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Topic: Acute Angles in N-gon (Read 4118 times) |
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Barukh
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Acute Angles in N-gon
« on: Nov 21st, 2003, 11:56am » |
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How many acute angles an N-gon can have?
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harpanet
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Re: Acute Angles in N-gon
« Reply #1 on: Nov 21st, 2003, 1:54pm » |
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:I'd say 3.
I don't know if this is the correct terminology but, what I shall call the external angle of a vertex, (the angle of turn from one edge to the next) is 180o - the internal angle. The sum of the external angles must be 360o. Therefore if the internal angle is acute the external angle must be at least 91o.
3 * 91 < 360, 4 * 91 > 360.
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Sir Col
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Re: Acute Angles in N-gon
« Reply #2 on: Nov 21st, 2003, 4:51pm » |
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That's a nice demonstration, Harpanet, for convex polygons, but consider a (concave) 'bow-tie' shaped hexagon formed by placing two equilateral triangles apex-to-apex and sliding them together; it would have four interior angles less than 90o.
[e] Edited because my reasoning below was flawed! 
I've removed the hidden and written the erroneous part in orange. [/e]
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Obviously a triangle can have three actute angles.
A quadrilateral arrow can have three acute angles.
It can be shown that all n-gons, for n>4, can have n–2 acute angles. I'm not sure if this can be improved.
I shall prove this by a form of geometrical induction.
We shall make repeated use of congruent trapeziums (Q1, Q2, Q3, ...) with two acute base angles, and two triangles. Both triangles have an acute angle at their apex, but the base of one triangle (T1) is equal to the shorter side (top) of the trapzium, and the base of the other triangle (T2) is equal to the longer (bottom) side of the trapezium.
Start with Q1 and add T1 to the shorter side; this will form a 5-gon with three acute angles.
By placing Q1 and Q2 short side to short side, we form a 6-gon with four acute angles.
Adding T2 to the long side of Q2, forms a 7-gon with five acute angles.
This process can be repeated, by adding trapeziums in alternating directions and adding T1 and T2 respectively.
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It is possible to establish a lower bound generalisation:
(3k+a)-gons, where a=0,1,2, will have k+2 acute angles.
Proof:
Consider a line segment 3 units in length.
If an equilateral triangle, length 1 unit, is added to the middle third of the segment, we introduce one acute angle and three more edges.
An equilateral triangle has has three acute angles. Using the above procedure will produce a 6-gon, with one extra acute angle. Continuing this process ad infinitum completes the proof for (3k)-gons.
This process can be applied to any given n-gon. If the n-gon has m acute angles, then an (n+3)-gon will have x+1 acute angles.
Hence the following generalisations can be made:
As a 3-gon has 3 acute angles, (3k)-gons have at least k+2 acute angles.
As a 4-gon and a 5-gon have 3 acute angles, (3k+1)-gons and (3k+2)-gons have at least k+2 acute angles.
However, a 10-gon (which should have 5 acute angles: 3x3+1=10, 3+2=5), can have 6 acute angles. Consider Q1Q2Q1 (from above) placed back to back.
Hmm... is this an easy problem?
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| « Last Edit: Nov 22nd, 2003, 3:22am by Sir Col » |
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Eigenray
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Re: Acute Angles in N-gon
« Reply #3 on: Nov 22nd, 2003, 2:33am » |
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I suppose it would be a bit of a hyperbole if I said N.
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towr
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Re: Acute Angles in N-gon
« Reply #4 on: Nov 22nd, 2003, 8:26am » |
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If N-gon doesn't imply none of the sides may cross, I'd also say N
So, is a pentagram a pentagon?
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Sir Col
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Re: Acute Angles in N-gon
« Reply #5 on: Nov 22nd, 2003, 12:08pm » |
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We're going to need clarity from Barukh on this. I would say that a pentagram is a 10-gon, as I count the number of external edges to determine the name of the polygon. The word polygon comes from the Greek, poly=many and gon=corners; so strictly speaking we should be counting the number of vertices on the external boundary.
