I don't buy from Amazon anymore. You can also buy it at Abe Books: http://www.abebooks.com/servlet/SearchResults?sts=t&tn=%22The+Big%2C+Big%2C+... Bob Baillie --- ed pegg wrote:

It's rare when a new puzzle book has a dozen puzzle types that I haven't seen before. I'm really jaded about puzzle books, but I keep buying them.

Here's the best puzzle book I've seen in years, with hundreds of new puzzles: http://www.amazon.com/Grabarchuk/dp/1402777280/

Highly recommended.

--Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

Fred and others, Sorry that I haven't brought you up-to-date. My messages to math-fun were being kicked back, but Rich has kindlily now fixed this. I will try to circulate the latest on the incentres problem, but note that it has been established that the 6 incentres of the triangles formed by the medians do NOT in general lie on a conic. Peter Moses has obtained several related results. R. On Tue, 13 Sep 2011, Fred lunnon wrote:

It has just dawned on me that I have been attempting to solve the wrong problem: the original version placed the fourth point at the centre-of-gravity, whereas I have been placing it at (or near) the in-centre.

Assuming that the 6 subsidiary in-centres lie on a conic when the fourth point is either the in-centre or the centre-of-gravity (which I have not investigated) but not in general, what is the locus of all fourth points for which such a conic exists?

[And what diabolically ingenious argument will be required to establish it?]

Fred Lunnon

On 9/13/11, Fred lunnon <fred.lunnon@gmail.com> wrote:

...

I'm guessing that the cryptic reference to a "problem debunked" was intended to relate to Richard Guy's Cevian variation, which I confirm does not stand up. This isn't very surprising --- if the outer triangle's in-centre is perturbed to a more general "Cevian" point, the subsidiary triangles' in-centres would surely also need to be perturbed in some compatible fashion.

The original in-centre statement certainly does appear to hold numerically, and I have to confess that I've not the foggiest notion how to prove it. But the following question occurs to me: is it feasible to bound the number of specific instances (independent in some appropriate manner) for which a claim like this must be (exactly!) verified numerically, in order for its truth to implied in general?

Fred Lunnon

On 9/9/11, Dan Asimov <dasimov@earthlink.net> wrote:

...

<< Twice again my messages have been bounced back, so math-fun don't know that the problem has been debunked by Stan Rabinowitz, using Geometer's Sketchpad. R.

On Thu, 8 Sep 2011, Dan Asimov wrote:

...

Richard Guy, who is having trouble posting to math-fun, sent this message:

-------------------------------------------------------------------------- Apologies for butting in, but my messages to math-fun are bounced back by a spam filter, so I don't know how to send. What I wanted to say was:

The following problem was given to me by Julian Salazar, a high school student, who said he got it second-hand from members of the US IMO team. There are several subsidiary problems.

The medians of a triangle divide it into six smaller triangles. Show that the six incentres lie on an ellipse.

1. Who originated the problem?

2. Did he/she have a proof?

3. Experiment suggests that if the three ``cevians'' are drawn through any point, not only the centroid, then the result is still true, except that if the point is outside the triangle, then the conic is a hyperbola, and if the point is on an edge of the triangle, the conic is a pair of straight lines.

4. In the original problem, is the centre of the ellipse among Clark Kimberling's three-thousand-odd triangle centres?

5. Same question as 4., but taking the defining point to be other of the more familiar triangle centres, besides the centroid.

6. If we apply Conway's extraversion to the problem, are there any selections of six excentres which lie on a conic?

I must be getting old, because any attempt at finding a proof bogs down in a morass between Euclidean and projective geometry. Or, just possibly, it isn't true?!?

Happy new academic year to all. R.

...

Sometimes the brain has a mind of its own.

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On 9/13/11, Fred lunnon <fred.lunnon@gmail.com> wrote:

Agreed; I speculate however that the 6 subsidiary incentres lie on a conic when the fourth point is the incentre, and not otherwise.

Presumably somebody has already looked at the 6 subsidiary centroids when the fourth point is the centroid, etc. --- maybe I should wait for RKG's report.

WFL

Or perhaps somebody hasn't ... let S be the fourth point within the original triangle T, the Cevian lines through S partitioning T into six subsidiary triangles. It looks to me as if not only (1) When S is the incentre of T, then the incentres of the six sub-triangles lie on a conic; but also (2) When S is the centroid of T, then the centroids of the six sub-triangles lie on a conic; suggesting analogous results lurking for other special points of T. Possibly these are all special cases of a result in which a general point S of T is put in correspondence with an "analogous" points of arbitrary triangles (including the six). Finally, why should its proposer have thought the "mixed" version mentioned by RKG would work at all? I'm perpetually amazed at my ability to select "random" geometric configurations apparently supporting some pet conjecture, despite the latter eventually proving to be utter wombat's do's. This canard is no exception. If T happens to be right-angled and S is its centroid, then the six incentres do apparently lie on a conic after all! Fred Lunnon

On 9/14/11, Eugene Salamin <gene_salamin@yahoo.com> wrote:

Fred Lunnon: ... let S be the fourth point within the original triangle T, the Cevian lines through S partitioning T into six subsidiary triangles. It looks to me as if not only

(1) When S is the incentre of T, then the incentres of the six sub-triangles lie on a conic;

but also

(2) When S is the centroid of T, then the centroids of the six sub-triangles lie on a conic. ----------

The proposition (2) is invariant under affine transformations. There exists an affine transformation that takes T into an equilateral triangle, in which case, the six centroids lie on a circle.

-- Gene

Neatly puncturing an earlier claim of mine that affine geometry was no use! But while this also gives a way to define corresponding Cevian points in distinct triangles, and preserves conics, of course it fails to preserve incentres and other special points. In the plane, the only transformations which preserve all lines are projective transformations; Moebius transformations preserve angle, and their intersection with the projectivities is just the Euclidean similarities. I don't actually know that other conformal plane transformations --- such as z -> exp(z) --- cannot ever be projectivities, but it looks plausible. However I've often wondered whether there exist more subtle plane transformations which preserve (say) some lines but not all, together with some angles but not all, which might attack the in-centre case in the same manner as Gene's one-liner for the centroid case? Fred Lunnon