1979 USAMO Problems/Problem 4
lies between the rays and . Find on and on collinear with so that is as large as possible.
There are two ways to solve this problem. The first is more subtle, and the second is just bashing.
Perform the inversion with center and radius Lines go to the circles passing through and the line cuts again at the inverses of Hence
Thus, it suffices to find the line through that maximizes the length of the segment If are the midpoints of i.e. the projections of onto then from the right trapezoid we deduce that Consequently, is the greatest possible length of which obviously occurs when is a rectangle. Hence, are the intersections of with the perpendicular to at
Solution (trig bash)
Let and Then and Using the Law of Sines on gives and using the Law of Sines on gives Note that and are given constants. Hence,
Clearly, this quantity is maximized when Because must be less than , is as large as possible when or when line is perpendicular to line .
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