Brain Teaser 2  Unusual challenge

01182016, 02:17 PM
(This post was last modified: 01192016 06:44 PM by Pekis.)
Post: #1




Brain Teaser 2  Unusual challenge  
01182016, 10:06 PM
Post: #2




RE: Brain Teaser 2  Unusual challenge
Well, it might involve a modicum of unnatural affection for an inanimate object, but if necessary, osculari circulos .
BEST! SlideRule 

01192016, 10:41 AM
Post: #3




RE: Brain Teaser 2  Unusual challenge
... Last call for this challenging challenge ... very far from being a simple translation ... Anyone to answer ?


01192016, 02:58 PM
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RE: Brain Teaser 2  Unusual challenge
(01182016 02:17 PM)Pekis Wrote: Hello, Good brain teaser! Do you want a solution or say the minimum distances solution? In either case, I am sure some of us will plunge into this shortly... Osculum anuli ? Best, Carl 

01192016, 03:26 PM
(This post was last modified: 01192016 06:08 PM by Pekis.)
Post: #5




RE: Brain Teaser 2  Unusual challenge
Hello,
I just want the solution around x = 2 as in the picture (and not the solution around x=12). I don't think there are other solutions satisfying equidistance (shortest) from the 3 figures simultaneously. Can you solve it, Carl ? Thanks 

01192016, 06:24 PM
(This post was last modified: 01192016 06:24 PM by CR Haeger.)
Post: #6




RE: Brain Teaser 2  Unusual challenge
(01192016 03:26 PM)Pekis Wrote: Hello, Ill try but it will take (me) a while. Others will probably knock this out in no time. 

01192016, 08:34 PM
(This post was last modified: 01192016 08:35 PM by fhub.)
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RE: Brain Teaser 2  Unusual challenge
(01182016 02:17 PM)Pekis Wrote: What are the coordinates of the point (xp,yp) on the image, equidistant (shortest) from the 3 figures ? Which distance ?Here are my results: Code:
Franz 

01202016, 08:50 AM
(This post was last modified: 01212016 07:09 AM by Pekis.)
Post: #8




RE: Brain Teaser 2  Unusual challenge
Hello Franz,
Well done ! Your solution is a fast & straight one ... and mine was rather unusual: I used the graphic intersection of the socalled parallel curves. Except in the case of a line or a circle, a parallel curve is a rather heavy polynomial => the parallel curve of a parabola is not a parabola ! But if the curves are expressed in parametric form, the world becomes simple: For a curve defined as (f(t),g(t)), the parallel curve at a given distance d (d for the other side) is (f(t)+d*g'(t)/sqrt(f'(t)^2+g'(t)^2),g(t)d*f'(t)/sqrt(f'(t)^2+g'(t)^2)) I had then to find the parametric form of the parallel curves of y=x^2+1, y=x2, x^2+y^2=2 and find the value of d (with same sign for all curves) which makes them intersect simultaneously. y=x^2+1: Original curve: (t/2,t^2/4+1) Parallel curve: (t/2+d*t/sqrt(1+t^2),t^2/4+1d/sqrt(1+t^2)) (d>0 towards the intersection) y=x2: Original curve: (2*t,2*t2) Parallel curve: (2*td/sqrt(2),2*t2+d/sqrt(2)) (d>0 towards the intersection, but had to change the sign of d) x^2+y^2=2: Original curve: (sqrt(2)*cos(t),sqrt(2)*sin(t)) Parallel curve: ((sqrt(2)+d)*cos(t),(sqrt(2)+d)*sin(t)) (d>0 towards the intersection) And here is the result on the impressive free online grapher desmos, with my graph I used the slider to change d and refined it with the appropriate zoom level, and voilà ! Still using the slider, one can easily find another solution around x=12 (distance around 10.6). Franz, how would you transform your system to also find it ? I had much fun on my way to the solution, like Steve McQueen on the way to San Mateo . You too ? Thanks to Carl & Franz 

01202016, 11:59 AM
(This post was last modified: 01202016 12:00 PM by fhub.)
Post: #9




RE: Brain Teaser 2  Unusual challenge
(01202016 08:50 AM)Pekis Wrote: Franz, how would you transform your system to also find it ? Well, I just need to add the condition xs<0 to my equations, and the numeric solver that I use now gives the following solutions: Code:
Franz 

01202016, 04:30 PM
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RE: Brain Teaser 2  Unusual challenge
(01202016 11:59 AM)fhub Wrote: BTW, I've used the ancient (more than 15 years old!) numeric solver 'Eureka 2.11', which is still a very good (DOS!) program for such tasks. I remember Eureka (and Mercury, too), from the 80's, very well. Anyone interested in it can still get a free copy from here Greetings, Massimo +×÷ ↔ left is right and right is wrong 

01212016, 12:17 PM
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RE: Brain Teaser 2  Unusual challenge
Very interesting challenge. I only miss what it has to do with HP calculators.


01212016, 02:09 PM
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RE: Brain Teaser 2  Unusual challenge
My Cotidiana oscula circulorum approach doesn't seem to intersect the three functions with the solution coordinates (see attached Png fie). Has anyone calculated the three intersection points as well as the tri(tangents)intersect point?
[attachment=3062] ps: isn't the minimum intersect point inscribed in the circle equation? BEST! SlideRule 

01212016, 02:43 PM
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RE: Brain Teaser 2  Unusual challenge
(01212016 02:09 PM)SlideRule Wrote: My Cotidiana oscula circulorum approach doesn't seem to intersect the three functions with the solution coordinates (see attached Png fie). Has anyone calculated the three intersection points as well as the tri(tangents)intersect point? You forgot to square the radius ... BTW, inverting d sign in the parallel curve of the circle, it seems there is also another solution at x0=0,168042723 y0=0,168093174 d=1,17652959 

01212016, 03:02 PM
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RE: Brain Teaser 2  Unusual challenge
Quote:You forgot to square the radius ... ARGGH!!! thanks! SlideRule 

01212016, 04:15 PM
(This post was last modified: 01222016 07:08 AM by Pekis.)
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RE: Brain Teaser 2  Unusual challenge  
01212016, 05:01 PM
Post: #16




RE: Brain Teaser 2  Unusual challenge
Quote:BTW, inverting d sign in the parallel curve of the circle, it seems there is also another solution at if interested see attached SlideRule [attachment=3063] 

01212016, 10:26 PM
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RE: Brain Teaser 2  Unusual challenge
(01212016 12:17 PM)RMollov Wrote: Very interesting challenge. I only miss what it has to do with HP calculators. It helps perhaps to read the description of the forum. Quote:General Forum Günter 

01252016, 09:28 AM
Post: #18




RE: Brain Teaser 2  Unusual challenge  
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