Aichi University of Education 2007: Sum of Line Segment Lengths
Problem
Fix point A(1, 0) on the coordinate plane, take point P on the line $${y=x+2}$$, and take point Q on the circle $${x^2+y^2=1}$$. At this time, find the minimum value of the sum of the line segment lengths AP+PQ, and the coordinates of points P and Q at that time.
This is a problem from Aichi University of Education in 2007. If you draw a diagram, it looks like this.
If 'Q were not a point on the circumference of a circle', it would be the common 'problem of going to the river to fetch water'.
In the following diagram, there is a house at point A and a field at B. This is a problem of finding the shortest path from the house, to the river (a straight line) to fetch water, and then to the field.
Assuming the field is on the opposite bank of the river (at a position B' symmetric with respect to the river), then PB=PB'. Since the path connecting point B' and A is a straight line, which is the shortest, you should fetch water at the intersection P with the river.
As shown in the diagram, it is the same thing if you take the symmetric point of the house instead of the symmetric point of the field.
An application of this problem is the Aichi University of Education problem. Such applications are also given as entrance exam problems involving a plane in space and two points.
Now, in the problem of going to the river to fetch water earlier, it was the same whether you took the symmetric point of A or the symmetric point of B.
Then, how about in this problem?
Let's consider both the case of taking the symmetric point of Q and the case of taking the symmetric point of A.
This is quite difficult. While A is a fixed point, Q is an arbitrary point on the circumference of the circle. If you take the symmetric point of Q, that point will also move every time you move Q, so it is difficult to think about. If it is the symmetric point of A, this point will not move even if Q moves, so you only need to consider A'Q.
The correct answer is $${\sqrt{5}-1}$$, which is approximately 1.236, but let's verify this while actually moving the points.
There is an interactively movable version at the following link.
The beginning is the problem diagram. Pressing the 'Auxiliary Diagram 1' button takes the symmetric point of Q, and pressing the 'Auxiliary Diagram 2' button takes the symmetric point of A. (The diagram shows the state after pressing 'Auxiliary Diagram 2')
Since the value of AP+PQ is displayed, you can move points Q and P to find the position where it is minimized.
※ The diagram is created with Cinderella (CindyJS).