Three cubes with edge lengths of 3, 2, and 1 are stacked in order with

Question

https://gmatclub.com/forum/download/file.php?id=58295 Three cubes with edge lengths of 3, 2, and 1 are stacked in order with their left and back faces aligned, as shown above. A line is drawn between opposite corners of the bottom face of the cube at the bottom and the top face of the cube at the top. What is the length of the line contained inside the middle cube? A. 2 B. \sqrt{6} C. \sqrt{7} D. 3 E. 4 Are You Up For the Challenge: 700 Level Questions Untitled.png

nick1816 · Answer

Untitled (4).png AC^2=AB^2+BC^2 AC^2 = 6^2+(3√2)^2 AC^2 =54 AC= 3√6 \frac{AB}{AC} = \frac{DF}{EG} \frac{6}{3√6} = \frac{2}{EG} EG = √6

IanStewart · Answer

The height of the stack of cubes is 6, and the diagonal along the base, a square with sides of length 3, is 3√2 (using Pythagoras, or 45-45-90 triangle ratios). The long sloping line cutting through all three boxes in the diagram is just the hypotenuse of a right triangle with those two short sides. So if the length of the long sloping line is d, we have

d^2 = 6^2 + (3√2)^2
d^2 = 54
d = 3√6

By similarity, 1/6 of that line will be in the top box, 2/6 will be in the middle box, and 3/6 of it will be in the bottom box. We want the part in the middle box, so the answer is (2/6)(3√6) = √6. Technically, you also need to confirm the entire blue line is contained in the middle box, since that's all the question asks for, but it is, which you can confirm by using slopes.

nick1816  -- my solution differs from yours; it appears you got 3√3 for the diagonal of the base of the bottom cube, instead of 3√2? edit: I see you already caught the mistake!

GMATGuruNY · Answer

A slightly different way to calculate the length of the line segment drawn through the three cubes:

Attachment:

diagonal of 3 cubes.png [ 178 KiB | Viewed 3756 times ]

AB constitutes the main diagonal of the 3 x 3 x 6 red rectangular solid shown above.
Since the formula for the main diagonal is \(d = \sqrt{l^2 + w^2 + h^2}\), we get:
\(AB = \sqrt{3^2 + 3^2 + 6^2} = \sqrt{3^2(1+1+2^2)} = 3\sqrt{6}\)

Three cubes with edge lengths of 3, 2, and 1 are stacked in order with

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Three cubes with edge lengths of 3, 2, and 1 are stacked in order with

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