Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is it possible to construct the midpoint of a segment in the hyperbolic plane using the set square only?

With the set square one can

  • draw the line through the given two points and
  • drop the perpendicular from the given point to the given line.

The following construction produce the point $X'$ which is centrally symmetric to the point $X$ with respect to point $O$.

  1. Draw line $(OX)$ and let $m$ be the line perpendicular to $(OX)$ through $O$.
  2. Draw yet two perpendicular lines $l$ and $l'$ through $O$.
  3. Find the foot point $Y$ of $X$ on $l$.
  4. Draw the line through $Y$ perpendicular to $m$ and let $Z$ be its intersection with $l'$.
  5. Finally, $X'$ is the footpoint of $Z$ on $l$.
share|improve this question

3 Answers 3

I can do it if I am allowed to cheat - in axiom one for the set square (which is the axiom for the straight-edge) I'll allow one of the points to be ideal: that is, on the Gromov boundary of $H^2$. Also, I assume that the set square produces infinite geodesic rays. In particular, this means, in the second axiom, that the given point can be on the given line.

Suppose the segment $[x,y]$ is given. Draw the geodesic ray $P$, based at $x$ and perpendicular to $[x,y]$. Do the same at $y$ to get the ray $Q$, on the same side of $[x,y]$. Connect the ideal endpoint of $P$ to $y$ and the ideal endpoint of $Q$ to $x$. These new rays cross at $z$. Drop a perpendicular from $z$ to $[x,y]$ and we are done.

share|improve this answer
up vote 1 down vote accepted

Finally, I realized that the answer is NO.

Let us think of hyperbolic plane as about subset of projective plane for $\mathbb R^{2,1}$.

Note that in this model, the set square tool is the same as hyperbolic-cross-product tool; i.e. for any two vectors $u$ and $v$ you can construct the vector $$w=J(u\times v),$$ where $$ J\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x\\y\\-z\end{pmatrix}$$

If the end points of the segment are given by rational vectors then so are all the hyperbolic-cross-products. On the other hand, it is easy to construct two rational vectors such that their hyperbolic bisector goes in an irrational directin. Hence the result follows.


In spherical geometry, the same question has also answer NO. It is easier to visualize since in spherical geometry the set square tool is equivalent to the cross-product tool.

share|improve this answer

Here's another construction that works in the Euclidean plane. Suppose we want the midpoint of $A$ and $B$. Take any $C$ between $A$ and $B$, and draw some $D$ off of $\overline{AB}$ such that $\overline{CD}$ and ${AB}$ are perpendicular. You can construct a line parallel to $\overline{AD}$ that goes through $B$ by drawing a perpendicular to $\overline{AD}$ and then another perpendicular to that new line. You can also draw a parallel to $\overline{BD}$ through $A$. These two parallels intersect at $E$. Then $ADBE$ forms a parallelogram, and you can intersect the diagonals to find the midpoint $F$ of $AB$.

Now, in the hyperbolic plane, the parallel you find isn't unique, so there are multiple ways to do this same construction. And when you do, $F$ is not going to be a midpoint. However, $F$ will be closer to the midpoint than $C$ (I believe). Hence, repeating the construction, you could construct maybe construct a sequence of points that converge to the midpoint. Obviously that sort of thing isn't allowed in a construction, but it's at least going in the right direction =).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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