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# 312

The Riddler's Cube (Puzzle) - October 6, 2010
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If you said 1+sqrt(2), then you are............. wrong!! Try again before looking at the solution.

1 + square root of 2 meters

2*sqrt( 1 * 1 + 0.5 * 0.5 ) = 2 * sqrt( 1,25 ) ~ 2.24

You could have made it simpler by making it root 5

It's the square root of 5 metres - think of the trip as the square root of (2^2 + 1^2) which is root 5

Square root of 5.
If you flatten the cube, they're on opposite corners of adjacent 1x2 squares. the diagonal of these is the the sqrt of 1^2 + 2^2, which is the square root of 5.

This is clever. I got the right answer, by doing this: http://www.wolframalpha.com/input/?i=minimize+sqrt(1%2Ba^2)+%2B+sqrt(1%2B(1-a)^2)
But you just did it by reasoning, which is way smarter. No calculus (or WolframAlpha) needed!

I'm pretty sure you can do it in sqrt(5)

Square Root of 3

well, that would be the shortest distance between them, but because they are trapped on the surface of the cube (ie, cannot cut across the diagonal), the shortest path would have to be sqrt(5).

This is one of my favorite puzzles - it has even stumped a few mathematicians!

I founded this problem in an exercise in the text to enter at Pisa University(i don't know if you know)but it was a little bit more intersting: you have a parallelepiped of volume 1, and you have to proof which is the best configuretion for the shortest path among the two opposite corner

Now I am wondering what happens if you change coordinate systems.

If they move towards each other, the answer is the golden ratio minus 0'5 :)

I'm gonna go with 1 metre across, 1 metre down, then 1 metre across...so 3 metres in total!!

c=√a^2+b^2
c=√1^2+2^2
c=√5

;)

Wait… the cube is only 1m on a side? How can they possibly be lonely? It'd be like talking to someone across the room! Make it 100m/side and I'll buy it.

I'm with those who can get there in square root of 5 meters, by cutting across one of the cube face to the middle of an opposite side, then continuing across the new face to Robin. It's the diagonal of a Knight's move in chess: according to the Pythagorean Theorem, the square root of (1^2 + 2^2), or sqroot(5) = ca. 2.24

that was easy.. Take the function F(x)=sqrt(1+x^2)+sqrt(1+(1-x)^2).... differentiate wrt to x ... for x=0.5, F(x) is minimized.. the value of F(x) at x=0.5 is "2*sqrt(1.25) "

There are six paths of length sqrt(5) m form Batman to Robin and no shorter one. Thus there is no one "shortest path" and the question can't be answered.

Ah, but they're all local geodesics!

Whoops, I should have put "a shortest path".

You could also just have said:
the smallest distance

Before calculating anything, I'd need to know what the metric on the space is ;-) Some of us care about such things, you know!

This was quite easy. First I saw that you could unfold the cube, then to check that answer I did a max min problem. Great puzzle.

Geodesic. sqrt(5)

given batman's height, i'm going to have to say about two steps

ooo... my math teacher asked us a more complicated version of this question once. (the side lengths were different, so you had to check a few possible unfolding patterns.)

Just unfold the box and trace a line from point a to point b. :)

Well, according to what I remember, i remember having glanced at a paper defining the shortest path on a cube. something like (m+n)! over m!n! so 2 i guess? can't imagine the pathway though

Awesome! You should definitely do more puzzles.

Haha I actually differentiated the distance function with respect to the point the path crosses from one face to the other, talk about overkill!

I was about to do the same thing, but I like to draw pictures before doing math. And then I was too lazy to draw a whole cube, so I started to draw just two sides of it unfolded, so it made it pretty obvious...

Anyway, really great puzzle. Fairly easy to get fooled by 1+sqrt(2) answer.

honest to god, i didnt think 1+sq(2) first, and i skipped past everyone else's

Why should Batman reach Robin and not viceversa?

2 stratimetres. 1 stratimetre is equivalent to half the shortest distance from batman to robin on this cube.

One of my freinds in Maths class a few years back found this in the text book as 1 + sqrt(2)

Well, I thought of it as two 1*.5 rectangles, found the diagonal of a triangle with sides .5 and 1, sqrt(1.25), and doubled that and got sqrt(5), so all in all, it worked out for me. Just my two cents.

