How many possible sets exist for $\frac1a + \frac1b + \frac1c = \frac12$? Sorry for the confusing title, but I don't know any other way to present the problem.
My problem is that in a competition math book, the problem is to find how many different sets of $3$ regular polygons can be placed around a point such that the three polygons surround the point entirely, basically their interior angles add to 360 degrees around the point.
I've did some algebra and found $\frac1a + \frac1b + \frac1c = \frac12$. 
(Sorry for the lack of formatting. I am in a rush at the moment. If somebody can help me format this, thanks!)
$a,b,$ and $c$ represent the number of sides each regular polygon has. I'm not going to explain what the equation above means and how it relates to the problem I am working on, as it's not important for this question I am asking, but I am seeking how many possibilities for the set $(a,b,c)$ exists if I give a random solution for the variable a, and I wish to solve for what b and c are.
So far, if $a = 3$, I've found two different sets of $(a,b,c)$, which are $(3,9,18)$ and $(3,12,12)$.
Please don't solve the math book question for me. My question is how do I know how many sets of $(a,b,c)$ exist for $\frac1a + \frac1b + \frac1c = \frac12$. Right now, I am giving a random solution for a and solving for b and c by guessing and checking. 
If anybody can help, that would be great. Thanks!
Also, it would be nice if you can tell me how to find the number of possibilities of the set $(a,b,c)$.
 A: This is related to Egyptian fractions.  For results of the form $4/n$ or $5/n$, it's actually an unsolved question as to where there is always a solution.   For 1/2 (or 4/8 or 5/10), a proof by exhaustion is possible.
Ten solutions. One hint is to let $a$ be the smallest number, and $b$ the second smallest.  Can $a$ be 7 or higher?
Unsolved Conjectures about Egyptian Fractions
A: Following the hints of others if you assume $a \le b \le c$ so $1/a \ge 1/b \ge 1/c$ we can conclude.
$1/2 > 1/a \ge 1/6$ so $2 < a \le 6$
then $1/b + 1/c = 1/2 - 1/a = (a-2)/2a$.
In all but $a = 5$ we have $(a-2)/2a = 1/d$ for some $d$.
So $1/b + 1/c = 1/d$.
Then $1/d > 1/b \ge 1/2b$ so $d < b < 2d$.
And $1/c = 1/d - 1/b = (b-d)/bd$  So $b-d|bd$  If $gcd(b-d,b) =1$ or $gcd(b-d,d)$ that doesn't happen.
That gives a strategy for trying to solve by brute force.
$a = 3$;
$d = 6$. $b=7,c=42;b=8,c=28;b=9,c=18;b=10,c=15;b=12,c=12$ are $5$ solutions.
$a =4$;
$d=4$.  $b=5,6,8$ and $c = 20,12,8$.  Three more solutions.
$a = 6$;
$d = 3$. $b = 6$ and $c = 6$.  One more solution.
$a = 5$ was the one where $(a-2)/2a = 3/10$ had no $d$.
$1/b + 1/c = 3/10$ so $3/10 > 1/b  \ge 3/20$ so $5\le b < 20/3 = 6 \frac 23$.  So $b = 5,6$.
so $1/6 + 1/c = 3/10$ so $1/c = 3/10 - 1/6 = 2/15$ so not a solution. 
If $b$ = 5, $1/5 + 1/c = 3/10$ the $1/c = 1/10$ and $c = 10$. Is a solution.
So I have 10 solutions if I didn't make a mistake.  (11 was a mistake.)
