Challenge Questions

Wow, this one must be difficult ;o(

Its currently Dec 16th here in Australia, and there are no replies, wish I had the time to try and work this one out, but the boss has decided to break up work on christmas eve ;o(

Its a quote from a cartoon (Snaketales.... SOLS.... www.snakecartoons.com)

Hmmm the end date is modified...

Look for the answer right here on December 23rd...

I guess we need to start with the basics. For the maximum case of R2, R2 = sqrt(2) * R1 will be the correct relation to match A2 being half the area of A1. I believe that selecting A2 at half of A1 or less is not arbitary because you can not have the A & B points go past N & S.

Agree. From there, R1 + R2 (or from W to the outer circumference of the inner circle) from W is used to establish the intersection at A and B. This will establish R3. No proof as of yet, but working on it.

This challenge question has part of the Grazing Goats and part of the Trisection challenges. The Goats challenge I understood. The Trisection I did not. I don't have a complete solution but I have the following insights.

The area enclosed by arc AB can be broken into two parts. These two parts are divided by a straight line of length 2h. If h along with R1 are treated as a known values, in the figure below:

a = sqrt(R1^2 – h^2),

b = R1 – a,

R3 = sqrt(h^2 + b^2),

θ1 = asin(h/R1),

θ3 = atan(h/b),

and the shaded area A2 = θ1*R1^2 + θ3*R3^2 –h*R1.

For the shaded area to be equal to R2, R2 = sqrt(A2/∏)

Using these equations, I get the following relationships between h, R1, R2, and R3. The relationships between h/R1 and R3/R1 versus R2/R1 are fairly linear and something here might provide insight into doing the compass and straight edge part of the challenge.

h/R1 and R3/R1 versus R2/R1 are almost straight lines. The average of the two is even closer to a straight line. Sorry about the poor graphic.

A linear fit through the origin has:

(h + R3)/(2*R2) = 1.486 or h + R3 = 2.9712*R2. If we use the approximation of h + R3 = 3* R2, we have a method to do a graphic iteration as follows:

Mark off on a straight line a distance equal to 3*R2. Divide this into two equal lengths of 1.5*R2. This is our initial quess of R3. Use the compass to mark off an arc with radius 1.5*R2 centered at E. Make a prependicular line to the EW line. This is the current guess for h. Make a line perpendicular to the line of length 3*R2. Somewhere on this line, make a prependicular line. Mark off the current guess of R3 and then add the current quess of h. These two will add up to something less than 3*R2. Connect the ends of these two line and extend these connecting lines until they intersect. Add a line from this intersection to through the point of the current quess of R3 to the first line. This provides a new guess for R3. Strike another arc at E, get a new guess of h and repeat until you can not see any need to continue.

Tonight when I've got time, I'll try to demonstrate that this is within 0.1 for all values of 0 < R2 < sqrt(2)/2*R1

Some of this might be on the right path but I'm reached a dead end on this part. Accuracy is not nearly good enough.

Darn

I do like the challenge though. Having to dust off part of my brain that hasn't been used in a while.

This one had me stumped. For some reason though, the idea that R3 = 2*R2 seems to fit, but I can not give any supporting argument.

I couldn't even understand the question. Does it mean equal as it says, or only approximately equal as implied later? If the latter, why should we consider that a test between 1/2 and 1/4 would demonstrate validity for the whole of region 0 ... 1/2? What is the relevance of the N/S and E/W quadrants?

(But I'm not certain I'd know where to start even if I did understand the question...)

I agree, it's difficult to make any sense out of this. Especially when its says "an inner circle of radius R2 which is half ... the area of the lager circle". How can a radius be half of an area? One is 1-dimensional, the other is 2-dimensional.

And I'm not quite sure what a 'lager circle' circle is but regardless, I prefers ales to lagers.

That was the one aspect where I had thought that the intention was clear.

I read it as: "an inner circle (of radius R2) which (circle)is less half .... the area of the larger circle".

But I'll admit even that could be wrong.

Yeah, actually, I thought that's what the presenter was trying to say.

