Area Calculations: CR4 Challenge (07/14/09)

I agree that it's only a little off topic, and so I gave him a GA.

I believe we have only 1 GA per challenge, (is this true? Did you just assume I had a GA left - or did you somehow know?) and so I tend to hoard mine to give to the one I think is the most deserving.

I always wondered if there was a rule about how many GAs I could give per challenge question. Usually I give one max but sometimes the challenge morphs into something different and then I may give another GA or at least I think I've given a second GA. I've never checked to see if the second GA counted or if the first GA was taken away. I've never given more than one GA to a single response though I think I've given the same person two GAs for the same challenge.

Is there some guidance?

Thanks,

Jim

Yes: you can give as many as you like. And, don't worry if you try to give the same post a good answer twice: CR4 remembers:-

Same thing with rating a thread, but, it doesn't explicitly warn you that you're wasting your time (this is my second attempt at rating this thread) :-

Thanks all, for the votes BTW. It's funny it seems so obvious when the penny drops, but, I needed SlideRuler to push me in the right direction.

You can only give one GA to a single posting (unless you have multiple aliases). But I know of no other constraint; you can certainly give multiple GA's in a single thread. If there are multiple answers that are good and truly different solutions I would happily give each one a GA (it doesn't often happen). On the other hand, I reserve my GAs for the first well-written answer of a particular solution (just my taste).

I too was quite taken in by the elegance of the triangle area solution, after having tried to play around with the figures in SlideRuler's post 129. Since I'm more comfortable with geometric interpretations of formulae, wherever possible, I tried to relate it to the standard (?) form for area of a triangle whose vertices are given as coordinates P₁ (x₁, y₁) etc.

The alternative version in my old school text book** can be written for convenience as:-

2A = x₁ (y₂ - y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )

The subscripts follow in cyclic order, hence easy to remember, and this is essentially a variation of your solution for the polygon abcde, by adding or subtracting rectangles instead of triangles. In this form the actual computation may be somewhat easier to do and cross-check.

If we put in some numerical values (all positive coordinates to start with) the procedure becomes visually apparent.

In the first figure the overlapping rectangles have opposite signs and hence those areas vanish on "addition" in the formula given above. If the vertices have any other arbitrary locations, the rectangles move around but the same pattern holds in principle. The picture becomes still clearer if the three points are in different quadrants, as in the second and third figures. It is also seen that the resulting area works out as negative or positive, depending on whether the points are numbered clockwise or anticlockwise.

I'll leave it to the CR4 geometricians to show whether the method can be extended to other straight-sided polygons. (Drawing those figures has left me exhausted.) I suppose there are many algebraic expressions which can be explained geometrically, but we seldom make the effort. =TeeSquare=

** Elements of Coordinate Geometry -- S L Loney (publ. 1895 !!), Article 25. Those ancient texts by Loney, Hall & Stevens, Hall & Knight, continued to be the mathematical bibles during my school days in the late nineteen fifties, and I find that local editions are still being reprinted in this corner of the globe!

aka

Well, okay....

But my diagram shows all the congruent triangles including the ones that are half-squares.

And I finally got to use those 'upper bound' and 'lower bound' terms from Calculus class ages ago.

If you gave me the GA, thanks!

-Kinsale

he he

I didn't give you the GA, but I shall now because it is a good diagram. I hadn't spotted the numbering first time round (), but that just adds to the proof that mine was even better with it's glorious colour coding .

Why not sign up to CR4 - that way you can collect all the GA's under a chosen user name like the rest of us ! The GA's aren't overly serious (and member titles certainly aren't), it's just that having a user name makes it easier to follow who posts what in a discussion.

I am signed up -- as 'Kinsale'. In fact, I've been signed up since April 2005 -- longer than anyone else here, or just about. My 'Guest' reply was sent from a different PC where I wasn't logged in.

I check in on the Challenge occassionally, but don't really have too much time for responses. This one came to mind quickly, so I posted it.

Yes your color coding was glorious! Had to view it wearing shades.

LOL - I probably should have checked the name ! Your relaxed approach to whether or not you collect a GA against your user-name is most praiseworthy .

