By Melvin Hausner

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To designate positive numbers (the “masses”). A mass-point (mass m located at point P) will continue to be designated by mP. * Suppose we have k mass-points m1P1, . . , mkPk. We have assumed that they uniquely determine a new mass-point mP, where m = m1 + · · · + mk and where P is their center of mass. We shall write mP = m1P1 + · · · + mkPk Thus m1P1 + · · · + mkPk is a shorthand way of writing, “The mass-point obtained when all of the masses of m1P1, . . ” We have seen in the examples that the center of mass can be obtained by taking the centers of two points at a time, and repeating the operation.

13 Exercises 1. In Fig. 14, AP = 2PB and QC = 2PQ. Compare AR and RC. 15 2. In Fig. 15, find the ratios BP/PQ and AP/PR. 3. In Fig. 16, the letters a, b, c, d, and e represent actual distances. Express QO in terms of them. (Hint: First ignore e, and use only the ratios a/b and c/d. 16 4. Menelaus’s theorem deals with the ratios into which an arbitrary transversal divides the three sides of a triangle. In Fig. 17, A1, B1, C1 are the points of intersection of a transversal with BC, CA, AB, respectively.

Find AR/RC and BQ/QR. The method is to take Q as the center of mass of a suitable physical system. Give A mass 1 and B mass 5. Then P is the location of the center of mass of 1A and 5B (see Fig. 9). By giving C mass 4 we obtain Q as the center of mass of 6P and 4C. The previous examples point the way. By observation, AR/RC = 4/1 and BQ/QR = 5/5 = 1. 9 EXAMPLE 4 (Ceva’s Theorem). Let P be inside ABC. Suppose AP is extended until it meets BC at A1, and that B1 and C1 are found similarly. Then (see Fig.