Wednesday, April 23, 2014

Geometry Problem 1007: Hexagon, Midpoint, Triangle, Median, Centroid, Concurrency

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1007.

Online Math: Geometry Problem 1007: Hexagon, Midpoint, Triangle, Median, Centroid, Concurrency


  1. Using mass point approach.

    Let all points in the figure have mass 1.


    Let O₁ and O₂ be the centroid of ΔM₁M₃M₅ and ΔM₂M₄M₆, thus
    3O₁=M₁+M₃+M₅, i.e. 6O₁=A+B+C+D+E+F
    3O₂=M₂+M₄+M₆, i.e. 6O₂=A+B+C+D+E+F

    Hence, O₁=O₂=O, which is the intersection of six medians.

  2. Let V(X) represent position vector of point X
    We have V(M1)= ½(V(A)+V(B))
    Similarly we also have V(M2) and V(M3)
    Since O is the centroid of triangle M1M3M5
    So V(O)=1/3(V(M1)+V(M3)+V(M5))= 1/6 ( V(A)+ V(B)+ V(C)+ V(D)+ V(C)+ V(F))
    Similarly we have V(M2), V(M4) and V(M6)
    And vector of centroid of triangle M2M4M6 will be the same as V(O)

  3. I really don’t like the notation in this question, so bear with me as I change the names of the points.
    Question (reworded): There is hexagon ABCDEF. L is midpoint of line AF, G is midpoint of line BA, H is midpoint of line CB, I is midpoint of line DC, J is midpoint of line ED, K is midpoint of line FE. Do the medians of triangle LHJ concur with medians of Triangle GIK?
    We use Barycentric Coordinates.
    Definition: Capital letters will denote vectors. P = x L + y H + z J will be written as P = (x, y, z). By definition, x + y + z = 1. AB means “length of AB” unless otherwise specified; most importantly, it does NOT mean “A times B”. If Z is midpoint of X, Y, then Z = (1/2)(X + Y).
    The intersection of medians is the centroid. That the centroid exists is trivially proven with Ceva’s theorem. The centroid of Triangle LHJ is (1/3, 1/3, 1/3); this is trivial to prove, but for those who want a proof: the equation of the median from L is y = z as this linear equation satisfies (1, 0, 0) and (0, ½, ½) which is midpoint of H, J. Similarly, the equation of the median from H is x = z. Given that x + y + z = 1, it’s obvious that the intersection occurs at (1/3, 1/3, 1/3).
    In other words, the centroid of LHJ is the average of L, H, J. Similarly, the centroid of Triangle GIK is (1/3)(G + I + K).
    As G, I, K are midpoints of line AB, line CD and line EF, it’s obvious that the centroid of Triangle GIK is (1/6)(A + B + C + D + E + F), which, when rearranged, makes (1/3)([A + F]/2 + [E + D]/2 + [C + B]/2). It’s known that L, K, J are midpoints of line AF, line BC and line DE. Hence, (1/3)([A + F]/2 + [E + D]/2 + [C + B]/2) = (1/3)(L + H + J) = (1/3, 1/3, 1/3).
    We have proved that the centroid of triangle LHJ is the same point as the centroid of triangle GIK. Hence, all the medians concur.