Question 1199136: If a polygonal n sides has (n/2)(n - 3) diagonals, how many sides will a polygon with 65 diagonals have? Is there a polygon with 80 diagonals?
Let me see.
I gotta set (n/2)(n - 3) equal to 65 and solve for n. The same with 80.
Yes?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52795)   (Show Source):
You can put this solution on YOUR website! .
Your first idea is right.
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| But in such problems, you ALWAYS must make one more (next) |
| step forward and to check, if integer solution does exist. |
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In case of 65 diagonals it does exist: the number of sides is n = 13.
In case of " 80 diagonals " integer solution DOES NOT exist.
You will easily detect it, when you apply the quadratic formula.
Not for every given " number of diagonals " the integer solution does exist.
So, very often such problems have a  hidden underwater stone (as a trap),
and the duty of a person who solves such problem is to detect it / (to recognize it),
if such trap does present in the hidden form in the problem.
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By the way, the formulation of the problem starts with the words
If a polygonal n sides has (n/2)(n - 3) diagonals . . .
The word " if " is excessive in this phrase : the number of diagonals
of any convex n-gon is  , for any n > 3, and without any " if ".
Answer by math_tutor2020(3817)  (Show Source):
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