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

Posted August 23, 2009 5:01 PM

This week's Challenge Question:

Given a rectangle CFKE of any given sizes of 1 to 10, and located in a right angle triangle ABC, such that the hypotenuse AB touches the corner of the rectangle at E. Show how to construct a triangle ABC using only a compass and straight edge, no measuring instrument to be used. The hypotenuse length being any combination of sizes taken from the rectangle, the chosen length for the hypotenuse AB being greater than the length that would be a right angle with the rectangle diagonal CE, and referring to the construct, show how to calculate angle CAB, such that using the formula (FE/Sine A ) + (EK/ Cos A) = AB, show that AB is correct to within three decimal places.

Special thanks to jdretired who supplied this week's challenge.

And the Answer is....

We will present two solutions to this problem. There may be more solutions given the fact that this is a geometric construction problem.

SOLUTION 1:

Step 1. The solution concept outlines a number of rules. If a circle is drawn around the triangle ABC, from point O at the centre of AB, such that it touches all points, and a line CE extended to intersect the circle at D, and D used to form a triangle ABD, the triangle ABD is a simular triangle to CEF, and if a rectangle EGDJ is drawn within the triangle ABD, then the triangle EDG is a simular triangle to ABC. Also if line CD is taken as the base line, then the heights of points F and G are the same, both equal (h = 2.4). And lastly if lines AB, FK and GJ are extended, they all intersect at L, the position of L varying with length AB.

Note: The following solution holds true for any rectangle who's side measure any where between 1 and 10. And angle CEA is equal to or greater than 90 deg.


With the above concept rules in mind we construct the triangle ABD on a base line equal to the chosen length of A,B, in STEP ONE the chosen length A,B = 11, A,D = 8.8, B,D = 6.6, D,W = 5.28 and B,W = 3.96. We then solve by a number of feed back loops, to locate point C. this is done by establishing a number of POINTS, the location of a POINT is referenced by X, the horizontal distance from B, and Y, the vertical distance from B.

Step 2. Draw line B,N, being a tangent of the small circle of radius 2.4, then draw a second line parallel to B,N off set by 2.4 to intersect line A,B at P1. With compass set to Radius 3, from centre P1 intersect line B,N at Q1. With compass set to radius 4, from centre Q1 intersect the circle at point T1, draw a line from B to T1, and a line from P1 to Q1 which is extend it to intersect line B,T1 at POINT2.


MATHS

POINT1 X1 = 0. Y1 = 0

Angle D,B,N = Sine 2.4/6.6 = 21.323686 deg.

Angle N,B,P1 = 53.1301 - 21.32369 = 31.806416 deg. B,P1 = 2.4/Sine 31.80642 = 4.553639.

By Cosine rule ( 3, 4.553639, Cos 21.323686 ) .B,Q1 = 2.069836.

By Cosine rule ( 5.5, 2.069836, Cos 31.806416 ) .Q1,O = 3.896799.

(Q1,M)^2 = 4^2 - (5.5-x)^2 and (Q1,M)^2 = 3.896798^2 - x^2 therefore x = 2.67591.

Using x, angle Q1,O,M = 46.6309 deg.

By Cosine rule (5.5,3.896798, Cos O = 2.0669836) Cos O = 16.25724 deg.

Therefore angle O,B,T1 = (180 - (16.25724+46.6309) )/2 = 58.55593 deg.

By Sine rule (Point2,B,P1) B,POINT2 = 1.68204.

X2 = 1.68204 * Cos 58.55594 = 0.87746, and Y2 = 1.68204 * Sine 58.55594 = 1.43503

POINT2 X2 = 0.87746, Y2 = 1.43503

Step 3. Draw a line B1,N1, being a tangent at N of the small circle of radius 2.4 and passing through POINT2, draw a second line parallel to B1,N1 off set by 2.4 to intersect line AB at P2.


MATHS

POINT2,G = 3.96 - X2 = 3.08236. D,G = 5.28 - Y2 = 3.84497.

Distance between POINT2 and D = square root of ( 3.08236^2 + 3.84497^2) = 4.927955.

Angle D,POINT2,G = Cos 3.08236/4.927955 = 51.282232 deg

Angle D,POINT2,N = Sine 2.4/4.92795 = 29.144736 deg

Angle G,POINT2,N and angle POINT2,L,P2 = 51.282188 - 29.144766 = 22.137496 deg

Angle B,POINT2,L = 180-((180-58.55592)+22.137496) = 36.418424 deg

Distance between L,P2 = 2.4/Sine22.137496 = 6.36891.

By Sine rule (1.68204/Sine 22.137496 = L,B,/Sine 36.418424) L,B = 2.64997.

Therefore B,P2 = 6.36891 - 2.64997 = 3.718941.


Step 4. With compass set to Radius 3, from centre P2 intersect line B1,N1 at Q2. With compass set to radius 4, from centre Q2 intersect the circle at point T2, draw line B,T2, and line P2 to Q2 which is extend it to intersect line B,T2 at POINT3.

MATHS

Angle B,P2,Q2 = 90 - ( 22.137496 + 36.8699 ) = 30.992606 deg.

O,Q2 By Cosine rule(3,(5.5 - 3.718941),Cos(180-30.992606)) = 4.618752.

(Q2,M)^2 = 4^2 - (5.5-x)^2 and (Q2,M)^2 = 4.618752^2 - x^2 therefore x = 3.234807.

Using x, Q2,O,M = 45.543799 deg.

By sine rule [ Sine B,O,Q2/3 = Sine((180-30.992606)/4.618752) B,O,Q2 = 19.539633 deg.

Therefore angle B,O,T2 = 19.539633+45.543799 = 65.083432 deg.

Angle T2,B,P2 = (180 - 65.083432)/2 = 57.458284 deg.

Angle B,POINT3,P2 = 180 - ( 57.458284 + 30.992606) = 88.450891.

By Sine rule ( B,POINT3/Sine30.992606 = 3.718941/Sine 88.450891) B,POINT3 = 1.915685.

X2 = 1.915207 * Cos57.458284 = 1.030473, and Y2 = 1.915685 * Sine57.458284 = 1.61492.

POINT3 X3 = 1.030473, Y3 = 1.61492.


Step 5. With compass set to Radius 3, from centre POINT3 intersect line A,B at P3, with compass set to radius 5 (rectangle diagonal, CE ), from centre P3 intersect the circle at C. The lengths A,C and B,C will form a right angle triangle how's hypotenuse will touch a 3 by 4 rectangle at E.


MATHS

O,P3 =5.5-[ Square root(3^2 - Y3^2) + X3] = 1.94128.

By Cosine rule (5,5.5,Cos C ,(O,P3)) Cos C = 20.605454 deg

By Sine rule (SineO/5 = SineC/1.94128) SineO = 65.017921 deg

Therefore angle C,A,B = 65.017921/2 = 32.508961 deg.

CAB = 32.508961 deg, therefore by formula AB = 11.000.



SOLUTION 2:

With compass set at the selected hypotenuse length scribe an ark from point P1 to Q1, to dissect a vertical line extended from the rectangle, draw a line from Q1 to the corner of the rectangle at E, and extend it to dissect a point P2 on a line extended horizontally from the rectangle. With compass still at the selected hypotenuse length scribe an ark from point P2 to dissect the vertical line at Q2, draw a line from Q2 to the corner of the rectangle at E, and extend it to dissect the horizontal line at P3, repeat this a number of times that is practical, then transfer the final lengths C,Q and C,P and draw up triangle ABC.

