# Right angled triangle to all acute angled triangles

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

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

## 2 Answers

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.

• Yes, @Jaap Sch...s, you got it correct. Good illustration too. Aug 26 '17 at 8:39
• Can you prove this is optimal, or is that really difficult? Sep 4 '17 at 13:10
• @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. Sep 4 '17 at 13:45

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.

• well that is the first impression. However, if you try, you can get the solution. Aug 26 '17 at 6:52
• 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 ! Aug 26 '17 at 9:09
• @MeaCulpaNay To be fair, that should have been included in the question.
– Rubio
Aug 26 '17 at 21:01