Parallelogram ABCD is the base of a pyramid SABCD. Planes determined by triangles ASC and BSD are mutually perpendicular. Find the area of the side ASD, if areas of sides ASB, BSC and CSD are equal to x, y and z, respectively.
Edit: I wish I could change the title of this post to Guess at solution
It wasn't at first apparent to me that a pyramid with the given info could exist where opposite sides aren't equal. So I played around and created one in Desmos3D.
If ASC and BSD are perpendicular planes, why not make them the XZ and YZ planes. Then the vertex can be on the Z axis.
I arbitrarily decided the base would be the plane .5x+y+z=1 and the vertex (0,0,4) giving
A=(1,0,0.5)
B=(0,1,0)
C=(-1,0,1.5)
D=(0,-1,2)
(double checking my idea: ABCD is indeed a parallelogram)
Then using Herons Formula, the four sides in order are
sqrt(117)/4
sqrt(93)/4
sqrt(45)/4
sqrt(69)/4
Edit: One of the above areas was wrong. I've since fixed it. Observe: 117+45 = 93+69
So my guess is the squared sums of areas of opposite sides are equal implying [ASD]=sqrt(x^2-y^2+z^2)
My picture: https://www.desmos.com/3d/4af1112404
Edited on January 27, 2024, 12:58 pm
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Posted by Jer
on 2024-01-27 12:16:08 |
experiment, no solution