MATHS

Base = Rectangle width

Start of Loop

Triangle height = square root ( Hypotenuse^2 - Base^2) .

Short height = Triangle height - rectangle height.

Top length = square root ( Short heigth^2 + rectangle width^2).

Bottom length = (rectangular height/short length) * Top length.

Length PQ = Top length + Bottom length.

Base = square root( Length PQ^2 - Triangle height^2).

With new Base size return to Start of loop four times,

Sine of angle CAB = Base/length PQ.

Hypotenuses check = (rectangle width/ Sine A) + ( rectangle height/ Cos A).


Examples

Width 4, height 3, length 11, CAB = 32.50575 deg. AB = 11.000.

Width 4.6, height 3.9, length 17, CAB = 21.024 deg. AB = 17.000

Width 8, height 3, length 33, CAB = 15.5263 deg. AB = 33.000

Width 5.8, height 9.2, length 30, CAB = 16.5144 deg. AB = 30.000

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#1

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

08/24/2009 8:04 PM

I don't have an answer but I do want to restate the challenge question a little to see if I really understand it. By "rectangle of any size of 1 to 10" I am assuming that both both distances FE and CF must be greater than or equal to 1 and no greater than 10 units. Non-integer lengths are permitted. First we pick two parameters n and m where the length of the desired distance AB is n*FE + m*CF. We are then going to find a distance AC (or CB) based on our solution technique and locate point A (or B) followed by drawing a line from point A (or B) through E to a line through CK (or CF). We then compare the actual distance AB as plotted to the value n*FE + m*CF. The difference should be no greater than 0.001.

Is this everyone else's understanding?

Thanks,

Jim

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#2
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/24/2009 8:55 PM

Okay. I looked a little harder at the figure. We can also us the diagonal distance CE. The length of AB is therefore something like n*FE + m*CF + o*CE.

Thanks,

Jim

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#3
In reply to #2

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

08/25/2009 12:39 AM

Guys don't over analyse the construction of the orange. Just peel it and eat it.

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#4
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/25/2009 5:19 AM

Oh! You're supposed to take the peel off those suckers

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#5

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

08/25/2009 7:21 AM

Compass? Straight Egde? I think i have seen those things in a museum just joking

This is really off topic, but i have a cartoon that has a little kid laying on the ground coloring and his older brother looking at him a jealous like and saying "Wow....real crayons on paper! I can only draw with a computer program."

Unfortuantly i think a lot of skill was lost when we stopped teaching how to draw something on paper........oh well such is the way of technology

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#6
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/25/2009 9:18 AM

Compass? Straight Egde? I think i have seen those things in a museum

How true. We put a man on the moon with a compass, square, and slide rule. I have every expectation there are few people under the age of 60 who are able to operate these ancient tools of calculation.

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#7
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/25/2009 9:46 AM

i will admit that i learned how to use the compass and the straight edge.....even a t-square....i learned to design on the board before i ever learned to design on the computer.....my first design job was on a board.....and i am only 43 but the slide rule, while i own one, i never learned to use it....it was a gift from my uncle, calculators are so much easier......

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

08/25/2009 10:37 PM

I am over 60, and I still keep my slide rule handy. There's nothing to beat it for certain tasks like finding the nearest gear/sprocket tooth number combinations for a given ratio. I was 'forced' by colleagues to graduate to a calculator about three decades ago! Most engineering calculations require multiplication, and powers or trig functions which slide rules provide with 'adequate' precision. Calculators give you umpteen useless distracting digits, which should be rounded off meaningfully at each stage (a lot of folk just don't bother!). Errors in entering or reading digits on a calculator are not so easily spotted, whereas with a slide rule the error is more likely to be a factor of ten -- which should be apparent to someone who knows what (s)he is doing.

Engineering situations do not generally require more than two or three digits of precision because of the uncertainties in the data itself. Design procedures invariably involve service factors or safety factors like 1.5 or 2 which make those lengthy calculator displays largely redundant! I think all the great engineering feats were performed by those who used just slide rules and log tables. After calculators and computers and 'software' (cad, FEM, CFD ...) came along, people have been mostly tinkering around the edges or achieving more complexity/ speed/ productivity, but little by way of conceptualisation. Nowadays hardly anyone has time to uderstand a problem from basics. You can get away with mouse clicks on a bewildering array of nested icons.

I did get used to the regular scientific calculators with 'non-algebraic' logic. The two-line display versions are really crazy in my opinion. I can't multiply a column of figures by a constant, say for converting inches to mm. Earlier it was just "25.4 xx" and then "2.25=", "5.5=" and so on, reading off the results at each step. The same for any other kind of constant calculation, including those with bracketed expressions. Now the experts at the calculator company advise me to write a program to do this! The earlier version allowed me to keep track of (and note down) the value of each term in a long expression by closing a bracket. Thus you get a feel for the relative magnitudes, and can spot errors more easily. Now you can only get the final result in one shot!

I guess technological development is all about doing simple things in increasingly complex ways! And becoming dependent on some commercial organisation for everything which could once be achieved in a self-sufficient and eco-friendly manner.

OK, I guess I'm old-fashioned, possibly senile, for ranting like this (in other CR4 challenge threads as well). It has become nearly impossible for me to communicate with today's 'engineers' -- they speak an altogether different language. Oh well, I guess I've had my day, and hugely enjoyed my professional life, though no one can accuse me of being a 'success' in material terms!

=TeeSquare=

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#9
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/25/2009 11:21 PM

I really liked TeeSquare's comments. I voted it as not off topic and unchecked this reply as off topic so some more might read it. I encourage you all to read this comments and then vote it off topic and maybe then we could vote it a GA.

I do think there are many things we can do with computers and calculators that can not be done, at least effeciently, on a slide rule. However, there is much common sense that can be gained from having to use a slide rule. Common sense does seem to be coming rather uncommon.

Thanks,

Jim

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

08/26/2009 8:47 PM

Thank you. =TeeSquare=

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#11
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/27/2009 2:41 AM

I liked this sufficiently to follow jim in trying to vote you back on topic.

That said, I deal with three of your issues by using a spreadsheet (either Calc or EXCEL). That leaves the original entry open to inspection and makes repeat calculations extremely easy; it allows you to display the answers to the appropriate number of significant digits.

BTW, in the days of slide rule and tables I needed two rules: a straight rule for quick calculations where confidence to 3SF was good enough, and a large cylindrical rule for up to 5SF.

I'm possibly being dim here, but I don't follow your comment on safety factors meaning that calculations don't need to be precise. The 'most obvious' counter-example being the length of a strut used as a brace; however, in a competitive situation on a large project even a fraction of a percent over what is required to meet safety fators can significantly affect profits.

Maybe I'll find time to think about the challenge question...

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

08/27/2009 2:52 AM

Maybe I'll find time to think about the challenge question...

I've been thinking about it, but it hasn't got me anywhere (yet) !

<Teesquares comments are very worth a read, and I've voted him up too.>

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

08/28/2009 5:15 AM

Thanks for your thoughtful comments.

I must admit to being stubbornly old-fashioned! Ready access to a computer is not automatic for me. I started using one primarily as a glorified typewriter, and hardly got beyond using Word97 with Win98 on an assembled contraption. Now I am struggling with Linux/Firefox/OpenOffice on a laptop. Keeping up with current software, hardware, inter-compatibility and anti-virus requirements is not an affordable option for me. I just muddle through somehow with a minimum of stuff on board, because asking for help might solve an immediate problem, but with strange effects elsewhere. No one can explain how or why things happen in a computer -- seems like mostly trial and error to me. I avoid adding new software for fear of unexpected consequences.

