2018 AMC 10B Problem 4

A box has three pairs of matching faces (top and bottom, front and back, left and right). So only three areas are really different: $24$, $48$ and $72$.

1. Each face is a rectangle whose sides are two of the edges. So the three areas are the products of the edges two at a time:
$$XY = 24,\qquad YZ = 48,\qquad XZ = 72.$$
2. Compare the areas to find how the edges relate. Going from $XY = 24$ to $YZ = 48$, the $Y$ stays and $X$ is replaced by $Z$, and the area doubles. So $Z$ is twice $X$.
3. Going from $XY = 24$ to $XZ = 72$, the $X$ stays and $Y$ is replaced by $Z$, and the area triples. So $Z$ is three times $Y$.
4. Try $Z = 12$. Then $X = 6$ and $Y = 4$. Check all three areas:
$$6 \times 4 = 24,\qquad 4 \times 12 = 48,\qquad 6 \times 12 = 72.$$
All three work.
5. So $X + Y + Z = 6 + 4 + 12 = 22$.

(Why start with $Z = 12$? If $Z = 3Y$ and $X = Z/2$, then $XY = \tfrac{Z}{2}\cdot\tfrac{Z}{3} = \tfrac{Z^2}{6} = 24$, so $Z^2 = 144$ and $Z = 12$.)

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

When you know the products of three numbers two at a time, comparing the products tells you the ratios between the numbers. One more equation then pins down the size. Always check the answer against all three products.

Multiply all three equations: $XY \cdot YZ \cdot XZ = (XYZ)^2$, and $24 \times 48 \times 72 = 82944 = 288^2$. So $XYZ = 288$. Divide by each area to get the missing edge: $Z = 288 \div 24 = 12$, $X = 288 \div 48 = 6$, $Y = 288 \div 72 = 4$. The sum is again $22$.