Box Loop

Geometry Level 5

A small lizard is kept inside a glass box of dimensions $14\ \text{in.} \times 21\ \text{in.} \times 28\ \text{in.}$ . It begins climbing the wall at point $A$ along the red dotted line. Whenever it reaches another face, it continues at the same angle as if the two connected faces lie on the same plane.

Its trail is indicated in the diagram, touching all 6 faces where the first and fourth face are parallel, before coming back to A.

What is the distance (in inches) that the lizard travels in 1 loop?

The answer is 89.095.

1 solution

Calvin Lin Staff
Nov 28, 2016

Label the vertices of the box in the following manner:

Let's consider the path of the lizard starting from edge $ef$ of plane $abef$ . Note that fixing the starting point doesn't change the answer of the length of the loop. We draw the net of the cuboid by expanding along the path of the lizard. Since the first and fourth faces are parallel, we are forced to use the following net:

Drawn to scale with vertices labelled

Hence, the distance travelled by the lizard is $\sqrt{ (14 + 28 + 21)^2 + ( 21 + 14 + 28)^2 } = 63 \sqrt{2} \approx 89.095$ .

We can make the following observations, which apply to the general case too:

The $\theta$ angle that the lizard makes must satisfy $\tan \theta = \frac{ 63}{63} = 1$ , or that $\theta = 45 ^ \circ$ .
If we want to transverse all 6 faces, there are a limited number of starting points. For example, if the starting point was much closer to $e$ , then we will miss face $dcgh$ .
The entire path that the lizard takes lies on the same plane! This normal of the plane is $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ ( up to orientation of the sides).
An example of a valid starting point is the midpoint of $ef$ . The plane of the entire path passes through the center of the cuboid.
In the comments, Maria points out that there exists different paths if we relax the "opposing faces" condition. I was shocked when I first saw it, so I encourage you to check it out!

I think I just found another solution. It is for $\tan(\theta)=\frac{5}{13}$ . The length is $\sqrt{91^2+35^2} \approx 97.5$ .

Maria Kozlowska - 4 years, 6 months ago

If the face that you start on is the first face, by the diagram given the fourth face that the lizard travels upon must be the parallel face. In your diagram, the 5th place is the opposite face so this is not the same route.

Dan Ley - 4 years, 6 months ago

Ah interesting configuration which works. It didn't occur to me that this net could also work. I had that thought at the back of my head, but didn't follow up on it.

Let me be explicit in the problem that the "fourth face that the lizard travels upon must be parallel to the first face".

Calvin Lin Staff - 4 years, 6 months ago

Is this not suggested by the diagram given in the question?

Dan Ley - 4 years, 6 months ago

@Dan Ley – It is suggested by the diagram. However, our images are not drawn to scale / accurate representation unless otherwise stated (esp for geometry diagrams). Hence I have to make the unstated assumption explicit.

Calvin Lin Staff - 4 years, 6 months ago

@Worranat Pakornrat @Dan Ley @Maria Kozlowska @Jon Haussmann Can you review this solution?

It explains why we must have $\tan \theta = 1$ , and so the condition of $\tan \theta = \frac{3}{4}$ is incorrect.

I kept track of the vertices on the net, so that it's easy to see which edge (and orientation) the lizard must end up on.

Calvin Lin Staff - 4 years, 6 months ago

As far as I can see, your solution is correct. What I like about the method is that this net can be reproduced infinitely as the lizard keeps "looping" around the cuboid, meaning that we can set the starting position to any edge, such as $ef$ (you have exemplified the case of starting on the edge of length 14). If we do so, we can simply continue the net so that it "loops" back to its new starting position.

On a side note, what did you use to create your 2 diagrams above? It would be nice to make some of my solutions on Brilliant a little more pleasing to the eye:)

Dan Ley - 4 years, 6 months ago

We use Figma for diagrams.

I also use OmniGraffle for quick images. (E.g. I was too lazy to do the cuboid image properly)

Calvin Lin Staff - 4 years, 6 months ago

@Calvin Lin – OmniGraffle looks great but quite pricey! Figma is already proving of much use, thanks for the suggestion.

Dan Ley - 4 years, 6 months ago

Can you find a starting point which will result in a path lying on one plane? (I think that was a condition, unless I misunderstood the wording of the problem)

Maria Kozlowska - 4 years, 6 months ago

The diagram is drawn to scale, and so the dotted line is an example of a path that passes through all 6 faces. If the starting point was much closer to $e$ than $f$ , we see that we will miss out on the plane $dcgh$ completely.

Calvin Lin Staff - 4 years, 6 months ago

Can you prove that a single plane can pass through all the points of the path created in this way? (I do not mean the net, but the actual 3D box)

Maria Kozlowska - 4 years, 6 months ago

@Maria Kozlowska – Ah, ic the confusion is about "as if it were on the same plane". What it means is that when 2 faces meet, then line continues as if the two faces were on the same plane. It does not mean that the entire path that the lizard walks on is on the same plane. Let me edit the problem for clarity.

However, having said that, the path of the lizard does indeed lie on the same plane. Depending on the orientation of the cuboid, the plane is normal to the vector $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ , which accounts for the $\tan \theta = 1$ that we observed.

Calvin Lin Staff - 4 years, 6 months ago

@Calvin Lin – That seems to be correct. I think there is one starting for which triangles on parallel faces are congruent and the common plane passes through the center of the box.

Maria Kozlowska - 4 years, 6 months ago

@Maria Kozlowska – Yes indeed! That's the plane that passes through the midpoints of those 6 sides. For a cube, it corresponds to the plane that gives us the largest regular hexagon that can be inscribed.

This discussion is great! Let me add these observations to the solution.

Calvin Lin Staff - 4 years, 6 months ago

@Calvin Lin – It might be that an extra condition needs to be added to the wording of the problem to ensure uniqueness of a solution value. Condition would be to ask for the shortest path. I am not entirely convinced that there are no other paths possible with different angles.

Maria Kozlowska - 4 years, 6 months ago