http://s205.photobucket.com/albums/bb192/falzoon/Blockage.jpg

In 2D land, this patient will soon experience angina. Rigid plaque build-up, represented

by the two arms of y = tan(x), is so great, that the rigid, circular red blood cell, of

diameter 3 units, will not be able to course through the coronary artery. (Unfortunately,

the patient also suffers from rapid onset rigidity, so red cell and artery do not bend.)

What are the coordinates of the centre of the red cell, when it finally comes to rest?

Pretty good Scythian, after a lot shorter time than I took. In the end, I did manage to

get 14 decimal place accuracy with a promising 15th, but it was tedious (and fun).

However, I’ll be happy with 6th place accuracy. I won’t be expecting anyone to set

out a method of attack, unless they really want to. It would be nice though, only

briefly, just for my (and anyone else’s) interest, to compare efficiencies of method,

and to see how other minds work.

I think you’ve done enough work Scythian. That will be close enough. Doesn’t look like

there’ll be many more answerers, but I’ll keep it open for a little longer.

To 6 rounded decimal places I get (2.561523, 1.625488).

The ‘method’ I used :

1. Draw y = tan(x) with MathGV(tm) FREEWARE Version 4 (http://www.mathgv.com)

2. Copy image to Microsoft PAINT, then draw circle with diameter = 3 units.

3. Move circle until it’s in the approximate correct position.

4. Call left-hand tangent point (p, tan(p)). Estimate p by eye (approx. 1.1), then take

limits either side, so I started with 1.0 ≤ p ≤ 1.2.

5. Distance from (h, k) to (p, tan(p)) is r (= 3/2) = √[(h – p)^2 + (k – tan(p))^2], but slope

of tan(x) at p is sec^2(p), so slope of radius line is -cos^2(p), which is also equal to

[k – tan(p)]/(h – p). Therefore, [k – tan(p)]^2 = (h – p)^2*cos^4(p). Substituting gives

r = √[(h – p)^2 + (h – p)^2*cos^4(p)], whence, h = p + r/√[1 + cos^4(p)], which is h, in

terms of p only.

6. From the slope equation above, once h is calculated, k = tan(p) + (p – h)cos^2(p),

or basically, k in terms of p only.

7. Then set up the other slope equation : -cos^2(q) = [k – tan(q)]/(h – q), where

(q, tan(q)) is the right-hand tangent point. From this, tan(q) + (q – h)cos^2(q) – k = 0.

8. Write a program that steps through p from 1.0 to 1.2 in steps of 0.01, each time

calculating h and k, then calculate q (from the last equation) using Newton’s iterative

formula. Fortunately, only one initial guess of q = 3.9 for each iteration is necessary.

Set up a variable, z = ABS{r – √[(h – q)^2 + (k – tan(q))^2]} to calculate z for each

h, k, q obtained. Run the program and from the list of 21 p values displayed, select

that p which gives the least z value. Then change the limits, to either side of that p

value, reset the step to 0.001, and re-run the program to refine all values. Then

continue until the limiting accuracy of the computer is reached.

I see what you mean – basically, a Rube Goldberg machine is one that performs a

very simple task in a very complex manner. But then, this task may not be very

simple, as I think, attested to, by a marked lack of answers.

It does seem a contortional way to do it, but on the bright side, once all of the stuff

is in the program, with a couple of additions, such as getting IT to find the least z,

then getting IT to change the limits and step size (which is not very difficult), the

result should pop out in a matter of seconds.

But yes, calling all mathematical prodigious savants – small problem for you to try.

You can be sure that I appreciate elegant solutions and I heartily endorse all your

words of wisdom. I have been mulling over this problem of a systematic method, but

so far, every time a ray of light shines through, it’s snuffed out by a cloud.

I will endeavour to think more on this. It’s amazing that nature knows eventually where

the cell is going to end up, apparently without any effort of calculation, while us poor

sods must work hard, to the same end. Then again, as WE are not so precise in

many of our answers, neither is nature, because there’s bound to be some jostling of

atoms at the tangent points, so in reality, the cell is never stable on any particular

point. So is there something wrong with us, in seeking absolute precision?

Well, it seems to be useful to us as we grope towards the answer to the universe(s).

Your sentence about non-exact answers intrigues me. I must ponder that, as well.

I can think of only one thing to do, which is much the same, but perhaps simpler, but

also, not elegant. Move the circle so that it is tangent to the left arm, of tan(x).

Keeping it always tangent on (p, tan(p)), shimmy it down until some part of the right

hemisphere of the circle goes beyond the right arm of tan(x). Knowing approximately

which part of the circle will hit would help in cutting down calculations. It’s somewhat

similar to the method I outlined, except, where I calculated the right-hand contact

point, (q, tan(q)), to begin with, and then tested how the calculated radius compared

with the given radius, this second method would require testing many points on the

right hemisphere for collision with tan(x). Once a collision has occurred, you would

have to step back a fraction and repeat, with refinements. Unless, of course, you

initially embed refinements to a given accuracy, which will increase running time.

