Mathematics

Mathematics

Problem

If GG is the centroid of the △ABC\triangle ABC and if G′G' is the centroid of another △A′B′C′\triangle A'B'C', then prove that AA′+BB′+CC′=3GG′AA' + BB' + CC' = 3GG'.
Intuitive explanation

The key insight is that the centroid is the average position of the three vertices—so the displacement of the centroid depends only on the average displacement of the vertices. Express both centroids using position vectors (G=A⃗+B⃗+C⃗3G = \frac{\vec{A} + \vec{B} + \vec{C}}{3} and similarly for G′G'), then compute GG′⃗\vec{GG'} and relate it to the individual displacements AA′⃗\vec{AA'}, BB′⃗\vec{BB'}, and CC′⃗\vec{CC'}. The algebra will reveal that these three displacement vectors sum to exactly three times the centroid displacement.

Intuitive explanation

The key insight is that the centroid is the average position of the three vertices—so the displacement of the centroid depends only on the average displacement of the vertices. Express both centroids using position vectors (G=A⃗+B⃗+C⃗3G = \frac{\vec{A} + \vec{B} + \vec{C}}{3} and similarly for G′G'), then compute GG′⃗\vec{GG'} and relate it to the individual displacements AA′⃗\vec{AA'}, BB′⃗\vec{BB'}, and CC′⃗\vec{CC'}. The algebra will reveal that these three displacement vectors sum to exactly three times the centroid displacement.

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