2020-수학-가형-홀수-27¶
문제¶
풀이¶
from sympy import *
x, z = symbols('x z')
a = Point(-4*cos(pi/6),0,0)
b = Point(0,-4*sin(pi/6),0)
c = -a
d = -b
m=(c+b)/2
n=(c+d)/2
p = Point(x,0,z)
s, t = symbols('s t')
q = s*a+(1-s)*m
r = s*a+(1-s)*n
solve((m-a).dot(b-q),s)
[2/7]
q=q.subs(s,Rational(2,7))
r=r.subs(s,Rational(2,7))
[q,r]
[Point3D(sqrt(3)/7, -5/7, 0), Point3D(sqrt(3)/7, 5/7, 0)]
x_sol=solve((p-q).dot(a-m),x)
x_sol
[2*sqrt(3)/9]
eqn = Eq(p.distance(q),b.distance(q)).subs(x,x_sol[0])
z_sol = solve(eqn,z)
z_sol
[-4*sqrt(6)/9, 4*sqrt(6)/9]
area_amn = Triangle(a, m, n).area
area_amn
\[\displaystyle 3 \sqrt{3}\]
p = (x_sol[0],0,z_sol[1])
p
(2*sqrt(3)/9, 0, 4*sqrt(6)/9)
normal_pam = Point(Plane(p,a,m).normal_vector)
normal_pam = normal_pam/(normal_pam.distance(Point(0,0,0)))
normal_pam
\[\displaystyle Point3D\left(- \frac{\sqrt{2}}{9}, - \frac{\sqrt{6}}{3}, \frac{5}{9}\right)\]
area_amn*normal_pam[2]
\[\displaystyle \frac{5 \sqrt{3}}{3}\]