Baggage Problem: $1.5$-meter-long sword onto a train

Question

This is the problem I came across reading the book The Art and Craft of Problem Solving. When I read this question I wasn't able to figure out the solution and I saw the solution after a while, but still I couldn't understand the solution as well.

Question:

Pat wants to take a $1.5$-meter-long sword onto a train, but the conductor won't allow it as carry-on luggage. And the baggage person won't take any item which greatest dimension exceeds $1$ meter. What should Pat do?

Solution:

This is unsolvable if we limit ourselves to two-dimensional space. Once liberated from Flatland, we get a nice solution : The sword fits into a $1 \times 1 \times 1$ -meter-box, with a long diagonal of $\sqrt{(1^2 + 1^2 + 1^2)}$ = $\sqrt{3} > 1.69$ meters.

Can anyone give me a clear explanation of this solution?

Answer 1

The sword fits in that 3D box, because the long diagonal is $1.69$ meter. The longest dimension of this box is 1 meter (length, width and height are all exactly 1 meter). The sword goes like the green line in the picture:

If Pat would use a flat case, it would be denied. In the most naive way, the box would be $1.5$ meter long and have a width of the sword itself. A more clever way to think is in the maximum dimensions, which are $1 \times 1$ meter. In that case the diagonal is $\sqrt{2} = 1.41$ meter long, which is not long enough. So the only option is to include the height into account.

Answer 2

Explanation

The solution given hinges on the fact that, while the sword itself is longer than the maximum allowed length, we can construct a volume in which the sword fits, that doesn't exceed the restrictions. Contradictory, we can work around the restrictions that were undoubtedly enforced to save space, by using more space!

I've actually run into similar limitations once when sending a parcel and I solved it in the exact same way.

Escalation

But a $1\text{m} \times 1\text{m} \times 1\text{m}$ box is a bit overkill. The diagonal is almost $20\text{cm}$ longer than needed!
So let's find a couple of smaller solutions.

Minimal height

Let's say we have a box with a base of $1\text{m} \times 1\text{m}$. What is the minimal height $h$ we need?

The length $L$ of the sword is $1.5 \text{m}$. The diagonal $D = \sqrt{1^2 + 1^2 + h^2}$. If we fit the sword precisely in the diagonal, $L=D$, so $1.5=\sqrt{1^2 + 1^2 + h^2}$. That means that $2.25 = 2 + h^2$ and so $h=0.5\text{m}$.

Minimal square

If we want a box of exactly $1\text{m}$ high with a square base, how long would the edges $x$ of that square be?

This means that $D = \sqrt{1^2 + x^2 + x^2}$, so $\tfrac{9}{4} = 1 + 2x^2$. This makes $x^2=\tfrac{5}{8}$ and $x=\tfrac{1}{4}\sqrt{10} \approx 0.79\text{m}$.

Minimal cube

Let's try to get the minimal cube with edges $x$ in which to fit the sword.

In this case, $D = \sqrt{x^2 + x^2 + x^2}$, so $\tfrac{3}{2} = \sqrt{3x^2}$, which means that $\tfrac{9}{12} = x^2$, which yields $x = \tfrac{1}{2}\sqrt{3} \approx 0.87\text{m}$.

Answer 3

The baggage person won't allow the sword, because it's 1.5m long. They will, however, allow a 1m x 1m x 1m box. If you put the sword running from one corner of the box to the opposite, it will fit (because the diagonal is ~1.7m) and be allowed on board.