3
$\begingroup$

Please help me to solve the following problem:

Prove that in hyperbolic geometry there is no triangle $ABC$ such that the length of a mid-segment is half the length of the corresponding segment.

In other words: If there exist a triangle $ABC$, such that the length of a mid-segment is half the length of the corresponding segment, then it is only possible in Euclid geometry.

My attempt:

I think the structure of the solution must be the following: We should suppose that in hyperbolic geometry there exists triangle with mentioned property and then we should conclude that axiom about parallel lines in hyperbolic geometry fails : so there exists the line $L$ and the point outside the line $P$, such that there are 0 or 1 lines through $P$ that are parallel to $L$ and no more, thus contradicting the definition of hyperbolic geometry. I also think that I should find $L$ and $P$ by looking at $ABC$. But I am not able to move forward.

Thanks a lot for your hints and answers!

Update:

There is another possible approach: it is enough to prove that in hyperbolic geometry the length of mid segment is always less than the length of the segment.

$\endgroup$
  • $\begingroup$ Here's a trigonometric solution: In $\triangle ABC$, writing $a_2$ and $b_2$ for $a/2$ and $b/2$, the hyperbolic Law of Cosines gives $$\cosh c = \cosh 2 a_2 \cosh 2b_2 - \sinh 2 a_2 \sinh 2 b_2 \cos C$$ $$\cosh m = \cosh a_2 \cosh b_2 - \sinh a_2 \sinh b_2 \cos C$$ where $m$ is the mid-segment corresponding to $c$. With a bit of work, we find $$\cosh c - \cosh 2m = \cosh c - ( 2 \cosh^2 m - 1 ) = 2\sinh^2 a_2 \sinh^2 b_2 \sin^2 C \geq 0$$ with "$=0$" in degenerate cases. We conclude that $c > 2 m$ for non-degenerate $\triangle ABC$. $\endgroup$ – Blue May 29 '17 at 16:23
  • $\begingroup$ I think it's more-appropriate for me to post a full solution here (which I will do shortly), and for you to delete your other question. $\endgroup$ – Blue May 29 '17 at 16:27
  • 1
    $\begingroup$ Ok, I've deleted another question, please post your answer. As long as I understand it and work thought it I will be happy to accept. $\endgroup$ – Hedgehog May 29 '17 at 16:38
1
$\begingroup$

Converting a comment into an answer, as requested:


Here's a trigonometric solution: In $\triangle ABC$, writing $a_2$ and $b_2$ for $a/2$ and $b/2$, the hyperbolic Law of Cosines gives $$\cosh c = \cosh 2 a_2 \cosh 2b_2 - \sinh 2 a_2 \sinh 2 b_2 \cos C$$ $$\cosh m = \cosh a_2 \cosh b_2 - \sinh a_2 \sinh b_2 \cos C$$ where $m$ is the mid-segment corresponding to $c$. With a bit of work (see below), we find $$\cosh c - \cosh 2m = \cosh c - ( 2 \cosh^2 m - 1 ) = 2\sinh^2 a_2 \sinh^2 b_2 \sin^2 C \geq 0 \tag{$\star$}$$ with "$=0$" in degenerate cases. We conclude that $c > 2 m$ for non-degenerate $\triangle ABC$. $\square$


Getting $(\star)$ is a matter of slogging-through some symbol manipulation. To save some space and typing effort, I'll abbreviate "$\cosh u$" as simply "$u$" and "$\sinh u$" as "$\widehat{u}$".

The double-angle formulas for hyperbolic sine and cosine give $$\cosh 2u = 2 \cosh^2 u - 1 \qquad\qquad \sinh 2u = 2 \sinh u \cosh u$$

So, from the above, we have $$\begin{align} \cosh c &= ( 2 a_2^2 - 1 )( 2 b_2^2 - 1 )- 4 a_2 b_2 \widehat{a_2}\widehat{b_2} \cos C \\ &= 4 a_2^2 b_2^2 - 2 a_2^2 - 2 b_2^2 + 1 - 4 a_2 b_2 \widehat{a_2}\widehat{b_2} \cos C \end{align}$$ Also, $$\begin{align} \cosh 2m &= 2 \cosh^2 m - 1 \\ &= 2 ( a_2 b_2 - \widehat{a_2}\widehat{b_2} \cos C)^2 - 1 \\ &= 2 a_2^2 b_2^2 + 2 \widehat{a_2}^2 \widehat{b_2}^2 \cos^2 C - 4 a_2 b_2 \widehat{a_2}\widehat{b_2} \cos C - 1 \end{align}$$ Therefore, $$\begin{align} \cosh c - \cosh 2m &= 2 a_2^2 b_2^2 - 2 a_2^2 -2 b_2^2 + 2 - 2 \widehat{a_2}^2 \widehat{b_2}^2\cos^2 C \\ &= 2 ( a_2^2 - 1 )( b_2^2 - 1 ) - 2 \widehat{a_2}^2 \widehat{b_2}^2\cos^2 C \\ &= 2 \widehat{a_2}^2 \widehat{b_2}^2 - 2 \widehat{a_2}^2 \widehat{b_2}^2\cos^2 C \\ &= 2 \widehat{a_2}^2 \widehat{b_2}^2 \left( 1- \cos^2 C \right) \\ &= 2 \sinh^2 a_2 \sinh^2 b_2 \sin^2 C \quad \square \end{align}$$

$\endgroup$
  • $\begingroup$ Terrific, thanks, accepted even without understanding it, but I will understand soon. I have second hyperbolic geometry unanswered question, it is also half-solved by me, can you please check? Thanks a lot again. $\endgroup$ – Hedgehog May 29 '17 at 16:50
  • $\begingroup$ I may not have time to get to your other question today. We'll see. $\endgroup$ – Blue May 29 '17 at 16:54

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.