Let ABCDEFGH be a unit cube where A is directly above E, B is directly above F, C is directly above G, and D is directly above H. Let X, Y, and Z be on AG, BH, and CE, respectively such that XG/XA=3/2, YH/YB=2, and ZE/ZC=3. Let O denote the center of the cube. Then find the surface area of tetrahedron OXYZ.
X,Y,Z are on main diagonals.
Each diagonal length is √3, and the 3 diagonals are all orthogonal to each other, just as 3 cartesian axes are. Point O is where all 3 diagonals meet.
So if we place the x,y,z axes appropriately, the tetrahedron has point O on the origin, and each of X,Y,Z on one of the axes.
(.5*√3,0,0), (0,.5*√3,0), (0,0,.5*√3) will be 3 of the vertices of the cube, specifically A,B,C.
Let L = Length of the diagonals, L = √3
XG/XA=3/2 means OX is 1/10 L.
YH/YB=2 means OY is 1/6 L.
ZE/ZC=3 means OZ is 1/4 L.
The 4 tetrahedron vertices are (in order O A B C):
(0,0,0), (L/10,0,0), (0,L/6,0), (0,0,L/4)
The edge lengths are (L = √3)
OA L/10
OB L/6
OC L/4
AB L√(1/100 + 1/36)
AC L√(1/100 + 1/16)
BC L√(1/36 + 1/16)
The surface area is the sum of the areas of the 4 triangles.
OAB (OA,OB,AB)
OAC (OA,OC,AC)
OBC (OB,OC,BC)
ABC (AB,AC,BC)
The four areas, using Heron's formula, in order:
[0.024999999999999988, 0.037499999999999985, 0.0625, 0.07705517503711223]
The sum; surface area of tetrahedron: 0.2020551750371122
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I initially misread the problem and figured the volume, so I'll include that:
Consider the base to be XOY; this has area (1/2)(1/10)(1/6)L^2 = L^2/120
and then the height is L/4.
V = (1/3) * base * height = (1/3) * L^2/120 * L/4
Volume = L^3/1440 = 3√3/1440
= approx 0.0036084391824351613
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def areaHeron(a,b,c):
""" input 3 float numbers, the 3 sides of a triangle
output the area """
if c >= a+b or b >= a+c or a >= b+c:
return 0
s = (a+b+c)/2 # semiperimeter
return (s*(s-a)*(s-b)*(s-c))**(1/2)
OA = 1/10
OB = 1/6
OC = 1/4
AB = (1/100 + 1/36) ** .5
AC = (1/100 + 1/16) ** .5
BC = (1/36 + 1/16) ** .5
# (each length needs to be multiplied by L = √3)
# (each area needs to be multiplied by L^2 = 3)
triareas = []
triareas.append(3*areaHeron(OA,OB,AB))
triareas.append(3*areaHeron(OA,OC,AC))
triareas.append(3*areaHeron(OB,OC,BC))
triareas.append(3*areaHeron(AB,AC,BC))
print(triareas)
print(sum(triareas))
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Posted by Larry
on 2024-12-26 14:14:37 |
Solution
