Problem 12007
(American Mathematical Monthly, Vol.124, December 2017) Proposed by K. Altintas (Turkey) and L. Giugiuc (Romania).
Let G be the centroid of triangle ABC, and let M be an interior point of ABC. Let D, E, and F be the centroids of subtriangles CM B, AM C, and BM A, respectively.
(a) Prove that the lines AD, BE, and CF are concurrent.
(b) Suppose that M 6= G and that P is the point of concurrency in part (a). Prove that G, P , and M are collinear, with P between G and M , and P M = 3P G.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We denote with A, B, and C the position vectors of the vertices of the given triangle ABC. Then the centroids G, D, E, and F satisfy the relations
G= A+ B + C
3 , D= C+ M + B
3 , E= A+ M + C
3 , F =B+ M + A
3 .
Let
P:= M+ 3G
4 =M + A + B + C
4 .
Since P can be obtained as a convex combination of A and D, of B and E, and of C and F , that is
P= 3D + A
4 =3E + B
4 =3F + A 4 , then the lines AD, BE, and CF are concurrent at P . This proves (a).
Moreover, since by definition P is a convex combination of M and G, then M , G, P are collinear and P lays between M and G. Finally
|P − M | =
M+ 3G
4 − M
=3 |G − M |
4 =
M + 3G
4 − G
= 3|P − G|
which proves (b).
