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Is there any way to prove that for any isosceles triangle, the volume of a solid created when that triangle is projected to a point determining the height above the angle opposite the hypotenuse is one third of the volume of a triangular prism of the same height and base shape?

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Here's a sketch. The projected solid can be thought of as one quarter of a square pyramid when N is 90 degrees. Is there a way to prove that the projected solid's volume is one third that of the triangular prism of same height and base values for any value of N?

Both intuitive and purely mathematical proofs are acceptable.

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  • $\begingroup$ Any and all productive posts greatly appreciated. $\endgroup$
    – David Ball
    Aug 16, 2014 at 0:59

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This holds also for non-isosceles triangles. The volume of a pyramid is one third of the product between the area of the base and the length of the relative height.

Just use Cavalieri's principle. If the area of the base is $B$, the area of the section parallel to the base at height $h_0$ is proportional to $(h-h_0)^2$, hence the result follows from:

$$\int_{0}^{1} x^2\,dx = \frac{1}{3}.$$

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    $\begingroup$ This is more or less the same argument Archimedes used to prove that the volume of the sphere is $\frac{4}{3}\pi R^3$, given the area of the parabolic segment. $\endgroup$ Aug 16, 2014 at 1:01
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    $\begingroup$ Not merely "non-isosceles triangles" (which implies that you still are stuck with triangles) -- Any Shape Imaginable does this. $\endgroup$ Aug 16, 2014 at 1:16
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    $\begingroup$ It would be nice with a version without any (explicit) integration. $\endgroup$ Sep 8, 2015 at 14:57
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The three dimensional figure you are referring to is called a tetrahedron. The volume is $V= \frac{1}{3}Ah$ where A is the area of the base and h is the height of the tetrahedron. See http://en.wikipedia.org/wiki/Tetrahedron#General_properties

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