20070329, 13:44  #1 
Dec 2005
2^{2}·7^{2} 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<x^2 we can draw two tangents to this curve, enclosing a certain area. Describe the set of points for which the area is constant. 
20070402, 08:36  #2 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×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 
20070402, 13:37  #3 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Without doing any work, my guess is y=x^{2}c
David 
20070403, 07:40  #4 
Aug 2004
7·19 Posts 

20070403, 08:06  #5 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
THX. So c = (3*area/2)^{(2/3)}
Last fiddled with by davieddy on 20070403 at 08:07 
20070403, 09:00  #6 
Dec 2005
2^{2}×7^{2} 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 
20070404, 09:58  #7 
Dec 2005
304_{8} 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. 
20070404, 12:04  #8 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
The tangents from (a,0) hit y=x^{2} at (0,0) and
(2a,4a^{2}). 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 20070404 at 12:18 
20070404, 12:27  #9 
"Lucan"
Dec 2006
England
194A_{16} Posts 
Yes, Since the sheared parabola passes through the origin,
the locus of its apex is obviously y=x^{2} 
20070405, 09:36  #10 
"Lucan"
Dec 2006
England
2×3×13×83 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 20070405 at 09:40 
20070405, 11:13  #11  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Quote:
To me its a waste of time. "Shearing transformation"? Ha! ha ! Ha! Mally 

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