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 Exercise [02.04] 
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Joined: 01 Aug 2009, 17:36
Posts: 2
Post Exercise [02.04]
Hi all,
I was writing a solution for this problem and while checking if it was right I've found this one on Penrose's home page:

http://www.roadsolutions.ox.ac.uk/solut ... ution2.jpg

Can I kindly ask you why: \frac{OP'}{OP} = \frac{SQ}{SP}
Mainly what's not so clear to me is the fraction \frac{SQ}{SP}, could you please explain me why this equation holds?

For those interested there's another published solution: http://downroadtoreality.blogspot.com/2 ... se-24.html that's IMHO more elegant than the first one, even if it's less clear.

Thanks for your help!

01 Aug 2009, 17:44

Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [02.04]
From the diagram given with the solution:

\triangle SOP is similar to \triangle QP^\prime P, since\angle QPP^\prime =\angle SPO and \angle SOP=\angle QP^\prime P.

So \frac{PQ}{SP}=\frac{PP^\prime}{OP}~~~~~~~~(1)

Also from the diagram:




Dividing both sides of (2) by SP and both sides of (3) by OP gives:




Using (1) and substituting shows that the RHS of (4) and the RHS of (5) are equal and so


02 Aug 2009, 06:53

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Exercise [02.04]
Maybe your confusion stems from not realizing that two letters here is a positive distance. So SQ is the distance from point S to point Q

02 Aug 2009, 07:20

Joined: 01 Aug 2009, 17:36
Posts: 2
Post Re: Exercise [02.04]
Thank you vasco, I was puzzled because I didn't understand how he came up with the idea of dividing (2) by SP and (3) by OP (alas I didn't have the same idea when trying this kind of demonstration :().
Thank you so much again, everything is clear now :)))))))))))

02 Aug 2009, 09:14
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