2018 AMC 10B Problem 4

Question

A three-dimensional rectangular box with dimensions

X,

Y,

and

Z

has faces whose surface areas are

24,

24,

48,

48,

72,

and

72

square units. What is

X + Y + Z?

18

22

24

30

36

The three distinct face areas are the pairwise products

XY, XZ, YZ,

equal to

24, 48, 72

Multiplying all three gives

(XYZ)^2;

take the square root, then divide by each face area

The three distinct face areas are the pairwise products

XY = 24,

XZ = 48,

YZ = 72

in some order. Multiply all three:

(XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944,

so

XYZ = 288.

Now divide by each face area. We get

Z = \frac{288}{24} = 12,

Y = \frac{288}{48} = 6,

and

X = \frac{288}{72} = 4,

so

X + Y + Z = 22.

Therefore, the answer is B.

Problem 4 in other years