Oldphil's really hard math problem. Takers?

[Fsiphskilm]

you guys are fr

[Scribs]

I brought this to school and worked on it in my spare time, all i got was 08:00:40 and i think it works, but one thing is certain-you guys are way beyond me mathwise!

[Azier the Swordsman]

I think we can officially dub this "The Geek Thread".

[Technomancer]

Think like this for a minute... I started by assuming that one time this might work is around 8:20, which (without min/hr hand advance) puts all 3 hands on the numbers. Now, we have to count for the fact that the hour hand is part of the way to 9, the min hand is just past the four, and the second hand is what done it to 'em.

...Prepair for major jump in logic...

So our function has to account for the number of seconds in a 12 hour period (60 * 60 * 12 = 43200). X is the number of seconds past 8:20 we are...

Our function, therefore, looks like this:

43200 = x + 60(20 + X) + 3600(8 + x)
43200 = x + 120 + 60x + 28800 + 3600x
14280 = 3661x
x = 14280/3661
x = 3.09005736 seconds.

Convinced?

Incidental Note: I'm very evil, so I may very well have done this just to show you "logic errors" people make.

I see several problems with this solution. First, there is no verification that the hand positions actually are separated by 120 degrees (it is simply assumed to be correct or approximately correct). In fact, this approach says nothing about the angular separation at all, leaving one to wonder what exactly is being demonstrated here. A further problem is that you are adding different units of time without converting them. For example, 8+x where x is in seconds, but 8 is in hours. To add them properly, you have to write 3600*8+x, not 3600*(8+x).