Stuart Anderson

Altitudes of a triangle concur at the orthocenter. Here is a proof involving 3 dimensional geometry. It applies to acute triangles only, and (as far as I can see) does not generalize to the case of obtuse triangles. The proof proceeds by building concurrency into a figure, and then demonstrating that the concurrent lines are altitudes.

Let triangle ABC be obtained by slicing a cube corner with a plane, and define G to be the foot of O on the plane ABC, so that OG \perp ABC. Consider the plane OAG: since OA lies on the x axis, while BC lies in the yz plane, OA\perp BC, and since OG\perp BC by construction, therefore the whole plane satisfies OAG\perp BC. Similarly, OBG\perp CA and OCG\perp AB. These planes intersect the plane ABC in the three lines AD, BE, CF, which are concurrent by construction, and perpendicular to the opposite sides of triangle ABC because they lie in planes that are perpendicular to those sides. This completes the proof.

Now we had assumed that triangle ABC was obtained by slicing a corner from a cube. Let us now show that an arbitrary acute triangle can be so obtained. Let the origin of coordinates O be at the cube corner, with the x, y, z axes running along the cube edges. Then as in the figure, three right triangles are formed, AOB, \quad BOC, \quad COA. Applying the Pythagorean theorem to each of these gives x^2 + y^2 = c^2, y^2 + z^2 = a^2, z^2 + x^2 = b^2. Adding the first two and subtracting the third yields y^2 = c^2 + a^2 - b^2, which can be solved whenever the right side is positive. This happens only for acute triangles. Similarly, one can solve for x and z. Thus an arbitrary acute triangle can be realized as in the figure.