I've seldom used Excel or Calc, though I know the benefits of quickly comparing a lot of alternative options. I just don't have the patience to reformulate my problem (equations, headings, intermediate results ...) in a 'computer-friendly' manner. If I need to do it regularly I guess I'll learn or be forced to. I've been doing manual 'spreadsheets' all along -- perhaps before Lotus was created -- and it gives me an insight into how the variables behave when I write down each step. Not so obvious when Excel or Calc displays everything instantly. Obviously I'm not efficient or productive, but those who are can easily miss important details. Making things easy has its downside. I just prefer to be wary of computers, and actually dislike staring at a screen (I don't possess a TV set).

I don't know what a cylindrical (slide?) rule is -- will search google when I find time. I suppose 3SF and 5SF are statistical terms for levels of accuracy/confidence (a subject I've never grasped).

I'll try to explain my comment on safety factors. Taking your example of a strut, the formula for column buckling presumes a perfect geometric shape and exactly known mechanical properties for the material. In practice there will be dimensional variations, initial bend or twist, eccentricity of loading, uncertain end conditions, etc. Different authors and codes of practice do not have a uniform view on how these should be accounted for. It is also common for items outside the tolerance limits to be used without getting detected. In a 'large' project with thousands of identical struts under identical load conditions, it may be prudent or cost-effective to test two or three pieces to failure before finalising the design. But my experience has really been limited to one-off machinery design, where economics and competitiveness were not the primary criteria. Sometimes wild imagination was necessary with just a hope that it would work reliably. There are generally no benchmarks for comparison.

A formal calculation may have to be submitted to meet a legal requirement. That does not necessarily make a design safe! Or unsafe. For instance a gear pair which is known to operate quite satisfactorily may very well 'fail' a calculation test according to AGMA or DIN standards or both. There are major uncertainties with regard to material properties and operating conditions. It is impossible for a design code to provide fine-tuned factors to yield precisely safe results under all conceivable situations. Some tables of factors involve choices or judgements which can make a lot of difference -- like deciding between severe and very severe duty. A designer can blindly follow a code and wash his hands off further responsibility. If he has the option (as I frequently had) he can apply his mind to the problem in a constructive way (which can possibly lead to delays and disputes). Pardon me if I've overstated my point of view. To put it briefly, I believe that insight or intuition is necessary when using a text book or code of practice or computer programme to solve an engineering problem. And it still remains an uncertain solution. I guess very few will agree.

The larger issue is the extent to which technology (more so medical technology) is being hijacked or controlled by the legal profession and insurance industry. Fortunately, in my country we are still disorganised enough for competent persons (and, sadly, charlatans as well) to to take independent decisions without much fear of legal consequences. My doctor does little more than use a stethoscope, look at my tongue/eyes, poke my ribs/stomach, and check my blood test values before declaring me fit every few years. If I went to one of these new 'corporate' hospitals for a 'general checkup', I'll be subjected to a battery of costly procedures, and the doctor who diagnoses the reports will surely discover a lot of ailments whose treatment will put me out on the street. But globalisation is making rapid progress here too, and I'm hoping to kick the bucket sooner rather than later!

Hope I haven't exhausted your patience =TeeSquare=

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

09/02/2009 1:47 PM

This post is mixed on and off topic. I'm still new here so I'm not sure how to handle that. I'll say it is off topic and see what shakes.

First; the off topic portion. I'm seventy-four so I agree with TeeSquare to a large extent; except that I embraced calculators and then computers. In using the old iterative methods for analyzing statically indeterminate systems, we pulled structures apart and found what made them tick, we asked them questions. I did worry that engineers raised on the computer would not have the benefit of that experience but enough of them do still play with the systems and gain a good understanding.

As to the point about safety factors that TeeSquare made, and on which Physicist? commented: If in a large production run, it is worth the effort to minimize an item's cost, if in a one-off job it is not. Structural members, pipes etc. are off the shelf, their strengths etc. go up in steps, so the accuracy needed is only such as to get in the range between two of the steps, and then select the next one up.

I have solutions to the problem although they, I think, walk around jdretired's intention in that, they are exact, and self-evidently so. With the diagram as shown, divide CK equally into 2n parts where n is the smallest number for the part to be shorter than CF. Draw a diagonal from E to the closest mark on CK. Set the length of AB equal to 1+2n times the length of that diagonal. Use the compass to mark out KB equal to one part of CK. Use the compass to mark A, 2n lengths CF above F. Connect A to B and the straight line will nick the corner of the rectangle.

In the case where CF < CK, the rectangle itself is the only cell required. Set AB=2CE, make KB=CK and FA=CF.

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#73
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 2:10 PM

You should read passingtongreen's comment (and vote to take it off "off-topic" and give it a GA). Assuming that "the chosen length for the hypotenuse" in the challenge means chosen by the solver and not someone trying to make the problem harder, this is the best solution so far. It is easy and exact.

Thanks,

Jim

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#80
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 12:34 AM

Yes, I totally agree. I have voted Passingtongreen's post #72 as not off topic and I will give a GA if I get to do so.

The original problem states "The hypotenuse length being any combination of sizes taken from the rectangle"

I initially, like most people, interpreted this to mean that the hypotenuse chosen by the solver must be a length AB = L*CE + M*CF + N*CK where L, M and N can be zero or positive integers. However, JDRetired did not explicitly state this.

Now, after reading Passingtongreen's post #72, I would say that by dividing down the longer side by factors of 2, he is legitimately taking a valid " combination of sizes from the rectangle".

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

09/03/2009 12:46 AM

In post #18 jdretired writes "The hypotenuse length was to be a practical length that can be determined with out using a measuring instrument, FE is the horizontal distance or width of the rectangle, CF is the vertical or height of the rectangle, CE is the diagonal of the rectangle, or any other dimension that can be determined from the rectangle is legitimate".

I bring attention to the words "or any other dimension that can be determined from the rectangle is legitimate".

I look forward to your reponse jd

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#82
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 1:52 AM

Hi MPM.

Yes I agree with your analysis, the spirit of the whole thing was to construct a triangle who's hypotenuse touches the corner of a rectangle, and the rectangle and hypotenuse sizes being of your own choice, the remark concerning the use of the rectangle as a reference was intended to be in keeping with not using a measuring device, how one arrived at the length AB was completely up to the individual, the end result being to have a hypotenuse who's length was known before starting the construct. The combinations shown where only there to get the ball rolling.

Regards JD.

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#90
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 5:16 PM

"The rectangle and hypotenuse sizes being of your own choice.."

That is a much easier problem than the one most of use were attempting. If we construct our own rectangle, passingtongreen's proposal together with jim's extension does the job perfectly (constructing the rectangle for ourselves allows us to ensure AB satisfies the length constraint without making any estimates or measurements).

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

09/03/2009 6:38 PM

I agree.

This is why I believe Passingtongreen's answer was the best answer given and deserves a GA.

You said in response to Jim how do we know the hypotenuse gets longer if we increase n - measurement?

Well, we already know how to construct the perpendicular to CE - surely if we increase n as per passingtongreen's method until the hypotenuse is visibly greater than this perpendicular, then we have an exact solution.

Or am I missing something else here?

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#96
In reply to #91

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

09/04/2009 5:25 PM

I think it's a matter of taste. I would assume that if greater-than and less-than measurements were forbidden, the equivalent estimates should also be forbidden. The ability to construct your rectangle and choose your length avoids this issue.

