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A kite is a quadrilateral with reflection symmetry across a diagonal. A kite with two opposite angles right is a right kite. If a pair of angles in a kite are equal and acute (obtuse), we may say the kite is acute(obtuse). A rhombus that isn’t a square is a kite that is both acute and obtuse.

Any triangle allows partition into 3 right kites that meet at its incentre. So any polygon allows partition into right kites.

Question: It is possible to form convex polygons that can be partitioned into kites that are all obtuse. But can any convex polygon - in particular, any triangle - be partitioned into finitely many obtuse kites? If not, how does one decide if a given convex polygon allows such a partition?

And What about partition into kites that are all acute? What if rhombuses cannot be used as pieces?

Note: we are looking only for convex kites as pieces. For a partition into acute kites with some pieces convex and some non convex kites (‘darts’), see Peter Taylor’s comment below.

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    $\begingroup$ A regular $2n$-gon can be partitioned into $n$ kites by cutting from the centre to every other corner. $\endgroup$ Commented Aug 26 at 5:49
  • $\begingroup$ Thanks. That settles the question on the existence of convex polygons that can be cut into acute kites. $\endgroup$ Commented Aug 26 at 5:57
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    $\begingroup$ Given a right kite, pick a point inside the kite on its axis of symmetry and cut from that point to the two vertices not on that axis of symmetry. With the exception of the degenerate case where this cuts it into two triangles, this cuts it into two acute kites, and there are at most two exceptional points where one of those acute kites is a rhombus. Therefore any triangle can be cut into six non-rhombus acute kites. $\endgroup$ Commented Aug 27 at 16:06
  • $\begingroup$ Thanks. This does give 2 acute kites from a right kite but one of them is non convex. I was looking for only convex kites as pieces but had missed saying so explicitly. Let me specify that in question. $\endgroup$ Commented Aug 27 at 17:16
  • $\begingroup$ For questions involving convex quadrilaterals, there is a natural sense based on ideas of Branko Grunbaum in which there are 21 different types of convex quadrilaterals. So one can extend the spirit of the question asked here. See this paper for a discussion and extensions of Grunbaum’s approach: jstor.org/stable/10.5951/mathteacher.106.7.0541 $\endgroup$ Commented Aug 29 at 19:17

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But can any convex polygon - in particular, any triangle - be partitioned into finitely many obtuse kites?

No!

It's impossible when a polygon has two consecutive acute angles:
enter image description here

Proof:

The horizontal line must be formed by a sequence of $n$ kites, sharing $n-1$ vertices along that line:
enter image description here

We'll say one of these kites claims a vertex if that kite has an obtuse angle at that vertex. (e.g. The shaded kite claims the orange vertex).
enter image description here

Note that each obtuse kite must claim at least one vertex. Also, each vertex can be claimed once, at most.

So, $n$ vertices are required, but only $n-1$ are available.

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    $\begingroup$ Really nice n intuitive. thanks! Hope u could also address the acute kite case. Marking the question solved anyways. $\endgroup$ Commented Aug 29 at 8:58
  • $\begingroup$ Is the convex polygon having 2 consecutive acute angles necessary? Your proof shows sufficiency $\endgroup$ Commented Sep 2 at 13:30
  • $\begingroup$ The proof can include right angles as well. So a square can't be partitioned. Beyond that, I don't know -- I'd suggest posting a new question. $\endgroup$ Commented Sep 2 at 13:55
  • $\begingroup$ Thanks again! I will do as u suggest $\endgroup$ Commented Sep 4 at 13:45

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