Please show the problem solving process, thank you. If you write on paper, pleas
ID: 3167448 • Letter: P
Question
Please show the problem solving process, thank you. If you write on paper, please write it clean and clear(Sometime some answer were hardly read, so just remind).
Suppose f and g are rigid motions which agree at two distinct points P and Q, i.e., f(P) = g(P) and f(Q)-g(Q). Finish the proof that f and g must agree at every point on the infinite straight line defined by P and Q. We already showed that f and g agree on any point on the straight line segment PO Thus you only need to handle the case when R is a point on line which is not in between P and Q WLOG you may assume that is on the straight line segment PR.Explanation / Answer
f and g are rigid motions, so these preserve distances.
So for any two points A and B, |A-B| = |f(A)-f(B)| = |g(A) - g(B)|. Here |A-B| denotes the distance of A and B.
Now without loss of generality, we assume Q is on the line segement PR. We need to show f(R) = g(R)
Suppose this is not true. We already know that f(P)= g(P) and f(Q)=g(Q)
Now note that considering the distance of PR, we have
|P-R| = |f(P) - f(R)| = |g(P) - g(R)|.
So, |f(P) - f(R)| = |g(P) - g(R)|
So we have either, f(P) - f(R) = g(P) - g(R), or f(P) - f(R) = -(g(P) - g(R))
The first one implies f(R)=g(R) since f(P)=g(P).
So the first one can't be true, since by assumption f(R) is not equal to g(R). So the second equality holds.
So, f(P) - f(R) = -(g(P) - g(R))
This implies (since f(P)=g(P) )
2f(P) = f(R)+g(R)
By a similar argument, considering the points Q and R, we will get
2f(Q) = f(R)+g(R)
Adding these two, we get
f(P) + f(Q) = f(R) + g(R)
So, f(P)-f(R) = g(R) - g(Q)
Taking absolute values we get
|f(P)-f(R)| = |g(Q)-g(R)|
So, |P-R| = |Q-R|
But this imples the distance between P and R is the same as distance between Q and R. This cannot be true as P,Q and R lie on the same line and P and Q are two distinct points.
So we have a contradiction and hence f(R)= g(R).