There's more.......

Not all of them, surely?

Fyz

I think this is equivalent to the version you originated

I was far too modest to say so myself !

The coincidence of finding 'Pick' was pretty funny. I haven't even read the new book and was just leafing thru it. Google churns out quite a few links on Pick's theorem, though I haven't read them. cnpower posted a fun geometry challenge some time back, and I found myself re-inventing the wheel in a similar fashion to thinking about this problem. It's a lot more fun/satisfying to learn these things by working thru a problem set on CR4. Reading maths texts on their own can be a bit 'dry', and I'm sure that school pupils would benefit from more homework that revolves around solving obscure problems like this.

It looks like my posting the name 'Pick's theorem' has killed off further input here ! Bit of a shame, because half the fun is looking at different approaches to solving. I guess it boils down to personal taste - whether one wants to know an answer to a question or enjoys the hunt more than the quarry.

Agreed: so far as I am concerned, I definitely find it best to work it out for myself. Great teachers provide the problem together with just enough clues to allow you to do this quickly; I was lucky to have two of these at high school and then have college tutors of similar bent.

Naturally this becomes difficult to arrange (even for the very quickest) if an entire topic has to be covered in a single session and the rate of progress of the class is excessively varied. (Beef over).

(Did I miss out on cnpower's geometry?)

Maybe Tangent to circle or ellipse

The problem is to draw a tangent to a circle from a point using only a straight edge (no compasses!).

Thanks.

Looking at the solutions - when you showed the solution for the ellipse giving both tangents you used different constructions for each tangent and you "broke" the line between the two tangents. I thought that the same construction found both tangents...

I hadn't realised that the one construction gave both tangents. I didn't "break" the line because I didn't twig that it was just one line.

I had a mentor who was much the same as your teachers. He'd chuck occasional random problems at me, somehow gauging that they were within my ability to solve. Never easy, but never impossible. He kept me busy on what, with reflection, are trivial problems, but I learnt a heck of a lot on the way to solving each specific problem. Being stubborn and willing to sacrifice trees in the process helped a lot. A lot of the youngsters I meet simply don't have any capacity to persevere. At the risk of sounding fuddy-duddy, I'd ascribe that to the immediate gratification/i-culture that we find ourselves in.

You're quite right that it would be hard to implement such methods in a class, though I think it's well suited to homework (with the proviso that marks are awarded for writing up all trains of thought followed in attempted solutions). cnpowers question, and the problems presented in the Challenge Questions, are excellent ways of developing creative thought and learning mathematical principles. As with most things in life, what people get out of a situation depends upon what the put in.

<clambers down from pulpit>

The brilliance of these two teachers was that half the class was able to develop the solution during the lesson before the teachers got to the conclusion. Being a fairly uniform group, it wasn't always the same half of the class, so it remained a continuous mental challenge and no-one felt left out or inadequate (SFIK).

[Amongst other things, we "invented" both trapezoidal integration and differentiation, including developing the appropriate power laws ("for ourselves") each during a separate double-maths class (90-minutes)]. The downside is that some of us became somewhat cavalier about homework...

The downside is that some of us became somewhat cavalier about homework...

LOL - A fair bit of mine was last-minute, just how last minute depended upon what temptation there were to wreak havoc on with a spanner/hammer etc. Then there was the lure of going off and doing stuff that was just plain naughty and/or fun. Oddly enough, I don't think much has changed !

once upon a time I spent many days trying to trisect an angle with a ruler and compass... (being a draftsman) and it was said to be impossible. I came close. the person who posed the problem to me showed me with trigonometry that my solution didn't solve it. but I used hundreds of sheets of paper...

one day I read (somewhere) that there is a solution to this embedded in the great pyramid..?

I would be curious about that. any ideas?

Chris

Trisection using straight edge and compass when you are only allowed to run the straight edge through pre-defined points is one of those rare cases to have been proved impossible. Either the Egyptian thing is a myth, the constraints were different, or the (much re-examined) mathematical 'proof' is invalid.

thank you.