Maybe someone versed in collision theory could make it happen efficiently.

An excellent job, Frst Grade Rocks!, and with a different, definitely easier method than

I could muster. I’m a little perplexed by both your’s and Scythian’s method, but they

seem to work fine. I wasn’t sure what to do with your results, so I averaged each of

the four red figures, and to 6 decimal places, they are exactly the same as Scythian’s.

Thanks for that description Frst Grade Rocks!. I fully understand your method now

and I like it a lot. It definitely has that ‘wow’ factor because of it’s beautiful simplicity.

I’m ready to close this question now.

If I’m correct, the answer should have been : (2.561523, 1.625488)

To 14 decimals, I got : (2.56152254940429, 1.62548774489490),

with p = 1.09377933632813 and q = 3.87453719045212

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June 25th, 2010 at 10:01 pm

Approximately it is (2.564, 1.630). I’ll try to refine this later, but first, a seder.

Edit: Well, but that’s the thing, I haven’t been able to figure out a systematic means of finding the answer numerically. An exact solution is hopeless. Let’s see if anyone else wants to try.

Edit 2: Increased precision: (2.5615231, 1.6254889)

I suppose that if I keep working on this, I might be able to approach your 15-digit precision.

Edit 3: Oh god, that looks painful. There’s got to be an easier way. This is what makes this problem interesting, is there any systematic way to do this, besides doing a "Rube Goldberg"?

Edit 4: I’ve been fiddling around with other ways of doing this more straightforwardly, but without much luck. While it’s true that we can’t expect calculus to give us exact answers every time, I just wonder if there is a systematic numerical method to these kinds of optimization problems.

On another note, when presented with a problem to solve, sometimes we enjoy the art of sidestepping brute force methods with an elegant (and presumably simpler) solution, which is something I like doing because that itself can be its own puzzle. On the other hand, finding systematic methods of solving certain kinds of problems, even via brute force, can be its own reward, because then those methods can be used with similar problems elsewhere. Because I keep running into this kind of problem (or at least you seem to enjoy posing such), it intrigues me whether or not there can be a general method for treating it, even if not one that delivers exact answers.

Edit 5: I see now FGR’s spreadsheet approach. My method is to look at the INVERSE functions ArcTan(x) and ArcTan(x) + π, and subtract the circle (of opposite signs) so that I end up looking at two approximate parabolas not quite touching 0. Then I successively minimize the gap in both of them. But what I’ve been wrestling with since is whether or not there can be a generral and "quick" numeric method of determining if a function has a negative value anywhere, which allows a more rapid iteration for increasing precision. There should be a way.

Edit 6: I think I’ll be posting a Y!A question on a general problem of this nature. Stay tuned.

June 25th, 2010 at 10:01 pm

Still open!??

I started working on this two days ago. I was unable to get an exact solution, but I confirmed that Scythian’s was close.

My methodology was to create two lines. Each line was 1.5 units away and perpendicular to their respective tan line. Both lines fit the general pattern:

y’ = f(x) = tan(x) +/- 1.5*sin(atan(cos ²x)

x’ = g(x) = x +/- 1.5* cos(atan(cos ²x)

I could not get a solution for x in terms of x’, (i.e, x = h(x’) — and I don’t think one exists) which would have allowed me to solve for y’ in terms of x’.

But I plotted it out y’ = f(x’) using an excel spreadsheet looking for where the two lines intersected. (This is center of the ball when the ball gets stuck)

I think it is easier my way. Here is a redo of my spreadsheet:

http://i278.photobucket.com/albums/kk114/Remo_Aviron/Book2.jpg

(Excel has limited sig figs, and I didn’t run it out to max precision, just to establish methodology. But I could get eight numbers past the decimal point in accuracy fairly easily)

************

What I did was plot out the center of the circle presuming that it was rolling along either the left hand tangent line or the right hand tangent line.

The equation for the left hand tangent line is

x’ = x – 1.5* cos(atan(cos ²x)

y’ = tan(x) + 1.5*sin(atan(cos ²x)

where x and tan (x) is the position point on the left hand tangent line.

The equation for rolling on the right hand tangent line is:

x’ = x + 1.5* cos(atan(cos ²x)

y’ = tan(x) – 1.5*sin(atan(cos ²x)

Note: The general position of x for the right hand side is +π compared with the left hand side.

Where the two lines cross, the ball will get stuck. I’ve done a free-hand drawing:http://i278.photobucket.com/albums/kk114/Remo_Aviron/20100402142526.jpg

As for the accuracy of the spreadsheet, you keep pluging in new more accurate "x" until you have practically indentical for the two lines. In theory you can whittle it down quickly.