Take a unit square of paper ABCD. Label the midpoints of BC and CD by E and F. Fold up along AE, EF, and AF so that B, C, and D come together and a tetrahedral pyramid is formed.
Find the height to the base AEF.
https://www.desmos.com/calculator/mh2ibfjuth
First we need the angle of elevation of the apex (call it Z) as seen from A. Its sine will be the height of Z above the base as AZ = 1.
Spherical Trig:
The easiest way I see is to use spherical trig on a small sphere centered on vertex A. The elevation angle sought is one leg of a spherical right triangle whose hypotenuse is arctan(1/2) -- that's angle ZAE cutting off that arc on the sphere. (Remember the sides of a spherical triangle are arcs of a great circle, measured in angular measure.)
The other leg is 45° - arctan(1/2).
By the spherical law of cosines,
cos(arctan(1/2)) = cos(x) * cos(45° - arctan(1/2))
+ another term which includes cos(90°) = 0 as a factor so can be ignored.
x = arccos( cos(arctan(1/2)) / cos(45° - arctan(1/2)) )
The answer is going to be the sine of the resulting angle, so we can skip over taking the arccos as we don't really need the angle:
height = sqrt(1 - ( cos(arctan(1/2)) / cos(45° - arctan(1/2)) )^2) = 1/3
Plane trig:
AC = sqrt(2)
From the midpoint, M, of EF to C is 1/(2*sqrt(2)), so AM is
AM = sqrt(2) - 1/(2*sqrt(2))
and this is the base of a vertical triangle AZM.
AZ = 1 and ZM = 1/(2*sqrt(2))
ZM ^ 2 = AM ^ 2 + AZ^2 - 2 * AM * AZ * cos(ZAM)
cos(ZAM) = (ZM^2 - AM^2 - AZ^2) / (2*AM*AZ)
cos(ZAM) = -(1/8 - (sqrt(2) - 1/(2*sqrt(2)))^2 - 1) / (2*(sqrt(2) - 1/(2*sqrt(2))))
This comes out to .942809041582065, the same, within rounding error of the .942809041582063 found during the spherical method.
The height is then sqrt(1 - .942809041582065^2) or 1/3.
Edited on July 30, 2021, 6:58 am
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Posted by Charlie
on 2021-07-29 11:32:35 |
solution
