2015 AMC 10B Problem 25

The condition is $abc = 2(ab + bc + ca)$. Dividing by $2abc$:
$$\frac1a + \frac1b + \frac1c = \frac12.$$

Since $a \le b \le c$, $\frac1a$ is the largest term, so $\frac{3}{a} \ge \frac12$ gives $a \le 6$; and $\frac1a < \frac12$ gives $a \ge 3$. Now for each $a$, let $r = \frac12 - \frac1a$. Then $\frac1b + \frac1c = r$ with $b \le c$ forces $\frac1b < r \le \frac2b$, so $\frac1r < b \le \frac2r$, and $c = \frac{1}{r - 1/b}$ must be an integer at least $b$.

• $a = 3$, $r = \frac16$, $6 < b \le 12$, $c = \frac{6b}{b-6}$: $b = 7, 8, 9, 10, 12$ give $c = 42, 24, 18, 15, 12$; $b = 11$ gives $\frac{66}{5}$. 5 triples.
• $a = 4$, $r = \frac14$, $4 < b \le 8$, $c = \frac{4b}{b-4}$: $b = 5, 6, 8$ give $c = 20, 12, 8$; $b = 7$ gives $\frac{28}{3}$. 3 triples.
• $a = 5$, $r = \frac{3}{10}$, $b \ge 5$ and $b \le \frac{20}{3}$, $c = \frac{10b}{3b-10}$: $b = 5$ gives $c = 10$; $b = 6$ gives $\frac{60}{8}$. 1 triple.
• $a = 6$, $r = \frac13$, $b \ge 6$ and $b \le 6$: $b = 6$, $c = 6$. 1 triple.

Total: $5 + 3 + 1 + 1 = 10$.

The answer is $\boxed{\textbf{(B)}\ 10}$.

Dividing by the product converts a symmetric three-variable equation into a sum of unit fractions, and the ordering $a \le b \le c$ then gives two-sided bounds on the smallest variable at every stage ($\frac{k}{a} \ge$ target $> \frac1a$). This "bound the smallest, recurse" pattern solves nearly every Egyptian-fraction problem. Each case leaves a single unknown, whose integrality is a quick check.

Simon's Favorite Factoring Trick per case: for $a = 3$, $\frac1b + \frac1c = \frac16$ becomes $(b-6)(c-6) = 36$, and factor pairs of $36$ with $b \le c$ give $5$ triples; for $a = 4$, $(b-4)(c-4) = 16$ gives $3$; for $a = 6$, $(b-3)(c-3) = 9$ with $b \ge 6$ gives $1$; for $a = 5$, $(3b-10)(3c-10) = 100$ with $b \ge 5$ gives only $(5, 10)$.

• Losing solutions by bounding b too tightly, or counting unordered triples with permutations (the condition a <= b <= c means each solution set counts once).
• Forgetting the case $a = 5$ (its factoring is ugly) or the boundary case $b = c$ such as $(3, 12, 12)$, $(4, 8, 8)$, $(6, 6, 6)$, each of which is valid.