I've been here before, I think during a nice little diversion with Fyz. It boils down to the impossibility of finding a cube root via Euclidean means. I don't have the relevant book to hand, but I'll dig it out this evening and post the proof.

Meantime.....only days ? I could happily spend months chasing the madness of trisection. Honest ! Not because of an expected solution, just because I enjoy exploring geometry. Have a read of this for a giggle. I feel a certain affinity with the gallant author

Check out this little gem !

Some of the waffle about mathematical properties of pyramids is not quite as it seems. Figures massaged a little, and so on. Undoubtedly they're fantastic structures, but stuff by the likes of Von Daniken has to some extent blurred peoples vision of their beauty. They've probably got a mass of interesting properties, but offering up a euclidean solution to trisection isn't one of them. I can see the attraction of investigating Pyramid geometry, but it would be easy to become lost in looking for dimensional relationships that weren't there.

"I could happily spend months chasing the madness of trisection. Honest !"

I'm speechless..

btw, is this you? part squirrel, part human, part dancing-over-the-abyss madman?

LOL - the thing in the tree has more poise !

Right, here we go;

We only need to show that one angle can't be trisected.

Consider the trigonometric identity....

Cos 3θ = 4Cos³θ - 3 Cosθ

Let's play with Cos3θ = 60^o = 1/2

We can insert that, and Cosθ = y/2 (the quantity we aim to find), into the identity.

With a bit of manipulation it becomes....

y³ -3y -1 =0

To solve such an equation involves solving the cube root of a number (there are several algebraic methods which do just that). Easy enough to find the square root of an arbitrary length (side length of a triangle), but impossible for a cube root. I've no idea what the proof of that is, Fyz probably does. In short, trisection would require finding the cube root of an arbitrary length.

As you probably appreciate, to my mind that is just pushing the boundary of the proof around.

At one time I could have reproduced the full proof. Rust Senility means that these days I have to look it up (just like anyone else).

BTW, the Wikipedia article uses the same result as you do, but expresses it differently: "Note that a number constructible in one step from a field K is a solution of a second-order polynomial"

Rigorous mathematical proof often looks impenetrable. The 'proof' quoted is a reasonable simplification (in as far as I'd want to satisfy myself about the problems impossibility), though I agree that it's tinkering with the margins. What's the Wiki on this ? The sauce source I regurgitated was the same as last time we chatted on it - Mathematics from the Birth of Time.

In the first link you give;

is not constructible, because has minimal polynomial of degree 3 over Q

<My underlining>

The quoted reduction to y³ -3y -1 =0 looks a bit more reasoned to me, and is it not 3rd order ? Your point is well taken, but you haven't offered up a better proof of Euclidean trisection impossibility. I've never read Wilkes proof about Fermat, but I suspect it is equally difficult to grasp (certainly on my level of understanding such stuff).

I'm sure that even the most capable mathematicians have to refer back to notes when writing up a proof. There's a neat analysis of solving cubic equations I was once shown. 'neat' in that I could actually comprehend it (!), and I haven't so far seen it on my web travels. I'm pretty sure that the bloke who showed me didn't claim it as original work, but I'll refrain from reproducing it until a good excuse arises and I've checked around for it's originator (or at least whatever name it goes by).

I see that the official answer to this question is now up. The GA distribution makes for an interesting read in itself. I'm not griping at all - but it perfectly illustrates why there's value to reading all posts in a thread. Benefit derived from these weekly bits of fun is directly proportional to how much one reads.

What I was intending to point up was that the critical thing to prove is that the appropriate fields will not produce cube roots - which is the subject of the references. Then there is the little matter of the correspondence between the geometric method and the operations under the field in question - after all, isn't angle trisection supposed to be possible using some scheme of marking and sliding the rule (equivalent to placing a tangent through a point by rotating until the edge touches the circle)?

Yes, correspondence is somewhat of a problem. You can rotate the marked ruler, but not with Euclid looking over your shoulder. It's no or valid than using a Tomahawk. Let's cut to the chase - what/where is the best (ie most easily understood by most people) proof the Euclidean trisection is impossible ?

