The surface distance between 2 points on the surface of a cube is the

To find the Shortest distance between the diagonal points of the cube, we need to unfold the cube completely.
Imagine the base and right side face of the cube into a flat rectangle of length = 8 cm and breadth = 4 cm

Shortest distance = path traveled along the diagonal of the above rectangle
--> \(Distance = \sqrt{8^2 + 4^2}\)
--> \(Distance = 4\sqrt{2^2 + 1^2}\)
--> \(Distance = 4\sqrt{5}\)

IMO Option B

Pls Hit kudos if you like the solution

VeritasKarishma generis

How can we get an idea that we have to open up the cube and then solve this question??

Dear IanStewart,

My students, myself included, fall into the trap of choice E.

My question is how to approach this problem?

Instinctively, I thought that the shortest route would to walk upward across the edge of the cube and then along diagonal of the upper face.

When solving such problem, how do I know that my instinct is wrong here?

varotkorn

The shortest path between 2 points is a straight line, if the surface is flat. In other words, choose the path in which a ray of light is supposed to travel, without any constraints or diffraction.

In the path you chose, light has to bend( which is obviously not a natural path of light). Though, i love to see IanStewart views on this.

nick1816
Isn't question asking about the greatest distance? From top corner of one face to the bottom corner on opposite side on opposite face.
The formula used for determining greatest distance in cube is Side √3
So why its Side√5 ?
Please explain.

Posted from my mobile device

I didn't know you had students - are you teaching GMAT? I don't have much to add to what Nick has said already. The shortest distance between two points is in a straight line. Here, we're confined to distances that lie on the surface of the cube. If we just flatten the cube out, the same way you might flatten a cardboard box, none of the surface distances change, and then it's easier to see how to make a straight line between our two points, and you can solve the problem using the rectangle diagrams that were posted already. It's really a 2-dimensional distance problem disguised as a 3-dimensional one, which is typical of the GMAT - problems that seem to be testing more advanced math (3-D geometry, say) are often testing something simpler.

yashikaaggarwal

√3*(side) would be the shortest path, if there were no constraints. In question, we do have one constraint. The shortest distance must be calculated along the surface.

Going through this comment, it is quite clear that, reaching the same diagonal point through sides.
And not through the cube.

Thank you.

Wow im not gonna lie, this is actually a great problem...

How do we know that we have to open the box ?

Flattening out the box just gives us better visualization and doesn't change the surface distance.
You can solve this question without opening the box tho. You just gotta apply the Pythagoras theorem twice.


Because of the symmetry
CD=DE = 2

\(AD^2 = 4^2+2^2\)

AD = \sqrt{20} = \(2\sqrt{5}\)

\(BD^2 = 4^2 +2^2\)

BD = \(\sqrt{20}\) = \(2\sqrt{5}\)

Total distance = AD + BD = 2√5 + 2√5 = 4√5

Solution:

Letting x = the surface distance between the lower left vertex on its front face and the upper right vertex on its back face; we can create the equation:

x^2 = 4^2 + 8^2

x^2 = 80

x = √80 = 4√5

Answer: B

Thanks for the explanation.
Just curious to know why can't I connect ACB directly?
In that case answer will be 4√2+4 i.e. option E.
Kindly suggest.

For the formula lovers:

The longest surface distance between two points on a cube = \(s*\sqrt{5}\) where s is the length of the side.

The longest distance between two points on a cube (the diagonal of a cube) = \(s*\sqrt{3}\) where s is the length of the side.

Just need to visualize.

Lay the Front Face and the Top Face flat, as if you were opening a cardboard box.

The shortest distance between any 2 points is a straight line. This will be the the diagonal through the now flat rectangle with:

Length = 4

And

Width = 8


(D)^2 = (4)^2 + (8)^2

(D)^2 = 80

D = 4 * sqrt(5)

(B)

Posted from my mobile device

The surface distance is the distance on the outer surface and cannot be distance passing inside the cube. It is like moving your finger or walking on the cube. We need to think of the cube as paper box and spread the relevant sides open. The unfold should happen in such way that the surface distance need not travel extra distance between two given points. So only unfold the sides on which two points lie (at max 2 faces). The final figure becomes a rectangle whose diagonal gives the shortest surface distance

Visualize a little...

How would you have joined these two points if you were having a pen in your hand

Cut open the cube and unfold it... make a 2-D problem

You are essentially left with two points on the diagonally opposite vertices of a rectangle (with sides 4*8)

= 4^2 + 8^2

= 80

=> √80 = 4√5

I know, I know