Description

Rui Qin
30874157

Part 1 The weirdest 3d Pythagoras yet
A:
Set u and v which are span the parallelogram
In[]:= u={u1,u2,u3} v={v1,v2,v3}
Plane P
We can calculate the area of P and we can get P^2 which is the right hand side of equation
In[]:= AreaP=Norm[Cross[u,v]]
In[]:= AreaP^2
Out[]= Abs[-u2v1+u1v2]2+Abs[u3v1-u1v3]2+Abs[-u3v2+u2v3]2
Plane O
In[]:= uo={0,u2,u3} vo={0,v2,v3}
Then we use the cross product of uo and vo calculate the area of O, then we can know the area O square
In[]:= AreaO=Norm[Cross[uo,vo]] AreaO^2
Out[]= Abs[-u3v2+u2v3]
Out[]= Abs[-u3v2+u2v3]2
Plane N
In[]:= un={u1,0,u3} vn={v1,0,v3}
Then we use the cross product of un and vn calculate the area of N, then we can know the area N square
In[]:= AreaN=Norm[Cross[un,vn]] AreaN^2
Out[]= Abs[u3v1-u1v3]
Out[]= Abs[u3v1-u1v3]2
Plane M
In[]:= um={u1,u2,0} vm={v1,v2,0}
Then we use the cross product of um and vm calculate the area of M, then we can know the area M square
In[]:= AreaM=Norm[Cross[um,vm]] AreaM^2
Out[]= Abs[-u2v1+u1v2]
Out[]= Abs[-u2v1+u1v2]2
Final result
We can make a equation and and let Mathematica to verify it is True
In[]:= AreaM^2+AreaN^2+AreaO^2⩵AreaP^2
Out[]= True
B:
Set the highest point of tetrahedron is point D, and other point is point A, point B and point C
Then we can get vectors: DA, DB, DC
The bottom triangle side vectors: CA, CB ,AB
Area of ADC, ADB and CDB
We can using the cross product to calculate the area, which is the left hand side
In[]:= TopArea = 1/2*(Norm[Cross[DA,DC]]+Norm[Cross[DB,DC]]+Norm[Cross[DA,DB]])
Out[]= (Norm[DA⨯DB]+Norm[DA⨯DC]+Norm[DB⨯DC])
Area of ABC
We also use cross product of two vectors to calculate the area of bottom
Norm[CA⨯CB]= Norm[(DA-DC)x(DB-DC)] = Norm[(DA-DC)x(DB-DC)]
= Norm[(DAxDB)+ -(DAxDC)+ -(DCxDB)+(DCxDC)]
= Norm[(DAxDB)+ -(DAxDC)+ -(DCxDB)]
= Norm[(DAxDB)]+Norm[(DAxDC)]+Norm[(DCxDB)] = Left hand side
C:
We can firstly get the area of plane, then get the area of bottom plane, and add up
In[]:= upper = 1/2*(2.5*3+3*4+2.5*4) ⩵ 3.75+6+5
Out[]= True
In[]:= upper = 1/2*(2.5*3+3*4+2.5*4)
Out[]= 14.75
In[]:= bottom = (3.75^2+6^2+5^2)^(1/2)
Out[]= 8.66386
In[]:= theSum = upper + bottom
Out[]= 23.4139

