Interesting Circle Problem

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Goistein

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Oct 8, 2006
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Suppose you are running on a circle track x^2+y^2=1. You spot a tennis ball on the interior of the circle at point (x,y), so you decide to run over and take it. However, now you want to move back to the circumference of the track at point (a,b) so that the distance moved is equal to the length of the arc that you would have moved had you not taken the ball. It should be safe to say you started at point (1,0).

Dist((1,0) (x,y))+ Dist((x,y) (a,b))=Arc((1,0) (a,b))

It was a real pain for me, as I used 6 equations, 6 unknowns, but are there any other simpler ways?
 
Yikes!
Perhaps trying somethig a bit easier would help you "see"; like:

Circle with center (5,5) is tangent to x-axis and y-axis (in top right quadrant).
Point A is on circle's circumference at(0,5), point P is inside circle at (4,7),
and point B is on circle's circumference at (x,y).
If length line AP + length line BP = length Arc AB (the one above circle's center), what is x and y ?
 
That would bring me to

D(AP)=2sqrt(5)
D(PB)=sqrt(x^2+y^2+65-8x-14y)
A(AB)=5theta=5arctan((y-5)/(x-5))

and (x-5)^2+(y-5)^2=25

Still 2 equations 2 unknown and I still cant solve without a calculator...
 
Goistein said:
That would bring me to
D(AP)=2sqrt(5)
D(PB)=sqrt(x^2+y^2+65-8x-14y)
A(AB)=5theta=5arctan((y-5)/(x-5))
and (x-5)^2+(y-5)^2=25
Still 2 equations 2 unknown and I still cant solve without a calculator...
Click to expand...
Agree. Doesn't help much.

(x-5)^2 + (y-5)^2 = 25
This leads to x = 5 +- SQRT[y(10 - y)] ; but substituting that is scary!

Hope someone else can come up with something magical :wink:
 
Using my example
"Circle with center (5,5) is tangent to x-axis and y-axis (in top right quadrant).
Point A is on circle's circumference at(0,5), point P is inside circle at (4,7),
and point B is on circle's circumference at (x,y)."
this works, but unable to see (yet) how it's arrived at (T = theta):

x = 5 - 5cos(T) , y = 5 + 5sin(T)
where
T = [2sqrt(5) + sqrt{30 - 10cos(T) - 20sin(T)}] / 5

Only thing I recognize is the 2sqrt(5) for AP.

Anyhoo, makes iteration easy; to 5digits accuracy:
T = 86.40225 degrees
Arc = 7.54002
x = 4.68624
y = 9.99014

Graphed it all and all OK!
 
So then there's no way to get an exact answer...

You'd think the answer to a question like that would be simple after minimal use of trig/algebra, but I guess not
 
Maybe I’m just too hammered right now, but if my interpretation of this problem is correct, I’d say this is basically just a problem of the sine and cosine of real numbers.

Let’s analyze this fascinating phenomenon through the eyes of someone who’s been “enchanted”.

Suppose you are running on a circle track x^2+y^2=1 (i.e., the unit circle, say in a counter clockwise direction, like in a positive radian angle). You (then) spot a tennis ball on the interior of the circle at point P(x,y), so you decide to run over and take it.

So far so good.

However, now you want to move back to the circumference of the track at point B(a,b) so that the distance moved is equal to the length of the arc that you would have moved had you not taken the ball. It should be safe to say you started at point A(1,0).

I see this as: You then see an opportunity to “compensate/cheat” by heading to any point B(a,b) on the circumference of the unit circle track which you reckon would “fairly” compensate you for the trouble of running after the tennis ball. To your amazement, however, the “unit circle universe” “corrects” you (if your reckoning is less or more than fair) by teleporting you to a point C(c,d) on the circumference whose arc length from point A(1,0) is, according to my fine fellow knight-errant Sir Denis, equal to length of line AP + length of line BP.
Thus, given the arc length AB, what then are the Cartesian coordinates of c and d?

