Ziehe eine horizontale Linie und setze darauf den Punkt O
(
0
;
0
)
{\displaystyle (0\quad ;\quad 0)}
Zeichne um O den Einheitskreis c1 mit Radius
r
{\displaystyle r}
1 sind
P
0
=
(
+
r
;
0
)
{\displaystyle P_{0}=(+r\quad ;\quad 0)}
Q
=
(
−
r
;
0
)
{\displaystyle Q=(-r\quad ;\quad 0)}
Konstruiere die vertikale Mittelachse vom Umkreis c1 des entstehenden 17-Ecks, Schnittpunkt mit c1 ist
A
=
(
0
;
r
)
{\displaystyle A=(0\quad ;\quad r)}
Halbiere den Radius OQ in
Q
′
=
(
−
r
2
;
0
)
{\displaystyle Q\prime =\left({\frac {-r}{2}}\quad ;\quad 0\right)}
Ziehe den Kreisbogen c2 mit dem Radius r2 = Q’P0 (
r
2
=
3
2
⋅
r
{\displaystyle r_{2}={\frac {3}{2}}\cdot r}
Errichte eine Senkrechte auf dem Radius OQ ab Q’, Schnittpunkt mit c2 ist
M
0
=
(
−
r
2
;
−
1
,
5
r
)
{\displaystyle M_{0}=\left({\frac {-r}{2}}\quad ;\quad -1{,}5r\right)}
Ziehe den Carlyle-Kreisbogen Cc1 um M0 durch A (mit
r
C
c
1
=
1
2
26
⋅
r
{\displaystyle r_{Cc1}={\frac {1}{2}}{\sqrt {26}}\cdot r}
1 zweimal trifft, Schnittpunkte sind
H
0
,
2
=
(
1
2
(
−
1
+
17
)
⋅
r
;
0
)
{\displaystyle H_{0,2}=\left({\frac {1}{2}}\left(-1+{\sqrt {17}}\right)\cdot r\quad ;\quad 0\right)}
H
1
,
2
=
(
1
2
(
−
1
−
17
)
⋅
r
;
0
)
{\displaystyle H_{1,2}=\left({\frac {1}{2}}\left(-1-{\sqrt {17}}\right)\cdot r\quad ;\quad 0\right)}
Halbiere die Strecke OH0,2 in
M
0
,
2
=
(
1
4
(
−
1
+
17
)
⋅
r
;
0
)
{\displaystyle M_{0,2}=\left({\frac {1}{4}}\left(-1+{\sqrt {17}}\right)\cdot r\quad ;\quad 0\right)}
Halbiere die Strecke OH1,2 in
M
1
,
2
=
(
1
4
(
−
1
−
17
)
⋅
r
;
0
)
{\displaystyle M_{1,2}=\left({\frac {1}{4}}\left(-1-{\sqrt {17}}\right)\cdot r\quad ;\quad 0\right)}
Ziehe den Carlyle-Kreisbogen Cc2 um M1,2 ab A bis auf die horizontale Mittelachse, Schnittpunkt ist
H
1
,
4
=
(
r
4
(
34
+
2
17
−
17
−
1
)
;
0
)
{\displaystyle H_{1,4}=\left({\frac {r}{4}}\left({\sqrt {34+2{\sqrt {17}}}}-{\sqrt {17}}-1\right)\quad ;\quad 0\right)}
Ziehe den Carlyle-Kreisbogen Cc3 um M0,2 ab A bis auf die horizontale Mittelachse, Schnittpunkt ist
H
0
,
4
=
(
r
4
(
34
−
2
17
+
17
−
1
)
;
0
)
{\displaystyle H_{0,4}=\left({\frac {r}{4}}\left({\sqrt {34-2{\sqrt {17}}}}+{\sqrt {17}}-1\right)\quad ;\quad 0\right)}
Trage OH1,4 von Punkt A aus auf der Geraden OA ab. Du erhältst Punkt
Y
=
(
0
;
r
4
(
34
+
2
17
−
17
+
3
)
)
{\displaystyle Y=\left(0\quad ;\quad {\frac {r}{4}}\left({\sqrt {34+2{\sqrt {17}}}}-{\sqrt {17}}+3\right)\right)}
Verbinde Y mit H0,4 .
