Let ABCDEF be a convex hexagon. Let A', B', C', D', E', F' be the centroids of triangles FAB, ABC, BCD, CDE, DEF, EFA, respectively. (a) Show that every pair of opposite sides in hexagon A'B'C'D'E'F' (namely A'B' and D'E' B'C' and E'F' and C'D' and F'A' are parallel and equal in length. (b) Show that triangles A'C'E' and B'D'F' have equal areas.

i know there are other solutions to this but i have read them and I don't understand any of them.

I've tried connecting the inner hexagon to some of the triangles created by the points to create some medians but it hasn't really worked and I don't know what to do.

If you could help me that would be great!

Guest Aug 5, 2023

#1**-3 **

(a) Let M be the midpoint of AB. Since A′ is the centroid of △FAB, the line through A′ and M is the median of △FAB, and therefore passes through B′. Similarly, the line through C′ and M passes through B′, and the line through D′ and M passes through C′. By SAS similarity, we have △A′MB′∼△BMC′, so A′B′=BC′. By the same argument, we can show that B′C′=CE′, C′D′=EF′, and D′E′=FA′.

(b) Let P be the intersection of the line through A′ and D′ and the line through C′ and E′. Since A′ and D′ are the centroids of △FAB and △DEF, respectively, the line through A′ and D′ is parallel to BC and the line through C′ and E′ is parallel to AB. Therefore, P is the midpoint of BC.

[asy] unitsize(1 cm);

pair A, B, C, D, E, F, M, P;

A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (2,2); F = (2,0); M = (2,1); P = (1,2);

draw(A--B--C--D--E--F--cycle); draw(A'--B'--C'--D'--E'--F'--cycle); draw(A'--D',dashed); draw(C'--E',dashed); draw(P--A',dashed); draw(P--C',dashed);

label("A", A, SW); label("B", B, SE); label("C", C, NE); label("D", D, NW); label("E", E, dir(0)); label("F", F, SW); label("A′", A', SW); label("B′", B', SE); label("C′", C', NE); label("D′", D', NW); label("E′", E', dir(0)); label("F′", F', SW); label("M", M, NW); label("P", P, SW); [/asy]

Since P is the midpoint of BC, we have AP=PB=BC/2. Also, A′B′=BC′=BC/2, so △A′PB∼△B′C′. Therefore, the areas of △A′PB and △B′C′ are equal.

Similarly, we can show that the areas of △A′C′E′ and △B′D′F′ are equal. Therefore, △A′C′E′ and △B′D′F′ have equal areas.

maximum Aug 5, 2023