Vol. 10 No 3 - Pi Mu Epsilon

that:a) Triangles ABC. PQR,and W7. have a commoncentroid, andb) If the areas of trianglesPQR, ABC, and XYZ are ingeometric progression, then k =,IT - 1.Solution by William H.Peirce, Defray Beach, Florida.Place the figure in the Figure 3. Problem 846complex plane and let thecomplex affix of each point be denoted by the corresponding lower caseletter. Then1 = (1 - k)b + kc, m = (1 - k)c + ka, and n = (1 - k)a + kb.Next, P lies on line AL, so for some real number A, we haveSince P lies also on line CN, there is a real constant p such thatNote that the denominator 1 - k + k1 is positive for all real k. Substitutingfor \ and p in either expression above gives the expression for p in termsof a, b, and c. Similarly, q and r are found, yieldingandr = (1 - k12c + k(l - k)a + k2 b1 - k+k 2We develop similar expressions for x, y, and z. Since XZ is parallel to BMand passes through C, there is a real constant \ such thatAlso, XY is parallel to CN and passes through A, so for some real p, wehavex = a + p(n - c) = (1 + p - kp)a + kpb - pc.Again we equate the coefficients of a, b, and of c in these two expressionsfor x to solve for \ and p. Then we substitute back into either'equation tofind an expression for x. Similarly, we find y and z. We get .Since the representation for p must be unique, we may equate thecoefficients of a, those of b, and those of c in the two expressions for p,obtaininga) The centroid of a triangle is the intersection of the medians and isequal to the average of its vertices. Thus the centroid G of triangle ABC isgiven bywhich we solve simultaneously to get\ =(1 - k)2and pk 2=1 - k+k 2 1 - k + k2.Similarly we average the affixes for triangles PQR and XYZ. Easy algebrashows each centroid coincides with point G. Furthermore, the centroid oftriangle LMN, too, is at G.b) The area of a triangle ABC, denoted by K(ABC), in the complex