Cuerpos platónicos truncados (página 2)

Partes: 1, 2

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5 3 b cos5.t=16.cos5t-20.cos3t+5.cost ^
sen5.t=16.sen5t-20.sen3t+5.sent ? t=72°, cost=x ^ sent=y
cos360°=16.x5-20.x3+5.x=1 ^ sen360°=16.y5-20.y3+5.y=0
16.x5-20.x3+5.x-1=0 ^ 16.y4-20.y2+5=0 ? 16.y4-20.y2+5=0 ?
y1-2-3-4=±v[20±v(400-320)]=±v(20±v80)=±v(20±4.v5)=±v(10±2.v5)=
32 32 32 16 ±v(10±2.v5) ? sen72°=v(10+2.v5) 4 4
(16.x -20.x +5.x-1):(x-1)=16.x4+16.x3-4.x2-4.x+1 ?
16.x4+16.x3-4.x2-4.x+1=0
16.x4+16.x3+4.x2-8.x2-4.x+1=(4.x2)2+2.4.x2.2.x+(2.x)2-2.(4.x2+2.x)+1=(4.x2+2.x)2-2.(4.x2+2.x).1+12=
[(4.x2+2.x)-1]2=0 ? 4.x2+2.x-1=0 ?
x1-2-3-4=-2±v(4+16)=-2±v20=-2±2.v5=-1±v5
? 8 8 8 4 cos72°=-1+v5 ? 4
h=b.v(10+2.v5).4=b.(-2-2.v5).v(10+2.v5)=b.v[(2+2.v5)2.(10+2.v5)]=b.v[(4+8.v5+20).(10+2.v5)]=
2.(-1+v5).4 4-20 16 16
b.v[(24+8.v5).(10+2.v5)]=b.v(240+48.v5+80.v5+80)=b.v(320+128.v5)=b.v[64.(5+2.v5)]=
16 16 16 16 =b.8.v(5+2.v5)=b.v(5+2.v5) 16 2
Atriángulo=b.b.v(5+2.v5)= b2.v(5+2.v5) 2.2 4
Adecágono=10. b2.v(5+2.v5)=5. b2.v(5+2.v5) 4 2
sen60°=h ? h=b.sen60° ? h=v3.b h b
Atriángulo=b.v3.b 4 2 60° Atriángulo=v3.b2 4
Adodecaedro truncado=12.5.b2.v(5+2.v5)+20.v3.b2 2 4 Adodecaedro
truncado=5.b2.[6.v(5+2.v5)+v3] Fórmula para el volumen Se
calcula el volumen de la pirámide. ar h y b y 9 30° b
2

b cos30°= b ? y= ? y= 2.b ? y=v3.b 2.y 2.cos30° 2.v3 3
ar=b.(-1+v5) ? ar2=h2+y2 ? h2=ar2-y2 ?
h2=[b.(-1+v5)]2-(v3.b)2=b2.(-1+v5)2-b2=b2.(1-2.v5+5)-b2= 2 2 3 4
3 4 3 b2.(6-2.v5)-b2=b2.(3-v5)-b2=b2.(7-v5)=b2.(7-6.v5) ?
h=b.v6.v(7-6.v5) 4 3 2 3 6 6 6
Vpirámide=1.v3.b2.b.v6.v(7-6.v5). ?
Vpirámide=b3.v2.v(7-6.v5) 3 4 6 24 Vdodecaedro
truncado=Vdodecaedro-20.Vpirámide ? Vdodecaedro
truncado=5.a3.v(47+21.v5)-20.b3.v2.v(7-6.v5) ? 2 24 a=b.v5
Vdodecaedro
truncado=5.(b.v5)3.v(47+21.v5)-5.b3.v2.v(7-6.v5)=25.b3.v5.v(47+21.v5)-5.b3.v2.v(7-6.v5)=
2 6 2 6 Vdodecaedro
truncado=5.b3.[15.v5.v(47+21.v5)-v2.v(7-6.v5)] 6 Icosaedro
truncado Construcción Se construyen 20 hexágonos
regulares y 12 pentágonos regulares que tengan la medida
b. b a a=3.b ? b=a b b b 60° b 3 10

