I taught a Euclidean geometry class last year, which I thought was a dream. It is a first course in proofs, yet is not axiomatic... something with which I'm still wrestling.

This summer, I'm trying to improve the approach (in my opinion). One important tool I want to introduce early on is the idea of an affine map. Many proofs greatly simplify and some of the surprising results become plain (yes, I'm a party pooper). The catch is that we prefer synthetic proofs to using coordinates. So, here's the problem:

Start with Euclid's axioms and perhaps some others that you deem suitable. Define an affine map of the plane to be a bijection that takes lines to lines (I believe this is equivalent). Note, this means parallel lines are preserved.

Suppose $A,B,C$ are points on a line and $T$ is an affine map.

Harder problem Prove that the directed segments are in proportion: $$AB:AC=T(A)T(B):T(A)T(C)$$

The harder problem quickly implies that ratios of areas are preserved.

Easier problem Prove that if $B$ is between $A$ and $C$, then $T(B)$ is between $T(A)$ and $T(C)$.

Progress This paper is interesting, although I still don't know what a g-reflection is. I expect a much more basic solution.

Once the easier problem is solved, I will have the harder problem as is shown below. Of course, the easier problem has to do with betweenness. We therefore at least need the notion, say $B$ is between $A$ and $C$ if $AB+BC=AC$. Now, we just need that affine maps (as defined above) preserve this notion. I can find no counterexamples.

Affine maps do not preserve circles, so it seems these problems are related to straightedge-only constructions, such as the coffin problem

You are given two parallel segments. Using a straightedge, divide one of them into six equal parts.

Actually, the solution to this problem is one of those things that will become unsurprising if we can get affine maps going.

To our advantage, we have another tool in addition to the straight edge. We can draw parallel lines through points, since affine maps preserve parallel lines. It is then easy to prove that midpoints map to midpoints by construction.

Finding the midpoint of segment BC

For any segment $BC$, construct a parallelogram $\square ABCD$. Draw a line parallel to the sides adjacent to $BC$ and through the intersection of the diagonals. It will intersect $BC$ at the midpoint $M$. Following the same construction for $T(A),T(B)$ we see that $T(M)$ must be the midpoint of $T(A)$ and $T(B)$.

Once we have midpoints, we can take midpoints of midpoints, etc. until we have all the ticks on our ruler coinciding through the affine map. I have trouble making the jump to third-way points, much less constructible numbers. The key is in solving the easier problem, which will allow some continuity type argument by repeatedly subdividing and choosing which halves.

  • $\begingroup$ Your midpont argument is essentially the answer to the hard problem for the special case when the proportion is $\frac12$ (or $2$). It generalizes to rational proportions (i.e., the denominator need not be a power of $2$): With the situation of your sketch, we can find $D_n$ on $CD$ such that $CD_n:CD=n$. Clearly, $D_0=C$, $D_1=D$. Given $D_{n-1}$ and $ D_n$, we find $D_{n+1}$ by letting $P$ be the intersection of $D_{n-1}M$ with $AB$ and then $D_{n+1}$ the intersection of $PE$ with $CD$. Let $Q=CA\cap D_nB$ and $R$ the intersection of $D_kQ$ with $AB$. Then $AQ:AB=k:n$. $\endgroup$ – Hagen von Eitzen Jul 11 '18 at 16:38
  • $\begingroup$ Thanks! Now we have all the rational points. I think $D_{n-1}M$ should be $D_nM$, no? $\endgroup$ – Aaron Goldsmith Jul 11 '18 at 17:58

When I was working at University, we did a similar thing for projective geometry. Essentially show that if lines map to lines (in that case including the line at infinity which might become finite, so parallels are not preserved), then harmonic ratios are preserved (which is similar to your midpoint property), and by repeated application of harmonic ratios you can get to every rational point (rational with respect to other points). I'm sure you can find geometric construction to (a) construct $p$ times a given length and (b) divide a given length into $q$ parts, so rational lengths $\frac pq$ should be feasible.

Then it indeed becomes tricky. You really need number theory there. Suppose you have $$\mathbb K=\mathbb Q[\sqrt2]=\{a+b\sqrt2\mid a,b\in\mathbb Q\}$$ as your coordinate space. Then in $\mathbb K^2$ you could have a map $$\begin{pmatrix}a+b\sqrt2\\c+d\sqrt2\end{pmatrix}\mapsto\begin{pmatrix}a-b\sqrt2\\c-d\sqrt2\end{pmatrix}$$ which is not an affine map but which does preserve lines. The general statement would be that any map preserving lines can be written as an affine map combined with an automorphism of the underlying field. If you ever work with complex coordinates, keep in mind that complex conjugation is an automorphism, too.

So you need a property of the reals that the field above does not have, and that the complex numbers don't have either. One key aspect here would be the fact that there exists an order of field elements. With that you can nest intervals and prove convergence of a sequence.

To better concentrate on this relationship between field and geometry, I'd focus on the transformation of a single line, with designated points $0$ and $1$. Once you have established an affine map on such a line, getting the result for the whole plane is easy since you can always form a parallelogram with coordinate axes or some such.

What does it mean for two points $a$ and $b$ on your designated line to satisfy $a>b$, in a way you can express with geometry? One approach would be using squares: $$a>b\quad\Leftrightarrow\quad\exists c:a=b+c^2\;.$$ This uses the fact that over the reals, squares are always non-negative. So if you can come up with a construction that squares a number, using nothing but straightedge and parallels, then you have the key ingredient here. If you don't find such a construction, I can add a figure for that later on.

The idea is taken from lecture notes of a lecture by Prof. Jürgen Richter-Gebert, given at TU München in 2014.

  • $\begingroup$ Thanks MvG! I think this is what I've been looking for. I'll work on distilling it to the real case. $\endgroup$ – Aaron Goldsmith Jul 12 '18 at 15:34

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.