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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.

See The Solution Submitted by Jer    
Rating: 5.0000 (1 votes)

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Solution Solved without peeking Comment 7 of 7 |
Without looking at previous solvers.
Pyramid is a stack of squares with side length 1 at the bottom and 0 at the top.
At height h, the side length is (1-h).
The total volume is integral from 0 to 1 of (1-h)^2 dh; equals 1/3.
Call the height at point E, capital H; the height of some horizontal slice lower case h.
Call the point along CV which is at height h, lower case e (the little brother of E).
As the plane slices through the various squares at various heights, the result is a rectangle one dimension of which is still (1-h), and the other dimension is the key to the solution.

If you look from the side, point B is at (0,0), point C is at (1,0), point V at (1/2, 1).
Where is point e?  It's height is h.  The width of it's "square" is (1-h) which means h/2 has been taken away from both the left and right sides (when the pyramid was manufactured).  
So the location of point e is ((1-h/2), h); and point E is ((1-H/2), H)

Where does plane ABEF intersect the horizontal square which is at height h?  The equation of plane ABEF is:
y = H/(1-H/2) x  and the plane of our square is y=h.
So, h = H/(1-H/2) x
x = (h/H)*(1-H/2) where the sloped plane meets the horizontal plane
Also x = (1 - h/2) for point e.
The width of the desired rectangle is the x value at point e, minus the x value where the planes intersect.
Width (at height h) = (1 - h/2) - (h/H)*(1-H/2), which simplifies to
Width = (1 - h/H)
Our rectangles have area:  (1 - h/H)*(1-h) = 1 - h - h/H + h^2/H
Integrate that from the bottom up to point E, or as h goes from 0 to H:
Volume (below sloped plane) = H/2 - H^2/6  which we need to be half of the total Volume (which is 1/3)

set H/2 - H^2/6 = 1/6 --->  H^2 - 3H + 1 = 0  which gives H = (3 - sqrt(5))/2.
The distance from E to F is the long side of that rectangle, which is 1-H

EF = (sqrt(5) - 1) / 2 = about 0.618 and which is the reciprocal of the Golden Ratio, hence the title
  Posted by Larry on 2020-10-31 11:48:49
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