A set of points of the plane is called obtuse-angled if every three of it's points are not collinear and every triangle with vertices inside the set has one angle $ >91^o$. Is it correct that every finite obtuse-angled set can be extended to an infinite obtuse-angled set? (UK)
Problem
Source: BMO Shortlist 2015 G3 (UK)
Tags: geometry, combinatorial geometry, angles
11.04.2022 17:12
parmenides51 wrote: Is it correct that every finite acute-angled set can be extended to an infinite obtuse-angled set? Did you want to write "Is it correct that every finite obtuse-angled set can be extended to an infinite obtuse-angled set?"? Because if set is not obtuse, then there are 3 points such that they form non-obtuse triangle and since these 3 points are in this set until end, the set can't be extended to an obtuse set. So probably you made typo?
11.04.2022 22:29
indeed I had a typo, thanks for noticing there was no acute in the wording, only obtuse, just corrected it
31.01.2024 14:29
The answer is yes. We'll show that it's always possible to add a point to an obtuse-angled set, which readily implies the extension of any finite obtuse-angled set to an infinite one by constructing infinitely many points all at once using the same idea. Let the points in a finite obtuse-angled set be $A_{1}, A_{2}, \ldots, A_{n}$. Construct a point $X$ such that $\angle XA_{1}A_{2} = \theta$ and $XA_{1} = \varepsilon$ for small enough $\theta, \varepsilon > 0$. Then $\triangle XA_{i}A_{j}$ has an angle larger than $91^{\circ}$ for $i>j>1$ because $\triangle A_{1}A_{i}A_{j}$ does and $\varepsilon$ is small enough. For the triangles $XA_{1}A_{i}$, notice that $\angle A_{1}XA_{2} \approx 180^{\circ}$ and for $i\neq 2$ we have that $\angle A_{1}A_{i}X \approx 0^{\circ}$ and $\angle XA_{1}A_{2} \not\in [89^{\circ}, 91^{\circ}]$ as $\triangle A_{i}A_{1}A_{2}$ is obtuse-angled (so $\angle A_{i}A_{1}A_{2} \not\in [89^{\circ}, 91^{\circ}]$). These two angle conditions imply that $\triangle XA_{1}A_{i}$ is also obtuse-angled, which completes the construction.