There are 4 solid spheres arranged so that each one is touching all of the others. The 3 bottom spheres touch the flat floor at points A, B and C. The top sphere has a radius of 12 centimeters. If it were replaced by a sphere with radius 25 cm, then its center would be 14, 15 and 16 cm further from from points A, B and C, respectively.
What is the radius of each sphere?
A way to solve this problem (probably not the best one) is the following:
Spheres (on the horizontal plane M): we call them T,S,R with radios=T,S,R respectively con T<S<R
Superior sphere: call r with radio=r (12 cm or 25 cm)
The plane N formed by the centres of the spheres T,S,R is the only one where T,S,R are tangential to each other. The centres of the spheres (points A,B,C) describe a triangle on that plane N. With the centre of the superior sphere r this points compose a tetrahedron A,B,C,D (D being the centre of the superior sphere r)
We can express all the edges of the tetrahedron in terms of T,S,R,r. This also means we can know every angle in the faces of the tetrahedron in terms of T,S,R,r.
The distance between the contacts of the spheres T,S,R (call them points A',B',C') in the horizontal plane M with the centre D of the superior sphere can be calculate based on the horizontal and vertical distances from A',B',C' to D.
/ (Plane N)
| sphere T (radio T)
===== (Plane M)
A'A=T (distance in sphere T between contact A' with plane M and the centre A)
AP (distance between the centre A (sphere T) and the vertical projection of the point D on plane N)
DP (distance between the vertical projection P of point D and point D)
To solve the problem we need AP, DP and the angle between them. Or better the angle "gamma" between AP and the plane M.
For AP: if we think at the tetrahedron as being made of paper and we try to open a face (for ex. ABD), we notice that the face turns on the axe AB so that the point D describes a circle in a perpendicular plane to AB. This allows as to know where is the projection of D on the plane N (opening two contiguous faces). We can also know the height (h) of point D over the plane N.
But we may want better not this projection (P'= perpendicular projection of D to plane N) but the projection of D on plane N but perpendicular to plane M (point P). To calculate it we need to know the angle "Tau" between planes N and M and the line of biggest slope between both planes, as the desired projection will take that direction and angle. With this data we can fix the position of P and so we have the length and direction of AP in plane N.
For the angle "gamma" between AP and the horizontal plane M We need it to get the horizontal distance we are looking for:
Dist (horiz)= AP*cos(gamma).
This angle "gamma" will depend on the orientation of AP in the plane N: its maximum value will be "Tau" (the angle between planes N and M) if AP is in the direction of the biggest slope, but could have any inferior value or also 0 if AP is perpendicular to the line of maximum slope. Once we have got "gamma" [the final formula here is: sin(gamma)=sin(Tau)*cos(Eta) -Eta being the angle between AP and PP'].
For DP the formula comes straight as it will be h/cos(Tau) (h being the height of D over the plain N, I mean the perpendicular projection of D to plane N: we have got it to calculate AP).
With this ulterior data we have come to know the vertical distance from A' to D, which is
Dist (vert)=T + AP*sin(gamma) + h/cos(Tau).
We are now able to know for each value of R,S,T,r the distance between point A' (contact of sphere T) and D (centre of the superior sphere).
But we need also the distances to D from B' (sphere S) and C' (sphere R). These are not much difficult as we can relay on the values of AP and "gamma" to calculate both distances. Just be aware that while AP gains height over M BP, and CP are probably losing height, because the centres of S and R are probably higher than point P.
Once we have the formulas for the distance B'D and C'D, we pass them to Excel (I do not own other resources) and there we give values to T,S,R.
It's quite quick to get the values of the radios fitting with A'D=14 B'D=15 C'D=16
What I did got after this long process is:
T=17,4811 S=22,7954 R=28,3342
(approx.. thousandth of millimetre)
Posted by armando
on 2019-02-25 04:02:56