A school warm-up exercise to occupy your time - page 6

[Aleksey Nikolayev]

Cramer's theorem is in this book (Appendix, page 102).

[Maxim Kuznetsov]

assuming that indeed if all nodes lie on a circle, the area is maximal (which is so similar to the truth that it seems to be true),

the maximal area does not depend on the order of sides (it is shown on the given pictures), for example, the maximal area 1-2-3-4 is equal to the maximal area 1-4-3-2

for 3- angle, the formula should be reduced to Heron's formula, for the square x-x-x-x reduce to x^2

it seems to be simple and obvious thing, but it somehow does not count

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damn it, and these people are looking for a grail in the financial markets :-)

[Aleksey Nikolayev]

Maxim Kuznetsov:

assuming that indeed if all nodes lie on a circle, then the area is maximal (which is so similar to the truth that it seems to be true),

the maximal area does not depend on the order of sides (it is shown on the given pictures), for example, the maximal area 1-2-3-4 is equal to the maximal area 1-4-3-2

for 3- angle, the formula should be reduced to Heron's formula, for the square x-x-x-x reduce to x^2

it seems to be simple and obvious thing, but it somehow does not count

---

damn it, and these people are looking for a grail in the financial markets :-)

Read about Brahmagupta formula (quadrilateral). With more sides it seems to be much sadder - there's a wiki about it.

Your "school" tasks are not school tasks at all)

[Aleksey Vyazmikin]

Would you like to use this method?

The idea is to choose the side of the square grid so that it (its nodes) is closest to all the sides of the polygon.

[Iurii Tokman]

Renat Akhtyamov:

let's say the price goes along a parabola.

try a polynomial, with different degrees

[Maxim Kuznetsov]

Aleksey Nikolayev:

Read about Brahmagupta's formula (quadrilateral). With more sides it seems to be much sadder - the wiki is about that.

Your "school" problems are not school problems at all)

If you rape Wolfram (or Maxima) if you have it handy,

then for A-B-C-D-.

s is the area of a single segment (isosceles triangle) of A, r is the radius of the circumcircle.

The radii of all segments are the same, they can be equated or make a system. The area of s in sum = the area of the figure... Sum of angles opposite sides is 360 degrees

But the idea goes further than that...

[Nikolai Semko]

Vladimir Simakov:

The above solution is only valid for polygons whose centre of the circumcircle lies inside the perimeter. Try triangle {2,2,3.9}

In general terms (approximation by precision double) it is solved like this:

Yes, you are correct. Did not take into account if the centre is outside the polygon.

[Nikolai Semko]

Aleksey Nikolayev:

3) MathSum()

s=6.0

Ah, so these are external libraries. So it's the same as I wrote there. To be encumbered by them only to replace one line of code:

for(int i=0; i<size; i++) sum+=arr[i];

I don't see any sense

[Aleksey Nikolayev]

Nikolai Semko:

Ah, so these are external libraries. So it's the same as what I wrote. Just to be surrounded by them just to replace one line of code:

I don't see the point.

Not external, standard) external is your i-canvas)

[Nikolai Semko]

Aleksey Nikolayev:

Not external, standard) external is your i-canvas)

Is it?

Not at the computer at the moment.

I think I've tried it. ME didn't recognise this function and didn't find it in help by F1

I understood that it is an AlgLib library