I do a lot of technical writing, and I've found that it doesn't really take that much time to proofread and find errors. If someone isn't good at it, he (she) should set the work aside for a day, then proofread it. With 'fresh' eyes the errors will be noticed. This Challenge seems to have more than the usual number of careless errors and ambiguous statements.

Given a circle of radius R1, who's area is divided into four quadrant, by lines running from North to South and East to West, has an inner circle of radius R2, the area of which can be exactly half, or less than half in area of the larger outer circle which has a radius of R1.

With the aid of a compass and straight edge only, (no measuring instrument), show how to draw an arc AB, at radius R3 from centre point E, and, the arc AB, enclosing an area that is approximately equal to the area of the smaller circle of radius of R2, within the following limits. Show that if R1 equals 10, and three sample areas of a half, one third and a quarter, that mathematically R3 is correct to within an error of plus or minus 0.01 units, and holds true for any random size of R2.

A measuring instrument may be used to establish R1 and R2, but can only be used to confirm R3. The position of the above circles is for illustration purposes only and may be repositioned as necessary, along with any additional circles and lines.

Yes I am the author of the above question. Above I have tried to address the points raised, I don't have any proof readers, you represent them. I though the intent was clear, I prefer lager to ale.

Regards JD.

Thank you. One final check: I assume that the sentence "A measuring instrument may be used to establish R1 and R2, but can only be used to confirm R3" is intended to be used with "successive approximation" methods, so that you know when you have got close enough to the theoretical value of R3?

One additional question: how would regard a method that used a "modestly decent" initial approximation together with a truncated polynomial (to enhance the local accuracy)?

Hi Fyz.

The use of a measuring device to establish R1 and R2, is to legitimise the use of Autocad as a means of solving the problem, and as you suggest, the checking of your method. The solution is a one pass solution.

As a hint, your astute remark regarding N,S,E and W, was a hand down from my first attempt to find R3, using only a compass, which as we know is 11.59 if R1 = 10. This can be done with just two adjustments of the compass, the third adjustment being R3. This was a solution for half the area only, my aim then was to extend this to solve any size, and thereby simplify the maths.

Where to start? well, if you wanted to draw a circle of half the area or one third of the area or a quater of the area of the larger cilcle, how would you go about it, without the use of a measuring instrument?

Regards JD.

Clearly, the radius of a circle with area m/n (m,n integers) times that of the original circle R1 is R2=R1.√(m/n) and this can readily be constructed using the exterior secant theorem (the length of the tangent). But I'm clearly being stupid, because that isn't helping me with R3.

PS #19/22 is the method I tried previously, but it didn't appear to work; then I found a typo in the EXCEL spreadsheet.
Note that the error of these arrangements is always zero at Area(C2)=Area(C1). If we abandon this as a requirement, the fit over the area range Area(C2) = (0...0.5)*Area(C1), could be further improved by adding a constant positional offset.

I tried the tangent method as described in post #19, but with the addition of a constant offset. I was amazed how good it could be.

The two interesting results are with Y placed West of 'W' as follows:
YW = 0.25*R1 +1.25*(R1-R2) gave:
. ChallengeErr(R3) < 0.02 unit and PkError(Area) < 0.5 sq.unit

YW = 0.3*R1 +(1+3/5/5)*(R1-R2) gave:
. ChallengeErr(R3) <0.01 unit and PkError(Area) < 0.22 sq unit

Even though the method of #22 was less fully optimised, for a given constructional complexity the fixed offset method generated slightly larger worst-case errors for the enclosed area - but both constructions perform well within the requirements of the challenge.

Hello Kinsale,

This Challenge seems to have more than the usual number of careless errors and ambiguous statements.

I agree totally with this statement! Totally misleading.

Take care and have a wonderful holiday........................

Hello Physicist?:

I am still thinking about it. But the N S E W quadrants could be just to allow you to judge where the other lines are, and work out the degrees?

What kind of gets me is, as you indicate, both an 'exact' and guess is implied. Maybe that is a red herring?.......Anyway, if you get any closer to understanding it, and exactly why the arc has to be 'approximately' where indicated? Sounds a bit hit or miss? Is there an end product and reason for this placing of the arc?

