## Orthocentricity

Last week while in Tel Aviv I had an interesting conversation over lunch with Leonid Polterovich and Yaron Ostrover. I happened to mention the following gem from the remarkable book A=B by Wilf-Zeilberger. The book contains the following Theorem and “proof”:

Theorem 1.4.2. For every triangle ABC, the angle bisectors intersect at one point

Proof. Verify this for the 64 triangles for which the angle at A and B are one of 10, 20, 30, $\cdots$, 80. Since the theorem is true in these cases it is always true.

We are asked the provocative question: is this proof acceptable? The philosophy of the W-Z method is illustrated by pointing out that this proof is acceptable if one adds for clarity the remark that the coordinates of the intersections of the pairs of angle bisectors are rational functions of degree at most 7 in the tangents of A/2 and B/2; hence if they agree at 64 points they agree everywhere.

Leonid countered with a personal anecdote. Recall that an altitude in a triangle is a line through one vertex which is perpendicular to the opposite edge. Leonid related that one day his geometry class (I forget the precise context) were given the problem of showing that the altitudes in a hyperbolic triangle (i.e. a triangle in the hyperbolic plane) meet at a single point — the orthocenter of the triangle. After the class had struggled with this for some time, the professor laconically informed them that the result obviously followed immediately from the corresponding fact for Euclidean triangles “by analytic continuation”. Philosophically speaking, this is not too far from the W-Z example, although the details are slightly more shaky — in particular, the class of Euclidean triangles are not Zariski dense in the class of triangles in constant curvature spaces, so a little more remains to be done.

Actually, one might even go back and rethink the W-Z example — how exactly are we to verify that the angular bisectors intersect at a point for the triangles in question without doing a calculation no less complicated that the general case? Let’s raise the stakes further. After some thought, we see that not only will the intersections of pairs of angle bisectors be given by rational functions of the tangents of A/2 and B/2, but the (algebraic) heights of the coefficients of these rational functions can be easily estimated, and one can therefore compute an effective lower bound on how far apart the intersections of the angle bisectors would be if they were not equal. We can then literally draw the triangles on a piece of physical paper using a protractor, and verify by eyesight that the angle bisectors appear to coincide to within the necessary accuracy. After rigorously estimating the experimental errors, we can write qed.

While I am off on a tangent, this reminds me of a discussion I once had with Michael Aschbacher about the status of arguments (in topology, say) using diagrams. This could be a computation of the fundamental group of a knot complement by Wirtinger’s algorithm, for example, or a proof that some topological 4-sphere is smoothly standard via Kirby moves. He took what I think is an extreme view, that such arguments are never mathematically valid. This is a bit of a fuzzy argument to have if one is not careful to define precisely what one means by a “diagram” — suppose (as is in fact the case) I draw a diagram by writing a (finite) .eps file in ASCII. Then a “diagram” can be taken to be a certain kind of string in a finite alphabet, and the kinds of reasoning about diagrams one is prepared to accept could also be precisely specified and formalized, and could presumably be shown to be consistent with ZFC. This shows (in some very weak sense) that it is possible to conceive of a theory of “reasoning by diagrams” which must be respectable even to Michael Aschbacher. However, in practice one “reasons using diagrams” (just as one reasons in every other context) by a combination of explicit formal rules and pre logical “leaps”: if I extend this line indefinitely, it will intersect that line here; or, if I pick up this strand and pull it behind the other strand, it will eliminate these three crossings and introduce a new crossing here. And so on. If one pursues this line of reasoning too far it starts to degenerate into questions about the reliability of short term memory, or the psychophysics of perception, which throw any kind of mathematical reasoning in question. But before reaching that point, one can argue (and Aschbacher did argue) that arguments involving diagrams are “special” because of the sheer quantity and sophistication of the pre logical leaps involved. Anyone who has seen how much effort is involved in translating e.g. the Jordan curve theorem into a formal proof system like HOL light might be prepared to concede that Aschbacher has a point.

Anyway, back to hyperbolic orthocenters. If one substitutes spherical for hyperbolic geometry, there is quite a cute proof of the existence of an orthocenter as follows. Let’s fix the unit sphere in 3-space, and let ABC be a Euclidean triangle in a plane $\pi$ tangent to the unit sphere and touching it exactly at the orthocenter O of ABC. Radial projection of the vertices determines a spherical triangle A’B’C’. I claim that the radial projection of the altitudes of ABC become altitudes of A’B’C’, and therefore these altitudes intersect in O, which turns out also to be the (spherical) orthocenter of the (spherical) triangle A’B’C’. To see the validity of the claim, observe first that the radial projection of a straight line in $\pi$ to the sphere is a great circle on the sphere; so if L is any straight line in $\pi$ through O, the radial projection L’ is a great circle through O. Second, note that if M is a straight line in $\pi$ perpendicular to L (as above), the radial projection M’ is a great circle perpendicular to L’; this follows by symmetry: reflection in the plane through the origin containing L takes M to itself and therefore M’ to itself, while fixing L’ pointwise. This proves the claim, and therefore that the (spherical) altitudes of A’B’C’ intersect at O’. By a dimension count, all spherical triangles arise in this way; qed. At this point the appeal to analytic continuation (from spherical to hyperbolic geometry) is more persuasive.

