Problem: Consider a circle $C$ centered at a point $O$, and the points $A$ and $B$ outside of it, such that $O$, $A$, $B$ are colinear; see the diagram above. Find the point $P$ on the circle so that the angle $APB$ is as large as possible.
This problem is posed as a question in planetary astronomy in Heinrich Dörrie, 100 Great Problems of Elementary Mathematics, Dover, 1965, p. 370: At what latitude circle of Saturn does the ring appear widest?
In the diagram above, the yellow disk represents Saturn, and the horizontal line represents its equatorial plane. The ring's cross-section extends from $A$ to $B$. The angle $AOP$ is the latitude of the point $P$. The angle $APB$ is the width of the ring section $AB$ as seen from $P$.
Dörrie writes $\phi$ for the latitude angle $AOP$, and $\psi$ for the view angle $APB$. Applying elementary trigonometry, he shows that the optimal $P$ we have: \[ \cos \phi = \frac{(b+a)r}{ab + r^2}, \qquad \sin \psi = \frac{(b-a)r}{ab - r^2}, \] where $r$ is the circle's radius, and $a$ and $b$ are the distances of the points $A$ and $B$ from the circle's center.
Here I will give a straightedge-and-compass solution to the problem.
The main observation is—and
Dörrie notes this too—
that this problem is closely related to the
the maximal angle problem.
With the same reasoning as there, we conclude that
the point $P$ is obtained by passing a circle through the points
$A$ and $B$ and making it tangent to the circle $C$, as shown
in the diagram below. The point of tangency, $P$,
is the optimal viewing position.
Thus, the original question is reduced to the problem of construction of a circle through the points $A$ and $B$ which is tangent to the circle $C$. The straightedge-and-compass construction is effected through circle inversion. See Inversion for a quick introduction to the concept.
We begin by constructing a circle (shown in green in the diagram below) centered at $B$ and radius $AB$. This will be our inversion circle. The inverse of the yellow disk is shown as a pink disk. The inverse of the gray circle is the straight line $AD$ because the gray circle goes through the center of inversion. Since inversion preserves tangency, and since the yellow and gray circles are tangent, then their inverses will be tangent as well. Therefore, the line $AD$ is tangent to the pink disk. This furnishes the necessary clues for the construction, which we summarize as the following steps:
The point of tangency, $P$, is the optimal position for viewing the line segment $AB$.
Remark: Step 4 of the construction is not really necessary for locating the point $P$. It suffices to note that $P$ and $P'$ are inverses of each other relative to the inversion circle.
This page was created on June 25, 2010 and was last updated in February 2020.