Altitudes of a triangle - MathOverflow

Altitudes of a triangle

2012-07-09T14:04:42Z

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?

Answer by Agol for Altitudes of a triangle

2012-07-10T23:41:48Z

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.

Answer by Vitali Kapovitch for Altitudes of a triangle

2012-07-11T01:22:58Z

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.