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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    
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Some Thoughts computer solution | Comment 2 of 6 |
The answer, shown at the bottom below the method of solution, is sqrt(2)/2. But it's suspicious from the start. The volume of the large pyramid should be 1/3, not 1/6. Perhaps the formula given in the reference is wrong, or restricted to specific types of pyramids, but it doesn't give the right answer here. Or I misinterpreted it.

The puzzle states that A, B, E and F are coplanar. A plane swinging around AB will intersect VC and VD at the same distance above the base plane, ABCD. The two volumes created by the division made by the new plane ABEF are both pentahedrons, but only ABEFV is a pyramid, while ABCDEF is topologically equivalent to a triangular prism.

The website has a paper describing finding the volume of a pentahedron, in particular the kind that is a pyramid (having 5 vertices), as opposed to one topologically equivalent to a triangular prism, via determinants of matrices formed from the coordinates of the five vertices.

For solution purposes, I put A at the origin, B at (0,1,0), C at (1,1,0), D at (1,0,0) and V at (1/2, 1/2, 1).

The program below, a MATLAB script, finds the volume (vol0) of the full pyramid, ABCDV, and satisfyingly finds it to be 1/6. The formula given in the paper mentioned above actually found -1/6, so I added the abs() function around its findings.

The program uses a binary search for the proportion of CE, relative to CV, that makes the volume of the smaller pyramid be half the volume of the larger. That's the same as the proportion of DF to DV due to symmetry mentioned before. The high and low values are a little confusing, as when E (or F) is too high, the volume of the sought upper pyramid is to low and vice versa.

a=[0 0 0];
b=[0 1 0];
c=[1 1 0];
d=[1 0 0];
v=[.5 .5 1];

vol0=abs(det([a;b;c])+det([a;c;d])+det([a;d;v]) + det([a;v;b])+det([b;c;d])+det([c;d;v]))/6

lowrat=0; ,hirat=1;
while hirat>lowrat
    vol=abs(det([a;b;e])+det([a;e;f])+det([a;f;v]) +det([a;v;b])+det([b;e;f])+det([e;f;v]))/6;
    if vol==vol0/2 
    if vol<vol0/2

which finds, 

first the volume of the original pyramid

second the coordinates of E

third the coordinates of F

fifth the found volume of the upper pyramid

sixth the distance between E and F (the difference of their y coordinates, as their x and z coordinates are the same).

vol0 =
e =
         0.853553390593274    0.853553390593274   0.292893218813453
f =
         0.853553390593274    0.146446609406726    0.292893218813453
vol =
ans =
which is recognizable as sqrt(2)/2         

Edited on October 7, 2020, 8:43 am

Edited on October 7, 2020, 9:05 am
  Posted by Charlie on 2020-10-07 06:59:46

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