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

  • $\begingroup$ you may want to explain more. Such as What do we cut actually? A4? $\endgroup$ – Oray Aug 26 '17 at 7:43
  • $\begingroup$ :-) @Oray, probably you can cut it diagonally and take one half and proceed with the problem ! $\endgroup$ – Mea Culpa Nay Aug 26 '17 at 8:52

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

  • $\begingroup$ Yes, @Jaap Sch...s, you got it correct. Good illustration too. $\endgroup$ – Mea Culpa Nay Aug 26 '17 at 8:39
  • $\begingroup$ Can you prove this is optimal, or is that really difficult? $\endgroup$ – greenturtle3141 Sep 4 '17 at 13:10
  • 1
    $\begingroup$ @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. $\endgroup$ – Jaap Scherphuis Sep 4 '17 at 13:45

The minimum number of cuts is:

You can't :P


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.


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

  • $\begingroup$ well that is the first impression. However, if you try, you can get the solution. $\endgroup$ – Mea Culpa Nay Aug 26 '17 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 ! $\endgroup$ – Mea Culpa Nay Aug 26 '17 at 9:09
  • $\begingroup$ @MeaCulpaNay To be fair, that should have been included in the question. $\endgroup$ – Rubio Aug 26 '17 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.