One reason I did not believe this to be the intention is that the calculation at the tail of the challenge bacomes irrelevant.

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#74
In reply to #72

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

09/02/2009 6:09 PM

You can indeed define lengths for which there is a neat and exact construct. Your have defined a set of such lengths. Unfortunately, it does not appear to be one of the lengths that was defined in the original challenge.

N.B. that as the challenge does not appear to define which of the possible diagonals AB should be, I think you could with justification swap CF and CK if the relative sizes were in the "wrong" order for your simpler proposal.

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#75
In reply to #74

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

09/02/2009 6:16 PM

That was me - I didn't realise that I was not logged in. I also forgot to welcome you to the Old Stager's club. Welcome

For what it is worth, my view is that you should not mark a post off-topic if there is something significant that is on-topic. If you feel that there is too much that is off-topic to leave it unmarked, it's probably best to separate the on-topic bit into a separate posting.

If only we were all as meticulous as that all the time! But the upside is that CR4 is very forgiving - especially if at least some of your contributions have value - as in your case.

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#76
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 7:25 PM

I don't like to rain on someone else's parade. So, if I've got a side issue like this one, and don't think it worth a new thread, I post off-topic. I think this is interesting, but I don't know how many others would.

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#83
In reply to #76

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

09/03/2009 3:58 AM

Agreed - it is well established (and accepted) to leave the separate posting in the thread that stimulated it

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#84
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 4:30 AM

Thank you PVP45 and Physicist for a very interesting exchange, the points you have raised I have taken on board, and hope should I post any future question? I get it right, but though I tried very hard this time, apparently I did not. It was once pointed out to me the purpose of these questions is to enter into the spirit of what the author intended. A positive or negative contribution is a valid comment.

Regards JD.

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#79
In reply to #74

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

09/02/2009 9:35 PM

I was reading the diagram quite literally, with the long direction of the rectangle, horizontal, and the angle above the diagonal of the rectangle to be greater than a right angle. If we, instead, say that the angle must not be ninety degrees, the simpler solution holds in either direction. Thank you for your welcome, in your following post.

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#98
In reply to #72

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

09/04/2009 8:46 PM

Gratifying to learn there are still folk around who can appreciate the old methods. In my case it was mostly constrained access to modern gadgets in the circumstances of my working life which dictated my dependence on simple tools and slow procedures. Maybe I just never grew out of that mindset.

My aversion to computers is because I see so much of misuse and dependence on them, which I suspect is commercially driven. Competitive pressures apparently make it impossible for today's organisations to provide technical staff (engineers, designers, draughtsmen ...) with the kind of workshop or field training , which was routine at one time.

Coping with developments in computers and internet is a constant nightmare for me -- especially now as I try to work with Linux and OpenOffice (instead of Win98+Word97). I just about manage a quick look at CR4 every other day or so and save the page(s) if possible for viewing later.

It was dumb of me not to figure out that physicist?'s 3SF and 5SF referred to significant figures (I invariably face difficulties with abbreviations and acronyms). I had no idea what a cylindrical slide rule was until I looked up wikipedia after seeing it mentioned here. (Haven't figured out the mechanical construction yet.) I can see that he needed the greater accuracy for scientific and mathematical calculations.

The point I had tried to emphasise about engineering calculations was that in most practical problems (mechanics, strength of materials, fluid flow, heat transfer, ...) the available data or assumptions may involve an uncertainty of anything from 1% to 10% if not more. As you mentioned, a lot of choices are already standardised in steps (bolt sizes, gear modules, pipe diameters, motor ratings ...) and one just chooses the next size. At the end of a calculation we should of course trace back and check that all the choices were mutually compatible. In this process any more digits than a slide rule can provide represents but imaginary precision. I am inclined to endorse the following quote from the wikipedia article on 'slide rule':

"The spatial, manual operation of slide rules cultivates in the user an intuition for numerical relationships and scale that people who've used only digital calculators often lack. Since you must explicitly note the order of magnitude at each step in order to interpret your results, you are less likely to make wildly wrong errors. You are forced to use common sense and an understanding of your subject as you calculate. Since order of magnitude gets the greatest prominence when using a slide rule, and precision is limited only to the few digits that are normally useful, users are less likely to make errors of false precision."

Concepts like safety factor or margin of ignorance are mostly used in a subjective way by engineers depending on their experience and intuition, familiarity with similar problems, risk perception, and so on. Text book examples and codes of practice may serve as guidelines (or be mandatory) in somewhat routine situations. In an open-ended problem it is unlikely that two persons will follow the same thought process. I think the problem with computer software is that even a bad solution can be presented in an attractive and convincing format with pretty coloured pictures. The bugs are hidden away somewhere. My grouse with some 3D modelling software I've encountered is the absence of dimensions -- I'm only comfortable with proper views and sections in which all the details are apparent. Guess I'm just ranting as usual.

It is quite embarrassing to see just my lone post#8 appearing as an almost GA, when there have been so many expert on-topic contributions which tax my understanding -- and I've not quite understood the question yet! I agree with MathsPhysicsManiac #38 about this being a stimulating challenge (strictly observer status for me, besides off-topic comments).

=TeeSquare=

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#99
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/05/2009 10:47 AM

I did know what the cylindrical slide rule was, but if we needed more accuracy we used "Smoley's", (Smoley's Four Combined Tables For Engineers, Architects and Students (Logarithms and Squares, Slopes and Rises, Logarithmic Trigonometric Tables, Segmental Functions)).

I did spend some time to see if there was some iterative way, along the lines of "Hardy Cross' Moment Distribution".

I think the lack of a fixed answer has made for more enjoyment.

Back to the problem.

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#13
In reply to #8

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

08/27/2009 5:19 PM

Agree completely with you... the ABSOLUTE value of a quantity is seldom an ABSOULTE necessity. So what if I calculate the volume of concrete required as 8,396.156 ft3, the mix plant sells it to me by the cubic yard!

There is a discussion going on Forum Thread Problems with Pi that bangs around this idea. There is, actually, some hair pulling going on.

Lots of off-topic posts on this challenge question... maybe I better have another look, and see if I can contribute something useful.

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#14

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

08/27/2009 7:00 PM

I haven't got anywhere with a solution - but I think I may have been "looking in the wrong place". I was looking for a "theoretically exact" construction, but on consideration there are clues that suggest we should perhaps be looking for an approximation. On the other hand, that could be a pure cop-out. I also found the description very confusing*, and may be completely misunderstanding the basic intent.

The possible clues are:
The minimum length defined for AB is not usually the shortest possible hypotenuse, so the construction is not required to handle the general case.
If EK<<FE (or EK>>FE), the minimum length set for the hypotenuse will be very much longer than any of the three lengths illustrated.
We are invited to prove that the final formula is a good approximation - i.e. not that it is exact. That suggests that the construction method is an approximation - if only on the basis that the formula should bear some relation to the construction method.

If it is an approximation that is desired, I have no doubt that we could come up with a method that would work to any specified level of accuracy and for any length that was greater than the specified CK3/FE/EK (so no need to otherwise relate it to the dimensions of the rectangle). Unfortunately, I shan't have time to look further before Sept 1st (and possibly not access to the internet, for that matter).

