# Surface distances on a cuboid: corner to corner

A usenet thread from alt.brain.teasers in December 1998 recovered using www.dejanews.com

Of the six responses, four were wrong, even though the question had the correct answer. (Some names have been changed to protect the guilty; all e-mail addresses have been removed.)

(A simple question containing the correct solution)

## Subject: 2*1*1 cube.

Author: om7
Date: 12/22/1998

I have the following problem.

I have a 2*1*1 cuboid. What is the shortest distance between two (furthest) corners? You cannot join the points directly to get the shortest distance, instead you have to go over the outside surface.

I think the shortest distance is 2*(square root of 2). But this is without proof.

Any help would be gratefully appreciated. Thanks.

(Confirmation that the solution is correct)

Author: Robert H. Olson
Date: 12/22/1998

Cut open the box and lay it flat.

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The @'s are one of several of the two fartherest points.
Using c^2 = a^2 + b^2, you get the square root of 8 or 2 x square root of 2
--
Robert H. Olson
Some people fish to forget their troubles. Some to escape reality. Some to get away from the complexities of modern life.   Me? I like worms ;-)

(a incorrect solution which is too long, possibly because of the error in comparing sqrt(10) with 2*sqrt(2))

Author: JJ
Date: 12/22/1998

to find the shortest distance, first you'll have to open up the cuboid:

so now you'll have a 3*1 rectangle draw a line on the diagonal whose lenght would be sqr(10). now when you construct the cuboid back again you'll find a line joining the two farthest corners. you'll also notice that sqr(10) is less than 2*sqr(2).

note: you don't have to open up the cuboid, but have to understand why you get a 3*1 rectangle.    3*1 = (2+1)*1

nice puzzle.

(agrees with the previous incorrect solution while pointing out why it is wrong)

Author: MW
Date:12/22/1998

I agree, it is sqr(10), but sqr(10) > 2*sqr(2), since 2*sqr(2) = sqr(8).

(a different incorrect solution which is too long, by ignoring the possibility of going across a face to a point which is not a corner)

Author: DS
Date: 12/22/1998

Which is less? 2 + SQRT(2) or 1 + SQRT(3)?  One of those is the shortest route.  Probably 1+SQRT(3) (go along a diagonal across a 1x2 side, then down the side one unit to the other corner).

(an incorrect solution which is too short, since it involves a route through the middle of the volume rather than on the surface)

Author: NT
Date: 12/23/1998

You have to use the Pythagorean Theorem twice.

First: use the P.T. to find the length of the diagonal of the base
the square root of(2 squared plus 1 squared) = the square root of 5

Second: use the P.T. with the square root of five and the height(1)
the square root of (5 + 1 squared) = the square root of 6

(another correct solution)

Author: Robin G Sharman
Date: 12/23/1998

Pythagoras confirms that it is indeed 2 * root 2 because (see diagram below)...

a^2 = 1^2 + 1^2  = 1 + 1= 2
thus a = root 2

similarly...
b^2 = 1^2 + 1^2 = 1 + 1 = 2
thus b = root 2

thus a + b = root 2 + root 2 = 2 * root 2

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