0018132人目の素数さん
2019/01/26(土) 06:12:08.46ID:9LgO+YCH2)
BP1 = p√2 とおくと
AP1^2 = (2+p)^2 + p^2 = 2 + 2(1+p)^2,
ところで、 1) より
AP1 = CP1 = (6+2√2 -2 -2)/2 = 1+√2,
これらより
p = √(1/2 + √2) - 1 = 0.38355107
DP1 = DO + OP1 = √2 + (1+p)√2 = √2 + √(1+2√2) = 3.37085025
3)
AP1 = CP1 = 1+√2 = a,
OP1 = (1+p)√2 = 1.95663668743 = b,
とおくと
A(-√2, 0) B(0, √2) C(√2, 0) D(0, -√2)
P1(0, b) P2(0, -b) Q1(a, 0) Q2(-a, 0)
Pの軌跡は楕円
(x/a)^2 + (y/b)^2 = 1,