Besides right angled triangles, is there any polygon I could use to cover any given (regular or not) polygon? It's clear that given a triangle, square, hexagon or rectangle you would other options. Can this be proven, if right? That is, that the right angled triangle is the only polygon able to $100\%$ cover any other polygon.
Asked
Active
Viewed 333 times
1
-
What do you mean by covering a polygon? How can right-angled triangles be used to covery any given polygon? – Vectornaut Apr 03 '16 at 18:18
-
@Vectornaut: I mean, imagine you can get any size of tiles (provided they are right angled triangles) and you have to cover a floor that is a polygon (or whatever form). – Pierre B Apr 03 '16 at 19:07
-
Covering here should mean the following: Given a polygon $P$, there is a finite collection of polygons $T_i$ from the given class, such that the union of $T_i$'s equals $P$ and the intersection $T_i\cap T_j$ has empty interior whenever $i\ne j$. Yes, indeed, any polygon, convex or not, can be covered by right angled triangles. Incidentally, the problem reduces to the case when $P$ is a triangle, since every polygon can be triangulated. – Moishe Kohan Apr 03 '16 at 23:07
-
@studiosus: do we have to prove that every polygon (regular or not) can be triangulated? Intuitively that makes sense, is that enough? – Pierre B Apr 03 '16 at 23:11
-
@PierreB: if you want to prove that every polygon can be triangulated, you could try it by induction, but there are quite a few subtle details that often go wrong. You could also choose any line not parallel to any of the sides of the polygon. Move this line and whenever it hits a vertex of the polygon, draw the line segment(s) within the polygon. This divides the polygon into trapezoids and triangles. Since trapezoids can be split into two triangles, you're done. – HSN Apr 05 '16 at 11:34
-
http://math.stackexchange.com/questions/83580/every-polygon-has-an-interior-diagonal/83596#83596 – Moishe Kohan Apr 05 '16 at 22:22
3 Answers
2
You'll want to look at Laczkovich, M. "Tilings of Polygons with Similar Triangles." Combinatorica 10, 281-306, 1990.
46 triangles of type $45^\circ-60^\circ-75^\circ$ can perfectly cover a square.
2
First, let me try to formulate your question properly. Let $C$ denote the set of congruence classes of all planar polygons; let $C'\subset C$ be a subset. For instance, $C'$ can consist of all right-angles triangles. Or, fix an angle $\alpha\in (0,\pi)$ and let $C'=T_{\alpha}$ consist of all triangles where one of the angles is $\alpha$. Another example is the class of all isosceles triangles. Or $C'$ can consist of all triangles similar to the given one (as in the Laczkovich's paper in the answer by Ed Pegg). Or you can take $C'=C$ (this is not very interesting, of course).
Definition. Fix a subset $C'\subset C$. We say that $C'$ is universal if every planar polygon $P$ can be subdivided into a finite collection of subpolygons which all belong to $C'$. The latter means that there is a finite collection of polygons $T_i\in C'$, such that the union of $T_i$'s equals $P$ and the intersection $T_i \cap T_j$ has empty interior whenever i≠j.
Theorem. The classes $T_{\pi/2}$ (all right angled triangles) and $T_{2\pi/3}$ are universal.
Proof. First of all, every planar polygon admits a triangulation, see here. Thus, to prove universality of some $C'$, it suffices to show that every triangle can be subdivided into a finite collection of polygons in $C'$.
Each triangle $ABC$ can be subdivided in two right-angled triangles: One of the altitudes of $ABC$ (say, the one from $A$) connects $A$ to a point $D\in BC$. (This is the case if the angle $\angle BAC$ is the largest angle of $ABC$.) Now, cut $ABC$ along $AD$.
The proof of universality of triangles of the angle $2\pi/3$ is a bit more complex. First, each triangle can be subdivided in two or three triangles $T$ such that each $T$ has at least one angle in the interval $(\pi/3, \pi/6)$. Then you show that such $T=ABC$ admits an interior point $D$ such that all three angles $\angle ADB, \angle BDC, \angle CDA$ are the same, equal to $2\pi/3$. qed
It is an interesting question which other classes $T_\alpha$ are universal. Or if the class of isosceles triangles is universal.
Another remark is that the class of all convex quadrilaterals is universal. To prove this, given a triangle $T$, pick its interior point $D$ and interior points $P, Q, R$ of the three edges of $T$. Now, connect $D$ to $P, Q, R$ by segments.
Moishe Kohan
- 97,719
-1
In FEA (Finite Element Analysis), meshing (specially triangle meshing) is required to perform the analysis.
Perhaps some insights of your proof can be found in Wikipedia, Types of Mesh: https://en.wikipedia.org/wiki/Types_of_mesh
CAGT
- 831
- 1
- 7
- 17