I keep wondering if the date, (the day before Xmas, is relevant?)

Take care and a wonderful holiday to you and all on CR4.

Hi BB.

Sorry for the confusion, what I am trying to say is that the result for R3 is correct to within 0.01 of giving a perfect area bounded by AB. For example the result for R3 for a smaller circle half the area of the larger circle is 11.589. this is within 0.01 of giving a perfect answer. As R3 is not perfect the resulting area is an approximation, in this case an error in area of 0.0003.

Regards JD.

Hello jdretired:

I was not criticizing you for your explanation.

More to the point, the explanation that goes with the question says # approximate, and at other times exact? And why use those particular arc definers?

No offence taken, I'm open to any constructive criticism at any time.

The remarks exact and approximate came about by my efforts to clarify the question wording, which I though conveyed my thoughts alright, but not to some. The area bounded by arc AB is a perfectly known area represented by a circle of radius R2, but as R3 has a flouting error within stated limits, the answer is a close first approximation with regard to the bounded area. I see no problem with this in regards to solving the problem, as it has been allowed for.

Regards JD.

Hello jdretired:

This is the piece I said didn't make sense to me. Saying "which is half, or less than half, of the lager circle".

How can it work out on several different sizes?

I just want to try and get my confusion on the above statement. OK?

has an inner circle of radius R2 which is half, or less than half, the area of the lager circle,

Take care and have a wonderful holiday....................

No problem happy to oblige. This means that you can nominate a circle of any area as long as it is not greater that half the area of the larger circle. The circle in the sketch as state at the end of the question is a representation and may be any size and located anywhere within the larger circle. The circle as draw is not literally the nominated circle as required to solve the problem, the circle can be any size and anywhere.

The position of the above circles is for illustration purposes only and may be repositioned as necessary, along with any addition circles and lines.

Regards JD.

I found a rather simple construction that gives a peak error of 0.018 units for the length R3 according to this test. That is not sufficient for the challenge, but I think it is a reasonable first step.

Construction: Extend EW to the west of W by a distance of (R1-R2) to create point X, draw the tangent from X to the circle C2*. This crosses the circle C1 twice, the Easterly crossing being the point A that defines the radius R3. We could make the error slightly smaller (0.15) by changing the extension to 1.95, but this is overly complex and still not adequate.

I will now look at the next level of correction (adding some fraction of H to the distance from W to X)

*Circle Cn is the circle with radious Rn

Hello Physicist?:,

This question has grabbed me! The explanation kind of put me off. I fell asleep for about three hours and I am back to see my mail. I do not know where they are as they 'have been tidied away'! But I am going to find my compass' and draw it myself. Until now I have nit been able to get my head round it but with your and jdretired's help I am slowly getting there. I do not have a rule so can't cheat!

Looks like you got pretty close?.

I am going to check my mail, get some sleep, and come back tonight.

Take care and have a wonderful holiday!......................

Do we need an old goat to solve this ?

Construction: Extend EW to the west of W by a distance of (R1-R2).√2 to create point Y, draw the tangent from Y to the circle C2*. This crosses the circle C1 twice, the Easterly crossing being the point H. The height of H above the line EW is h. Extend EY to the West by a distance h/5 from Y to create the point X.
Draw the tangent from X to the circle C2. This crosses the circle C1 twice, the Easterly crossing being the point A. R3 is the distance AE.

The maximum error of this construction according to the chosen measure is 0.041 units. This is by no means the limit for this specific construction, but is chosen to maintain reasonable simplicity of construction (see notes).

*Cn being the circle of radius Rn

Notes:

1) The √2 multiplier for (R1-R2) was chosen initially on the basis that the error in the area is second order for small values of the area C2. Although it is obviously not optimal, it was retained on the basis that the length is simple to generate (= the diagonal of an isosceles right triangle).
2) The value of h/5 was also chosen for simplicity. 3.h/14 reduces the measured error of of R3 to less than 0.015 units. Perhaps more significantly the error in the worst-case area [over the range of Area(C2)/Area(C1) = 0...0.5] is reduced from 0.9 areal units to 0.15 areal units (or from about 0.3% to about 0.05%).
3) Further improvement is clearly possible using basically the same construction. As previously stated, in the end there is no special merit in using the √2 multiplier for the construction that is used to generate h. In addition, we can replace the original position of Y with Y' based on a different multiplier of (R1-R2) as the starting point for finding X.
. So, if this was a practical problem that required greater precision it would be worth modifying these values to find an optimum fit. Given the capability to vary the shape of the correction function (the height h), we could expect to see at least a further order of magnitude improvement in the precision.