Halbiere die Strecke H0,4 Y in
M
0
,
4
=
(
r
1
8
(
34
−
2
17
+
17
−
1
)
;
r
1
8
(
34
+
2
17
−
17
+
3
)
)
{\displaystyle M_{0,4}=\left({\frac {r_{1}}{8}}\left({\sqrt {34-2{\sqrt {17}}}}+{\sqrt {17}}-1\right)\quad ;\quad {\frac {r_{1}}{8}}\left({\sqrt {34+2{\sqrt {17}}}}-{\sqrt {17}}+3\right)\right)}
Ziehe den Carlyle-Kreisbogen Cc4 um M0,4 ab A bis auf die horizontale Mittelachse, Schnittpunkt ist H0,8 .
Ziehe den Kreisbogen c3 mit dem Radius OP0 um H0,8 , Schnittpunkte mit dem Umkreis c1 sind die Eckpunkte P1 und P16 , somit ist die Strecke P0 P1 die erste Seite des gesuchten 17-Ecks.
Ein vierzehnmaliges Abtragen der Strecke P0 P1 auf dem Umkreis c1 , ab dem Eckpunkt P1 gegen dem Uhrzeigersinn, ergibt der Reihe nach die Eckpunkte P2 bis P16 .
Verbinde die so gefundenen Punkte P1 , P2 , …, P16 , P0 , dann ist das 17-Eck vollständig gezeichnet.
Ausgangspunkt: die Punkte
H
0
,
2
=
(
1
2
(
−
1
+
17
)
⋅
r
;
0
)
{\displaystyle H_{0,2}=\left({\frac {1}{2}}\left(-1+{\sqrt {17}}\right)\cdot r\quad ;\quad 0\right)}
H
1
,
2
=
(
1
2
(
−
1
−
17
)
⋅
r
;
0
)
{\displaystyle H_{1,2}=\left({\frac {1}{2}}\left(-1-{\sqrt {17}}\right)\cdot r\quad ;\quad 0\right)}
H
1
,
4
=
(
r
4
(
34
+
2
17
−
17
−
1
)
;
0
)
{\displaystyle H_{1,4}=\left({\frac {r}{4}}\left({\sqrt {34+2{\sqrt {17}}}}-{\sqrt {17}}-1\right)\quad ;\quad 0\right)}
H
0
,
4
=
(
r
4
(
34
−
2
17
+
17
−
1
)
;
0
)
{\displaystyle H_{0,4}=\left({\frac {r}{4}}\left({\sqrt {34-2{\sqrt {17}}}}+{\sqrt {17}}-1\right)\quad ;\quad 0\right)}
Substitution:
17
=
z
{\displaystyle {\sqrt {17}}=z}
17
=
z
2
{\displaystyle 17=z^{2}}
H
1
,
4
=
(
r
4
(
2
z
2
+
2
z
−
z
−
1
)
;
0
)
{\displaystyle H_{1,4}=\left({\frac {r}{4}}\left({\sqrt {2z^{2}+2z}}-z-1\right)\quad ;\quad 0\right)}
H
0
,
4
=
(
r
4
(
2
z
2
−
2
z
+
z
−
1
)
;
0
)
{\displaystyle H_{0,4}=\left({\frac {r}{4}}\left({\sqrt {2z^{2}-2z}}+z-1\right)\quad ;\quad 0\right)}
Y
=
(
0
;
r
4
(
2
z
2
+
2
z
−
z
+
3
)
)
{\displaystyle Y=\left(0\quad ;\quad {\frac {r}{4}}\left({\sqrt {2z^{2}+2z}}-z+3\right)\right)}
M
0
,
4
=
(
r
8
(
2
(
z
2
−
z
)
+
z
−
1
)
;
r
8
(
2
(
z
2
+
z
)
−
z
+
3
)
)
{\displaystyle M_{0,4}=\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\quad ;\quad {\frac {r}{8}}\left({\sqrt {2(z^{2}+z)}}-z+3\right)\right)}
Mit den Substitutionen:
r
8
(
2
(
z
2
−
z
)
+
z
−
1
)
=
s
