I was trying to prove that every convex $n$-gon can be decomposed into $2n$ right triangles. Here a sketch of the flawed proof:
$(i)$ Every convex $n$-gon with consecutive vertices $V_1, V_2, \dots, V_n$ and sides $s_1, s_2, \dots, s_n$ has some interior region $R$ with infinite points $P$ such that for every side $s_i$ of the $n$-gon there exists some straight line $l_i$ perpendicular to $s_i$ such that $P\in l_i$ (therefore, $P\in l_i \space \forall i$).
$(ii)$ From $(i)$, if we denote as $C_i$ the crossing point of $s_i$ and $l_i$, it is straightforward to note that $PV_{i}C_i$ and $PC_{i}V_{i+1}$ form two right triangles for every $i$ (take into account that $i$ is cyclic, and so $n+1=1$).
The problem comes with region $R$. This region can be constructed tracing perpendiculars to every side $s_i$ through each of its vertices (thus, every vertex will be contained in two of the perpendiculars, one for each of the sides it joins). Below you can find am example:
$R$" />Unfortunately, not every convex $n$-gon has a region $R$. If the interior angle formed at some $V_i$ by $s_i$ and lines from $V_i$ to each other vertex are all obtuse, then clearly for $n>4$ there will be at least some side $s_j$ whose perpendicular lines will not cross $s_i$'s perpendicular lines in the interior of the $n$-gon (indeed, if $s_i$ and $s_j$ are parallel, the perpendicular lines will never cross).
It seems then that a sufficient condition for some convex $n$-gon to be decomposable into $2n$ right triangles would be that at least one of the interior angles formed at some $V_i$ by $s_i$ and lines from $V_i$ to each other vertex is not obtuse.
I would like to know if (as most probable) this is a general result known in geometry, and a reference of where to find it; also, if the proof seems sound, how to prove the sufficiency condition (maybe I am mistaken, and is necessary but not sufficient), and in general how to improve it.
Thanks!
I assume by "rectangle triangle" you mean "right triangle".
Take a convex $n$-gon $A$ with consecutive vertices $V_i$ for $1\leq i\leq n$. Fix $P=V_n$. Form $n-2$ (in general non-right) triangles $B_j$ by the vertices $PV_jV_{j+1}$ for $1\leq j\leq n-2$. Each of these $n-2$ triangles can be subdivided into two right triangles by the altitude through the vertex with the largest angle - this produces $2(n-2)=2n-4$ right triangles subdividing $A$. By subdividing some of the right triangles further into smaller right triangles we can get to $2n$ right triangles.
One might ask further what the minimum number of right triangles is to subdivide a convex $n$-gon.
Every simple polygon (convex polygons are simple, but in general not vice versa) must have a minimum of $n-2$ (possibly non-right) triangles in a triangle decomposition. This can be proved by induction. It is more tricky than one thinks at first, because often for plane polygons one requires that a triangulation of a polygon only includes triangles whose vertices coincide with some vertices of the original polygon. Thus usual proofs for standard triangulations will not be general enough for the type of triangle decomposition you have in mind. It is however not too hard either.
Thus every decomposition of a convex (or even simple) $n$-gon must consist of a minimum of $n-2$ right triangles. Our procedure above shows, that in any case the number of minimum right triangles is at most $2\cdot(n-2)$. Obviously there is a gap here.
It is definitely true that some (convex) $n$-gons can be decomposed by $n-2$ right triangles. Look at a spiral made up by right triangles for inspiration: In the example in the link the polygon will however no longer be convex after attaching the 6th matching right triangle. But if you do not start with a triangle of side lengths $1$, $1$ and $\sqrt{2}$, but with a "flatter" right triangle you can successively attach more right triangles before becoming nonconvex. E.g. when one starts later in the spiral with side lengths $1, \sqrt{6}, \sqrt{7}$ only attaching the $11$th right triangle will produce something nonconvex. When this happens and you still want your $n$ to be larger: restart with an "even flatter" right triangle to begin with.
However there are $n=3$-gons (i.e. triangles) that need $2\cdot(n-2)=2$ right triangles in a decomposition. If you have a non-right triangle it obviously cannot be covered by $1$ right triangle you need at least $2=2\cdot (3-2)$.
I do currently not know, if for every $n$ there is a (convex) $n$-gon that cannot decomposed into less than $2\cdot(n-2)$ right triangles. I leave that as an open problem.