*I couldn't find a fitting interpretation of the phrase "of any given sizes of 1 to 10" in this context. The only constraining meaning I could ascribe was a rectangle was constructed from whole squares, of any number between 1 and 10 - but none of the exemplar lengths would then satisfy the length constraint for a rectangle built from 7 squares, so I have ignored this constraint for the time being. I wasn't certain whether the intention was to constrain us to just the three given examples - in the end I decided that, given that there was no apparently relevant relation between them, they were just examples. Then the first sentence of the description has the rectangle is already inscribed in triangle ABC, which would mean that all the work has been done before we start - so all we need to do is draw over it.

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#15
In reply to #14

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

08/27/2009 10:05 PM

Hi fyz.

To clear up some points any number between 1 and 10 includes fraction numbers, (3.4 by 5.6) or (9.2 by 5.8) or ( 7 by 4) etc. and the hypotenuse is greater than 90 deg with the diagonal as this would be an answer in its self. the reason for stipulating that the selected hypotenuse be multiple sizes taken from the rectangle was to omit the use of a measuring instrument, but in reality it can be any nominated size.

I have submitted two answers one that gives an answer of say 15.0001, or 20.0001 etc. as it would be very difficult to visually see such small tolerances with the naked eye, the first answer over comes that problem, the second answer is more precise, for thoughts that use spread sheets or write computer programmes, and that answer will give an answer like 15.00000000 or 20.00000000. so that should be a clue?

Good to see you involved regards JD.

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#16
In reply to #15

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

08/28/2009 1:36 AM

jd,

Are you confirming that the construction sought is an approximation ? Your reference to 15.0001 and 20.0001 suggests so, but is not explicit. Can the desired length(s) be placed as a hypotenuse ? - "yes" or "no".

Please also clarify the lettering in the diagrams for construction lengths used (hypotenuse = FE+FE+CF,,,,FE+FE+FE,,,CE+CE+FE ?) - it's a bit hard to read.

Thanks,

Kris

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#18
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/28/2009 3:16 AM

Hi Kris

15.0001 and 20.0001 are references to the hypotenuse as calculated when the angle CAB has been determined mathematically. If 15.0001 or 20.0001 is an approximation then yes, but 15 and 20 are original lengths we are trying to reproduce in our construct. The hypotenuse length was to be a practical length that that can be determined with out using a measuring instrument, FE is the horizontal distance or width of the rectangle, CF is the vertical or height of the rectangle, CE is the diagonal of the rectangle, or any other dimension that can be determined from the rectangle is legitimate. The problem is we start with a rectangle X wide by Y high and a hypotenuse Z in length that touches the rectangle at E, show how to place the hypotenuse such that when the check formula is applied that it is correct to N.0001.

Thank you Kris for your question if this is not clear please ask me to explain further.

REgards JD.

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#30
In reply to #18

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

08/28/2009 9:31 AM

Many thanks, JD. I shall have a play with this over the weekend. Even if I'm unable to come up with a decent solution proposal, I shall enjoy reading the thread as it develops.

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#17
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/28/2009 2:58 AM

To my eye, the phrase "Given a rectangle CFKE of any given sizes of 1 to 10" is still either meaningless or irrelevant. Is there something missing, or do you mean the relationship between the lengths of the sides (i.e. the value of FE/EK)?

I have a successive approximation method that depends on the nominal length AB being greater than the orthogonal to CE (or CE3/FE/EK - miswritten last time). It doesn't require any of your other constraints, however. I'll post it as a direct answer to the challenge.

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#20
In reply to #17

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

08/28/2009 3:23 AM

Yes you are correct, there is no relationship between the lengths, the reason I made this statement was to restrict the rectangle size, because if the the size was double digits the tolerance was reduced from 0.0001 to 0.001. as simple as that.

Regards JD.

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#21
In reply to #20

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

08/28/2009 3:27 AM

So is the size meant to be the perimeter, or FE+EK, or the longer of FE & EK, or? (I won't have time to do the calculations, unfortunately)

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#22
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/28/2009 3:46 AM

If you mean is the hypotenuse equal to the perimeter, No, do the sides of the rectangle have any relationship, No, the statement was meant to convey that one is free to determine the sizes, the end result was to show that the construct can be applied to any combination of sizes. A rectangle of any combination of lengths, a hypotenuse that is greater than 90 deg with the rectangle diagonal and touches the rectangle at E, so when the tolerance test is applied the constructed hypotenuse it is within certain limits. N.0001.

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#24
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/28/2009 4:48 AM

I meant exactly what I wrote. No more no less. Perhaps you think your description is standard???

The only legitimate "size" for a rectangle is its area. Any linear size you give needs further refinement. Amognst others, it can be the length of the longest size, or it can be the sum of the lengths of the two adjacent sides, or it can be the length of the perimeter, or it can be the length of the diagonal, or it can be the square-root of the area. You presumably know which one of these you meant. I still don't, even though your non-answer half-implies that you may have meant the diagonal.

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

08/28/2009 5:02 AM
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#37
In reply to #25

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

08/30/2009 10:30 PM

After giving it some thought, and though you have raised a good point, I stated sizes not size. an area is a size not sizes, a diagonal is a size not sizes and the sum of the sides is a size not sizes.

Regards JD.

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#41
In reply to #37

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

08/31/2009 4:37 PM

So you mean more than one of the sizes must obey the constraint. That still doesn't answer which. From what you now write I suppose I should infer that this was intended to include sides, area, and diagonal. Was it intended to include the perimeter as well?

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#45
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/31/2009 7:20 PM

I see you come back from your August bank holiday just a little bit sweeter.

Regards JD.

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#47
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 5:13 AM

Not a relevant comment (either way). I interpreted the avoidance of elucidation as obscurantism, and I did not anticipate it from your quarter. Grrr.

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#49
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 6:47 AM

Any and all dimension are included, as long as the vertical length and horizontal length and diagonal length can be determined and referenced, (required for the maths) and the rectangle can be dissected to find any other size you require. Hope you enjoyed your break.

Regards JD.

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#19

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

08/28/2009 3:19 AM

A simple successive approximation method:

The following will work whenever the intended length AB is longer than the orthogonal to CE (or CE3/FE/EK). None of the other constraints is necessary for the construction to work, although the rate of error reduction will vary. N.B. that if AB is too short, the method does not converge.

Construct the orthogonal line to CE passing through E to create* line A0B0. (The left hand of the diagrams in the challenge).

For r=0 to n-1 (n at choice):
Construct Ar+1Bn of intended length AB.
Draw Ar+1E, extended to Br+1

The final line "AB" is the AnBn. Accuracy depends only on your patience.

I shall be incommunicado for a few days, so unable to respond with the tolerance calculations (I can't do them yet because I still don't understand the constraints against which JD wishes them to be made)

*Note: all Ar will be on the line CF beyond F; and all Br will be on the line CK, beyond K

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#23
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/28/2009 4:29 AM

Yes, looks like the anticipated answer, and I agree that patience is required, but if I am right this will give an highly accurate answer when applied to a spread sheet. But my first answer does not require patience, and is easily understood.

Rgards JD.

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#46
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 5:07 AM

Other suggestions have (so far) used the successive approximation posted above, but have used different starting points. My reason for this choice of starting point was that I would intuitively expect convergence to be slowest when the orientation of the eventual line AB is closest to the the angle Θ at which d(AB)/dΘ is zero (Θ being the angle CAB. That makes this starting point the least-bad for the most critical lengths of AB (slightly greater than CE3/FE/FK).
It could of course be that this intuition is incorrect - I haven't had leisure to check in detail.

