Mean Geometry By Zachary Abel Averages on the Number-Line The average of 4 and 8 is 1 1 + = (4) (8) 6 2 2 A weighted average of 4 and 8, with weights of 3/4 and 1/4, is 3 1 + = (4) (8) 5 4 4 (the weights must add to 1) Averages of Points The average of two points A and B is the midpoint of segment AB, and is written as 1 1 + = A B C 2 2 The weighted average of two points A 3 1 and B, for example A + B , is the point 4 4 D that cuts segment AB into pieces that 3 1 are and of segment AB: 4 4 3 1 + = A B D 4 4 Averages of Figures To take the average of two figures A A A A and B B B B , simply 1 2 3 4 1 2 3 4 average each pair of corresponding points: 1 1 + = A A A A B B B B C C C C 1 2 3 4 1 2 3 4 1 2 3 4 2 2 Directly Similar Two figures are Similar if they have the same shape, i.e. one is a scaled version of the other. Two figures are Directly Similar if they also have the same orientation. The Fundamental Theorem of Directly Similar Figures The weighted average of two directly similar figures forms a figure that is directly similar to the first two. The average of two directly similar figures is another directly similar figure. Right-Isosceles Triangle On the outside of triangle ABC, erect two squares ABDE and ACFG, with centers at X and Y respectively. If M is the midpoint of side BC, prove that triangle XMY is a right-isosceles triangle. 1 1 What is DBA + ACF ? 2 2 1 1 + = D A X 2 2 1 1 + = B C M 1 1 2 2 + = A F Y 2 2 1 1 + = DBA ACF XMY 2 2 By the Fundamental Theorem, XMY is right-isosceles, as desired. Problem 2: Napoleon’s Theorem Napoleon Bonaparte If Equilateral Triangles BCP, CAQ, and ABR are erected externally on the sides of any triangle ABC, their centers X, Y, and Z form an equilateral triangle. Proof of Napoleon’s Theorem, Step 1 1 1 + = PCB BAR JKL 2 2