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Gold-in pyramid (Posted on 2020-10-06) Difficulty: 3 of 5
A right pyramid has a unit square base ABCD and vertex V. Its height is 1 unit.

Points E and F are on CV and DV respectively such that ABEF is a plane section that splits the pyramid into two pieces of equal volume.

Find the length EF.

  Submitted by Jer    
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Solution: (Hide)
The two pieces of pyramid are rather complicated, but the bottom piece is easy to deal with. Consider the two parallel plane sections through E and F, perpendicular to the base and segment EF. This cuts the bottom piece into three pieces. The middle piece is a (sideways) prism. The other two pieces can be pushed together to make a new pyramid with a rectangular base.

By similar triangles (seen in side view of the original pyramid). If EF=x, the height of the rectangle based pyramid is (1-x) and the new width C'D' is also (1-x).

The prism volume is then (1/2)(1-x)(x)
The pyramid volume is (1/3)(1-x)(1-x)
For a total volume -x^2/6 - x/6 + 1/3
Equate this to half the original pyramid (1/3)/2 = 1/6
and simplify to the quadratic x^2 + x + 1 = 0

The solution is (sqrt(5)-1)/2 which is the reciprocal of the golden ratio.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionBrian Smith2020-10-10 00:30:57
Hints/TipsSimpler solutions existJer2020-10-09 21:56:27
Solutionre(2): computer solutionCharlie2020-10-07 15:50:23
re: computer solutionCharlie2020-10-07 09:06:12
Some Thoughtscomputer solutionCharlie2020-10-07 06:59:46
some thoughtsSteven Lord2020-10-06 22:49:58
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