Given a line $L$ and points $A$ and $B$ on one side of it, find a point $P$ on $L$ such that the angle $APB$ is as large as possible.
A special case of this,
where the lines $AB$ and $L$ are orthogonal, was
posed
as a puzzle,
in the geometry.puzzles newsgroup on October 18, 1993,
in the form of a rugby problem:
What is the best scoring position of the kicker $P$ on the line $L$
relative to the goal posts $A$ and $B$?
The puzzle is equivalent to the Regiomontani Problem: At what point on the earth's surface does a vertically suspended rod appear longest? (I.e., at what point is the visual angle at a maximum?)
According to Heinrich Dörrie, 100 Great Problems of Elementary Mathematics, Dover, 1965, p. 369:
“This problem was posed in 1471 by the mathematician Johannes Müller, called Regiomontanus after his birthplace Königsburg in Franconia, to the Erfurt professor Christian Roder. This problem, which itself is not difficult, deserves special attention as the first extreme problem encountered in the history of mathematics since the days of antiquity.”
I have removed the orthogonality assumption, thus generalizing the problem somewhat.
The solution in brief: Construct a circle that goes through the points $A$ and $B$ and is tangent to the line $L$. The point of tangency is the optimal $P$.
Upon closer examination, we find out that there are two such circles. The diagram below shows the two circles and their tangency points labeled $P$ and $P'$. As the point $R$ slides along the line $L$, the angle $ARB$ achieves local maxima at $P$ and $P'$. The smaller of the two circles corresponds to the absolute maximum.
The reason for the optimality of the points $P$ and $P'$ is due to the well-known fact of plane geometry that the locus of points $P$ (not necessarily on the line $L$) such that the angle $APB$ remains a constant is an arc of a circle through the points $A$ and $B$. To maximize the angle, we need to minimize the size of the circle. The smallest circle that goes through $A$ and $B$ and still shares a point with the line $L$ is the circle that is tangent to $L$.
As we have seen above, the solution of the problem reduces to constructing a circle that passes through the points $A$ and $B$ and is tangent to the line $L$. Here is a classical straightedge-and-compass solution to this problem.
As we will see below, a circle through $A$, $B$, $P$ will be tangent to the line $L$ at $P$. Similarly, a circle through $A$, $B$, $P'$ will be tangent to the line $L$ at $P'$.
Proof: Let $P$ is the desired point, that is, $P$ is on $L$ and the circle through $A$, $B$, $P$ is tangent to $L$ at $P$. From the elementary properties of lines and circles we know that $QP^2 = QB \cdot QA = QB' \cdot QA$. On the other hand, in the right triangle $AHB'$ we have $QB' \cdot QA = QH^2$. It follows that $QP = QH$. similarly, $QP' = QH$. Therefore $P$ and $P'$ lie on the circle centered at $Q$ passing through $H$. QED
This page was created on November 23, 2001 and was last updated in February 2020.