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 Skip Navigation LinksMath Help > Math Puzzles > Triangle Partition Puzzle > Triangle Partition Puzzle

Partition a triangle into a finite number of acute triangles

Is it possible to dissect any triangle into a finite number of acute triangles?

Source: http://mathforum.org/wagon/spring02/p960.html, which credits
James Tanton, Math Horizons, April 2002.
In researching this question, I found Acute Square Triangulation, which gives a method of partitioning a square into eight acute triangles (diagram, right).  This website points out that a triangulation using triangles that are "more acute" (i.e. a smaller maximum angle) is possible with 14 triangles.
 
This diagram, left, is fascinating because it shows the partitioning of two 45�-45�-90� triangles into acute triangles -- not just acute, but "very" acute.  That means there's a lot of "give" in this triangulation, which would allow the 45�-45�-90� triangle to be "squashed" or "stretched" slightly without invalidating the triangulation.
The dashed circles above represent
"forbidden regions" in which one of
 the angles would be obtuse.
In this same web page, the curious comment is made, "Later, I and others showed that any polygon could be triangulated with non-obtuse triangles, and that in fact only O(n) non-obtuse triangles suffice to triangulate an n-sided polygon. However some polygons might require right angles or angles very close to 90 degrees."

If it were true that any triangle could be partitioned into acute triangles then any polygon that could be triangulated with non-obtuse triangles could be further partitioned into acute triangles.  So the fact this author chose to triangulate with "non-obtuse" triangles instead of making the stronger statement involving "acute" triangles leads me to speculate that some right triangles can't be partitioned into acute triangles.  Or maybe the reason for "non-obtuse" instead of "acute" is to keep the number of partitions down to O(n).
Enough speculation.  Let's see if we can just partition a triangle.

If the triangle is acute, we're done, so let's assume it's non-acute.  Let the angles be A ≤ B ≤ C.  So angle C is at least 90�, so the sum of angles A and B is at most 90�, so angle A is at most 45�.

If angle B is more than 45�, then the triangle can be partitioned by drawing line CD into an acute triangle BCD and an obtuse triangle ACD in which angle ACD is at most 45�.

(Demonstration:
First, note that angle C is less than 135� because angle B is at least 45� and angle A is greater than zero.
So C-45� is less than a right angle.
Now select point D so that angle C-45� < angle BCD < 90�
Angle BDC = 180� - angle B - angle BCD,
  < 180� - angle B - (C-45�),
  = 180� - (B-45�) - angle C,
  < 180� - angle C,
  < 90�
Since angles BDC, BCD, and B are all less than 90�, this is an acute triangle.)

The remaining triangle, triangle ADC has two angles that are at most 45� and one angle that is at least 90�.  Our partitioning problem boils down to the question of whether such a "flattened" obtuse triangle can be partitioned into a finite number of acute triangles.
The answer, in a word, is yes.

First, look at the diagram, then I'll explain the method.
First, place point D directly below, and close enough, to C so that angle FDH will be acute.  You'll see why the proximity of D to C is important in a minute.

Then draw line DE almost, but not quite horizontal, so angle CDE is less than 90�.  Similarly, draw line DG almost horizontal.

Then draw line EF almost vertical, so angle AFE is less than 90�.  Similarly, draw GH almost vertical.

That's all there is to it.  If your "almost" angle is small enough, and if segment CD is short enough, all the inner triangles will be acute.

Now you see what happens if line CD is too long:

The bottom triangle, FDH, is no longer acute.  This is never a problem if you make CD short enough, and there's no way to make it too short.
If the triangle is "flatter" you will need to make CD shorter so that triangles all sort of "bunch together" near C, and so every triangle is acute.  Also, the word "almost", when I use it in the instructions, above, refers to an angle that is very small compared to angle A.  So if the triangle is very flat (i.e. angle A is very small) then the inclination of DE, with respect to the horizontal base AB, must be "much less" than angle A.
So far, I've been dealing with isosceles (or nearly isosceles) triangles.  A new challenge emerges when angle A is very small compared to angle B.  Here is what that looks like (points B, E, and F are "off camera").  The problem is that angle FDH has a tendency to get too big when you follow these instructions.  I want you to see that angle D can be made smaller than 90� by simply shrinking the length of AD to the point that DH becomes "almost almost" vertical.  (I say "almost almost", because DG is almost horizontal, and DH can't become as vertical as DG is horizontal without making angle HDG bigger than 90�, but it can come arbitrarily close to 90�, which brings angle FDH back under 90�.  So "almost almost" is a smaller angle than just "almost"!) 
I'll admit this is far from a rigorous mathematical proof.  However, I'm quite confident that it could be turned into such a thing (at the risk of obscuring its logic), and I'm prepared to leave it as it is.  If you're still not satisfied, then I encourage you to take it to the next step of rigor, and express just how small AD needs to be, and just how small "almost" and "almost almost" need to be in terms of angles A, B, and C.  Then email it to me, or better yet, put it on the web and let me know where it is so I can link to it.

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