Ülesanne 3, TJ13. 
Kolmnurga ABC sees on vabalt valitud kaks
punkti D ja E. 
Ühendada punktid D ja E lühimat teed pidi, mis puudutab 
kõiki kolmnurga külgi.

Käesolevaga on ülesandele leitud kaks lahendit
1. Peegeldades punkti D küljest BC
2. Peegeldades punkt D küljest AB
Lahenditel puudub ühisosa.

A = (7.17, 13.45)
B = (5.16, 5.56)
C = (20.28, 5.29)
t1, c, a, b = Polygon(A, B, C)
D = (7.53, 8.26)
D'_{2} = Mirror(D, c)
P = Mirror(D'_{2}, a)
E = (15.93, 5.53)
E'_{2} = Mirror(E, b)
e = Segment(P, E'_{2})
g = Line(B, A)
D'_{1} = Mirror(D, a)
I = Mirror(D'_{1}, c)
E' = Mirror(E, b)
l = Segment(I, E')
h = Segment(E, E')
m = Segment(D, D'_{1})
J = Intersect(b, l)
n = Segment(J, E)
p = Segment(D'_{1}, I)
N = Intersect(g, l)
t2, n_1, i_1, d'_{1} = Polygon(I, D'_{1}, N)
K = Intersect(i_1, a)
k = Segment(D, K)
q = PolyLine(D, K, N, J, E)
f = Segment(E, E')
E'_{1} = Mirror(E, a)
F = Mirror(E'_{1}, b)
D' = Mirror(D, g)
i = Segment(D', F)
G = Intersect(g, c)
H = Intersect(i, g)
j = Segment(D, H)
L = Intersect(b, i)
r = Segment(E'_{1}, L)
s = Segment(F, E'_{1})
M = Intersect(r, a)
t = Segment(E, M)
d = PolyLine(E, M, L, H, D)
O = Intersect(e, b)
f_1 = Segment(E, O)
g_1 = Segment(E, E'_{2})
h_1 = Segment(D'_{2}, P)
Q = Intersect(e, a)
j_1 = Segment(D'_{2}, Q)
k_1 = Segment(D, D'_{2})
R = Intersect(g, c)
S = Intersect(j_1, g)
l_1 = Segment(D, S)
m_1 = PolyLine(D, S, Q, O, E)
T = Intersect(e, a, true)