Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

Do your triplets mean the lengths of the sides and the rectangle's hypotenuse? If so isn't 'play' an overstatement? The length of the "perpendicular hypotenuse" for a 3x4 rectangle is 5³/3/4=125/12 (~10.417) and for the other case is 85³/13/84/10 = 122825/2184 (~56.239)

I do mean the sides and the hypotenuse of the rectangle, but, while I suspected that such a formula existed, I had no idea what it was, or what to call it if I did want to look for it. I draw the diagrams and work with slopes, similar triangles and Mr. Pythagoras, they all work together and help me to understand thoroughly and to see all of the ramifications; the formula seems disembodied to me. I suspect TeeSquare will understand this preference.

My point though, was that in the original question, we needed to know the length of the perpendicular hypotenuse in order to select one that was longer, but we were not equipped with a measuring stick.

The formula is derived from Pharaoh Pythagoras, similar triangles, and a bit of simplification. Apart from the compactness, it has (for my money) the elegance of maintaining the diagram's symmetry between the CF and EF in the mathematics.

For CE perpendicular to AB, we have:
CE² =CF²+ EF²
AE = EF.(CE/CF)
EB = CF.(CE/EF)
AB = EF.(CE/CF) + CF.(CE/EF) = (EF².CE+CF².CE)/EF/CF = (EF² + CF²).CE/EF/CF
As EF² + CF² = CE², we have
AB = CE³/EF/CF

I hope you agree that this is not just embodied, but positively mummified

Very good. Thank you. It is easy of course to construct a line (an arc radius) of 11. The problem is locating the point on line CA where the arc would have to be centered in order to have line AB pass through point E. It is obvious that there is no exact solution without measurement. The question then turns to the best iteration process. I can see several, but they don't appear any better than some others proposed. The accuracy question is answerable, however.

In constructing an arc radius of 11, it is necessary to make three arc swings, each involving a line thickness. We generally accept that the thinnest pencil line we can make is around 0.005" (We could use an engraving scribe on metal, but I'll stick with paper for now). We can estimate that we will need perhaps a minimum of 6 swings at best and, even using something like √6 x 0.005" x 0.5 (to allow for "calibrated" eyeballs), we still end up with a resolution of 0.006" as the closest we can get. So, no, we would not get an error of +/- 0.0005". I actually think I would need 10 or 12 swings to do this, but I'm allowing for someone smarter than I to do it in much fewer.

I'll have to think a lot more about the best iteration. I don't know much about that sort of thing.

I suspect that this was a conceptual (limiting) accuracy.

Although I think you are being pessimistic about the eyes ability to estimate the middle of a line, I fear that our ability to position the compass point would have a similar effect. However, successive approximation only works if each stage reduces the error, so the errors in this case would not accumulative as you suggest.

In addition, I don't think we know the measure was inches. I suppose that the ideal unit (given the absolute size of our expected error) would be decimetres? Which would require a rather large sheet of paper...

Finally for now on this topic, the difference between systematic and random errors could be important. If the objective were to use drawing methods to find an angle or a length, we could in principle repeat the process as often as is required to reduce the random part of the error however far we wish - but using a fixed process the systematic constructional error would be the same on each occasion.

BTW, using my originally proposed method, I think I would be inclined to ascertain the "worst case" of rectangle dimensions and hypotenuse length as follows:
Set up a spread-sheet calculation that uses a fixed rectangle* shape and has a different outer hypotenuse length in each row. I would then work out a single iteration in each column, and use the sheet to discover the number of iterations needed for the worst-length outer hypotenuse. Then I would search (using multiple spreadsheets) in the region of the worst allowable shape. It should be possible to constrain the problem into at most 20 sheets - not too bad as the only one that will take significant time is the first; I might even get around to it later this week.

*The longer side would always be 10 units long

I am not so pessimistic as you might think about the eye's ability to estimate the middle of a line. I have worked as a draftsman, and have taught it, and have a good feel for what is possible. The 0.5 factor in my estimate was just that estimate.

