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Let $v_1, v_2, v_3$ and $w_1, w_2, w_3$ denote the vertices of two spherical triangles $\bigtriangleup_1, \bigtriangleup_2$ with the property, that $\|v_i - v_j \| = \|w_i - w_j \|$, e.g. their sides have the same length. Prove, that there exists a transformation $f \in O(3)$ that takes $v_i$ to the $w_i$ for $i=1,2,3$.

So $\bigtriangleup_1$ and $\bigtriangleup_2$ are congruent triangles and certainly the congruence transformation that takes one to the other has to be an orthogonal one, e.g. a transformation that fixes lengths and angles. But I don't see how to actually prove this theorem. Maybe that's because i can't picture how this transformation for two given triangles will work; the three vectors $v_i$ that define a triangle on $S^2$ are linearly independent and thus form a normed basis for $\mathbb{R}^3$, but it's not necessarily an orthonormal basis. So it's not as easy as taking one orthonormal basis to another... How can I prove this? Any help is appreciated.

EDIT: My progress so far: It's clear that $O(n+1)$ acts transitively on $S^n$. Thus there exists a rotation $f_1 \in O(3)$, such that (for example) $f_1(v_1) = w_1$. Since lengths and angles are invariant under orthogonal transformations $\bigtriangleup_1' := f_1(\bigtriangleup_1)$ is congruent to $\bigtriangleup_1$. Now $\bigtriangleup_1'$ and $\bigtriangleup_2$ share at least one vertex. Now we either have to rotate about the axis through $w_1$ and its antipodal point until $\bigtriangleup_1'$ and $\bigtriangleup_2$ align or, if $\bigtriangleup_1$ and $\bigtriangleup_2$ are mirror images of each other we have to reflect $\bigtriangleup_1'$ using some geodesic as a reflection line and then follow up with some combination of rotations (which are just translations on the sphere). In conclusion, I think the reason why such a transformation exists in $O(3)$ is the transitive nature of $O(n+1)$ acting on the sphere $S^n$. Thoughts?

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  • $\begingroup$ Yeah, but aren't A and B still congruent iff you can take A to B with a combination of rotations, translations and reflections (which reverse the orientation and have determinant -1)? $\endgroup$ – jazzinsilhouette Jul 14 '15 at 16:28
  • $\begingroup$ Sorry, silly mistake -- I've deleted the comment. $\endgroup$ – joriki Jul 14 '15 at 16:31

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