{\displaystyle {\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)=s}
und
r
8
(
2
(
z
2
+
z
)
−
z
+
3
)
=
t
{\displaystyle {\frac {r}{8}}\left({\sqrt {2(z^{2}+z)}}-z+3\right)=t}
gilt:
M
0
,
4
=
(
s
;
t
)
{\displaystyle M_{0,4}=(s\quad ;\quad t)}
Radius von
C
C
4
{\displaystyle C_{C4}}
M
0
,
4
A
¯
{\displaystyle {\overline {M_{0,4}A}}}
M
0
,
4
A
¯
=
(
x
[
M
0
,
4
]
−
x
[
A
]
)
2
+
(
y
[
M
0
,
4
]
−
y
[
A
]
)
2
{\displaystyle {\overline {M_{0,4}A}}={\sqrt {(x[M_{0,4}]-x[A])^{2}+(y[M_{0,4}]-y[A])^{2}}}}
M
0
,
4
A
¯
=
(
s
−
0
)
2
+
(
t
−
r
)
2
{\displaystyle {\overline {M_{0,4}A}}={\sqrt {(s-0)^{2}+(t-r)^{2}}}}
M
0
,
4
A
¯
=
r
c
c
4
=
s
2
+
t
2
−
2
t
r
+
r
2
{\displaystyle {\overline {M_{0,4}A}}=r_{cc4}={\sqrt {s^{2}+t^{2}-2tr+r^{2}}}}
Berechnen von
H
0
,
8
{\displaystyle H_{0,8}}
y
[
H
0
,
8
]
=
0
{\displaystyle y[H_{0,8}]=0}
x
[
H
0
,
8
]
=
x
[
M
0
,
4
]
+
r
c
c
4
2
−
y
[
M
0
,
4
]
2
{\displaystyle x[H_{0,8}]=x[M_{0,4}]+{\sqrt {r_{cc4}^{2}-y[M_{0,4}]^{2}}}}
x
[
H
0
,
8
]
=
s
+
(
s
2
+
t
2
−
2
t
r
+
r
2
)
−
t
2
{\displaystyle x[H_{0,8}]=s+{\sqrt {(s^{2}+t^{2}-2tr+r^{2})-t^{2}}}}
x
[
H
0
,
8
]
=
s
+
s
2
+
r
2
−
2
t
r
{\displaystyle x[H_{0,8}]=s+{\sqrt {s^{2}+r^{2}-2tr}}}
Berechnen von
P
1
{\displaystyle P_{1}}
x
[
P
1
]
=
1
2
x
[
H
0
,
8
]
{\displaystyle x[P_{1}]={\frac {1}{2}}x[H_{0,8}]}
x
[
P
1
]
=
1
2
(
s
+
s
2
+
r
2
−
2
t
r
)
{\displaystyle x[P_{1}]={\frac {1}{2}}\left(s+{\sqrt {s^{2}+r^{2}-2tr}}\right)}
y
[
P
1
]
=
r
2
−
x
[
P
1
]
2
{\displaystyle y[P_{1}]={\sqrt {r^{2}-x[P_{1}]^{2}}}}
Test
cos
360
∘
17
=
1
16
(
−
1
+
17
+
2
(
17
−
17
)
+
2
17
+
3
17
−
2
(
17
−
17
)
−
2
2
(
17
+
17
)
)
{\displaystyle \cos {\frac {360^{\circ }}{17}}={\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {2\left(17-{\sqrt {17}}\right)}}+2{\sqrt {17+3{\sqrt {17}}-{\sqrt {2\left(17-{\sqrt {17}}\right)}}-2{\sqrt {2\left(17+{\sqrt {17}}\right)}}}}\right)}
cos
360
∘
17
=
1
16
(
−
1
+
z
+
2
(
z
2
−
z
)
+
2
z
2
+
3
z
−
2
2
(
z
2
+
z
)
−
2
(
z
2
−
z
)
)
{\displaystyle \cos {\frac {360^{\circ }}{17}}={\frac {1}{16}}\left(-1+z+{\sqrt {2\left(z^{2}-z\right)}}+2{\sqrt {z^{2}+3z-2{\sqrt {2\left(z^{2}+z\right)}}-{\sqrt {2\left(z^{2}-z\right)}}}}\right)}
x
[
P
1
]
=
cos
μ
{\displaystyle x[P_{1}]=\cos {\mu }}
x
[
P
1
]
=
1
2
(
r
8
(
2
(
z
2
−
z
)
+
z
−
1
)
)
+
(
r
8
(
2
(
z
2
−
z
)
+
z
−
1
)
)
2
+
r
2
−
2
r
r
8
(
2
(
z
2
+
z
)
−
z
+
3
)
)
