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I ended up seeing 3 different solutions to the problem, a fast, a medium, and a slow solution. The solution that Bob is pointing to is the medium one. The slow one is just a twisted (or flipped) version of that one. The fast one directly uses areas in units of the area of the large triangle, and for me can be done in my head without need of paper. It is sort of the "Ah Hah!" solution. I didn't see it until I put in the height from C, i.e. after the other solutions. The problem with that solution is giving a hint. It's a little like getting someone to see one those stereograms, you want to say "Well, you just gotta see it". Not too helpful. I think the key hint is still angle B although maybe stated in not quite as contorted fashion. I hope classof2020 wasn't too put off by my response and made a stab at it, and if he or she is still following, good luck in what I assume is your geometry class.
On a side issue, I've never considered the sine, etc to be purely circular functions, or trig functions as classof2020 put it. I was (as in way way back) introduced to them as geometry concepts relating to similar right triangles, and it always has felt right tieing (tying?) them to similar triangles early on. If one uses that standard diagram with two rays forming an angle and a sequence of parallel perpendiculars set out from one ray to the other, then it is visually obvious that those are a sequence of similar triangles and can be extended indefinitely. Consequently the ratio of two corresponding sides of any two triangles is constant and consequently constant for the figure as a whole. Then it's easy to see that there can be 6 constants (3 possible sides * 2 remaining sides) which gives the 6 functions or constants for that angle. However, I don't know if that's how they present it these days.
Basically, the sine is just a ratio of triangle side lengths which then stays the same for similar (right) triangles. Look at the areas of the triangles by drawing in the triangle heights making sure that the angle B is opposite the height. You should notice that same similar triangle reappears in several places in the figure.
Use law of sines focusing on angle B. Actually, just use sines.
Okay. That illustration has the 9 point circle, but what of the circle that is tangent to it, i.e. one of the excircles or the incircle?
If needbe there is a procedure for recreating a triangle from for example the 9 point circle and a tangent excircle, but I don't think it's really necessary for setting up the illustration. All that is necessary in your illustration is to add one of those circles and then add the two oblique axises. (I would suggest using the incircle to make the illustration look somewhat like Blum's figure 1.) Once all of necessary elements are in the illustration, the extraneous items (extraneous with respect to the problem) might be deleted for a cleaner figure. However, that's more up to you as to what you want the figure to represent.
Notice how his figure two is with an excircle.
Does the above help?
-jeff
Thanks for the link. Hhmmm. He didn't make it easy did he? The diagram isn't next to the text so you have to keep flitting back and forth, many points aren't labelled and there's no triangle shown. So I'm really confused about this and don't know if I'll ever be able to follow what you want, let alone help with an answer.
Nope. Not easy. Not easy at all. His target audience was professional mathematicians. I'm not sure whether he even made it too confusing for them. Coxeter didn't complain about the article in his text. Like you it was confusing for me too initially. However I did make it as far as the equation for tangents before getting stuck. I think I would put the problem at either a fourth year undergraduate or first year graduate student level. I can certainly help out in clearing up confusion up to the point of the equation of tangents.
It would help if you could reproduce the diagram with some letters to identify what is what, and then maybe I can follow what you are seeking.
Hmmm, we may have to make do with pencil and paper and maybe a compass and straightedge here. I don't think I have it in me to get together the software tools (and familiarity with them) to put together graphics that in the end just mimic what was there in the paper. If you do have them convenient and want to make a stab at the figure then we could definitely use that. However in its absence we can use the diagram in the paper, either reproduced by hand or captured via a screen shot, for easy reference.
To possibly help with clearing up some of the confusion, let me give a little bit of overview. The xi and eta coordinates that he provides are tied to one of the common tangents of the 9 point circle and of any one of the excircles or the incircle, *and* a side of the triangle tangent to whichever incircle or excircle was chosen. His diagram to me seems to say that he went towards the incircle, he doesn't say. Anyway, a (somewhat more understandable) diagram can then be created by drawing two circles tangent to each other, making the tangent one of the axises(sp?) and filling in a triangle for that 9pt circle and tangent circle and using one of the sides of the triangle that intersect with the common tangent (now an axis) as the other axis. In his diagram eta references the common tangent and xi is one of sides of the triangle that intersects the eta axis. *Note* the two axises in all probability are not orthogonal. For this particular diagram, that is unavoidable.
At this point one then works his way through the sequence of key equations in the paper...until of course one gets stuck and goes begging for help. In any case I hope the above helps in clearing up a bit of the confusion.
-jeff
Hi Bob,
Regarding the old thread, I'm not sure that it will be necessary.
In any case, I'm not sure what you want me to post since there
are possibly many graphics scattered through the thread. (Odd
though that it displays okay for me on my tablet.)
Anyway, the pertinent animal to look at is:
https://books.google.com/books?id=N1t02 … CE8Q6AEwBw
Again, the confusing aspect is on page 2 regarding the equation of the tangent lines.
One confusion is whether that equation is a starting point that he
pulled from another source or a finishing point that summarizes two
equations of lines in one easy to provide form. The second confusion is
the exact form of the "polar form" term.
-jeff
In some ways this question is an extension of:
http://mathisfunforum.com/viewtopic.php?id=14945
My question arises from an exercise in Coxeter's Intro to Geometry,
pg 114 ex4. He references Richard Blum's paper (1958) that I was able
to track down in books.google.com.
On page 2 of the paper, Blum gives an equation of the tangents that I
have not been able to make sense of. I did derive the equation of the
tangent circle in his oblique coordinates. However the next step he gives
eludes me, i.e. the C(p,n)*C(p0,n0) - C^2(p,n | p0,n0) equation for the
tangent lines.
Although I could beat my head against the wall a bit longer, I figure it's
time to move. So I would appreciate any pointers to making sense of it
that anyone could provide. Thanks.
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