(ie most easily understood by most people).
Most people can't even understand the question.
_{Look out he's right behind you...}
Del

Most people can't even understand the question

You think I did ? Even if I'd typed 'more' instead of 'or' I'd still be flombusculated !

_{....he could be behind me, but he isn't necessarily right ! I'm rapidly getting left behind - yippee !! The perfect excuse for me to demonstrate my half-***** understanding}

I don't know of a proof that can be handled without substantial preparation, as (my recollection) the cube-root proof depends on lemmas that have themselves been developed through several stages.

We can regard the proof as three processes:
1) Demonstration that trisection is equivalent to taking a cube root (done)
2) Conversion of the Euclidean constructions into transformations in the equivalent Field (reasonably straightforward)
3) Proof that these transformations cannot effect a cube root (my recollection is that this is a term's maths course to get to the starting point)

Even more interesting (perhaps) is to project the sliding-marking constructs onto the Field. I think that this creates a set of transformations that extend the standard set that are defined for a Field - i.e. we move into a different mathematical realm (mathematicians being inveterate fiddlers, the consequences have probably been extensively investigated - just not in my field [PI] of view).

Much as it might make you shudder, can't I just stick with "you can't find the cube root of an arbitrary length" ? I'm going to anyway, because it will give me hours of fun trying to do it. Not !

Thanks for the breakdown of the stages, but I think I could wallow in trying to understand it all for a rather long time ! However, the posts you've made are duly noted for possible further study. I like to know where the doors are, even if I'm a bit too chicken to venture thru them right now !

Check out this little gem !

And Fyz' brilliant post #89 is still only nearly good!

LOL - I do believe I made exactly the same point in the last line of my #119.

If I'm not mistaken, there seems to be something of a theme here. Do not be alarmed, Chrisg288, Fyz does not (as far as I know) have almond eyes.

I rate this question as good because people of all ages/ability can play with the geometry of it and learn something.

If you reserve one corner for "start and finish", there should be 4 choices for the second point three for the third, two for the fourth and (the fifth is then fixed). That would be 24 pentagons, but half are duplicates - merely tracing in the reverse direction. So that should leave 12 different options. (Example: connect each of the bottom corners to the top, connect the outside corners together, and the two right-most corners to each other, and ditto the two left-most corners).
BTW, your rightmost drawing doesn't seem to use the same points as the others.

Thanks. I was only trying to determine the different 'kinds of sub-shape combinations' which could be generated by joining any five points (not necessarily the same locations each time!). The sequence you suggested still gives three triangles. Sorry I didn't state my problem clearly enough.

On the area formula, maybe someone can check whether AutoCad gives the result as 42 or 52 (or something else?) if only the outer vertices and path are specified. I suspect that it would be necessary to determine the coordinates of PQRST separately in the general case, and trace the path around the periphery. My mistrust of 'software' is due to possible misinterpretations of hidden procedures.

=TeeSquare=

Sorry - not reading carefully enough. I think you got all types.

CAD packages such as Autocad are intended to generate physical shapes rather than abstract or topological concepts. So they should not allow negative areas or double counting. On that basis, if you specify the vertices and the path you ought to get the area of the region enclosed by the outermost sections of the paths. The alternative would be to complain* that your definition was ambiguous.

*Software is generally good at complaining about such things - but usually without telling you what is at the heart of the problem.

But I could easily be being overoptimistic.

Thanks for your suggestions. Fact is I'm rather scared of computers -- more on this in the recent challenge thread (Triangle-Rectangle). Downloading and installing new software is something I won't do voluntarily.

I agree with you about the limitations of the Draw command in MSWord. Things get unstable after a while. Never tried drawing in Excel (and never will now, since I've moved to Linux & Open Office).

So far I have been making diagrams in Word97 to go with text, so the document is self-contained and not linked to some other file which may get moved or deleted. For putting figures into CR4 posts I make them in Word and save as jpg file. MSPaint was good enough for my limited image manipulation needs.

The Draw commands in Open Office Writer are more comprehensive than in Word but I'm not yet familiar with them. The GIMP programme in Linux is way too complex, and I wish there was a simpler alternative.

Despite teething troubles, I'm glad to get away from Word and Windows.

=TeeSquare=