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polygon

by shashank.ism » Tue Feb 09, 2010 12:56 pm
Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?

24
23
22
20
21
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by komal » Wed Feb 17, 2010 4:54 am
shashank.ism wrote:Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?

24
23
22
20
21
Each angle is 180(p-2)/p.
180-360/p=k So 360/p has to be an integer.
Factors of 360 = 23.32.51
So there are 4.3.2 solutions, but we exclude 1 and 2, because p>= 3 So , 24 -2 =22
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by harsh.champ » Thu Feb 18, 2010 2:28 am
komal wrote:
shashank.ism wrote:Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?

24
23
22
20
21
Each angle is 180(p-2)/p.
180-360/p=k So 360/p has to be an integer.
Factors of 360 = 23.32.51
So there are 4.3.2 solutions, but we exclude 1 and 2, because p>= 3 So , 24 -2 =22
Hey komal,
Each angle is 180(p-2)/p.
Is this a standard formula?Did you work this out or have you memorized it?

Also,I could not understand this explanation:-
Factors of 360 = 23.32.51
So there are 4.3.2 solutions
Can you clarify the soln. approach??
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