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 Gold-in pyramid (Posted on 2020-10-06)
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 No Rating 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.

 Subject Author Date Solution Brian Smith 2020-10-10 00:30:57 Simpler solutions exist Jer 2020-10-09 21:56:27 re(2): computer solution Charlie 2020-10-07 15:50:23 re: computer solution Charlie 2020-10-07 09:06:12 computer solution Charlie 2020-10-07 06:59:46 some thoughts Steven Lord 2020-10-06 22:49:58

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