The medial triangle of a triangle ABC is the triangle whose vertices are located at the midpoints of the sides AB, AC, and BC of triangle ABC. From an arbitrary point O that is not a vertex of triangle ABC, you may take it as a given fact that the location of the centroid of triangle ABC is the vector (vector OA + vector OB + vector OC)/3.

Question

The medial triangle of a triangle ABC is the triangle whose vertices are located at the midpoints of the sides AB, AC, and BC of triangle ABC. From an arbitrary point O that is not a vertex of triangle ABC, you may take it as a given fact that the location of the centroid of triangle ABC is the vector (vector OA + vector OB + vector OC)/3.

Task:

A. Use vector techniques to prove that a triangle and its medial triangle have the same centroid, stating each step of the proof.

1. Provide written justification for each step of your proof.

Answer 1

To prove that the triangle ABC and its medial triangle have the same centroid using vector techniques, we can follow these steps:

Step 1: Denote the coordinates of points A, B, and C as vectors OA, OB, and OC, respectively.

Step 2: To find the coordinates of the midpoints of the sides AB, AC, and BC, we need to find the average of the vectors OA and OB, the average of the vectors OA and OC, and the average of the vectors OB and OC.

Let's denote the midpoint of AB as vector M_AB, midpoint of AC as vector M_AC, and midpoint of BC as vector M_BC.

M_AB = (OA + OB) / 2

M_AC = (OA + OC) / 2

M_BC = (OB + OC) / 2

Step 3: The centroid of triangle ABC is the average of its three vertices. Let's denote the centroid as vector G.

G = (OA + OB + OC) / 3

Step 4: Now, we need to prove that G is also the centroid of the medial triangle formed by the midpoints.

To prove this, we need to show that the sum of vectors GM_AB, GM_AC, and GM_BC is equal to the zero vector.

GM_AB = M_AB - G = [(OA + OB) / 2] - [(OA + OB + OC) / 3]
= (3(OA + OB) - 2(OA + OB + OC)) / 6
= (3OA + 3OB - 2OA - 2OB - 2OC) / 6
= (OA + OB - 2OC) / 6

Similarly, you can calculate GM_AC and GM_BC.

Step 5: Now, let's calculate the sum of these vectors: GM_AB + GM_AC + GM_BC.

GM_AB + GM_AC + GM_BC = (OA + OB - 2OC) / 6 + (OA + OC - 2OB) / 6 + (OB + OC - 2OA) / 6
= (2OA + 2OB + 2OC - 2OA - 2OB - 2OC) / 6
= (0) / 6
= 0

Since the sum of the vectors GM_AB, GM_AC, and GM_BC is equal to the zero vector, it implies that the centroid of the medial triangle is indeed vector G, which is also the centroid of triangle ABC.

Hence, we have proved that the triangle ABC and its medial triangle have the same centroid.