Problem 61082. Slicing the area of a circle

Created by Luisa

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Given the area, A, of a square, consider a circle having the area, πA, and the radius, r.
For a given slicing number n>1, find the (n+1)×2 matrix, M = [A1/π a1; A2/π a2; ...; An/π an; A_r L], where
in the first row (i=1), A1 stands for the area of one slice (like a pizza slice), and a1 stands for the logical 1 if A1 is smaller than or reaches the square's area A or a1 stands for the logical 0 if A1 surpasses A;
in the second row (i=2), A2 stands for the area of two slices and a2 has the same previous false-true meaning relative to the areas A2 and A;
and so on, until last slice of the circle;
in the last row (i=n+1), A_r is the area of the rectangle, with dimensions L×r, which contains the maximum possible number of slices, such that have the true meaning of their sum has the area smaller or equal than the square's area, in their adjacent arrangement.
Hint: Compare with Problem 61081.
input: (A,n)
output:  M = [A1/π a1; A2/π a2; ...; An/π an; A_r L]

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Last Solution submitted on Nov 24, 2025

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Christian Schröder on 21 Nov 2025 at 17:34

I must admit I'm not fully understanding the last requirement. You're supposed to consider the maximum number of slices such that their combined area does not exceed A, arrange them with the "tips" pointing up and down alternatingly, and then report the area and width of the rectangle of height r containing them?

Also, it'd be nice if you could always use some tolerance when checking for numerical equality, for all elements of the matrix returned. isapprox() is a builtin that does this BTW.

Luisa on 22 Nov 2025 at 16:16

Yes, the rectangle is exactly as you described. It is like the rectangle of Problem 61081, but it contains slices of circle instead triangles.

Luisa on 22 Nov 2025 at 16:22

Thank you for the advice, @Christian Schröder. But I will need some extra time to fix it as a matrix, because
"Error using assert The condition input argument must be convertible to a scalar logical. Error in Test2 (line 5) assert(isapprox(slicing_circle(A,n),y_correct))".

Luisa on 22 Nov 2025 at 17:32

Sorry, I will be away all week. So, I will only be able to fix the tolerance tests as you claimed later.
Hoping that this does not prevent the resolution of the problem.

Christian Schröder on 22 Nov 2025 at 19:07

Thanks for the feedback on the rectangle, Luisa. In that case there's either a bug in my code or in the test cases.

isapprox(A, B) returns a logical matrix of the same size as A and B; you may want all(isapprox(...), "all").

William on 23 Nov 2025 at 15:07

Luisa -- I have it all working except test problem #2, which needs a tolerance.

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Problem 61082. Slicing the area of a circle

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