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I scratching my head over this problem from my projective geometry book (C. R. Wylie, Jr).

Given a triangle in the plane $z = 0$, is it possible to find a viewing point, $C$, from which the triangle will appear on the plane $y = 0$ as a right triangle.

What about for isosceles triangle? For equilateral triangle?

I have tried using algebra to find the dot product of the image vectors to find a solution of constructing right angle triangles, but it seems really messy. Is there a way to prove or disprove using elementary geometry?

Mark
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Let $m_1$ and $m_2$ be two sides of the original triangle that are not parallel to the image plane, meeting at $A$.

Consider the plane through $A$ parallel to the image plane and choose two lines $k_1$ and $k_2$ in it, which meet in a right angle at $A$, such that neither $k_1$ nor $k_2$ is parallel to the plane of the original triangle.

Let $P_1$ be the plane spanned by $m_1$ and $k_1$, and let $P_2$ be the plane spanned by $m_2$ and $k_2$. Clearly $P_1$ and $P_2$ intersect each other (because both contain $A$) but do not coincide (because then they would be the triangle plane, contradicting the choice of $k_1$ and $k_2$).

Chose $C$ on the intersection between $P_1$ and $P_2$, far enough form $A$ that the entire triangle projects onto the image plane in one piece.

Then the image of $m_1$ as seen from $C$ is the intersection between $P_1$ and the image plane (which exists because $m_1$ wasn't parallel to the image plane), and is therefore parallel to $k_1$. Similarly the image of $m_2$ is parallel to $k_2$. Therefore the image of the triangle is right.

This works whenever the triangle is not parallel to the image plane. In fact, by projecting back to the camera point from all three sides, you can get to choose all three angles of the image triangle and except for degenerate cases even its orientation. So it is also possible to get the image to be isosceles or equilateral (or for that matter right isosceles).

hmakholm left over Monica
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  • It seems to me there'd be a much simpler and shorter proof using point set topology and an appropriately chosen homeomorphism from z=0 onto y=0. But the OP specifically requested a proof based on elementary geometry,so......... – Mathemagician1234 Jan 19 '15 at 18:17
  • @Mathemagician1234: "Homeomorphism" alone won't do it; they don't necessarily preserve straight lines. – hmakholm left over Monica Jan 19 '15 at 18:20
  • suppose the object plane is the xy-plane and picture plane is the xz-plane, since $k_1$, $k_2$ touches A wouldn't the intersection of $P_1$ and $P_2$ be on the xy-plane? That would imply that the viewing point is on the xy-plane as well... Am I missing something here? – Mark Jan 19 '15 at 20:46
  • @Mark: No, $P_1$ intersects the object plane at $m_1$ only and $P_2$ intersects the object plane at $m_2$ only. Since $m_1$ and $m_2$ are different lines $P_1$ and $P_2$ cannot have their intersection in the object plane. – hmakholm left over Monica Jan 19 '15 at 20:53
  • so $m_1$, $m_2$ are projecting "onto" $k_1$ and $k_2$? and then you push it towards the picture plane along $P_1$ and $P_2$s' intersection to create a right triangle on the picture plane? So $k_1$ and $k_2$ are like the shadow of $m_1$ and $m_2$? – Mark Jan 19 '15 at 21:22
  • @Mark: Yes, that's one way of putting it. – hmakholm left over Monica Jan 19 '15 at 21:26