First, this deserves a Good Answer. I also think the Challenge deserves a Good Question.

I have done a check of the solution and agree. I do believe there is a typo in the calculation of L2. Where it says "The distance G'F' is the reference length L2 (=[R1-R2].√2)." I think the distance is really G'E'.

I think I could have worked on this for months and not done this well. Good work Fyz.

As you say, G'F' was a mistype - it should have read G'E'.

And there was another: "at R2 = 0.72" units should have read "at R2=7.2 units".

For me it was a good question, and I have rated it accordingly. But (other than yourself) there seems so far to have been a shortage of attempts at solution; was this the wrong forum, or are there too many people away on vacation?

P.S. (I though I wrote all this before - I must have forgotten to submit)
Quite a bit of luck was involved:
First, that one of my standard fitting methods was available: i.e. match the gradient at small area (R3=√2.R2), and also pass through the maximum possible value (R3=2*R2 at R2=R1).
Then, remarkably, there was a correction term (the height of H) available that didn't disturb the ends - and we could modify its shape.
There may be even more mileage available by moving away from an exact fit at small area, or we could generate explicit higher order correction terms using the exterior secant theorem - but don't expect that here...

Hi Fyz.

Would you agree that this is a correct representation of your construct?

This autocad drawing gives a result of 11.5906 for R3.

Regards JD.

I haven't read all this yet, but I'm giving you a GA for helpfulness (even though it's your Q !). Fyz too, just 'cos I'm sure the solution will be right. It's on my list of subjects to read when Christmas has driven me nuts.

That is reassuring - it both looks to be as I described it, and (within the presented resolution) the Autocad value is the same as my spreadsheet value.
The Autocad route is certainly easier to check than the spread-sheet... On the other hand, once the spreadsheet is correct (!) it is easily extended to cover the conditions of interest - and then adjusted to check different values of the parameters...

Did I really type "0.2175=1/5+1/5/16 "?
[No need to answer - I clearly did]

That would give 0.2125, which would not be close enough to meet the challenge. However 5/23 (~0.2174) is close enough. Moreover, it is a marginally simpler construction*, and gives a worst-case measure according the the challenge of 0.00358 units and a worst-case area error of 0.08-square units [again over the full range area(C2)=0...area(C1)].

*As here:

This is not the answer, but gives those still thinking about it some ideas? This only solves for a circle of radius R2 being half the area of the larger circle of radius R1, R3 = 11.589. The question is how to solve for any value of R2, when the smaller circle is half or less than half the area of the large circle? I think that Fyz has given a good alternative answer.

Regards JD.

Giving up about drawing solution,i tried solve first by equations to interpret geometrically later,i arrived to this equation that i still didn't check:

(half area case): B=SIN2B/(1-2(SINB)^2) ,being B the R3-E-W angle,no easy neither,and no way that helps for geometric construction.

Its not easy? But half area case, R1/cos(30.3432) will solve the length of R3 for any given value of R1.

Regards JD.

Sorry i mistake a figure in the the eq.however the angle what contents the arc is 2B,If i am not wrong (of course):A=(R1^2)(PI+2B(1-2(SINB)^2)-SIN2B);so then you can equal to some value like 1.57R1^2 and send to "solve" function from your calculator to work.You need several digits:A/R1^2=1.5 (should be 1.57..)for B=.31PI (=55,8 deg,not 1rad.)and is1,77 for B=.28pi. I am not going to do this mannually and this problem is interesting only if it may be solved by rule and compass.

Good approximations that fit over over wide ranges can be interesting in their own right; but this is the sort of problem where you might expect multiple solutions to exist (with differing complexity and precision).