Part 2 The Big Cube
Find Aa length and it vector
Let’s tell Mathematica about A and a, b. (a is A’ )
In[263]:= A={43.1162, 29.9541, 7.72383} a={42.3208,27.8349,0} b = {25.7441,19.8898,0}
and we make vector Aa
In[266]:= Aa = a-A
Out[266]= {-0.7954,-2.1192,-7.72383} and we figure out the length of Aa
In[166]:= Aa_length = Norm[Aa]
Out[166]= 8.04868
Find Plane ABCD equation
aA is vertical with plane ABCD, so we can set aA as normal vector of ABCD, and we put the coordinate of A in it to get the equation(A is on the plane)
In[169]:= ABCD = 0.7954x+2.1192y+7.72383z⩵157.431
Find the distance of Bb
We start with finding the distance of Bb
Find point p on the plane setting y and z =0
In[170]:= p = {157.431/0.7954,0,0}
Out[170]= {197.927,0,0}
In[171]:= v = b – p
Out[171]= {-172.183,19.8898,0}
In[172]:= distanceBb = Norm[Projection[v,n]]
Out[172]= 11.7788
Find point B
We set B as {Bx,By,Bz}
In[176]:= B = {Bx,By,Bz}
In[173]:= Bb = {25.7441-Bx,19.8898-By,-Bz}
In[174]:= Norm[Bb]==distanceBb
[25.7441`-Bx]2+[19.8898`-By]2+Abs[Bz]2 ⩵11.778788659188484`
We find AB and AB vertical Bb
In[193]:= AB = B-A
Dot[AB,Bb]⩵0
Out[194]= -16.2081(25.7441-Bx)-6.96297(19.8898-By)-3.57956Bz⩵0
Then we set up 3 equations for solving the coordinate of B
1. The B is on the plane ABCD
2. Norm[Bb]==11.77
3. Dot[AB,Bb]⩵0
In[195]:= Solve0.7954Bx+2.1192By+7.72383Bz⩵157.431&&
(25.7441-Bx)2+(19.8898-By)2+(Bz)2 ⩵11.7788&&(25.7441-Bx)(-43.1162+Bx)+
(19.8898-By)(-29.9541+By)-(-7.72383+Bz)Bz⩵0,{Bx,By,Bz}
Out[195]= {{Bx→26.9081,By→22.9911,Bz→11.3034}}
In[190]:= B = {26.90813027581234`,22.991133262610784`,11.303390101606857`}
In[196]:= Bb = b-B
Out[196]= {-1.16403,-3.10133,-11.3034}
In[210]:= pointD = {Dx,Dy,11.303390101606857`}
Out[210]= {Dx,Dy,11.3034}
In[211]:= AD = pointD-A
Out[211]= {-43.1162+Dx,-29.9541+Dy,3.57956}
We set up 2 equations for solving the coordinate of D
1. AB vertical AD
2. D on plane ABCD
In[212]:= Dot[AB,AD]⩵0
Out[212]= 12.8133-16.2081(-43.1162+Dx)-6.96297(-29.9541+Dy)⩵0
In[213]:= 0.7954Dx+2.1192Dy+7.72383*11.3034⩵157.431
Out[213]= 87.3055+0.7954Dx+2.1192Dy⩵157.431
In[214]:= Solve[0.7954Dx+2.1192Dy+7.72383*11.3034⩵157.431&&Dot[AB,AD]⩵0,{Dx,Dy}]
Out[214]= {{Dx→50.7409,Dy→14.0459}}
In[215]:= pointD = {50.740870303284865`, 14.045947404099287`,11.303390101606857`}
Out[215]= {50.7409,14.0459,11.3034}
Find point d
The point d in plane abcd, that mean dz = 0
In[219]:= pointd= {dx,dy,0}
In[220]:= Dd = pointd-pointD
Out[220]= {-50.7409+dx,-14.0459+dy,-11.3034}
We set up 2 equations for solving the coordinate of d
1. Aa//Dd
2. Dd vertical AD
In[232]:= AD = pointD-A
Solve[Norm[Cross[Dd,Aa]]⩵0&&Dot[Dd,AD]⩵0,{dx,dy}]
Out[232]= {7.62467,-15.9082,3.57956}
Out[233]= {{dx→49.5768,dy→10.9446}}
In[236]:= pointd = {49.57684784855689`,10.944616098272855`,0}
Out[236]= {49.5768,10.9446,0} Find point C
In[240]:= pointC = {Cx,Cy,Cz}
In[260]:= CD = pointD-pointC
CB = B – pointC
Out[260]= {50.7409-Cx,14.0459-Cy,11.3034-Cz}
Out[261]= {26.9081-Cx,22.9911-Cy,11.3034-Cz}
We set up 3 equations for solving the coordinate of B
1. CD vertical CB
2. The C is on the plane ABCD
3. CD//AB
In[262]:= Solve[Dot[CD,CB]⩵0&&0.7954Cx+2.1192Cy+7.72383Cz⩵157.431&& Norm[Cross[CD,AB]]⩵0,{Cx,Cy, Cz}]
Out[262]= {{Cx→50.7409,Cy→14.0459,Cz→11.3034},{Cx→34.5328,Cy→7.08298,Cz→14.8829}}
In[268]:= pointC = {34.532800339947386`,7.082980029533567`,14.882947880911194`}
Out[268]= {34.5328,7.08298,14.8829}
Find point c
In[237]:= c = {cx,cy,0}
In[287]:= cd = d-c cb = b-c
Cc = c-pointC
CD = pointD-pointC
Dd = d-pointD
Out[287]= {49.5768-cx,10.9446-cy,0}
Out[288]= {25.7441-cx,19.8898-cy,0}
Out[289]= {-34.5328+cx,-7.08298+cy,-14.8829}
Out[290]= {16.2081,6.96297,-3.57956}
Out[291]= {-1.16402,-3.10133,-11.3034}
We set up 2 equations for solving the coordinate of c
1. Cc vertical CD
2. Dd //Cc
In[292]:= Solve[Norm[Cross[Dd,Cc]]⩵0&&Dot[Cc,CD]⩵0,{cx,cy}]
Out[292]= {{cx→33.0001,cy→2.99952}}
In[295]:= c = {33.00014769626712`,2.9995154578492857`,0}
Out[295]= {33.0001,2.99952,0} Find side length
In[304]:= Cc = c-pointC cd = d-c
Out[304]= {-1.53265,-4.08346,-14.8829}
Out[305]= {16.5767,7.9451,0}
In[307]:= sidelength = Norm[AB]*4+Norm[Aa]+Norm[Cc]+Norm[Bb]*2+Norm[ab]*2+Norm[cd]*2
Out[307]= 192.644
Find size of plane abcd
abcd is not a square, but ab = ad, bc = dc so we can draw auxiliary lines ac and bd Two lines separate abcd into two triangle
In[309]:= ac = c-a bd = d-b
Out[309]= {-9.32065,-24.8354,0}
Out[310]= {23.8327,-8.94518,0}
In[311]:= abcd =1/2*(Norm[ac]*(1/2*Norm[bd]))*2
Out[311]= 337.635
Find volume
If we make a rotate of this cube and set it as cube 2, the combination of cube and cube 2 will look like a cuboid, and we just need to divide the volume of combination by 2.
In[315]:= areaOfABCD = Norm[AB]*Norm[BC]/2
Out[315]= 168.532
In[318]:= Height = Norm[Bb]*2
Out[318]= 23.5576
In[320]:= volume = areaOfABCD*Height/2
Out[320]= 1985.1
Final Result
In[336]:= A
B
C = pointC
D = pointD
a b c d Norm[Aa]
Norm[Bb]
Norm[Cc] Norm[Dd] sidelength abcd volume
Out[336]= {43.1162,29.9541,7.72383}
Out[337]= {26.9081,22.9911,11.3034}
Out[338]= {34.5328,7.08298,14.8829}
Out[339]= {50.7409,14.0459,11.3034}
Out[340]= {42.3208,27.8349,0}
Out[341]= {25.7441,19.8898,0}
Out[342]= {33.0001,2.99952,0}
Out[343]= {49.5768,10.9446,0}
Out[344]= 8.04868
Out[345]= 11.7788
Out[346]= 15.5089
Out[347]= 11.7788
Out[348]= 192.644
Out[349]= 337.635
Out[350]= 1985.1
In[296]:= r=1 Graphics3D[{
{Red,Sphere[A,r]},
{Blue,Sphere[B,r]},
{Green,Sphere[pointD,r]},
{Yellow,Sphere[pointC,r]},
Sphere[b,r],
Sphere[a,r],
Sphere[d,r],
Sphere[c,r],
Line[{A,a}],
Line[{B,b}],
Line[{pointD,d}],
Line[{pointC,c}]
}]

