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Moulton plane
The Moulton plane $M$ is an incidence structure defined on the Euclidean plane $\mathbb{E}^{2}$ as follows:

the points of $M$ are points of $\mathbb{E}^{2}$

the lines of $M$ are the following three kinds:
(a) vertical lines (with equations of the form $x=a$) of $\mathbb{E}^{2}$ are lines of $M$
(b) nonnegative sloped lines (with equations of the form $y=mx+b$ with $m\geq 0$) of $\mathbb{E}^{2}$ are lines of $M$
(c) sets of pairs $(x,y)$ where, for some $m,b\in\mathbb{R}$, with $m<0$,
$y=\left\{\begin{array}[]{ll}mx+b&\textrm{if }x<0,\\ 2mx+b&\textrm{otherwise,}\end{array}\right.$ are lines of $M$. These lines are called the bent lines, because they are bent at the $y$axis. We call $m$ and $b$ respectively the slope and the $y$intercept of the (bent) line.

incidence between points and lines of $M$ is $\in$, the usual set membership relation.
The following figure shows an example of a bent line:
\pspicture(4,2.75)(4,2.75) \pssetunit=25pt \psline¡¿(1.5,2.5)(1.5,2.5) \uput[r](1.5,2.4)$y$ \psline¡¿(4,1.5)(4,1.5) \uput[r](4,1.5)$x$ \psline¡(4,2)(1.5,1) \psline¿(1.5,1)(2.5,2.5)
For convenience, let us write $y=x*m*b$ to represent the bent line whose slope is $m$ and whose $y$intercept is $b$.
Proposition 1.
$M$ is an affine plane where the Desargues’ theorem fails.
Proof.
Verifying that the plane is affine is straightforward:

Given two distinct points $(s,t),(u,v)$, if $s=u$, then $x=s$ is the line passing through them. Otherwise, let $m=(tv)/(su)$. If $m\geq 0$, then $y=mx+b$ is a line passing through them, where $b=tms$. If $m<0$, we have three subcases: (we may assume that $s<u$)

$0\leq s<u$. Then the line passing through the points is given by the equation $y=x*(m/2)*b$.

$s<u\leq 0$. Then the line passing through the points is given by the equation $y=x*m*b$.

$s<0<u$. Then, by setting $m^{{\prime}}=(tv)/(s2u)$ and $b^{{\prime}}=tm^{{\prime}}s$, the line $y=x*m^{{\prime}}*b^{{\prime}}$ passes through the two points.
These lines are clearly uniquely determined by the two points.


Axiom 2: given a line and a point not on the line, there is exactly one line incident with the point that is parallel to the given line (Playfair axiom).
Given a line $\ell$, and a point $(s,t)$ not on $\ell$. If $\ell$ is of the form $x=a$, then $s\neq a$, and the line $x=s$ passes through $(s,t)$ and is parallel to $\ell$. If $\ell$ is of the form $y=mx+b$ with $m\geq 0$, then $t\neq ms+b$. By setting $c=tms\neq b$, the line $y=mx+c$ passes through $(s,t)$ and is parallel to $\ell$. Finally, if $\ell$ is a bent line given by $y=x*m*b$. There are two subcases:

If $s\leq 0$, then $t\neq ms+b$. By setting $c=tms\neq b$, we get a bent line $y=x*m*c$ through $(s,t)$ parallel to $\ell$.

If $s>0$, then $t\neq 2ms+b$. By setting $d=t2ms\neq b$, we get a bent line $y=x*m*d$ through $(s,t)$ parallel to $\ell$.
Again, all of the lines determined are unique.


