+0  
 
0
96
2
avatar

ABCD is a regular tetrahedron (right triangular pyramid). If M is the midpoint of CD, then what is cos ABM?

 Jul 29, 2022
 #1
avatar+302 
+1

Answer: \(\frac{\sqrt3}{3}\)

Solution:

In the orientation that I* would view this in, A would tbe the point on the 'top'. Side CD as well as point B are on the base of the tetrahedron. Since BCD is an equilateral triangle, the median from CD to B would pass through its center. Call the center O. AOB would be 90 degrees (if it isn't clear, angle ABO is the same as angle ABM). Because there is a right triangle with the desired angle in the right place 'SOH CAH TOA' can be applied, where AB is hypotenuse and BO is adjacent. BO is equal to 2/3 the length of the median MB, which itself is \(\frac{\sqrt3}{2}\) of AB. That makes BO \(\frac{\sqrt3}{3}\) of AB. BO/AB = cos ABM = \(\frac{\sqrt3}{3}\).

 

 

 

 

 

*Being bad at geometry, especially 3d geometry, runs in my family. This may not be correct (especially since the question has been asked before on this site and answered with 2/5; no work shown. The answer had -2 points, so it probably wasn't correct, but...)

 Jul 30, 2022
edited by WhyamIdoingthis  Jul 30, 2022
 #2
avatar+124676 
+1

For convenience   let   the side of the tetrahedron = 1....let

A  = ( -1/2 , 0 , 0)

B  = ( 1/2, 0, 0  )

C =  (0, sqrt (3)/2, 0)

D =  (0, sqrt (3)/4 , 6/sqrt (3) )

M =  (0 , (3/8) sqrt (3) , 3/sqrt (3) )  =  (0, (3/8)sqrt (3) , sqrt (3) )

 

AB  = 1

AM = BM   =  sqrt   [ (-1/2)^2  + (3sqrt (3) / 8)^2  + (sqrt 3)^2   =  sqrt [ 1/4 + 27/64 + 3 ]   = 

sqrt (235) / 8

 

By the Law of Cosines

 

AM^2  =  AB^2 + BM^2 - 2 (AB) (BM)cos (ABM)

 

0  = 1 - 2 ( 1) (sqrt (235) / 8) cos ABM

 

0 =  1 - [sqrt ( 235) / 8] cos ABM

 

-1 /  - [  sqrt (235) / 8]  =  cos ABM

 

8 / sqrt ( 235)  =  cos ABM

 

 

cool cool cool

 Jul 30, 2022

38 Online Users

avatar
avatar