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Iterations in Geometry, an exampleLet there be given a triangle ABC and a point P. First, point P moves towards A. But half way from its original position to A it makes a turn and continues towards B. However, half way from B, it turns towards C. Half way towards C, it turns towards A, and so on. Verify and try to prove that point P eventually settles into cycling between three successive positions irrespective of its initial position. These three positions form a triangle whose area is 1/7 that of the area of triangle ABC. (Click anywhere in the applet to start iterations.)
From its initial position point P moves to (P + A)/2. From here on the next step, it moves half distance towards B. It's new position is at We may now consider a function f of a point P in a plane defined by This function defines an iterative process: Pn+1 - Pn = f(Pn) - f(Pn-1) = (Pn + D)/8 - (Pn-1 + D)/8 = (Pn - Pn-1)/8. In other words, the distance between Pn+1 and Pn is 1/8 the distance between Pn and Pn-1. The latter is of course 1/8 the previous distance: Pn - Pn-1 = (Pn-1 - Pn-2)/8. Which means that Pn+1 - Pn = (Pn-1 - Pn-2)/82. Continuing in this manner we obtain that Pn+1 - Pn = (P1 - P0)/8n. Remembering the definitions of P0 and P1, this is the same as Pn+1 - Pn = (D - 7P)/8n+1. Write a sequence of similar identities for finitely many indices n and sum them up (let Pm - Pm-1 = (D - 7P)/8m Summing all of these up, we get Pm - Pn = (D - 7P)(1 + 2-1 + ... + 2-(m-n))·2-n Since the sum in parentheses is less then 2, we may claim that {Pn} is a Cauchy sequence. Since the plane along with the real line are complete metric spaces, the sequence {Pn} has a limit which we denote Q. From Pn+1 = f(Pn) = (Pn + D)/8 we obtain Q = (Q + D)/8, or Q = D/7 = (A + 2B + 4C)/7. Other vertices of the limiting triangle are found by cycling the symbols in that expression: There are several ways to prove that that triangle has the area 1/7 that of triangle ABC. The simplest is probably to recognize a construction we discussed on another occasion. One can also use the barycentric coordinates to identify the intersection A' of AD and BC. A' divides BC in a 2:1 ratio, which naturally leads to the same result. The above problem admits a two-fold generalization.
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