Axiom 3: there are at least three noncollinear points.
$(0,0),(1,0)$, and $(1,1)$ are three noncollinear points of $M$.
Lastly, to show that Desargues’ theorem fails in $M$, we need to find a pair of triangles $ABC$ and $A^{{\prime}}B^{{\prime}}C^{{\prime}}$, such that lines $AA^{{\prime}},BB^{{\prime}}$, and $CC^{{\prime}}$ are concurrent, but points $X:=AB\cap A^{{\prime}}B^{{\prime}}$, $Y:=BC\cap B^{{\prime}}C^{{\prime}}$ and $Z:=CA\cap C^{{\prime}}A^{{\prime}}$ fail to be collinear. Below is such an example. Note that $X,Y$ are on the $x$axis, but $Z$ is not. The dotted line merely shows that if the line is “unbent”, then the Desarguesian property is restored.
\pspicture(8,4.5)(6,8) \pssetunit=25pt \psline¡¿(0,4.5)(0,5) \uput[r](0,4.9)$y$ \psline¡¿(7,1.5)(7,1.5) \uput[r](7,1.5)$x$ \psline¡¿(6.5,4.5)(2.58824,5) \psline¡¿(3,4.5)(3,5) \psline¡¿(0.5,4.5)(3.29412,5) \psdots[linecolor=red,dotsize=5pt](4.6,0.114286) \uput[l](4.6,0.114286)$A$ \psdots[linecolor=blue,dotsize=5pt](6.25,3.89286) \uput[l](6.25,3.89286)$A^{{\prime}}$ \psdots[linecolor=red,dotsize=5pt](3,2.5) \uput[l](3,2.75)$B$ \psdots[linecolor=blue,dotsize=5pt](3,0.75) \uput[r](3,0.5)$B^{{\prime}}$ \psdots[linecolor=red,dotsize=5pt](1.75,0.25) \uput[r](1.75,0.05)$C$ \psdots[linecolor=blue,dotsize=5pt](1,2.8) \uput[r](1,3)$C^{{\prime}}$ \psline[linecolor=red,linewidth=2pt](4.6,0.114286)(3,2.5) \psline[linecolor=red,linewidth=2pt](3,2.5)(1.75,0.25) \psline[linecolor=red,linewidth=2pt](1.75,0.25)(4.6,0.114286) \psline[linecolor=blue,linewidth=2pt](6.25,3.89286)(3,0.75) \psline[linecolor=blue,linewidth=2pt](3,0.75)(1,2.8) \psline[linecolor=blue,linewidth=2pt](1,2.8)(6.25,3.89286) \psdots[linecolor=green,dotsize=8pt](3.67283,1.45) \uput[l](3.7,1.25)$X$ \psdots[linecolor=green,dotsize=8pt](2.38053,1.45) \uput[r](2.3,1.25)$Y$ \psdots[dotsize=8pt](6.304334,1.45) \psline[linestyle=dotted](1.75,0.25)(6.304334,1.45) \psline(1,2.8)(6.304334,1.45) \psline(1.75,0.25)(0,0.51073) \psline(0,0.51073)(4.089655,1.85932) \psdots[linecolor=green,dotsize=8pt](4.089655,1.85932) \uput[d](4.089655,1.95)$Z$
Using the above diagram as a guide, one may find the appropriate coordinates for the configuration that work. We invite the reader to complete the proof by finding these coordinates. ∎
Remarks.

The $2$ in the definition of bent lines is sometimes called the refraction index, borrowing from physics.

Another curious fact about the Moulton plane is that there exists a triangle such that the sum of its interior angles is greater than $\pi$. We invite the reader to find such a triangle.

Variations of the Moulton plane exist. Of course, one may replace $2$ by any other positive real number $r$ (nonpositive $r$ is not allowed, or else the plane fails to be affine entirely) other than $1$, and the resulting plane is still nonDesarguesian affine. Or, instead of the $y$axis as the “line of refraction”, one may choose a different line. Another popular choice is the $x$axis.

Using the Moulton plane, one can then construct a nonDesarguesian projective plane using the usual method of completion by adding a line of infinity, such that each point on the line (point of infinity) is the equivalent class of parallel affine lines (two bent lines are parallel iff they have the same slope).
References
 1 R. Artzy, Linear Geometry, AddisonWesley (1965)
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