March 23rd, 2014, 04:35 PM  #1 
Senior Member Joined: Nov 2013 Posts: 247 Thanks: 2  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 14 acute angles, 13 right angles, and 15 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? 
March 23rd, 2014, 07:05 PM  #2 
Senior Member Joined: Mar 2014 Posts: 112 Thanks: 8 
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. 
March 23rd, 2014, 07:13 PM  #3  
Senior Member Joined: Nov 2013 Posts: 247 Thanks: 2  Re: Quote:
Thats what I already said. I also said that pentagons can have 15 obtuse angles, 13 right angles, and 14 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.  
March 23rd, 2014, 08:40 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Re: Polygon types Code: A E B D C Whatcha ya gonna do with that one :P 
March 24th, 2014, 10:30 AM  #6  
Senior Member Joined: Nov 2013 Posts: 247 Thanks: 2  Re: Polygon types Quote:
 
April 1st, 2014, 12:15 PM  #7 
Senior Member Joined: Nov 2013 Posts: 247 Thanks: 2  Simpler Problem
Triangles can have 1 obtuse angle, 1 right angle, and 23 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? 
April 2nd, 2014, 06:45 PM  #8  
Senior Member Joined: Nov 2013 Posts: 247 Thanks: 2  Quote:
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 (n1)/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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