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Pythagoras' corner (Posted on 2017-06-16) Difficulty: 3 of 5
Slice off the corner of a right rectangular prism so that the result is a tetrahedron with three right triangular faces mutually perpendicular to each other. The fourth face is a triangle formed by the hypotenuses.

Prove: The sum of the squares of the areas of the three right triangles is equal to the square of the area of the fourth.

No Solution Yet Submitted by Jer    
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solution | Comment 1 of 3
let the vertices have coordinates
(0,0,0)
(x,0,0)
(0,y,0)
(0,0,z)

Then the right triangle faces have areas:
xy/2=P
xz/2=Q
yz/2=R

for the 4th face has side lengths a,b,c given by
a^2=x^2+y^2
b^2=x^2+z^2
c^2=y^2+z^2

let s=(a+b+c)/2
then from herons formula we get that the area A of the 4th face is given by
A^2=s*(s-a)*(s-b)*(s-c)
substitute, expand, and simplify to get
A^2=x^2y^2/4+x^2z^2/4+y^2z^2/4
A^2=(xy/2)^2+(xz/2)^2+(yz/2)^2
A^2=P^2+Q^2+R^2

Thus the square of the area of the 4th face is equal to the sum of the squares of the areas of the other 3 sides.

  Posted by Daniel on 2017-06-16 11:56:04
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