Pralinenschachtel mit maximalem Volumen
Author: Prof. Dr. Dörte Haftendorn

Achtung: die Lösung ist kein Polynom 3. Grades:

Ortslinie

Ha 2011

A = Intersect(xAxis, yAxis)
B = Point(xAxis)
Vieleck1, a, b, c, d, C, D = Polygon(A, B, 4.0)
E = Midpoint(A, D)
e = OrthogonalLine(E, d)
F = Point(e)
f = Segment(D, F)
g = Segment(A, F)
G = Midpoint(A, B)
h = OrthogonalLine(G, a)
D' = Mirror(D, h)
F' = Mirror(F, h)
f' = Segment(D', F')
A' = Mirror(A, h)
F'_2 = Mirror(F, h)
g' = Segment(A', F'_2)
H = Intersect(h, e)
F'_3 = Rotate(F, 90.0*°, H)
i = Segment(A, F'_3)
j = Segment(F'_3, B)
Vieleck2, f_1, f'_1, k, l, I, J = Polygon(F, F'_3, 4.0)
m = Segment(D, J)
n = Segment(J, D')
p = Circle(B, Segment[E, F])
K, L = Intersect(p, xAxis)
q = OrthogonalLine(L, xAxis)
r = Segment(A, D')
M = Intersect(r, k)
N = Point(xAxis)
s = Circle(N, k / 2.0)
O, P = Intersect(s, xAxis)
t = OrthogonalLine(N, xAxis)
a_1 = Segment(D', M)
c_1 = Circle(P, a_1)
Q, Spitze = Intersect(c_1, t)
Vieleck3, o, p_1, r_1 = Polygon(P, Spitze, O)
b_1 = Segment(N, Spitze)
V: (x(L), 1.0 / 3.0 * k^(2.0) * b_1)
d_1 = Segment(E, F)
e_1 = Segment(B, L)
g_1 = Segment(H, M)
h_1 = Segment(N, P)
Vieleck4, d', d_2, j_1 = Polygon(D, J, D')
Vieleck5, a_2, d_3, f_2 = Polygon(D, F, A)
Vieleck6, i_1, d'_1, b_2 = Polygon(D', B, I)
Vieleck7, a_3, b_3, f'_2 = Polygon(B, F'_3, A)
k_1 = Segment(V, L)
Ortslinie1 = Locus(V, F)
u = false