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.
(In reply to
re: computer solution by Charlie)
We can use a cross section of the pyramid to get angle and distance relations for the cutting plane and height of the smaller pyramid using plane trig.
Let alpha be the angle the cutting plane is inclined to the base of the pyramid.
The slant height, s, of the pyramid is sqrt(1/4 + 1).
The base angle of the original pyramid is atan(2).
The height of the smaller pyramid with the same vertex that's half the volume of the larger pyramid is s * (baseAngle  alpha).
The distance, d, along the slanted side of the large pyramid that the cutting plane intersects the side, measured from the edge of the base is sin(alpha)/sin(beta), where beta is the angle between the cutting plane and the face of the pyramid opposite AB, and is 180°  alpha  baseAngle.
The sought top edge, t, of the trapezoid that cuts the pyramid is (sd)/s as it's the proportion of the way down for the width to become 1.
The bottom of that trapezoid, edge AB, has length 1, top edge t, and its height, planeh is calculated by the plane law of cosines: sqrt(1 + d^2  d*cos(baseAngle)), making the area, A, equal sqrt(1+d^2  2*d*cos(baseAngle)) * (t+1)/2.
And V=A*h/3.
vol0=1/3;
s=sqrt(5/4);
baseAngle=atand(2);
high=atand(2); low=0;
while high>low
alpha=(high+low)/2;
beta=180alphaatand(2);
h=s*sind(baseAnglealpha);
d=sind(alpha)/sind(beta);
t=(sd)/s;
planeh=sqrt(1+d^2 2*d*cosd(baseAngle));
A=planeh*(t+1)/2;
V=h*A/3;
if V==vol0/2
break
end
if V>vol0/2
low=alpha;
end
if V<vol0/2
high=alpha;
end
end
This didn't actually stop; I had to stop it with high not very different from low  a good 15 decimal places. I have to learn better stopping conditions.
V =
0.166666666666667
d =
0.427050983124842
t =
0.618033988749895
s =
1.11803398874989
>> d/s
ans =
0.381966011250105
That is the plane cuts the opposite side of the pyramid about 38% up the side, a distance of .42705...
The EF distance sought is the t value (the top of the trapezoid), 0.618033988749895 approximately, and that explains the gold, and gives me confidence I'm right this time.

Posted by Charlie
on 20201007 15:50:23 