{\displaystyle x[P_{1}]={\frac {1}{2}}\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\right)+{\sqrt {\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\right)^{2}+r^{2}-2r{\frac {r}{8}}\left({\sqrt {2(z^{2}+z)}}-z+3\right)}})}
2
⋅
x
[
P
1
]
=
(
r
8
(
2
(
z
2
−
z
)
+
z
−
1
)
)
+
(
r
8
)
2
(
2
(
z
2
−
z
)
+
z
−
1
)
2
+
8
2
(
r
8
)
2
−
16
(
r
8
)
2
(
2
(
z
2
+
z
)
−
z
+
3
)
{\displaystyle 2\cdot x[P_{1}]=\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\right)+{\sqrt {\left({\frac {r}{8}}\right)^{2}\left({\sqrt {2(z^{2}-z)}}+z-1\right)^{2}+8^{2}\left({\frac {r}{8}}\right)^{2}-16\left({\frac {r}{8}}\right)^{2}\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}}
2
⋅
x
[
P
1
]
=
(
r
8
(
2
(
z
2
−
z
)
+
z
−
1
)
)
+
r
8
(
2
(
z
2
−
z
)
+
z
−
1
)
2
+
8
2
−
16
(
2
(
z
2
+
z
)
−
z
+
3
)
{\displaystyle 2\cdot x[P_{1}]=\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\right)+{\frac {r}{8}}{\sqrt {\left({\sqrt {2(z^{2}-z)}}+z-1\right)^{2}+8^{2}-16\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}}
x
[
P
1
]
=
r
16
(
(
2
(
z
2
−
z
)
+
z
−
1
)
+
(
2
(
z
2
−
z
)
+
z
−
1
)
2
+
8
2
−
16
(
2
(
z
2
+
z
)
−
z
+
3
)
)
{\displaystyle x[P_{1}]={\frac {r}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\sqrt {\left({\sqrt {2(z^{2}-z)}}+z-1\right)^{2}+8^{2}-16\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}\right)}
x
[
P
1
]
r
=
cos
μ
=
1
16
(
(
2
(
z
2
−
z
)
+
z
−
1
)
+
(
2
(
z
2
−
z
)
+
z
−
1
)
2
+
8
2
−
16
(
2
(
z
2
+
z
)
−
z
+
3
)
)
{\displaystyle {\frac {x[P_{1}]}{r}}=\cos \mu ={\frac {1}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\sqrt {{\color {red}\left({\sqrt {2(z^{2}-z)}}+z-1\right)^{2}}+8^{2}-16\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}\right)}
cos
μ
=
1
16
(
(
2
(
z
2
−
z
)
+
z
−
1
)
+
2
(
z
2
−
z
)
+
z
2
+
1
+
2
z
2
(
z
2
−
z
)
−
2
2
(
z
2
−
z
)
−
2
z
+
8
2
−
16
(
2
(
z
2
+
z
)
−
z
+
3
)
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\sqrt {{\color {red}2(z^{2}-z)+z^{2}+1+2z{\sqrt {2(z^{2}-z)}}-2{\sqrt {2(z^{2}-z)}}-2z}+8^{2}-{\color {blue}16\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}}\right)}
cos
μ
=
1
16
(
(
2
(
z
2
−
z
)
+
z
−
1
)
+
2
z
2
−
4
z
+
z
2
+
1
+
2
z
2
(
z
2
−
z
)
−
2
2
(
z
2
−
z
)
+
8
2
−
16
2
(
z
2
+
z
)
+
16
z
−
48
