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 Post subject: Travel fun
PostPosted: Wed, 10 Aug 2011 05:54:30 UTC 
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Code:
A(X@x)******************************>C<.....{65}.....(Y@y)B

A<**********************************(X@x)

A(X@x)***24***>D<...................(Y@y)

X and Y are at ends of straight path AB.
X's speed = x, Y's speed = y ; x > y.
They leave at same time and meet at point C.
X then turns around and travels back to A, Y keeps going towards A.
X reaches A, turns around, heads back East: meets Y at point D.
BC = 65 and AD = 24. How long is AB?

No computer help; can be solved directly...

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 Post subject: Re: Travel fun
PostPosted: Wed, 10 Aug 2011 09:35:07 UTC 
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Express AC/BC and AD/YD in terms of x and y, then express YD and AC in terms of AB minus known lengths.

Spoiler:
\displaystyle\\\frac{AC}{BC}=\frac{AD}{YD}=\frac{x}{y}\\\therefore\;AD.BC=YD.AC=(AB-2BC-AD)(AB-BC)\\\therefore\;24\cdot 65=(AB-154)(AB-65)\\\therefore\;AB=169,\;\cancel{54}


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 Post subject: Re: Travel fun
PostPosted: Wed, 10 Aug 2011 14:38:19 UTC 
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Agree; did mine slightly different: in terms of y / (x+y);
yours is more elegant (but only a little more!).

With a = AD and b = BC, mine is:
AB = [k +- SQRT(k^2 - 8b^2)] / 2 : where k = a + 3b
(which, of course, is same as yours...)

Btw, shouldn't your 2nd solution be 50, not 54?
I'm curious about this 2nd solution: obviously to be rejected, since BC = 65.
However, can it be made to work, like with Y going in reverse for a while?!
Any thoughts on that? Thanks.

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 Post subject: Re: Travel fun
PostPosted: Wed, 10 Aug 2011 19:22:00 UTC 
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Hold on, are you telling me they did all this without a lunch or potty break? And what about the time it took X to turn around? This problem seems ill-posed to me.

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 Post subject: Re: Travel fun
PostPosted: Wed, 10 Aug 2011 20:22:11 UTC 
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Shadow wrote:
Hold on, are you telling me they did all this without a lunch or potty break? And what about the time it took X to turn around? This problem seems ill-posed to me.

Geesshhhhh...you should be old enough to realise that X and Y are robots...

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 Post subject: Re: Travel fun
PostPosted: Wed, 10 Aug 2011 20:26:03 UTC 
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Of course, how foolish of me.

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 Post subject: Re: Travel fun
PostPosted: Wed, 10 Aug 2011 22:27:05 UTC 
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You're forgiven.

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 Post subject: Re: Travel fun
PostPosted: Thu, 11 Aug 2011 14:47:10 UTC 
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Denis wrote:
Btw, shouldn't your 2nd solution be 50, not 54?


Yes, I misread my own scribble. Also, I didn't explain that the point Y is the position of Y when X is back at A.

Denis wrote:
I'm curious about this 2nd solution: obviously to be rejected, since BC = 65.
However, can it be made to work, like with Y going in reverse for a while?!
Any thoughts on that? Thanks.

It implies YD and AC are negative distances, and the question requires X and Y to be travelling towards each other; if they are heading in the same direction, then C is outside AB, but then AD cannot be less than BC.

If you assume that Y changes direction the moment X arrives back at A, then you get 24*65 = (2BC+AD-AB)(AB-BC) = (154-AB)(AB-65), which makes AB = 89 or 130. To flip the other factor you need AB<BC and it's too big a change to the question.


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 Post subject: Re: Travel fun
PostPosted: Thu, 11 Aug 2011 16:58:29 UTC 
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Clear. Merci beaucoup.

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