I see how that works and that may be a simpler way to deal with more complex forms of this (such as a object with 1000 sides; note, the sides would not be square then)... but then if you factor in irregularity into the shapes, the other way would be easier.

The shortest path is still sqrt(3) it is just that Batman cannot take that path as he is confined to the surface and hence would take some other path over the surface of Cube.

Holey torus Batman! We need to think outside the box on this one.

This is fun! I have a nice generalization for you folks:
What's the solution for n dimensions?
In 2d it's trivial 2 = 1+1, because one can only walk the sides (and there is only one path modulo direction),
In 3d is $\sqrt(2^2 + 1^2)$.
4d ? 5d ? n-d?
Have fun with it!

At first glance, the length keeps increasing--so if the Riddler can just access a high enough dimension, the time for Batman to reach Robin can be arbitrarily long. On the other hand, as dimension increases, so does the number of possible paths, so Riddler cannot hope to keep them apart by blocking all possible paths.

I like Sonobox's question. It's easy to see that the edge path for an n-cube is simply n. For 1 and 2 the shortest path is also this. If n = 3 then the answer is 5^.5. My question is now this - is Batman constricted to 2 dimensions? Or for higher dimensions are we allowing hyper-surfaces. A tessaract, for instance, has 4 cubes on its "surface". I should technically be allowed to try that surface, which gives a very different answer to the 2d-soln.

8 cubes... sorry 'bout that.

I think you should see it as a stratifold, and Batman can use all of them - like in 3d, he can use 2d, 1d and 0d objects which are on the surface.

Good job.

The solution is pretty simple and i don't think that any body found the answer.So Batman can move anywhere on the surface of the cube, so the answer is 0.5*sqrt(0.5)+1+0.5*sqrt(0.5).I't preetty close to sqrt(5) i don't know of it is sqrt(5) because we are playing with irrational numbers.
on paper this is the shortes version: Batman goes diagonal to the right till the center of the cube surface, then straight until the half way to the next surface then it goes diagonal right to reach Robin.

Watch the solution... Your solution is wrong: The length you described is 1 + sqrt(2) > sqrt(5).

2sqrt(5/4) = 2sqrt(5)/sqrt(4) = 2sqrt(5)/2 = sqrt(5)

Seems that most of those who got the correct answer did so by unfolding the cube.

My reasoning went: Since Batman must transverse two planar surfaces of the cube in order to reach robin (assuming that his route is the shortest distance), and since all planar surfaces of a given cube are spatially congruent, it follows that he must cover half the route in transversing one planar surface.

Furthermore, since his two-surface route is bisected by a vertex of the cube, and since his distance covered on the first surface is equal to his distance covered on the second surface, it follows that he will cross the vertex at its midpoint.

Insert appropriate numbers and solve algebraically.

What is the Batmobile's instantaneous velocity at the point where it crosses the vertex?

hmmm, now what I am most curious about now is how batman (roughly a 2m tall cylinder of 0.5m diameter) misses robin who is only less tall, but about the same radius)... and then how did they even fit on the cube without causing all sorts of weird stuff to happen ;)

A similar question was actually asked at the Flemish Mathematical Olympiad many years ago. (it was on the surface of an icosahedron).
Not surprisingly, 85% of the respondents chose the logical wrong answer.
The question (in Dutch, but understandable by the figure) can be found here: (question 28)
http://www.vwo.be/vwo/files/2r2001.pdf

Isn't the answer sqrt(5) BS? Because that would hold only in a cube that has been flattened. The cube in question is a cubical cube. Things happen differently in 3D and 2D. What an illogical puzzle.

I don't agree with this solution, seeing as the space that constrains movement can be folded the shortest possible distance is equal to 0m, unless it can only be folded at the edges, then the shortest distance is 1m

Nice .. It helps to open out the cube into "one"(intact) sheet of surface and then work on it.. Gr8 site ..!!

For sure, sqrt(3)

2 * sqrt(1,25).. which happens to be the same as sqrt(5)

if you look at the cube like an optical illusion, then it looks like an open 3-sided cube and the shortest distance would be 1+sqrt2

What if they are at the centers of the two opposite walls.. what will be the shortest distance then...??????
Please answer this question i hv been looking for it for many days...

Well, it's Batman. So instead of square root of 5, it is zero distance, because Batman is awesome.

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