■等面単体の体積(その257)

 (その253)をやり直し.

  x^2+y^2+z^2=4

  (x−3/2)^2+(y−(√5)/2)^2+(z−(√10)/2)^2=4

  (x−1)^2+(y−√5)^2+z^2=6

[2]x=−t/2+1,y=(√5)t/2+√5,z=(√10)t/2

   t^2/4+5t^2/4+10t^2/4=6,t^2=3/2

   (t−2)^2/4+5(t+2)^2/4+10t^2/4=4

   (t+1)^2/4+5(t+1)^2/4+10t^2/4=6

   (t−2)^2+5(t+2)^2+10t^2=16

   (t+1)^2+5(t+1)^2+10t^2=24

   6t^2+10t+24=16

   6t^2+12t+6=24・・・(NG)

===================================