2018 AMC 10B Problem 4

Opposite faces of a box are congruent, so the three different face areas are the three pairwise products of the edges. Label them
$$XY = 24,\qquad YZ = 48,\qquad XZ = 72.$$
Multiplying all three equations gives $(XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944 = 288^2$, so $XYZ = 288$. Dividing by each face area isolates the remaining edge:
$$Z = \frac{288}{24} = 12,\qquad X = \frac{288}{48} = 6,\qquad Y = \frac{288}{72} = 4.$$
Check: $6 \cdot 4 = 24$, $4 \cdot 12 = 48$, $6 \cdot 12 = 72$. Hence $X + Y + Z = 6 + 4 + 12 = 22$.

The answer is $\boxed{\textbf{(B)}\ 22}$.

A symmetric system of pairwise products is solved by multiplying everything together: the product of the three equations is the square of $XYZ$, and dividing back out recovers each variable in one step. The same trick handles $ab, bc, ca$ in any setting.

Ratios: $XZ/XY = Z/Y = 3$ and $YZ/XY = Z/X = 2$, so with $Y = t$ we get $Z = 3t$, $X = 3t/2$, and $XY = 3t^2/2 = 24$ gives $t = 4$. Or simply spot $4 \cdot 6 = 24$, $4 \cdot 12 = 48$, $6 \cdot 12 = 72$.

• Guessing edge lengths from one factor pair without checking all three products, or treating the six listed areas as six different equations.
• Forgetting the square root and using $XYZ = 82944$, or summing $24 + 48 + 72$ and looking for something to do with $144$.