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\sqrt {{\color {red}2z^{2}-4z+z^{2}+1+2z{\sqrt {2(z^{2}-z)}}-2{\sqrt {2(z^{2}-z)}}}+8^{2}-{\color {blue}16{\sqrt {2(z^{2}+z)}}+16z-48}}}\right)}
1
+
8
2
−
48
=
17
=
z
2
{\displaystyle 1+8^{2}-48=17=z^{2}}
cos
μ
=
1
16
(
(
2
(
z
2
−
z
)
+
z
−
1
)
+
4
z
2
−
4
z
+
2
z
2
(
z
2
−
z
)
−
2
2
(
z
2
−
z
)
−
16
2
(
z
2
+
z
)
+
16
z
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\sqrt {{\color {red}4z^{2}-4z+2z{\sqrt {{\color {green}2}(z^{2}-z)}}-2{\sqrt {{\color {green}2}(z^{2}-z)}}}-{\color {blue}16{\sqrt {2(z^{2}+z)}}+16z}}}\right)}
cos
μ
=
1
16
(
(
2
(
z
2
−
z
)
+
z
−
1
)
+
4
z
2
+
4
⋅
3
z
+
2
z
4
2
(
z
2
−
z
)
−
2
4
2
(
z
2
−
z
)
−
4
⋅
4
2
(
z
2
+
z
)
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\sqrt {{\color {red}4z^{2}+4\cdot 3z+2z{\sqrt {{\color {green}{\frac {4}{2}}}(z^{2}-z)}}-2{\sqrt {{\color {green}{\frac {4}{2}}}(z^{2}-z)}}}-{\color {blue}4\cdot 4{\sqrt {2(z^{2}+z)}}}}}\right)}
cos
μ
=
1
16
(
(
2
(
z
2
−
z
)
+
z
−
1
)
+
2
z
2
+
3
z
+
z
1
2
(
z
2
−
z
)
−
1
2
(
z
2
−
z
)
−
4
2
(
z
2
+
z
)
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\color {green}2}{\sqrt {{\color {red}z^{2}+3z+z{\sqrt {{\frac {1}{2}}(z^{2}-z)}}-{\sqrt {{\frac {1}{2}}(z^{2}-z)}}}-{\color {blue}4{\sqrt {2(z^{2}+z)}}}}}\right)}
Rücksubstitution
z
=
17
{\displaystyle z={\sqrt {17}}}
cos
μ
=
1
16
(
−
1
+
17
+
2
(
17
−
17
)
+
2
(
17
−
17
)
+
17
+
1
+
2
2
(
17
−
17
)
17
−
2
2
(
17
−
17
)
−
2
17
+
8
2
−
16
(
2
17
+
17
−
17
+
3
)
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {2(17-{\sqrt {17}})}}{\color {cyan}+}{\sqrt {{\color {red}2(17-{\sqrt {17}})}+{\color {green}17}+{\color {blue}1}+{\color {mulberry}2{\sqrt {2(17-{\sqrt {17}})}}{\sqrt {17}}}-{\color {dandelion}2{\sqrt {2(17-{\sqrt {17}})}}}-{\color {purple}2{\sqrt {17}}}+8^{2}-16\left(2{\sqrt {17+{\sqrt {17}}}}-{\sqrt {17}}+3\right)}}\right)}
cos
μ
=
1
16
(
−
1
+
17
+
2
(
17
−
17
)
+
4
(
17
−
17
)
+
4
17
−
17
⋅
17
−
4
17
−
17
−
4
⋅
17
2
−
16
(
2
17
+
17
−
17
+
3
)
+
4
⋅
∗
41
2
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {2(17-{\sqrt {17}})}}+{\sqrt {4(17-{\sqrt {17}})+4{\sqrt {17-{\sqrt {17}}}}\cdot {\sqrt {17}}-4{\sqrt {17-{\sqrt {17}}}}-4\cdot {\frac {\sqrt {17}}{2}}-16\left(2{\sqrt {17+{\sqrt {17}}}}-{\sqrt {17}}+3\right)+4\cdot *{\frac {41}{2}}}}\right)}
cos
μ
=
1
16
(
−
1
+
17
+
2
(
17
−
17
)
+
2
(
17
−
17
)
+
17
−
17
⋅
17
−
17
−
17
−
17
2
−
4
(
2
17
+
17
−
