Length Of The String

"Open up" the box, (i.e., deconstruct the box into a two-dimensional representation), so that the $4$ side faces lie in a row with $C$ at the lower right corner and $D$ at the upper left corner.

Then the shortest distance between $C$ and $D$ is the straight line joining them, which will be the hypotenuse of a right triangle with sides length $4$ and $1$ . Thus the shortest possible length of the string is $\sqrt{1^{2} + 4^{2}} = \boxed{\sqrt{17}}$ .

there's a logical way to think about this too. The string wraps around four walls so at each wall, the string increases its height a bit. The height of the cube is 1, thus the string increases its height in increments of 1/4. Using triangles and Pythagorean theorem, we can find one length of the string on one side and multiply by 4.

My solution is same as Brian's.

A cube has 6 walls. The shortest length is 3.6.

Not as elegant as the other answer, but I looked at the individual triangles formed, with the adjacent side = 1, the obtuse side = x, and the hypotenuse = y. The problem is essentially asking for 4y. We know x = 1/4 cm (given that there are four turns to reach 1 cm), and can then solve for y.

$y\quad =\sqrt { 1+0.25 } cm\\ 4y=\sqrt [ 4 ]{ 1+0.25 } cm$

SQRT [(1 + 1+ 1 + 1)^2 + (1)^2] = SQRT (17) = 4.1231056256176605498214098559741

I also got the answer by detaching the faces

Having seen solutions to problems very similar to this, it was immediately obvious (≈ 5 sec), and I don't figure I can claim any "Brilliance" for this.

After getting the answer, I spent multiples of that time trying to find a "catch," but there was none.

open the box and join c and d we get a right angled triangle of sides 4 and 1 and the hypotenuse is the length so it is root 17 = 4.12

Once I realized that the "slope" was constant, I also pictured a NET of the cube, so my solution was the same as Brian's.