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rmsgrey
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Re: Acute Angles in N-gon
« Reply #6 on: Nov 22nd, 2003, 9:14pm » |
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assuming non-intersecting polygons, confined to a Euclidean (sorry Eigenray) plane:
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As harpanet said, the sum of external angles must be 360. Allowing for reflex (>180) internal angles, external angles must lie in the open interval (-180,180), with those corresponding to acute internal angles lying in (90,180). If the intervals were closed, it would be possible to get 4 acute angles plus 2 per reflex angle. As they are not, the (arbitrarily small) errors have to be accounted for - where possible by making a reflex angle more so; where not, sacrificing an otherwise acute angle.
So, for an N-gon (assuming my calculations are right - I'm falling asleep as I type), the maximum number of acute angles is:
floor((2N+2)/3)+1
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The same result applies to spherical geometry (longer side lengths give bigger angles, but for small enough side lengths, things become arbitrarily close to the Euclidean case)
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Barukh
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Re: Acute Angles in N-gon
« Reply #7 on: Nov 23rd, 2003, 12:39am » |
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on Nov 21st, 2003, 4:51pm, Sir Col wrote:| Hmm... is this an easy problem? |
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Actually, when I posted this problem, I had in mind the convex polygon. Hence, the easy section. And Harpanet gave an elegant answer to that question.
Nevertheless, my mistake did produce some additional thinking, which is good!
on Nov 22nd, 2003, 8:26am, towr wrote:| So, is a pentagram a pentagon? |
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Yes, I think pentagram is a pentagon. Moreover, it’s a regular polygon of a star type. Look, for instance, here.
Anyway, sorry for misleading…
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Sir Col
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Re: Acute Angles in N-gon
« Reply #8 on: Nov 23rd, 2003, 12:39pm » |
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No problem, it was an interesting problem to contemplate. I'm not convinced of your formula, rmsgrey. Not even considering larger N, your formula gives a maximum of 4 acute angles for a 4-gon.
Barukh, I checked my mathematical encyclopaedias and wikipedia.org (if you want to check online), and it seems that a simple polygon has no intersecting lines; otherwise it's called a complex polygon. If we're considering the outline, a pentagram would be a simple concave decagon: five acute and five reflex angles; otherwise, it's a complex pentagon.
Ask Dr. Math at the MathForum addressed my question too:
http://mathforum.org/library/drmath/view/57854.html
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towr
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Re: Acute Angles in N-gon
« Reply #9 on: Nov 24th, 2003, 1:25am » |
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on Nov 23rd, 2003, 12:39pm, Sir Col wrote:| No problem, it was an interesting problem to contemplate. I'm not convinced of your formula, rmsgrey. Not even considering larger N, your formula gives a maximum of 4 acute angles for a 4-gon. |
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He's not talking about convex polygons though.. so a 4-gon can have 4 acute angles..
it's easy to see that for every n=2k you can get n acute angles, just make a saw-tooth out of n-2 edges, and finish the polygon off with the other two..
actually for odd n you can do something similar..
Just as long as the edges aren't of a given length.. In some cases that will make it impossible.. (but that's so even for a simple triangle..)
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| « Last Edit: Nov 24th, 2003, 1:28am by towr » |
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James Fingas
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Re: Acute Angles in N-gon
« Reply #10 on: Nov 25th, 2003, 6:32am » |
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N for complex polygons
[smiley=lfloor.gif]2N/3[smiley=rfloor.gif]+1 for simple polygons
min(N, 3) for simple convex polygons
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All I did for the simple non-convex case was to start generating such polygons. The best I could do with just one reflex angle was 5 acute angles, and it's evident that the shape in the limit goes to two acute angles per reflex angle.
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| « Last Edit: Nov 25th, 2003, 6:47am by James Fingas » |
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rmsgrey
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Re: Acute Angles in N-gon
« Reply #11 on: Nov 26th, 2003, 4:34am » |
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I did say I was falling asleep - redoing my sums while awake gets the same answer as James Fingas - with rather more work!
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