Three balls of radius r lie on the lower base of a regular triangular prism, and each of them touches two other balls and two side faces of the prism. On these balls lies the fourth ball, which touches all the side faces and the upper base of the prism. Find the height of the prism.
For simplicity r=1, call the radius of the fourth ball R.
The side lengths of the equilateral triangle base are easily found as 2+2sqrt(3). In terms of R this length is also Rsqrt(3), so R=(3+sqrt(3))/3 = 1.577
The centers of the four balls are the vertices of a pyramid with an equilateral triangle base with lengths 2, an edge height of 1+R, a slant height of l, and a height h.
The slant height: 1+l^2 = (1+R)^2, l^2=(10+4sqrt(3))/3
(I didn't need to do it this way, but l=2.375)
Height: (sqrt(3)/3)^2+h^2 = l^2, h^2=9+4sqrt(3))/3
h = sqrt(27+12sqrt(3))/3 = 2.304
Height of the prism = 1 + R + h
= (6+sqrt(3)+sqrt(17+12sqrt(3))/3 = 4.882
Edit to add a Desmos3D picture
https://www.desmos.com/3d/ueiezsnsry
Edited on July 5, 2024, 10:27 am
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Posted by Jer
on 2024-07-05 10:00:52 |
Solution
