2015 AMC 10B Problem 25

Step 1: write the condition. The volume is $abc$. The surface area is $2ab + 2bc + 2ca$. They are equal:
$$abc = 2(ab + bc + ca).$$

Step 2: make it simpler. Divide both sides by $2abc$. Each term on the right loses two of its letters:
$$\frac12 = \frac1c + \frac1a + \frac1b.$$
So we need three fractions $\frac1a, \frac1b, \frac1c$ that add to $\frac12$.

Step 3: limit $a$. Since $a$ is the smallest, $\frac1a$ is the biggest of the three fractions.
- It must be less than $\frac12$ (the other two are positive), so $a \ge 3$.
- It must be at least a third of the total, $\frac16$ (it is the biggest of three), so $a \le 6$.

So $a$ is $3$, $4$, $5$ or $6$.

Step 4: try each $a$. For each $a$, work out what $\frac1b + \frac1c$ must equal. Since $b \le c$, $\frac1b$ is at least half of that amount, which limits $b$. Then solve for $c$ and keep only whole numbers with $c \ge b$.

• $a = 3$: $\frac1b + \frac1c = \frac12 - \frac13 = \frac16$. So $b > 6$, and $\frac1b \ge \frac1{12}$ gives $b \le 12$.

That is $5$ boxes.
- $a = 4$: $\frac1b + \frac1c = \frac12 - \frac14 = \frac14$. So $4 < b \le 8$. Then $b = 5, 6, 7, 8$ give $\frac1c = \frac1{20}, \frac1{12}, \frac{3}{28}, \frac18$, so $c = 20, 12$, (not whole), $8$. That is $3$ boxes.
- $a = 5$: $\frac1b + \frac1c = \frac12 - \frac15 = \frac{3}{10}$. We need $b \ge 5$ and $\frac1b \ge \frac{3}{20}$, so $b \le 6$. Then $b = 5$ gives $\frac1c = \frac{3}{10} - \frac15 = \frac1{10}$, so $c = 10$. And $b = 6$ gives $\frac1c = \frac{3}{10} - \frac16 = \frac{2}{15}$, not a whole-number $c$. That is $1$ box.
- $a = 6$: $\frac1b + \frac1c = \frac12 - \frac16 = \frac13$. We need $b \ge 6$ and $\frac1b \ge \frac16$, so $b = 6$ and then $c = 6$. That is $1$ box (a cube).

Step 5: add up. $5 + 3 + 1 + 1 = 10$.

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

Dividing by $abc$ turns a messy equation into a sum of fractions like $\frac1a$. The ordering $a \le b \le c$ then does the rest: the biggest fraction must be at least its fair share of the total but less than the whole total, which traps the smallest number between two limits. After fixing $a$, the same trick traps $b$, and $c$ is then fixed. This "limit the smallest, then repeat" plan works on most problems with sums of fractions like $\frac1a$.

For each $a$, clear fractions and factor. For $a = 3$: $\frac1b + \frac1c = \frac16$ becomes $bc - 6b - 6c = 0$, which is $(b - 6)(c - 6) = 36$. The factor pairs of $36$ with the first no bigger than the second are $1 \cdot 36$, $2 \cdot 18$, $3 \cdot 12$, $4 \cdot 9$, $6 \cdot 6$, giving $5$ boxes. In the same way $a = 4$ gives $(b - 4)(c - 4) = 16$ with pairs $1 \cdot 16$, $2 \cdot 8$, $4 \cdot 4$, which is $3$ boxes; $a = 5$ gives $(3b - 10)(3c - 10) = 100$, where only $5 \cdot 20$ works, giving $(5, 5, 10)$; and $a = 6$ gives $(b - 3)(c - 3) = 9$ with $b \ge 6$, so only $3 \cdot 3$, giving $(6, 6, 6)$.

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 $\boxed{\textbf{(B)}\ 10}$.