4)√2 + (15)√3 + (-33/4)√6 ≈ 7.862569

BTW, for such an interesting question, I'm surprised at the number of stars so far.

EDIT: Here's a list of all 12 in descending area order -

(15, 90, 60, 45, 30, 120): Area ≈ 7.862569

(15, 45, 60, 90, 30, 120): Area ≈ 7.049387

(15, 90, 60, 30, 45, 120): Area ≈ 3.630452

(15, 30, 60, 90, 45, 120): Area ≈ 2.657675

(15, 60, 30, 45, 120, 90): Area ≈ 1.869781

(15, 45, 120, 90, 30, 60): Area ≈ 1.676400

(15, 60, 45, 30, 120, 90): Area ≈ 0.474067

(15, 30, 120, 90, 45, 60): Area ≈ 0.347041

(15, 45, 60, 30, 90, 120): Area ≈ 0.319805

(15, 30, 60, 45, 90, 120): Area ≈ 0.261120

(15, 45, 120, 30, 90, 60): Area ≈ 0.188298

(15, 30, 120, 45, 90, 60): Area ≈ 0.153745

EDIT: To obtain an EXACT answer to a given set of arcs, I had to -

1. Find the EXACT points at the ends of the arcs.

2. Find the equations to the 3 triangle lines with EXACT slopes & EXACT y-intercepts.

3. Find the EXACT (x,y) points of the 3 intersecting lines after equating to each other.

4. Find the EXACT area using the area equation for 3 points.

Scythian, each one takes quite a lot of time, even using WolframAlpha, so I was

interested in knowing if you had a quicker method, such as software that can

manipulate radicals and output radical answers (is there such a thing?), or do you

have to do the hard slog too?

EDIT: Thanks for the explanation, Scythian, In the end, once I had ironed out all

the mistakes, the maximum popped out easily, and like you, all I had to do was

calculate that one exactly. It's a pity I had to work through those others though,

when I didn't really need to. "Opposite arcs adding to 180º" didn't occur to me.

Instead, I used the perpendicular slopes formula. Having calculated 3 of the 12

exactly, it looks like your answer format may be true for the other 9, but I'm not

about to look into that soon.

I'm still working on it, but I'm down to 12 cases. To be cont'd...

• 6 different arcs can be arranged in (6-1)! = 120 different ways around a circle, but the arrangements come in pairs of mirror images. So really at most "only" 60 arrangements need looked at.

• Let O be the center of the circle. For a given arrangement, label the arc endpoints P(1), ..., P(6) in, say, clockwise order, and for all n ∈ ℤ let P(n) = P(n mod 6). Then a chord must connect endpoints P(n) and P(n+3) for each n. Let A(n) := ∡(P(n),O,P(n+1)). So by hypothesis

(1): { A(n) | n ∈ ℤ } = { τ/3, τ/4, τ/6, τ/8, τ/12, τ/24 },

where τ := 2π.

• The exterior (turning) angle between a pair of adjacent sides of a polygon is equal to the angle by which a line perpendicular to one of those sides must be turned in order to become perpendicular to the other. So in the case of a polygon formed by chords of a circle, this will be the same as the angle subtended by the chords' radial bisectors. Consequently, the exterior angle between chord (P(n),P(n+3)) and chord (P(n+1),P(n+4)) is equal to

(2): ∡(n) := (1/2)(A(n) + A(n+3)).

By hypothesis our polygon is a right triangle, so WLOG ∡(1) = τ/2 - τ/4 = τ/4. The only way this is possible, given (1) and (2), is if { A(1), A(4) } = { τ/3, τ/6 }. So WLOG let

(A(1), A(4)) = (τ/3, τ/6).

{ A(1), A(3), A(5), A(6) } can be matched with {τ/4, τ/8, τ/12, τ/24} in 4! = 24 different ways. But half of them are mirror images of the other half. So just 12 cases remain.

We must realize that. in order to be left out piece after the three cuts has to be a right triangle. This makes it necessary that two cuts have to be perpendicular to each other. next the second perpendicular cut cannot be after 15 degree cut because the remaining will be 360 -15, a odd number so the other cuts will be fractional. For sane reason it cannot be 30 36- -30 is not divisible by 4. So let us try with 60.and let perpendicular cut be the diameter.