It is well known that, as regards moment of inertia about any line in its plane, a uniform triangular lamina may be replaced by a set of three particles, each with mass equal to one-third of the mass of the triangle, and placed at the midpoints of the sides. The usual proofs given for this elegant result are rather cumbersome. Now it takes only a few seconds to verify that the set of particles has the same moment of inertia as the triangle about each of the three sides (using the h2/6 formula for the square of the radius of gyration). It is tempting to ask whether the equivalence of particles and lamina may not be deducible from this. In other words, starting from the fact that the equivalence holds for three particular lines in the plane, can it be deduced for all lines in the plane?