 Ruudu ABCD tipud on vastupäeva ühendatud vastaskülgede 
keskpunktidega nii, nagu on näidatud kõrvalloleval joonisel. 
Tulemusena on saadud väiksem ruut EFGH. Tõestame, et 
ruutude EFGH ja ABCD pindalade suhe on 1/5.

\frac{S_{EFGH}}{S_{ABCD}}=\frac{1}{5}

A = (0, 0)
D = (0, 4)
C = (4, 4)
B = (4.08, 0)
hulknurk1, a, b, c, d = Polygon(A, D, C, B)
I = Midpoint(A, D)
J = Midpoint(D, C)
K = Midpoint(C, B)
L = Midpoint(B, A)
f = Segment(D, L)
g = Segment(A, K)
h = Segment(I, C)
i = Segment(J, B)
H = Intersect(h, f)
G = Intersect(i, h)
F = Intersect(g, i)
E = Intersect(g, f)
hulknurk2, e, f_1, g_1, h_1 = Polygon(E, H, G, F)
n = 1 (slider from 1 to 5, step 1)
α = 0 rad (slider from 0 to 180°, step °)
F' = Rotate(F, α, K)
B' = Rotate(B, α, K)
G' = Rotate(G, α, J)
C' = Rotate(C, α, J)
H' = Rotate(H, α, I)
D' = Rotate(D, α, I)
A' = Rotate(A, α, L)
E' = Rotate(E, α, L)
hulknurk7, c', j, g' = Polygon(J, G', C')
hulknurk8, d', i_1, h' = Polygon(I, H', D')
hulknurk9, a', l, e' = Polygon(L, E', A')
hulknurk10, b', k, f' = Polygon(K, F', B')
Q_1: n > 3
Q_4: n > 2
Export_1 = (-1.6458507399577065, 5.970146094036651)
Export_2 = (14.5, -1)