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 2007-03-29, 13:44 #1 Kees     Dec 2005 110001002 Posts Analysis puzzle Here is a puzzle I found on some mathforum. I have not found a neat solution other than brute analysis (which, although some may consider this neat, is not really elegant). Consider the equation y=x^2 For points (x,y) in the plane with y
 2007-04-02, 08:36 #2 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22·33·19 Posts No takers! Since no one has attempted your problem, kindly give us your solution, no matter how 'brutal' it may be. I for one give up (though it can be worked out by calculus). I have resurrected this thread so that others who may have missed it will consider it and give a solution. However it will be interesting to know your way of solving it. In the meantime I will consult my books on analysis as Im now out of touch with it. Mally
 2007-04-02, 13:37 #3 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts Without doing any work, my guess is y=x2-c David
2007-04-03, 07:40   #4
Chris Card

Aug 2004

2058 Posts

Quote:
 Originally Posted by davieddy Without doing any work, my guess is y=x2-c David
Your guess is correct I reckon. I won't show the gory working, but I calculate the area in question (for a point (x,y), x^2 > y) to be

2/3(x2 - y)(3/2)

Chris

 2007-04-03, 08:06 #5 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts THX. So c = (3*area/2)(2/3) Last fiddled with by davieddy on 2007-04-03 at 08:07
 2007-04-03, 09:00 #6 Kees     Dec 2005 22×72 Posts Not going to show the gory details either. It is relatively simple to calculate the tangents to y=x^2 from a point (a,b) for which a>b^2, then integration and some triangle substraction will give an answer. But after finding the answer, which is just a translation of the y=x^2, I felt somewhat stupid and was certain that there was an easy geometrical explanation. Have not found one yet so I just continue to feel silly, Kees
 2007-04-04, 09:58 #7 Kees     Dec 2005 22·72 Posts Finally I found something. Well, to be more precise, Archimedes found something. Take two points A, B on a parabola and let the tangents to the parabola at A and B intersect at S. Then the area enclosed by the chord AB and the parabola is 2/3 of the area of the triangle ABS. The area which we are considering is the complentary 1/3 part of the triangle. So the question can transformed to the question of finding the set of all such triangles with constant area. Have not worked out the details yet, but this seems a good way to start.
 2007-04-04, 12:04 #8 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts The tangents from (a,0) hit y=x2 at (0,0) and (2a,4a2). Shearing the diagram keeping the line x=a fixed maintains areas, tangents and the shape of the parabola. It just shifts the apex. I think the locus of the apex offers a promising approach. David PS Perhaps starting with the tangents from (0,-c) is even simpler. (Shear keeping y axis fixed) Last fiddled with by davieddy on 2007-04-04 at 12:18
 2007-04-04, 12:27 #9 davieddy     "Lucan" Dec 2006 England 194A16 Posts Yes, Since the sheared parabola passes through the origin, the locus of its apex is obviously y=-x2
 2007-04-05, 09:36 #10 davieddy     "Lucan" Dec 2006 England 194A16 Posts I know the whole point of this thread is to avoid "gory" details, but formally, the "shearing" transformation I am considering is: x' = x y' = y+kx where k is a constant real number. Are you paying attention Mally? David Last fiddled with by davieddy on 2007-04-05 at 09:40
2007-04-05, 11:13   #11
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts

Quote:
 Originally Posted by davieddy I know the whole point of this thread is to avoid "gory" details, but formally, the "shearing" transformation I am considering is: x' = x y' = y+kx where k is a constant real number. Are you paying attention Mally? David
The problem is not put properly. The methods of attack are even worse.
To me its a waste of time.
"Shearing transformation"? Ha! ha ! Ha!
Mally

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