Iterations in Geometry, an example

Let 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.

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Explanation

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 [(P + A)/2 + B]/2 = (P + A + 2B)/4. After that it moves half way in the direction of C to [(P + A + 2B)/4 + C]/2 = (P + A + 2B + 4C)/8. From here the process starts again: half way towards A, half way towards B, half way towards C, and so on.

We may now consider a function f of a point P in a plane defined by f(P) = (P + A + 2B + 4C)/8, or f(P) = (P + D)/8, where D = (A + 2B + 4C)/8.

This function defines an iterative process: P_n+1 = f(P_n), n = 0, 1, 2, ..., where P₀ = P, the initial position of the given point. The distance between successive iterations shrinks by a factor of 8 in the following sense:

P_n+1 - P_n = f(P_n) - f(P_n-1) = (P_n + D)/8 - (P_n-1 + D)/8 = (P_n - P_n-1)/8.

In other words, the distance between P_n+1 and P_n is 1/8 the distance between P_n and P_n-1. The latter is of course 1/8 the previous distance:

P_n - P_n-1 = (P_n-1 - P_n-2)/8.

Which means that P_n+1 - P_n = (P_n-1 - P_n-2)/8². Continuing in this manner we obtain that

P_n+1 - P_n = (P₁ - P₀)/8ⁿ.

Remembering the definitions of P₀ and P₁, this is the same as

P_n+1 - P_n = (D - 7P)/8ⁿ⁺¹.

Write a sequence of similar identities for finitely many indices n and sum them up (let m > n):

P_m - P_m-1 = (D - 7P)/8^m
P_m-1 - P_m-2 = (D - 7P)/8^m-1
...
P_n+2 - P_n+1 = (D - 7P)/8ⁿ⁺²
P_n+1 - P_n = (D - 7P)/8ⁿ⁺¹

Summing all of these up, we get

P_m - P_n = (D - 7P)(1 + 2^-1 + ... + 2^-(m-n))·2^-n

Since the sum in parentheses is less then 2, we may claim that {P_n} is a Cauchy sequence. Since the plane along with the real line are complete metric spaces, the sequence {P_n} has a limit which we denote Q. From

P_n+1 = f(P_n) = (P_n + 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: (B + 2C + 4A)/7 and (C + 2A + 4B)/7.

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.

Limits in Geometry

Two Circles and a Limit
A Geometric Limit
Iterations in Geometry, an example
- Iterations in Geometry, a generalization
Iterated Function Systems

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