Let AA', BB', and CC' be the altitudes of
acute triangle ABC.

Prove that the feet of the perpendiculars
from A' to AB, BB', CC', and AC are collinear.

Generalisation of the problem.

1.Let P1, P2, P3, denote similar convex polygons of successively smaller size, nested together at a common vertex, V.
(a)  Now P1,P2,P3 share a common diagonal VV1, the vertices V2, V3 lying somewhere along VV1. Denote some other diagonal of P1 as D1; trivially, the corresponding diagonals D2 of P2 and D3 of P3 will be parallel to this.
(b) For convenience, denote the vertex of P3 at V, X.
2. Let VV2 +VV3 = VV1.
3. Now slide P3 along VV1 until V3 lies at V2, then rotate P3 through 180 degrees about  V2: D3 remains parallel to D1, D2 after this rotation; it follows, from 2, that X has now been translated to V1; and the diagonal V3X remains on VV1.
4. If the whole lengths VV2 +VV3 = VV1, then the partial lengths between the intersections: (D2, VV2) and (D3, V3X) sum to (D1, VV1), hence the diagonal D3 is not only parallel to, but lies on, D1.

Construction.

5. Construct lines AB'AC' and C'B'. The altitudes of an acute triangle are coincident within the triangle. Denote their intersection as O.

Proof

6. Consider the polygons A-C'-O-B'; A-AB'-A'-AC'; and O-BB'-A'-CC':
(a) These polygons are clearly similar;
(b) AO and OA' lie on AA'; and
(c) AO+OA'= AA'.
7. It follows at once from 1. to 5. above that AB', BB', CC' and AC' are collinear.

For good measure, the same proof also holds if ABC is obtuse, while the RHT case is straightforward.

Proof if ABC is obtuse.

8. If ABC is obtuse, O falls outside ABC. Without loss of generality, assume ABC is the obtuse angle.
9. Construct the triangle AOB. The point A' now lies at the intersection of the extension of CB and AO. The perpendiculars corresponding to those in the problem are from A' to AC, A' to AB, A' to BO, and A' to CO.
10. This construction is now exactly the same as for the acute case, except that points B and O have exchanged with each other. The same proof follows immediately.

I would like to add that I thought this was an excellent problem; thanks Bractals!

 

Edited on September 26, 2010, 12:20 pm

Posted by broll on 2010-09-26 12:07:58