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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.

 See The Solution Submitted by Jer No Rating

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 some thoughts | Comment 1 of 6
Since ABEF are in a plane, E and F must be the same height (H) above the base. Horizontal slices at height h of the lower shape (call it a wedge) will be trapezoids.  Preparing expressions for the lengths and separation of the parallel lines of each trapezoid (one found on the CDV face and the other with endpoints on the BCV and the ADV faces) as functions of h will give the trapezoidal area at each height. The areas can be integrated over h from 0 to H to get the wedge volume. Setting the result to 1/2 the volume of the pyramid will yield H and then EF.

Edited on October 7, 2020, 4:51 am
 Posted by Steven Lord on 2020-10-06 22:49:58

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