Drag the point A with the mouse. Press r to reset.
Requirement: Point A must stay to the right of the vertical line.
Problem: 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.
The construction is shown in the figure above. Here are the details:
It can be shown that as the point P slides along the line L, the angle APB achieves local maxima at P1 and P2.
Proof: First note that based on the construction above, B'HA is a right triangle therefore we have QH2 = B'Q . QA = QB . QA. We will need this later.
Let P be an arbitrary point on the line L. Consider the circle C that passes through the points A, B, and P. The sine of the angle APB equals (length of AB)/(diameter of C). In general, the sine of the angle subtending a chord of a circle equals (length of the chord)/(the diameter).
We conclude that with A, B, L fixed and P sliding on L, the angle APB achieves its maximum when the diameter of C achieves its minimum.
Since C is constrained to pass through A and B, then the smallest such C is achieved when it is tangent to the line L. Thus the problem of finding the maximum angle APB reduces to the problem of passing a circle through the points A and B such that it is tangent to the line L. The point of tangency, P, is the desired point
Let the point Q be as described in the construction. If the circumscribing circle of the points A, B and P (with P on the line L) is tangent to the line L, then we have QP2 = QB . QA. Recalling the previous remark about QH2 = QB . QA, we conclude that QP = QH, thus the circle centered at Q through H intersects the line L at P. Actually there are two such intersections, P1, and P2. Both correspond to local maxima of the angle APB. The one corresponding to the smaller circle C corresponds to the absolute maximum.
This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on November 23, 2001. It was revised on May 20, 2008 based on feedback received from Szczepan Holyszewski.