Re: Riddle for the bright minds here
From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 06/09/04
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Date: Wed, 09 Jun 2004 04:55:06 GMT
"César Sirvent" <8umucsxySPA@M_MAPSterraReMoVeThIs.es> wrote in message
news:p6pxc.847483$A6.3278159@telenews.teleline.es...
| OFF TOPIC:
|
| Another nice riddle for the bright minds here:
|
| There are 3 Ancient-Greek philosophers, which are discussing the Pitagoras
| theorem at open air. Suddenly, a bird flies near them and praise them with
a
| corresponding excrement in the head, to all of them. However, no one of
them
| notices because they are wearing hats.
|
| But obviously, they see each other's stain in the hat, so they start
| laughing, unaware of their own dirty hat but aware of the companion dirty
| hat.
|
| Until one of them, the most intelligent, suddenly stops laughing. How did
he
| know that he also was wearing a dirty hat? ( just reasoning, I mean ... )
|
| Cesar
He sees A laughing at B, and B laughing at A. If A has a clean hat, B would
not be laughing at A. However, A or B would soon realize that they could
only be laughing at each other, and one (or both) would stop. Therefore A
and B are laughing at C also, and C stops.
Here's a little ON topic for you.
"If we place x'=x+vt, it is clear that a point at rest in the system k must
have a system of values x', y, z, independent of time. We first define tau
as a function of x', y, z, and t. To do this we have to express in equations
that tau is nothing else than the summary of the data of clocks at rest in
system k, which have been synchronized according to the rule given in § 1.
>From the origin of system k let a ray be emitted at the time tau0 along the
X-axis to x', and at the time tau1 be reflected thence to the origin of the
co-ordinates, arriving there at the time tau2; we then must have
˝(tau0+tau2) =tau1, or, by inserting the arguments of the function tau and
applying the principle of the constancy of the velocity of light in the
stationary system:-
˝[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c+v))"
It's too tough for you, though.
Androcles
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