The three altitudes of a triangle are concurrent -- this is true in all three constant curvature geometries (Euclidean, hyperbolic, spherical), but, as far as I know, the proofs are different in the three cases. Is there a "uniform" proof?
$\begingroup$
$\endgroup$
4
asked Jul 9, 2012 at 14:04
-
1$\begingroup$ I don't think the altitudes of a triangle in the hyperbolic plane necessarily intersect. $\endgroup$– Allan EdmondsCommented Jul 9, 2012 at 14:32
-
3$\begingroup$ They always form a pencil, soothe intersection can be beyond infinity, in principle... $\endgroup$– Igor RivinCommented Jul 9, 2012 at 16:01
-
$\begingroup$ What do you mean by a pencil? $\endgroup$– Deane YangCommented Jul 9, 2012 at 21:43
-
1$\begingroup$ Pencil is a set of line going through the same point in the extended plane, so the intersection can be ideal (which means the lines are all parallel) or hyper ideal, which means that the lines all have a common perpendicular. $\endgroup$– Igor RivinCommented Jul 9, 2012 at 21:45
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
The spherical and hyperbolic versions may be proved in a uniform way.
Consider the cross product $\times$ on $\mathbb{R^3}$ or on $\mathbb{R}^{2,1}$. If the vertices of the triangle are $a,b,c$ thought of as vectors in the unit sphere or hyperboloid, then the line through $a,b$ is perpendicular to $a\times b$, etc. The altitude of $c$ to $\overline{ab}$ is the line through $c$ and $a\times b$, which is perpendicular to $c\times (a\times b)$. The intersection of two altitudes is therefore perpendicular to $c\times (a\times b)$ and $a\times (b\times c)$, which is therefore parallel to $(c\times (a\times b))\times (a\times (b\times c)))$. But by the Jacobi identity, $a\times (b\times c) = -c\times (a\times b) -(b\times (c\times a))$, so this is parallel to $-(c\times (a\times b)) \times (b\times (c\times a))$, which is parallel to the intersection of two other altitudes, so the three altitudes intersect.
The Euclidean case is a limit of the spherical or hyperbolic cases by shrinking triangles down to zero diameter, so I think this gives a uniform proof.
Addendum: There are some degenerate spherical cases, when $a\times(b\times c)=0$. This happens when there are two right angles at the corners $b$ and $c$. In this case, two altitudes will be the interval $\overline{bc}$, and the other can be any geodesic going through $a$. If all three angles are right angles, then all three cross products are zero, and altitudes don't necessarily meet (although certainly there are triples of altitudes which intersect at any point on the sphere).
In the hyperbolic case, the orthocenter might lie outside of hyperbolic space, or at its boundary. See jc's links in the comments for a discussion.
answered Jul 10, 2012 at 23:41
-
3$\begingroup$ "by Jacobi": there's a very nice paper by V.I. Arnold on the hyperbolic case sciencedirect.com/science/article/pii/S0393044004001172 see also jstor.org/stable/10.4169/amer.math.monthly.118.01.041 $\endgroup$– j.c.Commented Jul 11, 2012 at 0:53
-
$\begingroup$ Thanks jc, I didn't know about these references. $\endgroup$– Ian AgolCommented Jul 11, 2012 at 2:39
-
$\begingroup$
$\endgroup$
5
I think this (and other similar facts) can be derived uniformly using elementary analyticity arguments. First you prove it for the round unit sphere. by rescaling this implies that it's true for the round sphere of any radius. now look at the cosine law in the simply connected space form of constant curvature $k$. since this formula is analytic in $k$, the "size" of the potential failure of the altitudes to intersect at the same point (measured in any reasonable way) will also be analytic in $k$ and since it's constantly zero for $k>0$ it must be constantly zero for all $k$. It should not be hard to make the above into a rigorous argument.
answered Jul 11, 2012 at 1:22
-
$\begingroup$ Well, such an argument might work, given care, but I am not sure it is enlightening (the analyticity thing is what someone we both know has described as "proof by proof"). Also notice that there is some possible failure of analyticity in the hyperbolic case, since you might have to divide by zero at some point, so, again, one must pick one's coordinates carefully if analyticity have any hope of existing... $\endgroup$ Commented Jul 11, 2012 at 15:24
-
$\begingroup$ @Igor, Yes, I realize that this is not quite a proof yet in the hyperbolic case (I didn't try to make it precise anyway) but I don't see any issues making it work there too. you can first prove it for triangles where the altitudes actually do intersect in the space itself and no division by zero is needed and then (again using analyticity in, say Poincare model) extend it to the general case. $\endgroup$ Commented Jul 11, 2012 at 15:47
-
3$\begingroup$ Nothing wrong with proof-theoretic arguments (or "proof by proof", if you wish), in my opinion :-). One could also use algebraic geometry instead of analytic continuation as the underlying proof theory; once one complexifies, spherical and hyperbolic geometry are equivalent as algebraic geometries (Wick rotation). The space of real spherical triangles is Zariski dense in this complexification, so any algebraic geometry statement which holds for those triangles, in fact holds for all complex spherical triangles, and in particular for real hyperbolic triangles. $\endgroup$ Commented Jul 11, 2012 at 16:03
-
1$\begingroup$ ... admittedly, there is the issue that sometimes, the point of intersection will only lie in the complexification (and projectivisation) of the original geometry, but this problem was already noted earlier. $\endgroup$ Commented Jul 11, 2012 at 16:11
-
1$\begingroup$ ... and "all" triangles should be replaced with a "Zariski open set of triangles", since one also has to avoid the degenerate cases. Sorry! $\endgroup$ Commented Jul 11, 2012 at 16:15