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What is the minimum number of cuts needed to dissect a right angled triangle into acute-angled triangles ?

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  • $\begingroup$ you may want to explain more. Such as What do we cut actually? A4? $\endgroup$
    – Oray
    Commented Aug 26, 2017 at 7:43
  • $\begingroup$ :-) @Oray, probably you can cut it diagonally and take one half and proceed with the problem ! $\endgroup$ Commented Aug 26, 2017 at 8:52

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It depends very much on what you consider to be a cut. I have assumed every cut is like a straight line segment, and that an endpoint of a cut is allowed to be in the interior of the figure.

7 cuts as follows: Cut off two acute triangles so that you are left with a pentagon with four obtuse angles and one right angle. Then make 5 cuts from the corners to the centre of that pentagon.
enter image description here

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  • $\begingroup$ Yes, @Jaap Sch...s, you got it correct. Good illustration too. $\endgroup$ Commented Aug 26, 2017 at 8:39
  • $\begingroup$ Can you prove this is optimal, or is that really difficult? $\endgroup$ Commented Sep 4, 2017 at 13:10
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    $\begingroup$ @greenturtle3141 See boboquacks answer for why there must be at least one interior point where some cuts meet. At such a point there must be at least 5 cuts meeting there, otherwise the average angle size at that point is not acute. So that gives you 5 radial cuts. Where cuts meet an edge of the main triangle you get an obtuse angle unless two cuts meet there. I think that means you need at least two more cuts, because I think only one of the radial cuts can go to a corner of the main triangle. $\endgroup$ Commented Sep 4, 2017 at 13:45
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The minimum number of cuts is:

You can't :P

Since:

Suppose the minimum number of cuts is $m$ with $\triangle ABC$, which has its non-acute angle at $A$. Then some cut $K$ must pass through $\angle A$ else it would be part of a non-acute-angled triangle. Let that cut meet $BC$ at $D$. WLOG $\angle ADB\geq\angle ADC$, then, since they sum to $180^\circ$, $\angle ADB\geq 90^\circ$. So then $\triangle ADB$ must be partitioned into acute-angled triangles in $\leq m-1$ cuts (because $K$ won't help cut up the triangle), contradicting the minimality of $m$. So such a dissection is impossible.

Note:

I am assuming cuts go all the way through the triangle.

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  • $\begingroup$ well that is the first impression. However, if you try, you can get the solution. $\endgroup$ Commented Aug 26, 2017 at 6:52
  • $\begingroup$ Well, I agree that it depends upon how a cut is defined / interpreted as. It need not go through entirely (from a vertex to an opposite side altogether). It can stop in between ! $\endgroup$ Commented Aug 26, 2017 at 9:09
  • $\begingroup$ @MeaCulpaNay To be fair, that should have been included in the question. $\endgroup$
    – Rubio
    Commented Aug 26, 2017 at 21:01

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