My homework question is what is the product of rotations through opposite angles α,−α about two distinct points. The answer is clearly a translation, but I'm not sure how to prove it. My idea on how to prove this is to draw a triangle, ABC, and its two rotations, A'B'C' and A"B"C", and then show AA", BB", CC" are parallel, however I don't know how to get to this point.

  • $\begingroup$ Is this in $\mathbb{R}^3$? $\endgroup$ Oct 10, 2013 at 1:48
  • $\begingroup$ What level of math class is this? High school geometry, or something in college? (Or, said another way: how rigorous of a proof are they looking for?) $\endgroup$
    – apnorton
    Oct 10, 2013 at 1:48
  • $\begingroup$ @ncmathsadist this is in R2 $\endgroup$ Oct 10, 2013 at 1:54
  • $\begingroup$ @anorton This is a college level geometry cource $\endgroup$ Oct 10, 2013 at 1:54

1 Answer 1


Let $A$ and $B$ be the centers of the two rotations, and let $R$ be the rotation about $A$ by the angle $\alpha$. Of course then $R^{-1}$ is the rotation about $A$ by $-\alpha$, but we want the rotation by $-\alpha$ about $B$ instead. Let $T$ be the translation that sends $B$ to $A$. Then the rotation about $B$ by $-\alpha$ is $T^{-1}R^{-1}T$, i.e., first move $B$ to $A$ by $T$, then do the rotation about $A$, and finally move $A$ back to $B$. (The composite operation sends $B$ to itself and rotates all directions by $-\alpha$, since $T$ doesn't affect directions. So the composite is the desired rotation around $B$.) So the combination "first rotate around $A$ by $\alpha$ and then rotate around $B$ by $-\alpha$" is the composition $T^{-1}R^{-1}TR$. Now $R^{-1}TR$, being a conjugate of a translation, is itself a translation $T'$, so we have the composite of two translations, $T^{-1}T'$, and this is a translation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.