I'm working through the section in baby Hartshorne on Archimedian Neutral Geometry, and I've been stuck for a couple of days trying to (synthetically) prove the following: In a Hilbert plane satisfying Archimedes' axiom where the parallel postulate is false, given an angle $\epsilon \gt 0$, show that there exists a triangle with angles $\alpha, \beta, \gamma$, all three smaller than $\epsilon$.

There was even a hint to use an earlier exercise (which I have already proved), namely: For any angle $\alpha$, however small, there exists a line $l$ entirely contained in the inside of the angle. Also, we have already proved Saccheri-Legendre. I remain stuck. Any assistance would be welcome. Thanks.


I think I've answered my own question. I'll provide a very rough sketch of proof here in case it's of use to someone. We use the third fact below from Hartshorne's exercises (easily established, for example, with the help of some solutions here https://math.berkeley.edu/~serganov/130_2014/sol12.pdf ).


In the latter diagram, let $\alpha \lt \varepsilon$. Choose $B^\prime$ further along the ray $AB$, on the side opposite $A$, so that $\angle AB^\prime P \lt \varepsilon$ (It is readily observed that if we choose a point $X$ at a distance $BP$ further down the ray from $B$ then $\triangle B X P$ is isosceles and then $\angle A X P \leq \frac{\angle A B P}{2}$ by the (semi-hyperbolic) exterior angle theorem. We can repeat choosing new points and halving the angle until we arrive at a suitable $B^\prime$. See for example baby Hartshorne p. 322). Similarly, we can choose $C^\prime$ further along the ray $AC$ so that $\angle AC^\prime P \lt \varepsilon$. Then in $\triangle A B^\prime C^\prime$, $\angle B^\prime A C^\prime \lt \varepsilon$ by construction, $\angle A B^\prime C^\prime \lt \angle A B^\prime P \lt \varepsilon$ and $\angle A C^\prime B^\prime \lt \angle A C^\prime P \lt \varepsilon$, by the third exercise, giving us the desired triangle.


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