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#50
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 6:56 AM

Yes so I see, I included a progressive solution in my answers as it requires no selected rectangle or hypotenuse sizes to be stated. And the maths only take a few lines to show how to calculate angle CAB. I included it because I thought the maths where interesting, the other answer relies on a number of applied rules. Have fun.

Regards JD.

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#26

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

08/28/2009 5:08 AM

Oops! Much embarrassed by all the superfluous publicity over a patently off-topic post. Perhaps I asked for it. But I was just responding to a couple of remarks which struck a chord in me.

I should apologise to jdretired for further diverting attention away from the challenge question.

The funny thing is that my few GAs were awarded mainly for off-topic limericks and such, in the 'fixing the pump' thread. For me the certificate from jim35848 and comments by physicist?, Kris and Doorman are worth more than a crop of GAs. At least I'm satisfied that there are people on this(?) planet who do not treat ALL my outlandish notions as complete rubbish. There's just the niggling doubt about where CR4 is located. (Some asylum in cyberspace?)

Now this is clearly off-topic. =TeeSquare=

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#28
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/28/2009 5:25 AM

Apologises not necessary, I was one who voted that you where not off topic. I enjoy physicist involvement and Kris and doorman and jim35848 who first responded and yourself, I'm an outside the box thinker and and roll with positive and negative responses, just keep them coming.

Regards JD.

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#29

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

08/28/2009 7:13 AM

It seems to me that we're looking for a circle which comes close to following the pink hyperbola y=ab/x in this spread sheet:-

a and b are the sides of the original rectangle and L is the length of the required hypotenuse.

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#31
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/28/2009 12:35 PM

Nicely put!

By setting the equations for circle and hyperbola equal, you will get two real solutions at the curve intersections, one which will be for the example you have shown. Unfortunately the solution is a 4th order polynomial, which makes the values very difficult (impossible?) to solve and construct. I came up with a similar solution in a much more roundabout (and messy) manner, but was also limited by the 4th order solution. I have been looking for other ways to simplify the solution down to 2nd order, which can then be solved and constructed exactly. However, it may be necessary to find a way to approximate the 4th order solution - thus the 0.0001 estimation accuracy.

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#32

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

08/28/2009 8:13 PM

There are actually two exact solutions. One is for the center of AB to between A and E. The other is for the center of AB to between E and B. My solution is for the center of AB between E and B.

This approach is iterative. First, draw a line from C to E and construct a perpendicular to this line at E. Extend this line to intersect the extensions of line segments CF (y-axis) and CK (x-axis). The length of this line segment is the shortest we can use. Lay out some combination of distances CF, FE, and CE until the total greater than the shortest we can use. Call this distance L.

Extend a straight line from F through E a distance L. Using the compass, mark off an arc centered at E and the pencil point and the far right end of this line (length L – FE). Intersect the positive x-axis. This is the first approximation to point B which we can call B'. Draw a straight line from B' through E to the y-axis. Along this line, mark off a point a distance L from B'. Draw a straight line from this intersection to the y-axis. Where this line intersects the y-axis is the first approximation of A called A'. From A', mark off an arc of length L that again intersects the positive x-axis. This intersection is our next approximation of B. We now repeat the following steps:

1) Draw a straight line from our current estimate of B through E to the Y-axis

2) Mark off a length L from the current guess of B along our current guess for L

3) Draw a line from this intersection to the y-axis getting a new approximation for A

4) Draw an arc of length L from our current approximation for A to the positive X-axis. This be becomes our current approximation for B.

You can get an approximation to the other solution if you draw your first line from K through E.

Iterate a few times. If I get some time this weekend, I'll try to determine how many iterations are required to meet the challenge criteria.

I'm guessing there is a a better solution out there but sometimes the most elegant solution is one that lets you get onto the next problem the quickest.

Thanks,

Jim

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#34
In reply to #32

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

08/30/2009 2:32 PM

Using the above technique, I tried the various combinations of rectangle widths and heights with integer values (1 by 1, 1 by 2, … ,3 by 5, …, 10 by 9, and 10 by 10) assuming these to representative of all combinations of real values for width and height between 1 and 10. I also always used my selected length of L to be some multiple of the rectangles diagonal. Actually, I chose the smallest multiple that would be larger than the minimum required. If Li is the ith iteration approximation for L, I found that at most five iterations were required for L – Li to be less than 0.001, six iterations for L – Li to be less than 0.0001, and seven iterations for L – Li to be less than 0.00001.

Thanks,

Jim

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#39
In reply to #32

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

08/31/2009 4:01 PM

"There are actually two exact solutions".

Perhaps it would be good to clarify what you mean:

I think you mean that there are two possible locations for the line AB. But you don't appear to give an exact solution to the challenge.

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#43
In reply to #39

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

08/31/2009 5:16 PM

Maybe the following figure will help. For this case, the rectangle was 3 units wide and 4 units tall with a diagonal of 5 units. I picked the length of L to be 3*5 = 15 units. The values for A & B are 14.414 and 4.152 for one case and 5.080 and 14.114 for the other case. L is 15 for both cases.

Thanks,

Jim

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#44
In reply to #43

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

08/31/2009 6:41 PM

Those are indeed the two possible theoretical locations for the 15-cm line. But they are not solutions to the challenge (as written, this would require an exact construction using straight-edge and compass). Nor is it likely that a general solution is possible.

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#33

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

08/30/2009 10:01 AM

I know I get a little picky about challenge questions, but here I go again. The challenge says "...using only a compass and straight edge, no measuring instrument to be used." That means the compass cannot be used to transfer distances directly. We have to use an idealized compass that collapses the moment we lift it from the paper. There's a workaround, but it limits construction to transferring a line and prohibits comparing line lengths.

All that sounds a little pedantic, but without the "rules" I can trisect an agle exactly using nothing but a standard straightedge and compass. For this problem, if you let me transfer distances directly for this challenge problem, I'll just make a slide rule.

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#35
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/30/2009 2:36 PM

When I read compass, I envsioned an adjustable compass with a point on end and a pencil point on the other. With that, I can draw straight lines and arcs of lengths than can be measured on the paper.

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Jim

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#36
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/30/2009 4:06 PM

The "fly in the ointment" is that if it stays open on its own, it's also a divider, which is a measuring instrument. In traditional Euclidean geometry, dividers are a no-no.

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#40
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

08/31/2009 4:32 PM

You are of course theoretically correct. But I was under the impression that the reason modern geometers have resorted to a fixable compass it that you can use a Euclidean compass to do anything that a fixable compass will do - it's just the operation is a lot more complex. (Dividers will only measure out integers - just about the easiest thing you can do with a true Euclidean compass and a straight edge)

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#48
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 6:42 AM

Not exactly. A Euclidean compass can only transfer a line by swinging the arc. You can indeed limit your fixable compass to that task and you can do things like the familiar construction of a perpendicular. But, one of the things you absolutely cannot do is use the compass to mark (or compare) a distance alongside a straightedge. That creates a measuring instrument. And, you can use dividers for much more than integers.

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#53
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 4:29 PM

All that is apparently missing from the collapsing compass is the ability to create a circle centred on an arbitrary point of length equal L to a predrawn line XY

But in fact that can be done as follows:

You state that you know how to draw a perpendicular to a line that passes through a defined point. Use two such perpendiculars to creates OS parallel to XY and another two to XT parallel to OX. The intersection Z of these two lines is a distance L (=XY) from O

Set compass and draw circle.