2 Icosaedro a Icosaedro truncado b Fórmula para el
área Se calcula el área de un hexágono
regular y un pentágono regular. sen60°=h ?
h=b.sen60° ? h=v3.b b Atriángulo=b.v3.b=v3.b 2 2.2 4
Ahexágono=6.v3.b2=3.v3.b2 b 4 2 h 60° b tan54°=2.h
? h=b.tan54°=b.sen54° b 2 2.cos54° h 54° b 2
cos5.t=16.cos5t-20.cos3t+5.cost ^ sen5.t=16.sen5t-20.sen3t+5.sent
? t=54°, cost=x ^ sent=y cos270°=16.x5-20.x3+5.x=0 ^
sen270°=16.y5-20.y3+5.y=-1 11

4 2 2 3 3 16.x4-20.x2+5=0 ^ 16.y5-20.y3+5.y+1=0 ?
(16.y5-20.y3+5.y+1):(y+1)=16.y4-16.y3-4.y2+4.y+1 ?
16.y4-16.y3-4.y2+4.y+1=0
16.y4-16.y3+4.y2-8.y2+4.y+1=(4.y2)2-2.4.y2.2.y+(2.y)2-2.(4.y2-2.y)+1=(4.y2-2.y)2-2.(4.y2-2.y).1+12=
[(4.y2-2.y)-1]2=0 ? 4.y2-2.y-1=0 ?
y1-2-3-4=2±v(4+16)=2±v20=2±2.v5=1±v5
? sen54°=1+v5 8 8 8 4 4 16.x -20.x +5=0 ?
x1-2-3-4=±v[20±v(400-320)]=±v(20±v80)=±v(20±4.v5)=±v(10±2.v5)=
32 32 32 16 ±v(10±2.v5) ? cos54°=v(10-2.v5) 4 4
h= b.(1+v5).4
=b.(1+v5).v(10+2.v5)=b.v[(1+v5)2.(10+2.v5)]=b.v5.v[(1+2.v5+5).(10+2.v5)]=
2.v(10-2.v5).4 2.v80 8.v5 40
b.v5.v[(6+2.v5).(10+2.v5)]=b.v5.v(60+12.v5+20.v5+20)=b.v5.v(80+32.v5)=b.v5.v[16.(5+2.v5)]=
40 40 40 40 b.v5.4.v(5+2.v5)= b.v5.v(5+2.v5) 40 10
Atriángulo=b.b.v5.v(5+2.v5)=b2.v5.v(5+2.v5) 2.10 20
Apentágono=5. b2.v5.v(5+2.v5)=b2.v5.v(5+2.v5) 20 4
Aicosaedro truncado=20.3.v3.b2+12. b2.v5.v(5+2.v5) 2 4 Aicosaedro
truncado=3.b2.[10.v3+v5.v(5+2.v5)] Fórmula para el volumen
Se calcula el volumen de la pirámide. ap h x x b 54° b
2 cos54°= b ? x= b ? x= b.4
=b.2.v(10+2.v5)=b.2.v(10+2.v5)=b.v5.v(10+2.v5) 2.x 2.cos54°
2.v(10-2.v5) v80 4.v5 10 ap=b ? ap2=h2+x2 ? h2=ap2-x2 ?
h2=b2-[b.v5.v(10+2.v5)]2=b2-b2.(10+2.v5)=b2-b2.2.(5+v5)=
b2.(1-1-v5)= b2.(5-v5) ? h=b.v10.v(5-v5) 2 10 10 10 10 20 20
Vpirámide=1. b .v5.v(5+2.v5).
b.v10.v(5-v5)=b3.v2.v[(5+2.v5).(5-v5)]=b3.v2.v(25-5.v5+10.v5-10)=
3 4 10 24 24 b .v2.v(15+5.v5)=b .v10.v(3+v5) 24 24 12

Vicosaedro truncado=Vicosaedro-12.Vpiramide ? Vicosaedro
truncado=5.a3.v2.v(7+3.v5)-12. b3.v10.v(3+v5) ? 12 a=3.b ?
Vicosaedro truncado=5.27.b3.v2.v(7+3.v5)-b3.v10.v(3+v5) 24 12 12
Vicosaedro truncado=b3.[135.v2.v(7+3.v5)-v10.v(3+v5)] 12
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