Round friends

[BMAD]

Suppose two cameras were installed x distance apart from each other in a hallway. They were placed directly across from each other in circular casings. The first camera is in a small casing that is about 3 times as small (1/3 the radius) as the bigger casing for the other camera which limits its viewing angle. The distance between the two cameras is such that the max they can see horzontally is the widest point on the other camera's casing. If you were to draw lines of the max viewing area from each camera to the casings of the other camera, the lines would intersect their own casings. Which camera would have the greater distance between its own intersections (intersection to intersection) at its casings.

[bonanova]

They are equal.
Join the centers and draw radii to the upper points of tangency.
The right triangles made by the half-chord and opposite radius are similar. The sines of the acute angles equate the ratios of the half-chords and corresponding radii with the opposite radii and the line joining the centers. Equating similar ratios for the two triangles leads quickly to the equality of the half-chords (and thus the chords themselves.)

[BMAD]

[BMAD]

We are trying to show which chord is bigger.