There is a legitimate question as to whether successive approximations decrease or increase (stacking) line width errors. Without some sort of error correction scheme (which I am not capable of devising), I think the errors will stack, the only question being how to do that. I thought I was optimistic in taking the quare root, but...

Do we get to use spread sheets? If I can have one of those, I can easily do this to 0.001 (whatever). I was assuming I could only do such math as can be done graphically.

As to units, I made the assumption we would be doing this on a somewhat ordinary piece of paper. Thus, the dimensions would be something close to inches. If we use paper like what we call "D" (or your A1), things get much better.

I'll consider the random errors first:

50% of line width corresponds to the edge of the line being used instead of the centre - to my mind this would correspond to my visual resolution being much appreciably worse than the line width allows (true in my case, but that's what spectacles are for...). However, positioning the compass point in two dimensions is another matter, so I'm content to accept your +/- 0.002" (say 0.0012" RMS) for a single construction.

Consider successive approximation without any random errors. It can only tend towards zero error if the each preceding error is divided by some number greater than unity; that same division must also apply to random error. If we now consider the case where it takes 12 steps to reduce the error from 0.1 to 0.005 you can see that this corresponds approximately to dividing by 1.5 each time. The RMS addition then gives:
√(0.00012².(1+1/1.5²+1/1.5⁴+1/1.5⁴+...)) = 0.002" RMS. Of course, (depending on the error probability that you accept, this could still correspond to a peak error of about 0.006" - as you wrote. But my checks so far (see belowThis either requires that we use the average of previous constructions, a larger unit, or scribing on metal.

Returning to the purely theoretical requirements - i.e your question: are we allowed to use a spread-sheet to determine whether successive approximation will be adequate?
Once we use successive approximation, we require some means of knowing whether it has converged to the required limit. The challenge forbids measurement on the diagram, so we need to find a way of proving that our method will give answers that are accurate to the required level regardless of the dimensions of the rectangle (within the permitted range) or the chosen length of the diagonal. We can't do this using drawing, which means we have to use arithmetic or algebra. The spread-sheet does nothing that we couldn't do by hand - it merely speeds it up a bit, so I'd say it was a valid method of proving the technique. However, I'd have to emphasize that we have show that we have covered the worst case, as (in the absence of measurements/estimation) we have no way of selecting different procedures (or numbers of iterations) for different case.

This is one of the better challenges that we should have more than a week to look at.

As nobody has even attempted the trivial cases I have shown a sketch below of the two simplest constructs I can find, maybe they will trigger others more complex.

A1B1 is a hypotenuse constructed by simply setting the compass point at E and pencil at C and drawing circle C1. I believe this is the smallest hypotenuse achievable and is equal to twice the length of CE.

The second right-angled hypotenuse to CE is achieved by drawing circle C1, with centre at E and radius CE. Then extend the line CE cutting through C1 at X. Draw two identical circles C2 and C3, centered at C and x respectively, their radius arbitrary but greater than CE. A line drawn through their overlaps bisects the line CX at E and forms the hypotenuse A2B2.

Still thinking about more complex ones...

Agreed* that 2.CE is the shortest possible length for AB. But JD constrains the length as greater than CE³/FE/EK, which can be much longer. I'm not certain of the reason...

I don't believe that an exact method exists (Oh, to be shown to be wrong here).
The successive approximation method that has been proposed would of course work (eventually) for any valid length.

*And can prove it - the easiest method to follow uses calculus - strictly post-Euclidean, of course...

Quick note! I have written to amichelen to ask if he can delay the answer to this challenge.

The first method is .... frightening. It'll be a few weeks before I have time to address it seriously - but I will try to understand it, and also to look at the tolerancing of the second method in more detail.

In the answer I used the sizes 2.4, 3 and 4 this may cause some confusion, these are not universal sizes, but sizes taken from the sample rectangle that I used for the maths.

