**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)*

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: 12/22/1998Cut 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: 12/22/1998to 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:12/22/1998I 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: 12/22/1998Which 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: 12/23/1998You 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 5Second: 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: 12/23/1998Pythagoras confirms that it is indeed 2 * root 2 because (see diagram below)...

a^2 = 1^2 + 1^2 = 1 + 1= 2

thus a = root 2similarly...

b^2 = 1^2 + 1^2 = 1 + 1 = 2

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

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Other pages: Usenet thread "Box and ribbon" Surface distances on a cube or cuboid