# Polygons Help!

• Jan 6th 2007, 05:14 AM
Ruler of Hell
Polygons Help!
Okay, I've posted after a long time, I know... But I'm here with another doubt.
The ratio between the sides of 2 regular polygons is 4 : 5 and the ratio between the interior angles is 15 : 16. Find the number of sides.

Thanks!
• Jan 6th 2007, 05:34 AM
CaptainBlack
Quote:

Originally Posted by Ruler of Hell
Okay, I've posted after a long time, I know... But I'm here with another doubt.
The ratio between the sides of 2 regular polygons is 4 : 5 and the ratio between the interior angles is 15 : 16. Find the number of sides.

Thanks!

Let the number of sides of the polygons be $\displaystyle n$ and $\displaystyle m$, then the first condition tell you that:

$\displaystyle \frac{n}{m}=\frac{4}{5}\ \ \ \dots(1)$,

and as the interior angle of a ploygon with $\displaystyle n$ sides is $\displaystyle 180(n-2)/n$, the second condition tells us that:

$\displaystyle \frac{n-2}{n}\,\frac{m}{m-2}=\frac{15}{16}$

which may be simplified as we know $\displaystyle n/m=4/5$ to:

$\displaystyle \frac{n-2}{m-2}=\frac{3}{4}\ \ \ \dots(2)$

Now we can proceed to solve $\displaystyle (1)$ and $\displaystyle (2)$, or we can just guess, when we find $\displaystyle n=8$, $\displaystyle m=10$.

RonL
• Jan 6th 2007, 05:50 AM
Ruler of Hell
Wait, I'm still a li'l confused. How did you solve equation 1 and 2? Sorry....
• Jan 6th 2007, 05:53 AM
CaptainBlack
Quote:

Originally Posted by Ruler of Hell
Wait, I'm still a li'l confused. How did you solve equation 1 and 2? Sorry....

I guessed, n and m have to be integers so I tried n=4, m=5 and that didn't
work so then I tried n=8, m=10 which did work. Also the solution is obviously
unique, so n=8, m=10 is the only solution.

RonL
• Jan 6th 2007, 06:06 AM
Ruler of Hell
Got it! Yay, thanks a tonne Captain Black... (Though I used 2 different methods, yours and another one) Nonetheless, thanks!