Shortest Path between Two Points in the Unit Disk Reflecting off the Circumference

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Given two points within the unit disk labeled (in green) and (in red), what is the shortest path between the two points that reflects off the unit circle? You can drag the two points.

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The central angle between the two points is . The angle from the axis to is . The shortest path that touches the unit circle is at angle .

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Contributed by: Jingang Shi and Aaron T. Becker (September 2017)
(University of Houston)
Open content licensed under CC BY-NC-SA


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Details

The shortest path between two points in the unit disk that reflects off the circumference is composed of two straight line segments. The problem can be simplified by choosing the coordinate system carefully. We define the axis along the position of the starting point: , and define the point of intersection by the angle from the axis , and the final point by a radius and angle , . Then define a symmetry point about of line named .

Then the length of the two line segments is

,

which is minimized by choosing an appropriate value. This equation can be simplified to

.

The length of the two line segments as a function of is drawn in the right plot.

There are several simple solutions. If is 1 or is 0 or is 0, the optimal angle is 0. If is 1 or is 0, the optimal angle is .

Label the origin . The optimal solution shows that the angle (from the origin to to ) is the same as the angle (from the origin to to ).

We name these angles . This can be proved by drawing an ellipse whose foci are and . When the ellipse is tangent to the circle, the point of tangency is exactly .

Since the distance from the origin to is always 1, we can set up three equalities using the law of sines:

From triangle : .

From triangle : .

If we mirror the point about the axis and label this point , from triangle : .

Simplifying this system of equations results in: .

Solving this last equation results in a quartic solution that has a closed-form solution with four roots, each of which can be either a clockwise or a counterclockwise rotation , depending on the sign of , with . We evaluate each and select the solution that results in the shortest length path.

Note that the optimal path satisfies the law of reflection off the unit circle, with angle of incidence equal to angle of reflection.



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