Question 1120642
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The number of diagonals in a polygon of *[tex \Large n] sides is given by *[tex \Large \frac{n(n\ -\ 3)}{2}].


So you have the following information about two polygons, one with *[tex \Large n] sides and one with *[tex \Large m] sides.  The sum of their sides is 13 and the sum of their diagonals is 25, which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ +\ m\ =\ 13]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n(n\ -\ 3)}{2}\ +\ \frac{m(m\ -\ 3)}{2}\ =\ 25]


Fractions suck, so get rid of them:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n(n\ -\ 3)\ +\ m(m\ -\ 3)\ =\ 50]


Multiply and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n^2\ -\ 3n\ +\ m^2\ -\ 3m\ =\ 50]


But since *[tex \Large n\ +\ m\ =\ 13], we know that *[tex \Large m\ =\ 13\ -\ n].  So substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n^2\ -\ 3n\ +\ (13\ -\ n)^2\ -\ 3(13\ -\ n)\ =\ 50]


Multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n^2\ -\ 3n\ +\ 169\ -\ 26n\ +\ n^2\ -\ 39\ +\ 3n\ =\ 50]


Collect like terms in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2n^2\ -\ 26n\ +\ 80\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n^2\ -\ 13n\ +\ 40\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (n\ -\ 5)(n\ -\ 8)\ =\ 0]


So *[tex \Large n\ =\ 5] in which case *[tex \Large m\ =\ 8]


or


*[tex \Large n\ =\ 8] in which case *[tex \Large m\ =\ 5]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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