Given 3 non-collinear points A, B, C, how can the bisector of angle ABC be constructed using neither a compass nor a ruler (e.g. any form of length measurement), just pair of perpendicular straightedges.

Construction is thus limited to the following techniques:

* A new point at the intersection of two lines.
* A new line thru 2 existing points. (This requires a straightedge, but not a ruler.)
* A new line perpendicular to an existing line through an existing point.

Given these three initial constructs, it is trivial to develop the following additional constructs, so assume they are availble as well.

* A new line parallel to an existing line through an existing point.
* The midpoint of two existing points.

Though a bit less trivial, I already know how to develop these additional constructs from the others, so assume they are available too if that helps, though other readers may benefit from deriving these tools in your response if they are necessary.

* The orthocenter of 3 non-collinear points.
* The circumcenter of 3 non-collinear points.
* The centroid of 3 non-collinear points.

If it's not obvious, my end goal is to find the incenter of triangle ABC, which is simply the intersection af any two angle bisectors of that triangle. Conversely, if the incenter can be found, it bisects all three angles so it answers the initial question.

I know that one point that bisects angle A is located the on the line thru the orthocenter of ABC and the midpoint of segment BC where it intersects the circumcircle of triangle ABC, but that doesn't help in this case because the circumcircle of triangle ABC can't be generated without a compass.

I found a couple of related question at Alternative compass and straightedge construction and How can we construct the incenter of a triangle using a compass alone?, but they don't exactly address my question.


It can't be done.

Suppose you start with the triangle $(0,0),(1,0),(0,1)$. It can be proven by induction that any point you can mark has rational coordinates (henceforth called rational points) and any line you can draw can be written in the form $ax+by=c$ with $a,b,c$ rational (call such a line rational line). Since the target incenter has irrational coordinates, it's impossible to mark.

To elucidate on the induction, suppose after $n$ steps we have only rational points and rational lines drawn. At the $n+1$-th step, we can do one of these three things:

  • Mark the intersection of two rational lines. Since this amounts to solving a linear system with rational coefficients, the resulting point will have rational coordinates.
  • Draw the line through two rational points. This amounts to solving the system $y_1=mx_1+c,y_2=mx_2+c$ for $m,c$ given $y_1,y_2,x_1,x_2$ rational (there's a special case when $x_1=x_2$, but that's easy to handle), but this clearly has a rational solution.
  • Draw a line perpendicular to $y=Mx+c$ through a rational point $(a,b)$, where $M,c$ are rational. Now this line will of the form $y=-x/M+c_1$, where $-1/M$ is clearly rational, and $c_1$ can be determined by solving the equation $b=-a/M+c_1$, which will again be rational. Again, there's an edge case to consider when $M=0$, but that can be dealt with similarly.

Note that the key idea used in this proof is that given an uniquely solvable linear system of equations with rational coefficients, the solution will be rational. This can be easily seen, from Cramer's rule, for example.

| cite | improve this answer | |
  • 1
    $\begingroup$ Thanks for the clear concise answer. I agree that the incenter you describe is irrational. It's 1-(sqrt(2)/2). It seems reasonable that any point I can mark has to be rational, but I'm not sure exactly why. Can you point me to the proof that you mention? In the case you mention, one of the side lengths is srqt(2) which is also the only irrational component of the incenter. While I haven't tried working thru your example yet, it's a great basis for a general explanation. I want to prove to myself that I can't mark sqrt(2) on the X axis with these constraints. If that's impossible, I'll agree. $\endgroup$ – David Mar 10 at 10:44
  • $\begingroup$ @David I have added some more details. Let me know if something is still unclear. $\endgroup$ – Ankoganit Mar 10 at 11:08
  • 1
    $\begingroup$ Again, thanks for clearing it up for me. The proof you added is perfect. Guess that answers my question. I can quit trying. ;-) $\endgroup$ – David Mar 10 at 11:15

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.