SOLUTION: 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

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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(52797)   (Show Source): You can put this solution on YOUR website!
.

Your first idea is right.


    +-----------------------------------------------------------------+
    |    But in such problems, you ALWAYS must make one more (next)   |
    |    step forward and to check,  if integer solution does exist.  |
    +-----------------------------------------------------------------+


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.


////////////////


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): You can put this solution on YOUR website!

n = number of sides, such that n > 3.
d = number of diagonals of a polygon with n sides












Apply the quadratic formula to solve for n.








or

or

or

Or you can factor like so




or

or
Ignore the negative value of n.
It doesn't make sense to have a negative number of sides.

Therefore, a polygon with n = 13 sides will have d = 65 diagonals.

---------------------------------------------------------------------------------

If you were to plug in d = 80, then,










Now use the quadratic formula






Each decimal value from here on out is approximate.

or

or

or
We unfortunately do not get a positive integer solution in either case.
This means it's impossible to have a polygon with exactly 80 diagonals.


We can see this by constructing a table of values
n = number of sides
d = number of diagonals
n456789101112131415
d259142027354454657790

Example scratch work calculation for n = 10:
d = (n*(n-3))/2 = (10*(10-3))/2 = 35

The value d = 80 does not exist in the table. The closest two values are d = 77 and d = 90 (for n = 14 and n = 15 respectively).
Note the solution found earlier is between n = 14 and n = 15.

Another way to see that we won't have a positive integer solution is to compute the discriminant







This result is not a perfect square, which means isn't an integer.
Consequently it causes both roots to be non-integer values.

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