In STEP ONE the rectangle CFKE measures 3 units vertically, and 4 units horizontally and the chosen length A,B = 11, therefore A,D = 8.8, B,D = 6.6, D,W = 5.28 and B,W = 3.96, and the rectangle diagonal height h = 2.4.

Regards JD.

For a square, the problem reduces to second order, and can therefore be solved exactly (albeit tediously) by geometric construction.

For a general rectangle and hypotenuse the problem can be reduced to third order (sillke, for example, gives a substitution that does this). But even with rectangular boxes there will be specific values for the lengths where the solution reduces to second order; however, you have to know the exact values for the lengths before you can use anything like this (which I think contravenes the spirit of the exercise?).

I have a solution, but it is a little difficult to present.

I drew the diagram shown in post #97. I constructed the hypotenuse, perpendicular to the diagonal of the rectangle, mark A₁ at the side and B₁ at the base. I extended the hypotenuse below the baseline. I used the compass to measure 11" up the side and then to measure 11" down the hypotenuse from A₁. A short length of it hangs below the base, call that the "tail". Use the compass to carry the tail and plot it above mark A₁ to mark A₂. draw a line from A₂ through the corner of the rectangle, E. Measured the new tail and repeated this process from A₂ to A_n when the length of the tail became insignificant.

The length of the tail diminishes rapidly so that the lines overlap and obscure each other; it would require a scale several multiples above full size for accurate measurement. I used a spreadsheet to calculate where the lines would land, if this protocol were executed perfectly.

FA : Tail

5.33 : 0.583

5.917 : 0.237

6.154 : 0.083

6.236 : 0.027

6.272 : 0.003

6.275 : 0.001

The final position of A is 9.276" above the base. B is 5.912" to the left of C.

I don't see any point in doing more calculation, the angle is defined by the lengths of the sides, so it would be circular to then use the angle to find the length of the sides.

Regards JD.

This does work, but it requires two compass stages for each step, compared with one for the method of post#19. The convergence is also slower than in post#19 - a limiting factor of 1/3.12 per step compared with 1/5.15. I also noticed that you omitted a step during copying (easily done).
For the 4x3 rectangle, theoretical errors at each stage should be as follows:

Post#115 . Post#19
0.5833 . . . 0.5833
0.2368 . . . 0.1496
0.0826 . . . 0.0308
0.0272 . . . 0.0606
0.0088 . . . 0.0012
0.0028 . . . 0.0002
0.0009

If we interchange the sides (a 3x4 rectangle), the step factors are 1/8.21 and 1/15.37 respectively.

If the rectangle dimensions are constrained only so that the maximum side length is 10 units and the minimum is 1 unit, and the outer diagonal AB can be any integral combination of lengths derived from the rectangle, then for the worst-case theoretical situation we would need 46 successive approximation steps using the method of post #19. The 46 steps are needed whenever the required construction for the eventual length of the diagonal is
AB = 2xCE + N.(CF - CE), where N is 2 or more.
All other constructions of AB will require fewer steps (most of them much fewer); the required number may be calculated on a case-by-case basis.
Note that the fact that we are not allowed to measure any lengths means that we cannot use observations of the size or the shape of the rectangle to reduce the number of steps (length comparison using the compass and estimation by eye both counting as the use of measuring instruments).
We should also note that the error is only reduced by a factor of about 1.05 for each stage, which means that the cumulative error due to inaccuracies in the physical construction will be about 10x the single-stage error.

As expected, things are worst when the rectangle is nearly square and the desired length is close to the length of the hypotenuse. For the specified "error < 0.001 units" the worst situation seems to be for a rectangle with provided dimensions CF=10, EF=10-(0.0033)/N and where the specified construction for the external diagonal is
AB = 2.CE + N(CF - EF) .
The accuracy gets worse as N is increased, but the number of required steps reaches its maximum of 46 when N=2. For N=1 the required number of steps is 45.
Interestingly (?) the error multiplier takes a long time to improve close to its final value. Even at the 45th step has only reached 1/1.045, compared with its eventual value of 1.054