 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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 Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 6 of 6 | I started by adding some more points:
Point M is the midpoint of AB.
Point N is the midpoint of CD.
Point P is the midpoint of FE.
Point O is the center of the base, which is also the foot of the altitude of the pyramid.
Point R is the point on MN such that PR is perpendicular to MN, PR will then be how far EF is above base ABCD.

Given from the problem: AB=BC=CD=DA=VO=1.

Let the length of EF be x.

Then AM=MB=CN=ND=MO=ON=1/2 and EP=PF=x/2.

The equations that follow come from looking at the plane containing triangle VCD (also includes points E,F,N,P) and the plane containing triangle VMN (also includes points O,P,R).

From right triangle VON, VO^2 + ON^2 = VN^2 which makes 1^2 + (1/2)^2 = VN^2.  Then VN=sqrt/2.

Triangles VND and VPF are similar, VP/VN = PF/ND which makes VP/(sqrt/2) = (x/2)/(1/2).  Then VP=(sqrt/2)*x.

P lies on VN thus VN = VP + PN; which makes sqrt/2 = (sqrt/2)*x + PN.  Then PN=(sqrt/2)*(1-x).

PRN is similar to VON thus PN/VN = PR/VO = RN/ON; which makes (sqrt/2)*(1-x)/(sqrt/2) = PR/1 = RN/(1/2).  Then PR = 1-x and RN=(1-x)/2.

R lies on ON thus ON = OR + RN; which makes 1/2 = OR + (1-x)/2.  Then OR = x/2.

This is enough to place the pyramid and points E and F on a coordinate system.  Let point O be at the origin.  Then
`V = (   1,    0,   0)A = (-1/2,  1/2,   0)B = (-1/2, -1/2,   0)C = ( 1/2, -1/2,   0)D = ( 1/2,  1/2,   0)E = ( x/2, -x/2, 1-x)F = ( x/2,  x/2, 1-x)`

The volume of pyramid ABEFV can be found by dividing they pyramid into two tetrahedrons and using the vertex determinant formula for a tetrahedron's volume on each half.  https://mathworld.wolfram.com/Tetrahedron.html  Note that the determinant gives a signed area, which maybe positive or negative depending on the order of the points entered in the determinant.

Without loss of generality, split ABEFV into VAEB and VAEF.  Because B and F are on opposite sides of VAE, their signs will be opposites so to find the total volume the two determinants need to be subtracted.

Now since the determinants differ in only the last row they can be combined by subtracting the elements in the last row and preserving the other rows.  Then the single determinant (V,A,E,F-B) becomes:
`               |       0        0    1  1 |Volume = (1/6)*|    -1/2      1/2    0  1 |               |     x/2     -x/2  1-x  1 |               | (x+1)/2  (x+1)/2  1-x  1 |`

Using row operations -R1+R2->R2, -R1+R3->R3, -R1+R4->R4, the determinant reduces to
`               |       0        0   1  1 |              |     -1/2      1/2  -1 |Volume = (1/6)*|    -1/2      1/2  -1  0 | = (1/6)*(-1)*|      x/2     -x/2  -x |               |     x/2     -x/2  -x  0 |              |  (x+1)/2  (x+1)/2  -x |               | (x+1)/2  (x+1)/2  -x  0 |`

Then row operations -x*R1+R2->R2 and -x*R1+R3->R3, the determinant reduces to
`                    |   -1/2  1/2  -1 |                    |     x   -x |Volume = (1/6)*(-1)*|      x   -x   0 | = (1/6)*(-1)*(-1)* | x+1/2  1/2 |                    |  x+1/2  1/2   0 |`

Then the determinant can be evaluated into Volume ABEFV = (x^2+x)/6.

The original pyramid ABCDV has a volume of 1/3.  Then 1/6 = (x^2+x)/6, which has one real root (sqrt-1)/2, or the golden ratio minus one.

 Posted by Brian Smith on 2020-10-10 00:30:57 Please log in:

 Search: Search body:
Forums (7)