If this is indeed the case scenario, and if t = length of line AP + length of line BP, then

\(\displaystyle t = \sqrt {\left( {1 - x} \right)^2 + \left( {0 - y} \right)^2 } + \sqrt {\left( {x - a} \right)^2 + \left( {y - b} \right)^2 }\)

and

\(\displaystyle \begin{array}{l} c = \cos t = \cos \left[ {\sqrt {\left( {1 - x} \right)^2 + \left( {0 - y} \right)^2 } + \sqrt {\left( {x - a} \right)^2 + \left( {y - b} \right)^2 } } \right] \\ d = \sin t = \sin \left[ {\sqrt {\left( {1 - x} \right)^2 + \left( {0 - y} \right)^2 } + \sqrt {\left( {x - a} \right)^2 + \left( {y - b} \right)^2 } } \right] \\ \end{array}\)

Or maybe I’m really seeing things due to my present heavy state of “enchantment”.
 
But the problem assumes that you (like me) do not have any common sense to figure out B. So (a,b) is (c,d) and t, c, and d are still in terms of c and d.

If you could get B accurately, that would mean we already have a point close enough to C.

However it did give me an idea:

Suppose you do just take an arbitrary B on the circle. Then you can repeatedly repeat those formulae to get (c,d), and each time hopefully closer to the true (c,d). Although it won't be exact, it will keep getting closer, and when you're down to an inch difference, nobody will care if you're an inch off.

Now just to see if it will converge like Euler's Method should...

Just for fun, a variation would be use a normal high school/college track with equations:
y=1
y=-1
(x-1)^2+y^2=1
(x+1)^2+y^2=1

Same problem but this time it has the actual straightaways and curves. Just in case someone's bored.
 
Goistein said:
However it did give me an idea:
Suppose you do just take an arbitrary B on the circle. Then you can repeatedly repeat those formulae to get (c,d), and each time hopefully closer to the true (c,d). Although it won't be exact, it will keep getting closer, and when you're down to an inch difference, nobody will care if you're an inch off.
Click to expand...
Yep. That's exactly what I meant by the term "iteration" in my previous post. "Hit and miss!".
With careful programming, you can get an answer to any precision you wish...
 
Suppose you are running on a circle track x^2+y^2=1. You spot a tennis ball on the interior of the circle at point (x,y), so you decide to run over and take it. However, now you want to move back to the circumference of the track at point (a,b) so that the distance moved is equal to the length of the arc that you would have moved had you not taken the ball. It should be safe to say you started at point (1,0).

Dist((1,0) (x,y))+ Dist((x,y) (a,b))=Arc((1,0) (a,b))

It was a real pain for me, as I used 6 equations, 6 unknowns, but are there any other simpler ways?

Having taken a route similar to jonah, I ended up with the following.

I went down a road similar to ??? and ended up with the following.

Given a circle of radius 1.0 and center O.
Point A on the circle at (1,0)
Point P is the ball location at (x,y).
Point B is the point of return to the circle of radius 1 along the radius (not necessarily the best way).
OP = sqrt(x^2 + y^2).
BP = 1 - sqrt(x^2 + y^2).
AP = sqrt[(1 - x)^2 + y^2].
Arc AB = 1(arctan(y/x)) = a

Test case:
a = 60º = 1.047197 rad
Arc AB =(1) 60(Pi)/180 = 1.047197 units.

Est.x = .9(.5( = .45
Est. y = .9(.866) = .7794
AP + PB = 1 - sqrt(x^2 + y^2) + sqrt](1 - x)^2 + y^2] = 1.05394

Est. x = .95(.5) = .475
Est. y = .95(.866) = .8227
AP + PB = 1.02596

Extrapolating, x = .456031 and y = .789845 making AP + PB = 1.047197 units.

<------------------------------------------------------------->

Considering a = 90º = 1.57079 rad
Arc AB = (1)90(pi)/180 = 1.57079 units

AP = sqrt( 1^2 + y^2)
PB = (1 - y)
AP + PB = sqrt(y^2 + 1) + (1 - y)

Est. y = .5
AP + PB = 1.61803

Est. y = .6
AP + PB = 1.566190

Extrapolating, y = .591126 making AP + PB = 1.57079 units.