Part 3 An amazing property of unit cubes
A:
First we set up the vectors (xyw, xyv and xyu means the orthogonal projection of vector on xy-plane
)
In[]:= xyw = {w1,w2,0} xyv ={v1,v2,0} xyu={u1,u2,0} w = {w1,w2,w3} v ={v1,v2,v3} u={u1,u2,u3}
and we can find the area of orthogonal projection of the cube onto the xy-plane
In[]:= area = Norm[Cross[xyw,xyv]]+Norm[Cross[xyw,xyu]]+Norm[Cross[xyu,xyv]]

Rui Qin_30874157_A1.nb
Out[]= Abs[-u2v1+u1v2]+Abs[u2w1-u1w2]+Abs[v2w1-v1w2]
Then we start calculate with the vector SN
In[]:= SN = Cross[v,u]+Cross[w,v]+Cross[w,u]
Out[]= {u3v2-u2v3+u3w2+v3w2-u2w3-v2w3,
-u3v1+u1v3-u3w1-v3w1+u1w3+v1w3,u2v1-u1v2+u2w1+v2w1-u1w2-v1w2}
Find the length of its orthogonal projection onto the z-axis
In[]:= z = {0,0,1}
Out[]= {0,0,1}
In[]:= Norm[Projection[SN,z]]
Out[]= Abs[u2v1-u1v2+u2w1+v2w1-u1w2-v1w2]
Then we find Norm[Projection[SN,z]] can be in form:
=Abs[u2 v1-u1 v2]+Abs[u2 w1-u1 w2]+Abs[v2 w1-v1 w2]=area So Norm[SN] = area B:
Set the start of vector u is S and the end of u is S’
Set the other side of N is N’
The minimum area of projection in xy-plane can deduced by the projection of SS’NN’plane on zcoordinates

The minimum length on z-coordinates is while SS’ // z, this is because after the cube roll over SS’// z, the Norm[SN] is not equal to the area on xy-plane anymore, it turn to Norm[S’N’] = area. While the SS’ // z, the shape on xy-plane is a square, it is the minimum area
In[]:= minimum = 1*1
Out[]= 1
Rui Qin_30874157_A1.nb
Maximum happen while SN // z, that means SN show it longest length on z-coordinates we can use Pythagorean theorem to find the Norm[SN]
In[]:= maximum = Square[Square[1^2+1^2]+1^2]
Out[]= (1+2) C:
If we change the side length, the relationship will not be change