01-18-2016, 02:17 PM

01-18-2016, 02:17 PM

01-18-2016, 10:06 PM

Well, it might involve a modicum of unnatural affection for an inanimate object, but if necessary, osculari circulos .

BEST!

SlideRule

BEST!

SlideRule

01-19-2016, 10:41 AM

... Last call for this challenging challenge ... very far from being a simple translation ... Anyone to answer ?

01-19-2016, 02:58 PM

(01-18-2016 02:17 PM)Pekis Wrote: [ -> ]Hello,

Thanks to CR Haeger for initial inspiration for this challenge:

What are the coordinates of the point (xp,yp) on the image, equidistant from the 3 figures ? Which distance ?

How did you get this ? Please provide maximum precision. Good luck !

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

01-19-2016, 03:26 PM

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

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

01-19-2016, 06:24 PM

(01-19-2016 03:26 PM)Pekis Wrote: [ -> ]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

Ill try but it will take (me) a while. Others will probably knock this out in no time.

01-19-2016, 08:34 PM

(01-18-2016 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:

How did you get this ? Please provide maximum precision. Good luck !

Code:

;distance to line: dl=abs(x0-y0-2)/sqrt(2)

;distance to circle: dc=sqrt(x0^2+y0^2)-sqrt(2)

;distance to parabola: dp=sqrt((x0-xs)^2+(y0-xs^2-1)^2)

;condition for xs (x-coord of intersection point with parabola):

;2*xs^3-(2*y0-3)*xs-x0=0

;equation system: dl=dc && dl=dp && equation for xs

abs(x0-y0-2)/sqrt(2)=sqrt(x0^2+y0^2)-sqrt(2)

abs(x0-y0-2)/sqrt(2)=sqrt((x0-xs)^2+(y0-xs^2-1)^2)

2*xs^3-(2*y0-3)*xs-x0=0

;distance

dl=abs(x0-y0-2)/sqrt(2)

Solution:

---------

Variables:

x0 = +1.9852721593222333

y0 = +1.529294579719819

xs = +1.007327946198892

dl = +1.0917887237671993

Franz

01-20-2016, 08:50 AM

Hello Franz,

Well done !

Your solution is a fast & straight one ... and mine was rather unusual:

I used the graphic intersection of the so-called 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=x-2, 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+1-d/sqrt(1+t^2))

(d>0 towards the intersection)

y=x-2:

Original curve: (2*t,2*t-2)

Parallel curve: (2*t-d/sqrt(2),2*t-2+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

Well done !

Your solution is a fast & straight one ... and mine was rather unusual:

I used the graphic intersection of the so-called 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=x-2, 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+1-d/sqrt(1+t^2))

(d>0 towards the intersection)

y=x-2:

Original curve: (2*t,2*t-2)

Parallel curve: (2*t-d/sqrt(2),2*t-2+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

01-20-2016, 11:59 AM

(01-20-2016 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:

x0 = -12.281421214881970

y0 = +1.16511414476357

xs = -1.77028039267528

dl = +10.922349898643146

Franz

01-20-2016, 04:30 PM

(01-20-2016 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.

Franz

I remember Eureka (and Mercury, too), from the 80's, very well.

Anyone interested in it can still get a free copy from here

01-21-2016, 12:17 PM

Very interesting challenge. I only miss what it has to do with HP calculators.

01-21-2016, 02:09 PM

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

[attachment=3062]

ps: isn't the minimum intersect point inscribed in the circle equation?

BEST!

SlideRule

01-21-2016, 02:43 PM

(01-21-2016 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?

ps: isn't the minimum intersect point inscribed in the circle equation?

BEST!

SlideRule

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

01-21-2016, 03:02 PM

Quote:You forgot to square the radius ...

ARGGH!!!

thanks!

SlideRule

01-21-2016, 04:15 PM

01-21-2016, 05:01 PM

Quote: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[/size]

if interested see attached

SlideRule

[attachment=3063]

01-21-2016, 10:26 PM

(01-21-2016 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

Including all HP calculators except the HP Prime. Also HP news, general math and science etc. (Essentially this is the original HP forum.)

Günter