Pythagoras-Puzzle
Author: Peter Hetterich
Date: 2007

Verändere die Lage der Vielecke so, dass sie im Quadrat Platz finden!

s = 6.2 (slider from 2 to 10, step 0.1)
A = (2.2666666666666666, -6.366666666666666)
B = Point(Circle[A, s])
c = Segment(A, B)
g_2 = OrthogonalLine(B, c)
g_1 = OrthogonalLine(A, c)
d = Circle(A, B)
e = Circle(B, A)
D_1, D = Intersect(d, g_1)
C_1, C = Intersect(e, g_2)
P, s_1, S_2, s_3, s_4 = Polygon(A, B, C, D)
E = Midpoint(D, C)
f = Circle(E, C)
Q = Point(f)
a = Segment(D, Q)
b = Segment(Q, C)
F = Point(Circle[A, a / b c])
G_1, G = Intersect(Circle[A, a], Circle[F, a a / b])
R, g, a_1, f_1 = Polygon(A, F, G)
h = Line(D, Q)
k = Circle(D, b - a)
H, I = Intersect(k, h)
i = OrthogonalLine(I, h)
J = Intersect(i, s_4)
K_1, K_2 = Intersect(Circle[I, a], h)
S, a_2, k_1, i_1, j = Polygon(A, K_1, I, J)
D' = Point(Circle[I, b - a])
Intersect(Circle[I, (b - a) / b a], Circle[D_1, (b - a) / b c])
Intersect(Circle[I, (b - a) / b a], Circle[D_1, (b - a) / b c])
p = Circle(D', (b - a) / b c)
q = Circle(I, (b - a) / b a)
K, L = Intersect(p, q)
T, d', i_2, k_2 = Polygon(I, K, D')
C_2 = Point(Circle[D', c])
r = Circle(C_2, b)
v = Circle(D', a)
M, N = Intersect(r, v)
U, c_2, d'_1, m = Polygon(D', M, C_2)
B_3 = Point(Circle[C_2, c])
l = Segment(C_2, B_3)
n = OrthogonalLine(B_3, l)
t = Circle(B_3, (b - a) / b c)
O, V = Intersect(t, n)
b_1 = Segment(B_3, O)
c_1 = Circle(C_2, b)
d_1 = Circle(O, c c / b - a)
W, Y = Intersect(c_1, d_1)
Z, c_3, y, o, b_3 = Polygon(C_2, Y, O, B_3)
P_1, d_2, q_1, c_4 = Polygon(Q, C, D)