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March 23rd, 2014, 04:35 PM   #1
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Polygon types

Equilateral pentagon: all sides are equal and all sets of 3 adjacent angles are in a range of angles(doesn't have to be regular meaning it is also equiangular)

Pentagons can have 1-4 acute angles, 1-3 right angles, and 1-5 obtuse angles.
They all have to add up to 540
with a 4th right angle the last angle would be 180 degrees which is always a problem with pentagons. Thus a pentagon cannot have a 4th right angle.
Is there a way to figure out all the different angles possible in a pentagon with these 2 sentences about angles? And is this a sum of permutations or a sum of combinations?
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March 23rd, 2014, 07:05 PM   #2
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Of course it has to add up to (5 - 2) × 180° = 540°. Also all regular pentagons are equilateral pentagons, but not vice versa.
Again if it's a regular pentagon then divide by 5 and we get angle with 108°.

Last edited by skipjack; April 1st, 2014 at 06:52 PM.
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March 23rd, 2014, 07:13 PM   #3
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Re:

Quote:
Originally Posted by the john
Of course it has to add up to (5 - 2) × 180° = 540°. Also all regular pentagons are equilateral pentagons, but not vice versa.
Again if it's a regular pentagon then divide by 5 and we get angle with 108°.

Thats what I already said.
I also said that pentagons can have 1-5 obtuse angles, 1-3 right angles, and 1-4 acute angles.

I was asking how I can figure out all the possible angles assuming that all degree measures are integers between 0 and 180 but not including 0 and 180 itself and if this is a sum of permutations or a sum of combinations.

Last edited by skipjack; April 1st, 2014 at 06:53 PM.
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March 23rd, 2014, 07:17 PM   #4
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Regardless of repeating what you said, this is the best information I could find.
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March 23rd, 2014, 08:40 PM   #5
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Re: Polygon types

Code:
              A




           E     B

D                           C
Thassa ABCDE, a cute concave pentagon, angles A,C,D = 30 degrees each, angles B,E = 225 degrees each.

Whatcha ya gonna do with that one :P
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March 24th, 2014, 10:30 AM   #6
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Re: Polygon types

Quote:
Originally Posted by Denis
Code:
              A




           E     B

D                           C
Thassa ABCDE, a cute concave pentagon, angles A,C,D = 30 degrees each, angles B,E = 225 degrees each.

Whatcha ya gonna do with that one :P
I didn't know pentagons could have reflex angles. It makes sense though because not all shapes are convex and the ones that aren't convex have reflex angles.
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April 1st, 2014, 12:15 PM   #7
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Simpler Problem

Triangles can have 1 obtuse angle, 1 right angle, and 2-3 acute angles.

The angles are integers between 0 and 180 but not including 0 and 180.

How many triangles are there as far as angles and how many of them are obtuse, how many of them are right, and how many of them are acute?
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April 2nd, 2014, 06:45 PM   #8
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Quote:
Originally Posted by caters View Post
Triangles can have 1 obtuse angle, 1 right angle, and 2-3 acute angles.

The angles are integers between 0 and 180 but not including 0 and 180.

How many triangles are there as far as angles and how many of them are obtuse, how many of them are right, and how many of them are acute?
I calculated the number of each this way. I was first thinking “Okay Obtuse Triangles. Lets see for each angle there are 2 less obtuse triangles starting with the isosceles.” There were 88 with 1 and 0 with 89 and each angle in degrees subtracts 2 triangles from the last so I had 88+86+84+82….+4+2+0. I added them all up and I got 1,980 total obtuse triangles. I got 89 right triangles. The same thing happened when I was adding the acute triangles. I did the integer sum of Xn = 2*1+2*2…+2*45 but than I didn’t half either the right or the acute like I should of.

Is there something wrong with just straight halving the acute and right and if it can't be cut evenly in half adding 1 to (n-1)/2 like this: 1980 obtuse + 88/2 + 1 right + 3916/2 acute = 3060 total triangles given what I said in the previous post about the same problem?

Last edited by caters; April 2nd, 2014 at 06:47 PM.
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