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### 12 Responses to Orthocentricity

1. Ian Agol says:

There is a degenerate spherical case of an all-right spherical triangle, which has the property that there are infinitely many altitudes, and different choices might not simultaneously intersect. You might add your answer here: http://mathoverflow.net/questions/101776/altitudes-of-a-triangle/101901#101901

• Danny Calegari says:

Hi Ian – thanks for pointing out the degenerate example. I looked at the mathoverflow question you pointed to, but it looks like the question was already adequately answered by someone (anyway, I understand that by Bousfield localization all morphisms can be inverted, so your linking this blog post to the mathoverflow question is good enough).

2. eppstein says:

Have you seen Hong’s paper “Proving by example and gap theorems” (FOCS 1986)? It gives a general method by which one can rigorously verify the truth of geometry propositions like 1.4.2 by constructing a well-chosen example to sufficient accuracy.

As for why the angle bisectors coincide: rather than calculating, I prefer to observe that the Voronoi diagram of the triangle’s edges must have a point where its three regions meet. But I don’t know of an analogous explanation for orthocenters.

3. jclf says:

That W-Z proof is dreadful, from so many perspectives: it’s at once ugly, impractical, unstable (relying on perfect construction 64 times), and lazy (all that work is “left to the reader”); it provides no insight (e.g. the point is the incenter, and also defines an inscribed circle); it relies on a relationship with mathematics that is inaccessible to the young mathematical mind; it relies purely on construction and (dubious, unreliable) foreknowledge, and exercises no thinking.

On the other hand, in the classic geometric proof, it’s sufficient to draw only one example triangle, it’s not error prone because it doesn’t require an accurate construction, it uses simple principles available to an unsophisticated mathematical mind, and the existence of the incircle about the incenter is a trivial deduction. Also requiring proof, certainly: but much cheaper in terms of effort than the W-Z horror.

But acceptable? Only if it’s used to illustrate the difference between ugly and elegant. I would only use it comparatively, to show what can be lost or gained; otherwise, not at all.

• Danny Calegari says:

Dear jclf – let me recommend to you Thurston’s well-known article “On proof and progress in mathematics”, available on the arXiv as http://arxiv.org/pdf/math/9404236v1.pdf

Thurston touches on many different issues in his article. One of them is the question of what a proof in mathematics is supposed to accomplish. Thurston makes the point that one of the functions of a proof is to achieve insight, and different kinds of proofs or ways of thinking about mathematical objects can convey different *kinds* of insights. I am sorry that the W-Z proof does not communicate an insight about mathematics to you different in kind from that communicated by the classic geometric proof; that’s too bad. Perhaps if you find it so unpleasant to contemplate you should just ignore it?

4. Ben Golub says:

Thanks for this thought-provoking post! Just a quick question about 64 points: don’t we need agreement on an 8-by-8 grid, X \times Y, rather than an arbitrary 64 points, to conclude “they agree everywhere”?

(Of course, in this case, with the family of triangles considered, we do have a grid.)

• Danny Calegari says:

Hi Ben – yes absolutely, the points need to be “generic” enough. For instance, you wouldn’t want them to all lie on the zero set of a low degree polynomial P, or else they wouldn’t be able to distinguish rational functions which differ by a multiple of P.

• Ben Golub says:

Thanks for clarifying!

5. Sang-hyun Kim says:

It is a very interesting post. I am also intrigued by that several people independently mused about orthocenters recently.

Genevieve Walsh and I found yet another proof of the existence of the orthocenter for an acute spherical triangle, purely as an accidental corollary in our paper (Corollary 4.3). At the time of our writing we did not know of this post or the mathoverflow link, but you might notice the similarity. As the proof uses more hyperbolic geometry, let me describe it here.

By a hyperbolic reflection cube, we mean a hyperbolic polyhedron P such that (i) P is combinatorially a cube and (ii) P is invariant under a $Z_2\times Z_2\times Z_2$ action (through three “mid-planes”). In particular, the eight links of P are isometric acute spherical triangles. It is a useful fact that every acute spherical triangle arises as the link of a unique hyperbolic reflection cube. The claim is, the four diagonals of P intersect the eight links at their orthocenters; in particular, orthocenters do exist.

One can see this by the symmetry of P. The hyperplane containing an opposite pair of edges of P intersect two faces of P perpendicularly. There are six such hyperplanes and these hyperplanes intersect the link triangles at the altitudes.

Bottema describes a proof of the existence of the orthocenter without using the parallel postulate (due to Gudermann, 1835). So there should be a uniform proof for elliptic and Euclidean geometry. In the same book (Topics in Elementary Geometry) he says “additional thought is required” for a hyperbolic case.

– Sam

• Danny Calegari says:

Very nice!

6. Justin says:

Why does the statement “the coordinates … are rational functions of degree at most 7” imply that 64 points are needed to show agreement?

First, where does the degree 7 come from?
Second, if I wanted to see if ABx (the x-coordinate of A intersect B) equals ACx (the x-coordinate of A intersect C, where ABx = P/Q (a rational function of degree at most 7) and ACx = R/S ( a rational function of degree at most 7), then I can check if P*S = R*Q, where each side of the equation is polynomial of at most degree 14. So why is it not 15 points (degree of polynomial + 1)?