Using this method, every time that you wish to use a collapsing compass to transfer a length you need to draw a parallelogram. It's tedious, but it does the same job as a fixable compass.

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#54
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 4:38 PM

No, I understand that and easily concede that is OK to do. But, you cannot hold a collapsing compass alongside a straightedge. It may seem a pedantic, picky point, but it makes all the difference in solving equations.

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#56
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 5:12 PM

"There's a workaround, but it limits construction to transferring a line and prohibits comparing line lengths."

Clearly, the construction I described will allow us to say which of two lines is longer - which is what I would mean by "comparing line lengths". So your meaning must be different; can you help with some detail?

I agree that the collapsible compass cannot be used as a sliding measure - but I don't believe that either dividers or a fixed compass will do that anyway.

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#57
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/01/2009 6:10 PM

Don't parse too much. I've seen you show your great familiarity with Euclid. I think I remember you even having a book of the postulates.

You may, of course, check lines for an inequality. You may not compare lines in any sense that equates to measuring.

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#59
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 4:59 AM

You write: "You may not compare lines in any sense that equates to measuring."
I fear that I would regard that statement as misleading in this context. It is of course true that Euclid was only concerned with relationships that were exact, which means that any approximation in measurement is outside his terms of reference. But you what you previously wrote was that the tools (collapsing compass and unmarked straight-edge) are incapable of performing what we would regard as measurement. This is demonstrably false.
All that you need do to measure the relative lengths of two lines is to construct a fixed "ruler" based on repeating an integral fraction of the length of one of the lines, and then transfer the length of the other line onto the ruler with one end on one of the markings. You can in principle achieve unlimited accuracy/resolution by repeatedly multiplying any residue.

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#60
In reply to #59

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

09/02/2009 7:29 AM

And, again, I believe you are familiar enough with Euclid to know that you cannot in any sense create rulers. It is thought by some that other geometers permitted rulers, but Euclid did not (or so we think). Some geometers will go so far as to insist that you may only hold one instrument at any one time - either the compass or the straightedge, never both.

I agree that once you allow "marking off", there is no limit to accuracy/resolution (in principle). That indeed is the foundation of any measurement system. And, once I can mark off distances, there doesn't seem to be any limit to my computational possibilities; certainly, I can then build an analog computer. Even a transcendental equation becomes simple to solve.

For example, one of the two exact solutions I know for trisecting an angle requires that I hold the compass in one hand while turning the straightedge with the other. It is exact. It does violate Euclid.

Why do I care? I think geometry one of the important bodies of knowledge and one that is rapidly vanishing. I don't suppose I can do anything about that, but I will, in my bastardization of Thomas, "Rage, rage, against the dying of the light.".

Cheers.

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#61
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 7:46 AM

I think geometry one of the important bodies of knowledge and one that is rapidly vanishing

How I would love to give you a GA for that ! If somehow it were possible to allow students the time to play with Euclidean construction, their knowledge would be so much better. They might find themselves 're-inventing the wheel', or pursuing impossible goals, but play-with-it-and-see is surely the best learning method. Reading endless tomes on theory has it's value, but time spent with blank paper and and abstract problem is never wasted.

You're more than likely correct in your assumption of what Fyz knows of Euclidean construction, but is not the greater point here learning ? Personally I feel that JD's initial description might have been a little clearer, but it really doesn't matter. JD has clarified, and it's up to individuals to draw (sorry for the pun) what learning experience they can. I've not come up with anything I consider worth contributing to the question, but I'm still enjoying it. The value of questions such as this one are not in the finding of a solution, but in the process of thinking.

Debating the finer points of Euclidean construction is quite valid and topical, but it's not the be-all and end-all. Much thanks to you all for an interesting read.

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#63
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 8:26 AM

I agree about the larger point. That's why I'm posting "off-topic".

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#64
In reply to #61

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

09/02/2009 8:27 AM

Based on this discussion, it would also be good if they were to experience the difference between Euclidean construction methods and the more narrowly defined (provable) Euclidean geometry.

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#62
In reply to #60

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

09/02/2009 8:25 AM

To misquote: "yes you can" (create rulers). Abandon your preconceptions and look at the evidence - better still, try it using Euclidean construction techniques (realising that this is not the same as Euclidean geometry - see below) and tell me if-and-where you find a difficulty.

I do agree that trisecting an angle in the way you mention uses the tools in a way that Euclid would have regarded as unsound. But (to labour the point) measurement in the manner I described can be done while using the tools in precisely the way that Euclid would have envisaged.

Of course you cannot move the marked ruler about in any convenient wholesale fashion, or combine its use with the compass - as this latter would destroy the setting of your (fragile) compass. But you can readily create a marked line ("ruler") that is any rational multiple of a pre-existing length. The most convenient way to use this would be at a fixed position on your papyrus, and then copy the length-under-consideration to your ruler.

The reason that this would not appear in Euclid's work is that the work only concerns relationships that are "provable" as exact. Anything involving measurement of any kind is by definition approximate.
N.B that I can't see how your computer would be analogue - the best I could manage using measurement is equivalent to a successive approximation digital method.

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#65
In reply to #62

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

09/02/2009 8:29 AM

An impasse perhaps? Since the answer has been provided, perhaps another place, another time?

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#66
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 10:01 AM

perhaps another place...

please, please , please.... don't openly discusss where with any degree of exactitude !

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#68
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 10:17 AM

I think he means the House of Lords.

Fortunately, I'm not qualified in that respect, so I'll try once more to see if TVP is willing to resolve it here - so at least neither of us will ever mislead anyone on this topic again.

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#70
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 11:24 AM

That's a very charitable definition - perhaps Parliament should move with the times and adopt the phrase "...another planet". I suspect that if you and TVP continue much longer, we'll shortly be hearing the phrase, " I refer the honorable gentleman..."

Yes, this is one occasion on when I'm quite happy to assert that neither of you are qualified in that respect - dissection of a Challenge Question is a much more tricky game than our glorious leaders could play ! I'm not privy to knowledge on JD's tonsorial endowment, but if he feels like I do right now he'll be checking the coefficient of friction up top. That is ,of course, presuming he's not brushing brick-dust from the forehead ! I jest too much - an interesting bit of debate between all. The question, and JD's solution are going to take me some time to get to grips with. Not that I can use the excuse of other tasks (though it's true enough), there are many subtle points at issue to ponder. Personally, I think that JD has posed a very good problem for us. It's not easy to come up with a concise proposal that answers it (in my case no proposal at all !).Putting forward a question is often a much greater art than solving one.

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#67
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 10:14 AM

Can you first clarify your position?

Are you saying that measurement against a scale is not possible using Euclidean construction techniques? or
Are you saying that Euclid explicitly disallows the use of this sort of measurement?

If it's the former, basic integrity would suggest that you either check where the technique I propose breaks down or admit that you aren't as certain as appears.
If it's the latter, I have looked (again) at a direct translation of Euclid's Elements, and can't find anything to that effect*. That said, the content of the Elements indicates that accurate but inexact measurements are irrelevant to the intent of the work - i.e. proof by example is not valid in this context, and only exact constructions are of interest.
*It is possible I've missed something you know about - if so, please give a reference. (Book#, Definition# or Proposition#, etc)

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#69
In reply to #36

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

09/02/2009 11:06 AM

I don't agree. Euclid used a compass and a straight edge to construct a perpendicular bisector of a line segment. This requires making arcs of equal radius on both ends of the line. He also constructed a parallel line through a point which required setting the compass distance from the intersection of an arc and two lines. None of the construction techniques used here to solve the challenge question require anymore than 1) drawing straight lines through the intersection of lines and 2) drawing arcs of "measured" lengths centered at the intersection of lines.

