Pentomino mit vier Symmetrieachsen - Typ 12
Author: Mag. G. Wengler

Dieser Typ ist 4-fach symmetrisch. Überlege, wie viele unterschiedliche Positionen es deshalb gibt.

hat 4 Achsen => ein Zentrum

4-fach symmetrisches Pentomino

A = (0, 1)
a = 0
b = -1
c = -1
d = -1
e = -1
f = -2
g = 0
h = -2
i = 0
j = -3
k = 1
l = -3
m = 2
n = -1
o = 1
p = -1
q = 1
r = 0
t = 0
u = -2
v = -1
w_5 = -2
z = -1
k_1 = -3
l_1 = -2
m_1 = -3
n_1 = -2
o_1 = -2
p_1 = -3
q_1 = -2
r_1 = -3
s_1 = -1
t_1 = -1
u_1 = -1
v_1 = -1
w_1 = 0
z_1 = 0
a_2 = -1
b_2 = -1
c_2 = -1
d_2 = -1
e_2 = -2
f_2 = 0
g_2 = -2
h_2 = 0
i_2 = -3
j_2 = 2
k_2 = -3
l_2 = 2
w_2 = -2
z_2 = 1
a_3 = -2
b_3 = 1
c_3 = 0
i_3 = -1
j_3 = 0
k_3 = -1
l_3 = 2
m_3 = 0
n_3 = 2
o_3 = 0
p_3 = 3
q_3 = 1
r_3 = 3
s_3 = 1
t_3 = 2
u_3 = 2
v_3 = 2
d_4 = 2
e_4 = 1
f_4 = 0
g_4 = 1
S_1 = (1.5, 0.5)
n_5 = false
s = 0
w = -1
B: (s + x(A), w + y(A))
a_1 = -1
b_1 = -1
C: (a_1 + x(A), b_1 + y(A))
c_1 = -1
d_1 = -2
D: (c_1 + x(A), d_1 + y(A))
e_1 = 0
f_1 = -2
E: (e_1 + x(A), f_1 + y(A))
g_1 = 0
h_1 = -3
F: (g_1 + x(A), h_1 + y(A))
i_1 = 1
j_1 = -3
G: (i_1 + x(A), j_1 + y(A))
m_2 = 1
n_2 = -2
H: (m_2 + x(A), n_2 + y(A))
o_2 = 2
p_2 = -2
I: (o_2 + x(A), p_2 + y(A))
q_2 = 2
r_2 = -1
J: (q_2 + x(A), r_2 + y(A))
s_2 = 1
t_2 = -1
T_1: (s_2 + x(A), t_2 + y(A))
u_2 = 1
v_2 = 0
U_1: (u_2 + x(A), v_2 + y(A))
Vieleck1, a_4, b_4, c_4, d_3, e_3, f_3, g_3, h_3, i_4, j_4, t_4, u_4 = Polygon(A, B, C, D, E, F, G, H, I, J, T_1, U_1)
h_4 = Segment((0.5, -3), (0.5, 2))
k_4 = Segment((-2, -0.5), (3, -0.5))
V_1 = (-1, 1)
W_1 = (2, -2)
l_4 = Segment(V_1, W_1)
Z_1 = (-1, -2)
A_2 = (2, 1)
m_4 = Segment(Z_1, A_2)
B_2 = Intersect(h_4, k_4)
C_2 = Intersect(h_4, k_4)