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Here's a problem that I'm stuck on:Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?

In the solution, it says that the apex of the pyramid is directly above the orthocenter of $\triangle{ABC}$, which I don't get. This is really obvious if the tetrahedron was regular, because then the projection would be directly above the center of the triangle which includes the orthocenter.

Any explaining would be great. Thanks.

TheLeogend
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    The edges of the midpoint triangle are parallel to the edges of the original triangle. A vertex of the triangle, folded about the corresponding midsegment, necessarily lies in a plane perpendicular to that midsegment. Thus, the projection of that folded vertex lies somewhere on the altitude to the opposite side. – Blue Sep 01 '18 at 14:53
  • Well thanks... if you could provide a good diagram that would be awesome – TheLeogend Sep 01 '18 at 15:00
  • I'm afraid I don't have time for that right now. Perhaps someone else will oblige. – Blue Sep 01 '18 at 15:03
  • So I called the apex A'. How do we know that the plane perpendicular to the midsegment, goes through A and A' will also go through the projection?Actually I think I might understand why... – TheLeogend Sep 01 '18 at 15:20
  • Once you've established that the foot of the altitude of the pyramid is the orthocenter of the original triangle, you can find $(\cot \alpha + \cot \beta) \cot \gamma =1/2$, where $\gamma$ is the angle opposite to the longest side of that triangle. This means that the orthocenter is the midpoint of the altitude to that side, therefore the pyramid is right-angled. Then finding the height of the pyramid is easy. – Maxim Sep 07 '18 at 02:02

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