Solution of Corrected Problem 3.1

Corrected Problem 3.1

There was an error in the original formulation of this problem. You will be given points for any reasonable answer, so you do not need to worry on that score. However, it is important that everyone get the mathematics out of this problem that was intended, so here is a correct statement of the problem with a solution.

3.1 Point Symmetry on a Line (10 points)

Definition: Given two points A and P, the point reflection of P in A is the point P' such that A is the midpoint of P and P' (in the special case that A = P, we set P' = P). Thus point reflection with center A can be thought of as a function taking point P to point P'.

Problem: Given 3 points A, B and P. Let P' be the point reflection of P in A and let P'' be the point reflection of P' in B. Now suppose that A, B and P are all on a line m with a given ruler. (a) If the ruler numbers of A, B and P are a, b, x, find the ruler numbers y of P' and z of P''. In other words, find a formula for P' and P'' in terms of a, b, x. (b) Compute the distance from P to P''. How does this distance change when P changes? How does this distance depend on the distance from A to B?

Solution:

It was proved in class that if F and G are points on a line with ruler numbers f and g, then the midpoint M of FG has ruler number (f+g)/2. We use this to find y and z.

First, P' is the reflection of P in A, if A is the midpoint of PP'. From the formula, this means that a = (x+y)/2, or y = 2a - x.

For the same reason, if P'' is the reflection of P' in B, then z = 2b - y.

Substituting for z in terms of x, we have z = 2b - (2a - x) = 2(b - a) + x.

From the ruler axiom, the distance PP'' = |z - x| = 2|b-a| = 2AB. This is twice the distance AB and does not change as P changes.

Comment:

Actually, we can say a little more than the distance statement. To get the coordinate (ruler number) z from x, we add the constant 2(b-a). This means that we shift the whole line the distance 2|b-a| in the direction from A to B.