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, 2013

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:
A bug is on one exterior vertex of solid $S$ , a $3 \times 3 \times 3$ cube that has its center $1 \times 1 \times 1$ cube removed, and wishes to travel to the opposite exterior vertex. Let $O$ denote the outer surface of $S$ (formed by the surface of the $3 \times 3 \times 3$ cube). Let $L(S)$ denote the length of the shortest path through $S$ . (Note that such a path cannot pass through the missing center cube, which is empty space.) Let $L(O)$ denote the length of the shortest path through $O$ . What is the ratio $\frac{L(S)}{L(O)}$ ?

Solution

Solution:
$\boxed{\frac{\sqrt{29}}{3 \sqrt{5}}}$ OR $\frac{\sqrt{145}}{15}$

By $\left(^*\right)$ , the shortest route in $O$ has length $2 \sqrt{1.5^2+3^2} = 3 \sqrt{5}$ .

By $\left(^{**}\right)$ , the shortest route overall (in $S$ ) has length $2 \sqrt{1.5^2+1^2+2^2} = \sqrt{3^2+2^2+4^2} = \sqrt{29}$ .

Therefore the desired ratio is $\frac{\sqrt{29}}{3 \sqrt{5}} = \frac{\sqrt{145}}{15}$ .

$\left(^*\right)$ Suppose we're trying to get from $(0,0,0)$ to $(3,3,3)$ through $O$ . Then one minimal-length path through $O$ is $(0,0,0) \rightarrow (1.5,0,3) \rightarrow (3,3,3)$ .

$\left(^{**}\right)$ Suppose we're trying to get from $(0,0,0)$ to $(3,3,3)$ through $S$ . Then the inner hole is $[1,2] \times [1,2] \times [1,2]$ , and one minimal-length path is $(0,0,0) \rightarrow (1.5,1,2) \rightarrow (3,3,3)$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.