1957 AMC 12 Problem 26

Question

From a point within a triangle, line segments are drawn to the vertices. A necessary and sufficient condition that the three triangles thus formed have equal areas is that the point be:

the center of the inscribed circle

the center of the circumscribed circle

such that the three angles formed at the point each be

120^\circ

the intersection of the altitudes of the triangle

the intersection of the medians of the triangle

Answer: E

Concepts:centroid triangle area median (geometry)

Difficulty rating: 1340

Small hint:

Recall how the medians partition a triangle’s area

Big hint:

The intersection of the medians divides the triangle into six small triangles paired into three equal-area regions

Solution:

The three medians meet at the centroid and divide the triangle into six small triangles of equal area. Each of the three triangles having the centroid and one side of the original triangle consists of two of those small triangles, so their areas are equal. Conversely, equal areas give equal barycentric coordinates, which uniquely locate the centroid.

Thus, the necessary and sufficient point is the intersection of the medians, and the correct answer is E.