3305.11 – Diagonals of Two Regular Polygons

Consider two regular polygons. (Lots of math problems start by asking you to "consider" something.) The total number of angles between them is thirteen, and the total number of diagonals is twenty-five. How many angles are in each polygon?

Solution

We have two distinct polygons: an m-gon and an n-gon, where m+n=13. The number of distinct diagonals, D, in a regular polygon with x sides can be found with the following formula:

D=x2(x−3)D=\frac{x}{2}(x-3).

You could do this by simply making a big chart, possibly on a spreadsheet, where you write a list of regular polygons from triangle up to, um decagon and next to it a list of the polygons from decagon down to triangle. Then use the formula to calculate the various numbers of diagonals and see which pair of polygons fits the requirement.

Or you can go the algebraic route, as follows:

m2(m−3)+n2(n−3)=25\frac{m}{2}(m-3)+\frac{n}{2}(n-3)=25

We now can simply substitute m=13−nm=13-n and solve for nn:

13−n2((13−n)−3)+n2(n−3)=2513−n2(10−n)+n2−3n2=2510(13−n)2−n(13−n)2+n2−3n2=2565−5n−(6.5n−0.5n2)+0.5n2−1.5n=25n2−13n+65=25n2−13n+40=0(n−5)(n−8)=0 \begin{aligned} \frac{13-n}{2}((13-n)-3)+\frac{n}{2}(n-3) &=& 25\\ \frac{13-n}{2}(10-n)+\frac{n^2-3n}{2} &=& 25\\ \frac{10(13-n)}{2}-\frac{n(13-n)}{2}+\frac{n^2-3n}{2} &=& 25\\ 65-5n-(6.5n-0.5n^2 )+0.5n^2-1.5n &=& 25\\ n^2-13n+65 &=& 25\\ n^2-13n+40 &=& 0\\ (n-5)(n-8) &=& 0 \end{aligned}

So, n=5n=5 or n=8n=8.

Thus, since nn can equal five or eight, we see that our two regular polygons are a pentagon and an octagon.

Which solution method do you prefer? Both are completely honorable--as simply guessing and checking is too.