come on people - are you only capable of hiding behind math that none of have studued and therefore have no hope of understaning. It's not that hard (honest), it just involves a touch of lateral thinking.
Originally posted by belgianfreak find the connection between these words:
mug elf kid low wag age hop eat tag
- they are all on the same line
- they are all writen by belgianfreak
- they all consist out of three letters (I personaly want to thank TheMaster37 for discovering that)
and last but not least
- they are all annoying me cause I can't see which connection is ment
come on people - are you only capable of hiding behind math that none of have studued and therefore have no hope of understaning. It's not that hard (honest), it just involves a touch of lateral thinking.
There are at least two people here who study math...
Originally posted by royalchicken And who might they be? If you're one, get a star and get in the Cult of Maths!
The Master and I both study math. (at least that's what we should do here instead of posting in the forums 🙂 )
Perhaps now is a good time to become a member. Who doesn't want to be in the Cult of Maths. I haven't got a creditcard nor a us bank account though, so it might take some while.
Originally posted by royalchicken What areas of math interest you in particular?
I'm interested in tha areas of financial mathematics, which is for the most part about stochastics. But also geometry interests me.
Perhaps you know this. I've got a triangle ABC. On the line AB I take a point D, on BC I take E and on CA F. Show that if AE, BF, CD intersect in one point, AD/DB * BE/EC * CF/CA = 1.
It is easy to prove for some special cases, but I can't do it in general.
Originally posted by Fiathahel I'm interested in tha areas of financial mathematics, which is for the most part about stochastics. But also geometry interests me.
Perhaps you know this. I've got a triangle ABC. On the line AB I take a point D, on BC I take E and on CA F. Show that if AE, BF, CD intersect in one point, AD/DB * BE/EC * CF/CA = 1.
It is easy to prove for some special cases, but I can't do it in general.
I don't think that's true in the general case. If D's near B, E's near C and F's near A, then AE and CD intersect near C, and AE and BF intersect near A.
Originally posted by richjohnson I don't think that's true in the general case. If D's near B, E's near C and F's near A, then AE and CD intersect near C, and AE and BF intersect near A.
But then AD/DB * BE/EC * CF/FA =/= 1 .
(I notice now I mistyped CF/FA the first time)
word games
22 Aug '03 12:52
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