Seeing as the going

claim is that there are only 36 ways to paint a

tetrahedron using four

unique colors, I'll

endeavor to

count them. Below I have '

drawn' a

tetrahedron for visualization purposes:

,|,
,7``\'VA,
,7` |, `'VA,
,7` `\ `'VA,
,7` |, `'VA,
,7` `\ `'VA,
,7` |, `'VA,
,7` `\ ,..ooOOTK`
,7` |,.ooOOT''` AV
,7` ,..ooOOT`\` /7
,7` ,..ooOOT''` |, AV
,T,..ooOOT''` `\ /7
`'TTs., |, AV
`'TTs., `\ /7
`'TTs., |, AV
`'TTs., `\ /7
`'TTs., |, AV
`'TTs.,\/7
`'T`

There are two ways to paint a T (tetrahedron) with a unique color for each

side; one is the

mirror of the other.

There are 24 ways to paint a T with three unique colors. Four

possibilities for the two that are the same, three for the first unique, and two for the second unique = 4*3*2 or 24. It is not

possible to create

duplicates, since duplicates are taken care of by

successively reducing the number of possible colors for the next

consecutive side.

There are 12 ways to paint a T where three sides are the same: 4 possibilities for the three same sides, three remaining possibilities for the remaining side for a total of 4*3=12. Again, duplicates do not exist.

There are four ways to paint a T where all the sides are the same (if it is

imagined that four is the total amount of colors available).

Now, to total the

results: 2+24+12+4=42. Hmm. I don't count 36.

Theory and

reality not quite in

sync, or are my

figures

wrong? Just a

side note, if

mirror images aren't

allowed, this doesn't quite cut the numbers in

half, but close. They would become 1+12+12+4=29.

Update: 15:51 Tue Aug 22 2000

In reply to mblase, umm, I still have to disagree, and even add some more to my list. Remember, a mirror image is one such that no

rotation can

achieve it while still within the same

dimension (

i.e. a left hand cannot be rotated into a right hand while still in the

third dimension). So my original count for three unique colors is still

accurate. For two unique colors, I

concede the

point. So that's another six onto the original, or 48, or another six onto the secondary, or 35. I have done the work on paper and I believe it stands as I have stated it.

Update again.

Now I see the

light. It took quite a bit of imagination (something I have trouble with . . .). The count is 36. So much for counting accuracy . . . The operative case here is a tetrahedron viewed on an

edge. Two unique colors are on the

faces visible, a third unique color is on the faces that aren't. The two visible

faces can be rotated either way 180 degrees to produce a mirror image (so much for my mirror "

logic"

above), so there are only 12 instead of 24, and the count is 36 (or 35 if mirrors aren't allowed . . .;).

Final note: I don't

intend to "

fix" the

mistakes above . .

rather I wish this to represent a

process of

thought that may (or equally well may not) be useful to someone wishing to

understand. My mistakes above are

legitimate; I really did think I was right for quite some time (an

hour at least . . .).