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.
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| 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 45º-45º-90º 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 45º-45º-90º triangle
to be "squashed" or "stretched" slightly
without invalidating the triangulation. |
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The dashed circles above represent
"forbidden regions" in which one of
the angles would be obtuse. |
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| 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 90º,
so the sum of angles A and B is at most 90º, so angle A is at most 45º.
If angle B is more than 45º, 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 45º.
(Demonstration:
First, note that angle C is less than 135º because angle B is at least
45º and angle A is greater than zero.
So C-45º is less than a right angle.
Now select point D so that angle C-45º < angle BCD < 90º
Angle BDC = 180º - angle B - angle BCD,
< 180º - angle B - (C-45º),
= 180º - (B-45º) - angle C,
< 180º - angle C,
< 90º
Since angles BDC, BCD, and B are all less than 90º, this is an acute
triangle.)
The remaining triangle, triangle ADC has two angles that are at most
45º and one angle that is at least 90º. 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.
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| 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 90º. Similarly, draw line DG almost horizontal.
 Then
draw line EF almost vertical, so angle AFE is less than 90º.
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 90º 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 90º, but it can come arbitrarily close to 90º, which brings
angle FDH back under 90º. 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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