17
+
3
)
+
41
2
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {2(17-{\sqrt {17}})}}+2{\sqrt {(17-{\sqrt {17}})+{\sqrt {17-{\sqrt {17}}}}\cdot {\sqrt {17}}-{\sqrt {17-{\sqrt {17}}}}-{\frac {\sqrt {17}}{2}}-4\left(2{\sqrt {17+{\sqrt {17}}}}-{\sqrt {17}}+3\right)+{\frac {41}{2}}}}\right)}
cos
μ
=
1
16
(
−
1
+
17
+
2
(
17
−
17
)
+
2
17
−
17
+
17
⋅
(
17
−
17
)
−
17
−
17
−
17
2
−
8
17
+
17
+
4
17
−
12
+
41
2
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {2(17-{\sqrt {17}})}}+2{\sqrt {17-{\sqrt {17}}+{\sqrt {17\cdot (17-{\sqrt {17}})}}-{\sqrt {17-{\sqrt {17}}}}-{\frac {\sqrt {17}}{2}}-8{\sqrt {17+{\sqrt {17}}}}+4{\sqrt {17}}-12+{\frac {41}{2}}}}\right)}
cos
μ
=
1
16
(
−
1
+
17
+
2
(
17
−
17
)
+
2
17
+
3
17
−
2
2
(
17
+
17
)
+
17
⋅
(
17
−
17
)
−
17
−
17
−
17
2
+
17
2
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {2(17-{\sqrt {17}})}}+2{\sqrt {17+3{\sqrt {17}}-2{\sqrt {2(17+{\sqrt {17}})}}+{\sqrt {17\cdot (17-{\sqrt {17}})}}-{\sqrt {17-{\sqrt {17}}}}-{\frac {\sqrt {17}}{2}}+{\frac {17}{2}}}}\right)}
cos
μ
=
1
16
(
−
1
+
17
+
2
(
17
−
17
)
+
2
17
+
3
17
−
2
2
(
17
+
17
)
+
17
17
−
17
−
17
−
17
+
17
−
17
2
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {2(17-{\sqrt {17}})}}+2{\sqrt {17+3{\sqrt {17}}-2{\sqrt {2(17+{\sqrt {17}})}}+{\sqrt {17}}{\sqrt {17-{\sqrt {17}}}}-{\sqrt {17-{\sqrt {17}}}}+{\frac {17-{\sqrt {17}}}{2}}}}\right)}
cos
μ
=
1
16
(
−
1
+
17
+
2
(
17
−
17
)
+
2
17
+
3
17
−
2
2
(
17
+
17
)
+
17
17
−
17
−
17
−
17
+
1
2
17
−
17
⋅
17
−
17
)
{\displaystyle \cos \mu ={\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {2(17-{\sqrt {17}})}}+2{\sqrt {17+3{\sqrt {17}}-2{\sqrt {2(17+{\sqrt {17}})}}+{\sqrt {17}}{\sqrt {17-{\sqrt {17}}}}-{\sqrt {17-{\sqrt {17}}}}+{\frac {1}{2}}{\sqrt {17-{\sqrt {17}}}}\cdot {\sqrt {17-{\sqrt {17}}}}}}\right)}
![{\displaystyle {\overline {M_{0,4}A}}={\sqrt {(x[M_{0,4}]-x[A])^{2}+(y[M_{0,4}]-y[A])^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6d198a39db86d182dc17137d884a9b4dac6cd5a)
![{\displaystyle y[H_{0,8}]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78b575c08558e0432c6bbaec7444aeb725c0b279)
![{\displaystyle x[H_{0,8}]=x[M_{0,4}]+{\sqrt {r_{cc4}^{2}-y[M_{0,4}]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f08b603002417ed403a38fbcfabb033afd76b9cd)
