mersenneforum.org  

Go Back   mersenneforum.org > Fun Stuff > Puzzles

Reply
 
Thread Tools
Old 2007-03-29, 13:44   #1
Kees
 
Kees's Avatar
 
Dec 2005

22·72 Posts
Default 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.
Kees is offline   Reply With Quote
Old 2007-04-02, 08:36   #2
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

205210 Posts
Exclamation 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
mfgoode is offline   Reply With Quote
Old 2007-04-02, 13:37   #3
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default

Without doing any work, my guess is y=x2-c

David
davieddy is offline   Reply With Quote
Old 2007-04-03, 07:40   #4
Chris Card
 
Chris Card's Avatar
 
Aug 2004

7·19 Posts
Default

Quote:
Originally Posted by davieddy View Post
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
Chris Card is offline   Reply With Quote
Old 2007-04-03, 08:06   #5
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2×3×13×83 Posts
Default

THX. So c = (3*area/2)(2/3)

Last fiddled with by davieddy on 2007-04-03 at 08:07
davieddy is offline   Reply With Quote
Old 2007-04-03, 09:00   #6
Kees
 
Kees's Avatar
 
Dec 2005

22·72 Posts
Default

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
Kees is offline   Reply With Quote
Old 2007-04-04, 09:58   #7
Kees
 
Kees's Avatar
 
Dec 2005

19610 Posts
Default

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.
Kees is offline   Reply With Quote
Old 2007-04-04, 12:04   #8
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default

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
davieddy is offline   Reply With Quote
Old 2007-04-04, 12:27   #9
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2×3×13×83 Posts
Default

Yes, Since the sheared parabola passes through the origin,
the locus of its apex is obviously y=-x2
davieddy is offline   Reply With Quote
Old 2007-04-05, 09:36   #10
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2×3×13×83 Posts
Default

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
davieddy is offline   Reply With Quote
Old 2007-04-05, 11:13   #11
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

22×33×19 Posts
Thumbs down

Quote:
Originally Posted by davieddy View Post
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
mfgoode is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Complexity analysis of 3 tests kurtulmehtap Math 10 2013-03-20 14:15
CLOCK_WATCHDOG_TIMEOUT (101) analysis TObject Software 4 2013-02-05 23:53
Dimensional analysis davieddy Puzzles 9 2011-08-02 09:59
Analysis of testing ranges for k's < 300 gd_barnes Riesel Prime Search 7 2007-11-17 05:48
mersenne analysis troels munkner Miscellaneous Math 2 2006-07-17 03:18

All times are UTC. The time now is 06:08.


Tue Jan 31 06:08:13 UTC 2023 up 166 days, 3:36, 0 users, load averages: 1.15, 1.12, 1.11

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2023, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