Clearly, there are an infinite number of solutions

Arc AB = 1.57079 rad
-------------------------------------------------------------->

Refering back to the a = 60º where P(x,y) = (.456031, .789845) satisfies th requirements of the problem, it should be noted that the distance of P inside the circle is a mere .087958 units. It would appear to be easier, and more efficient to stay on the track from point A until a point C is reached, where the distance from point C (just prior to point B) to the derived point P is equal to the distance from point C to point B. Once the ball is retrieved, the runner returns to the track along the mirror image of the path CP on the opposite side of the radial line OB.

There is a solution for every angle a assumed.

The bottom line is what determines the runners choice of when he departs from the circular path to retrieve the ball?

Any thoughts? I did not have much time to spend on this and may have taken a wrong path somehere.
 
Well, as far as I'm concerned:

if radius, departure point coordinates and ball coordinates are given: one solution, of course;

it really does not matter what is chosen as the departure coordinates, since the object
is not to find the "shortest" path inside the circle, but simply to end up with an inside
path equalling the "unjogged arc".

To use an easier to see scenario: circle track = 5 miles;
jogger "departs" after jogging 1 mile;
distance jogged "inside" plus length of remaining arc once back on the circle track
must total 4 miles.

Of course, if jogger's aim is "once around the track", then departure point will require
a "less than" range.

An interesting problem could be:
jogger decides to jog a circular track circmference = t ;
jogger takes off at point (a,b) ;
at point (c,d), jogger veers and heads for point (e,f) inside the circular track ;
when point (e,f) is reached, jogger then heads straight for point (a,b) ;
total distance travelled = t ;
get my drift?
 
I guess I really was way too much “enchanted” when I read your first post…so much so that I somehow managed to draw an analogy with a time travel feature I came across a few years ago on Discovery Channel. The show essentially focused on how the universe “punishes” a person who somehow managed to travel back in time (by travelling faster than light) by transporting him/her to the past of a parallel universe. In this sense, no paradox is created if he/she interferes with his/her ancestors business that will result in the prevention of his/her birth. But I digress…

In short order, what you want then was already in your possession. You simply need a restatement of your original expression. Thus

Dist((1,0) (x,y))+ Dist((x,y) (a,b))=Arc((1,0) (a,b))

\(\displaystyle \begin{array}{l} \Leftrightarrow \\ \sqrt {\left( {1 - x} \right)^2 + \left( {0 - y} \right)^2 } + \sqrt {\left( {x - \cos t} \right)^2 + \left( {y - \sin t} \right)^2 } = t \\ \end{array}\)

Using the unit circle equivalent of Sir Denis’ “easier” example, where x = 1/5 and y = 2/5, we get
t ? 1.508003662…
which completely agrees with Sir Denis’ result. Accordingly,

\(\displaystyle \frac{t}{\pi } = \frac{T}{{180}} \Leftrightarrow T \approx 86.40224533...^ \circ\)

Now how about indulging us by posting your so called painful “6 equations, 6 unknowns”.
 
Too many decimals...

Anyway, sorry it took a while to get back, but you asked for my earlier approach so here it is:

a^2+b^2=1
t=Theta(radian)
Theta=Arctan(b/a)
r+s=t
r=sqrt[(x-1)^2+y^2]
s=sqrt[(x-a)^2+(y-b)^2]

There's my 6 equations, and it would be too much to explain how many substitutions I tried...

The first obvious step is to redure it to 3 equations, but then it's all jumbled up.

Tchr: Yes, you probably could do that, but it would make for a much simpler solution, and nobody wants that, right? :wink:

Denis: Won't every interior point (e,f) make D[(a,b)(e,f)]+D[(c,d)(e,f)]<A[(c,d)(a,b)] because cutting through the track like that is always a shortcut? But it should work if (e,f) is an exterior point.
 
...still running around in circles? :wink:

WHY not leave the story stuff (tennis ball, circle track, jogger....) out?

Givens:
coordinates of points A and B on circumference of circle
radius of circle

Question:
Point P is somewhere inside circle such that AP + BP = arcAB;
what are P's coordinates?
 
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