![{\displaystyle x[H_{0,8}]=s+{\sqrt {(s^{2}+t^{2}-2tr+r^{2})-t^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fa8a6bfe3c336b2c908d8c81b92c2d65eabaa35)
![{\displaystyle x[H_{0,8}]=s+{\sqrt {s^{2}+r^{2}-2tr}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e47a9e2332ee77ca1b2ad2d56b397734cd2de018)
![{\displaystyle x[P_{1}]={\frac {1}{2}}x[H_{0,8}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf9e82cdb6aa2f2e05ef6d13d23072ccb05ce603)
![{\displaystyle x[P_{1}]={\frac {1}{2}}\left(s+{\sqrt {s^{2}+r^{2}-2tr}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/152129dd56420245e193d462889d3b5b11375783)
![{\displaystyle y[P_{1}]={\sqrt {r^{2}-x[P_{1}]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b9675ff5b86163351cc47bc6c4bc8c259a79179)
![{\displaystyle x[P_{1}]=\cos {\mu }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45180c5a150840822de7e2170935dd1f82c79d21)
![{\displaystyle x[P_{1}]={\frac {1}{2}}\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\right)+{\sqrt {\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\right)^{2}+r^{2}-2r{\frac {r}{8}}\left({\sqrt {2(z^{2}+z)}}-z+3\right)}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e611e18fc7e01b44fe28562c8db6366fb7ef35ea)
![{\displaystyle 2\cdot x[P_{1}]=\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\right)+{\sqrt {\left({\frac {r}{8}}\right)^{2}\left({\sqrt {2(z^{2}-z)}}+z-1\right)^{2}+8^{2}\left({\frac {r}{8}}\right)^{2}-16\left({\frac {r}{8}}\right)^{2}\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef47c4503963128477a758e276bbd54fdfe99b83)
![{\displaystyle 2\cdot x[P_{1}]=\left({\frac {r}{8}}\left({\sqrt {2(z^{2}-z)}}+z-1\right)\right)+{\frac {r}{8}}{\sqrt {\left({\sqrt {2(z^{2}-z)}}+z-1\right)^{2}+8^{2}-16\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a2f811e0c1289b4810454784f2248dd9518cb06)
![{\displaystyle x[P_{1}]={\frac {r}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\sqrt {\left({\sqrt {2(z^{2}-z)}}+z-1\right)^{2}+8^{2}-16\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5475672125e2b3ed8d35f92de95d5626b5d5a620)
![{\displaystyle {\frac {x[P_{1}]}{r}}=\cos \mu ={\frac {1}{16}}\left(\left({\sqrt {2(z^{2}-z)}}+z-1\right)+{\sqrt {{\color {red}\left({\sqrt {2(z^{2}-z)}}+z-1\right)^{2}}+8^{2}-16\left({\sqrt {2(z^{2}+z)}}-z+3\right)}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49af5494ba1fca0a40c4aa9af2bc372b727b2d06)
