\textbf{Zadání:}\ Určete\ vzdálenost\ bodu\ A\ pravidelného\ čtyřbokého\ jehlanu\ ABCDV\
od\ přímky\ CV,\ je-li\ |AB|=a, |AV|=b.

A

C

S

V

Potřebujeme\ znát \ |SV|\ (použijeme\ rovnost\ obsahů)

A

S

V

|AS|=\frac\sqrt22a\ (polovina\ úhlopříčky\ podstavy)\
{|AV|}^2-|AS|^2=|SV|^2\
b^2-\ \frac12a^2=|SV|^2\
\sqrt{b^2-\ \frac12a^2}=|SV|

Použijeme\ rovnost\ obsahů:\
\frac{|AC|\cdot|SV|}{2}=\frac{|CV|\cdot|PA|}{2}\ |\cdot 2 \
\sqrt2a\cdot\sqrt{b^2-\ \frac12a^2}=b^2\cdot|PA|\
\frac{a\cdot\sqrt{2\left(b^2-\ \frac12a^2\right)}}{b^2}=|PA|\
\frac{a\cdot\sqrt{2b^2-a^2}}{b^2}=|PA|

krok

zpět

V\ \Delta ACV\ hledáme\ výšku\ na\ CV.

\ \ \ \ \ \ \ \ \ \ \ \ \ b^2>\frac12a^2\
\left(b>\frac\sqrt22a\right)∨\left(a<\sqrt2b\right)

podmínky

vynuluj

udělej

velikost jehlanu

P

a = 2.1 (slider from 0.1 to 9, step 0.1)
b = 3 (slider from 0.1 to 9, step 0.1)
C = (-2.3763439363408176, 0.014629494757647521, 0)
D = PointIn(Sphere[C, a])
g = Segment(C, D)
c = Circle(C, a, xOyPlane)
f = Tangent(D, c)
d = Circle(D, a, xOyPlane)
A, B_1 = Intersect(f, d)
h = Tangent(A, d)
B, D_1 = Intersect(h, c)
mnohoúhelník1, a_1, f_1, e, c_1 = Polygon(A, D, C, B)
i, V, stena1, stena2, stena3, stena4, hrana1, hrana2, hrana3, hrana4 = Pyramid(mnohoúhelník1, (-sqrt(b^(2) - (1 / 2 * a^(2)))))
j = Segment(D, B)
F = (-3.84, -1.16)
G = Point(Circle[F, (sqrt(2) * a)])
k = Segment(F, G)
l = LineBisector(k)
H = Intersect(l, k)
p = Circle(H, sqrt(b^(2) - (1 / 2 * a^(2))))
I = Intersect(p, l, 2)
mnohoúhelník3, i_1, f_2, g_1 = Polygon(F, G, I)
m = OrthogonalLine(F, f_2)
P = Intersect(m, f_2)
n = Segment(F, P)
q = Segment(H, I)
E = (0.8654545454545466, -1.4109090909090862)
K = (2.7654545454545456, -1.3909090909090867)
E' = Rotate(E, (-(90°)), K)
α = Angle(E', K, E)
r = Ray(K, E')
L = Point(r)
mnohoúhelník4, l_1, e_1, k_1 = Polygon(E, K, L)
o: 0
β = Angle(F, P, G)
u: false
v = false
mnohoúhelník2, c_2, a_2, v_1 = Polygon(A, V, C)
J = Midpoint(v_1)
s = Segment(B, D)
t = Circle(J, (sqrt(2) / 2 * a), s)
M, N = Intersect(a_2, t)
h_1 = Segment(N, A)
j_1 = Segment(V, J)