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I have a problem with a special construction.

Is it possible to construct the angle bisector of two lines with only a right-angled ruler?

Here with a right-angled ruler one can performe the following constructions:

  • draw line through two points
  • draw perpendicular from any point to any line

I already know that with such a ruler one can divide a segment into half, double a segment, but I couldn’t construct an angle bisector. I am starting to believe that it is not possible. I am familiar for example with the impossibility of halving a segment with a ruler only, but I couldn’t generalize it. Any idea?

Blue
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user173628192
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  • I am assuming that the two lines are either infinite in length, or each represent a line segment, where the relative lengths of each line segment are unknown. Assuming so, then my guess (which could be wrong) is that you are right. Any idea that I might have requires that you are able to measure equal lengths from the two lines/line-segments. I don't see how a right angled ruler will work, assuming that you are not allowed to use any marking on the ruler, or the fixed length of the ruler. – user2661923 Apr 07 '21 at 22:51
  • On the other hand, the normal tools for construction are a straight edge and compass. The fact that you are permitted to use a right angled ruler, rather than a straight edge is intriguing. – user2661923 Apr 07 '21 at 22:54
  • Yes, the ruler is considered to be Euclidean, i.e. it has infinitely long legs. You can not mark segments on the ruler. Thus no fixed length can be used. – user173628192 Apr 07 '21 at 22:55
  • Have you even been able to construct a 45° angle? – Greg Martin Apr 07 '21 at 22:55
  • A bit of cheating allows the following construction step: Given points $A$, $B$ and a line $\ell$, move the right-angled ruler until one leg passes through $A$, one passes through $B$, and the apex is on $\ell$; then mark off the apex point. If such a step is allowed, your task is readily done (as you already know how to double a line segment) – Hagen von Eitzen Apr 07 '21 at 22:55
  • @HagenvonEitzen yes it is interesting, however illegal. – user173628192 Apr 07 '21 at 22:57
  • @LeoGardner Keeping in mind that I am generally ignorant in this area, could you please explain why it is cheating? Also, what legal methods (that you think might be pertinent) can be employed with a right angled ruler that a straight edge can't do? – user2661923 Apr 07 '21 at 22:59
  • @GregMartin actually no. The only tricky way I could bisect an angle if its measure is useful: for example if its 48 degrees, then if you reflect the lines over and over, the 8th time it will make a 8x48=384=24=48/2 angle. – user173628192 Apr 07 '21 at 22:59
  • @user2661923 it is considered cheating, because the construction step is not among the allowed basic constructions. Only drawing a line through two points and drawing a perpendicular is allowed. – user173628192 Apr 07 '21 at 23:01
  • But who determines those axioms? Shouldn’t the axioms be the ones that allow all the constructions to be made? – James Cleveland-Tran Sep 06 '23 at 01:11

1 Answers1

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Identify the plane with $\Bbb R^2$ and let $Q$ be a subfield of $\Bbb R$. We call a point $(x,y)$ nice if $x\in Q$ and $y\in Q$. We call a line nice if it passes through two nice points.

Claim. If all given points and lines are nice then all points and lines that are constructible with the right-angled ruler from these givens are also nice.

Proof: (By induction) We need to show that every construction step produces only nice objects from nice givens.

  • drawing the line through two nice points: Okay by the definition of nice line
  • drawing a line perpendicular to a nice line (through nice points $(x_A,y_A)$ and $(x_B,y_B)$) and through a nice point $(x_C,y_C)$: This line passes through the tow nice points $(x_C,y_C)$ and $(x_C+y_B-y_A,y_C+x_A-x_B)$.
  • intersecting two constructed lines: The coordinates of the intersection point are found by solving a system of linear equations based on the coordinates of some nice points. The solutions are in the same firld $Q$, hence the intersection point is nice
  • "Pick any point" (that does not alter the final outcome): An often overlooked construction step. But as $Q$ is dense in $\Bbb R$, we may assume that a picked point is nice
  • "Pick any point on a nice line": Again, points from $F\times F$ are dense on nice lines so that we may assume that a nice point is picked

I think those are all allowed construction steps, hence the claim follows. $\square$

The points $A=(1,0)$, $B=(0,0)$, $C(1,1)$ are nice if we let $Q=\Bbb Q$. Let $\ell$ be the bisector of $\angle ABC$ (which is an angle of $45^\circ$). As $\tan 22\frac12^\circ = \frac{1}{1+\sqrt 2}$ is irrational, we see that $(0,0)$ is the only nice point on $\ell$, i.e., $\ell$ is not nice. According to the claim, the bisector cannot be constructed.

Hagen von Eitzen
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