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#71
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 11:39 AM

I think TVP has agreed that we can transfer lengths using parallelograms and circles.

But to create the parallelogram requires arcs of equal radius without a transfer method - fortunately, we can create perpendiculars through a point by extending the line and marking an arc of a circle to cut the line, and then we can use the distance between the arcs to set the radius of the compass used for the perpendicular bisector.

I managed to start on your first method - as I understand it, this too starts with a successive approximation; so it's even more painful than the second - and still not an exact construction.

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#77
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 7:29 PM

I didn't challenge any of the solutions. I pointed out what I think is a legitimate obstacle to solving the question. I am slowly working my way through some of the solutions, but am not challenging them.

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#78
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/02/2009 8:18 PM

No. I think I have determined that none of the solutions satisfy the challenge. Perhaps I misunderstood the challenge, but I believe the conditions to be:

AB must be a combination of the sides and diagonals of the rectangle.

AB must be longer than the perpendicular of CE.

No measuring instrument may be used.

Are those incorrect?

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#85
In reply to #78

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

09/03/2009 4:59 AM

If that is what you were saying, I agree:
The challenge as posed has not been solved. In fact, it cannot be solved, as it asks for a line of any of those specific defined lengths passing through E - so measurement instruments would not help.

Rewording the challenge to demand a solution that give AB to a precision of 0.001 units, and does not require confirmation, I also agree that no solution has been proposed solutions is sufficient - more about that later. If we were to rephrase "no measuring instrument" as "no measurement may be made other than those that use the compass and the unmarked straight edge as separate stages", we would still require the extra step of dividing the side by 10000 and drawing a circle to show that the error was less than this (it is a mater of semantics whether a single larger/smaller should classify as a measurement?).

To return to the question of a solution to the re-posed (soluble) "accurate to 0.001 without confirmation": the successive approximation approach (first given in post #19) can be the basis for this. As implied there, the additional piece of information that would be needed is the number of iterations required. That would have to cover the worst case, so it either requires an analytic solution (very tough) or (more likely) a spread-sheet demonstration of the worst region (very tedious).

But perhaps we could now say that passingtongreen has shown a route to a solution to the challenge as posed - exploit the ambiguity of the presentation and choose the length so that it is one that you can fit. (Unfortunately, his proposal does not necessarily result in a length that is longer than the orthogonal to CE).

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#86
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 5:35 AM

It would have been nice if I was right, but alas its not the case, the challenge as proposed has not been solved, I did not state how wide the rectangle was, or high it was, or how long the hypotenuse was, how can there be an answer? such abstract questions, one has to be able to apply one knowledge, not come up with logical conclusions? of inferred facts, such depth of perception has to be admired.

Regards JD.

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#88
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 8:05 AM

I realised that the puzzle does not have a fixed answer, so I read it as a request for an iterative method that moved closer and closer to an exact answer, hence my saying that I thought I had "walked around" your real question. I was a little disappointed when I found, or thought I found, an exact solution.

It may have been more interesting because there is no single solution.

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#87
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 7:18 AM

I too found that for certain rectangles, "his proposal does not necessarily result in a length that is longer than the orthogonal to CE". However, if we increase the the value of n to be larger, we can still get an exact solution. Each time we increase n, we make the slope of L steeper and therefore longer.

Thanks,

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#89
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 5:03 PM

Agreed - but how do we determine whether the length is longer? Measurement?

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#92
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 6:53 PM

You write:-

"AB must be a combination of the sides and diagonals of the rectangle."

but the problem wording actually states:-

"The hypotenuse length being any combination of sizes taken from the rectangle"

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#93
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 7:07 PM

Yes, I know, but what other size could there be?

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#94
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/03/2009 8:40 PM

Hi TVP45

The concept of the challenge was to solve how to calculate the angle CAB, and test that the angle was correct to a given tolerance, the rectangle sizes and hypotenuse length being arbitrary, the challenge was about a construct that aided that endeavour, it was not intended to be a geometric proof, and the rules that go along with it, though that was unavoidable.

I would be interested to see if you can give me a calculated solution without the aid of the above constructs. A step by step mathematical solution? If you can I'll owe you a Guinness.

Regards JD.

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#95
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/04/2009 12:23 PM

I'm game. Can you briefly run over the challenge again? Several posters implied I didn't fully understand the original.

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#97
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/04/2009 8:12 PM

Question.

Given a rectangle CFKE measuring 3 units vertically, and 4 units horizontally, is located in a right angle corner, Show how to construct a triangle ABC using only a compass and un-incremented straight edge. Such that the hypotenuse AB being 11 units in length touches the corner of the rectangle at E, and referring to the construct, show how to calculate angle CAB, such that using the formula (4/Sine A ) + (3/ Cos A) = AB, show that AB = 11.000, correct to within three decimal places.

Regards JD

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#100
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/05/2009 11:15 AM

I'll have a go at this, I suspect it will be as difficult as you appear to think.

In the meantime, I was looking at the original problem again when it dawned on me that there was no quick way of finding the length of the perpendicular hypotenuse and thus, pick a longer one. I played with two Pythagorean triples. 3, 4 & 5, and 1.3, 8.4 & 8.5 (one tenth of 13,84 & 85) with very different results.

I don't consider your post off-topic and voted to remove that appellation.

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#101
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/05/2009 9:05 PM

I don't consider your post off-topic and voted to remove that appellation.

Ditto. Although giving specific dimensions doesn't alter the nature of the problem, it does gives us all something more specific to chew on. JD has worked to clarify all questions, but that post does seem to sum it up better. Doesn't make it any easier, but is more succinct. As (I think Fyz) pointed out, the question boils down to how many steps of approximation are required for desired accuracy. Being a complete gull for topics such as angular trisection, I've followed this with interest. The inevitable 'steps-to accuracy' is not my bag, but it's fun all the same.

Personally I'd have avoided the 3/4 rectangle in the example (exactly because it suggests Pythagorean proofs etc), but it captures the problem a little better than the original. Somewhere on CR4 is a similar problem concerning a ladder against a wall ( in which thread I missed a stupidly obvious point - hence my vague recollection).

Somewhat annoyingly, this particular problem keeps coming into my mind. If a question can keep the grey matter bubbling, then that's got to be a fair measure that it's a good question (at least by the measure of my own brains inability to solve !)

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#102
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/05/2009 10:34 PM

Hi Kris this should jog your memory.

Regards JD.

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#103
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/05/2009 11:23 PM

LOL- know where you're going with the 'vitruvian man', and iterative sequence always hold a fascination for me. I'll leave it to others to take that approach. Quite sure it leads to an answer (), but it's a case of ever decreasing circles. It's a good example of where to proceed on this question. Bitter experience has taught me the possibilities of disapeaering up yer own **** ! Yeah, have I been there !! A good point to make, and very valid in regard to 'solving' this querstion.

The 'ladder against wall' question I mentioned is out there somewhere. I'll check it later.

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#104
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/05/2009 11:24 PM

Oops, sorry, that was me !

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#105
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/06/2009 12:05 AM

The ladder is in the memory jog above, post #8.

Regards JD.

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#106
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Re: Triangle With Rectangle Insert: CR4 Challenge (08/25/09)

09/06/2009 6:38 AM

LOL - I really should check my own checking at times ! Cheers for pointing it out...simpers off muttering 'doh'

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