Polygonal Drama

There are $10$ such triples; proof as follows:

First note that (as polygons), each of $a,b,c$ is greater than $2$ .

Since $a \le b \le c$ , $\frac12=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}$

so that $a\le 6$ .

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Case $a=3$ : $\frac{1}{b}+\frac{1}{c}=\frac12-\frac13=\frac16$

We clearly need $b>6$ . By the same logic as above, we find $b\le 12$ .

Rearranging the equation, we have $c=\frac{6b}{b-6}$

It's easy to check that the following $b$ values all give valid solutions: $7,8,9,10,12$ .

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Case $a=4$ : $\frac{1}{b}+\frac{1}{c}=\frac14$

As above we find $4<b \le 8$ , with integer solutions for $c$ when $b$ is any of $5,6,8$

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Case $a=5$ : The only solution is $b=5$ and $c=10$ .

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Case $a=6$ : The only solution is $b=c=6$ .

In total, there are $10$ solutions above.

$a \leq b \leq c \\ \therefore \frac 12 = \frac 1a + \frac 1b + \frac 1c \leq \frac 1a + \frac 1a + \frac 1a \\ \therefore \frac 12 \leq \frac 3a \\ a \leq 6$

• $a = 3$ $\\ \frac 12 - \frac 1a = \frac 1b + \frac 1c \leq \frac 1b + \frac 1b \\ \therefore \frac 12 - \frac 13 \leq \frac 2b \\ \therefore b \leq 12 \\$ But when $b = 11$ , $c = \frac {66}5$ is not a integer. And when $b = 6$ , $c = \infty$ . Finally we can get $\\ b \in \{6, 7, 8 ,9 ,10, 12\}$

• $a = 4$ similarly we can get $\\ b \in \{4, 5, 6, 8\}$

• $a = 5$ there is only one solution $b = 5, c = 10$

• $a = 6$ there is only one solution $b = 6, c = 6$

So there are totally 12 solutions as below: $\\ (3, 6, \infty) \\ (3, 7, 42) \\ (3, 8, 24) \\ (3, 9, 18) \\ (3, 10, 15) \\ (3, 12, 12) \\ (4, 4, \infty) \\ (4, 5, 20) \\ (4, 6, 12) \\ (4, 8, 8) \\ (5, 5, 10) \\ (6, 6, 6)$