292 CHAPTER 8 GEOMETRY FOR AMERICANS
Phantom Points and Concurrent Lines
You have seen the method of phantom points before, where we invent a point with a
desired property or location. We can also use phantom points explicitly in a proof
by-contradiction argument, where the phantom plays the role of a straw man whose
demise yields the desired conclusion. Here is a beautiful and instructive example that
uses two phantom points to show that three lines meet in a point.
Example 8.4.6 Place three circles COl, ID2, COJ in such a way that each pair of circles
intersects in two points, as shown below. These pairs of points are the endpoints
of three chords (that are each common to two circles). Prove that these chords are
concurrent, in other words, that chords AB, CD, and EF below intersect in a single
point.Solution: This is a beautiful result; both surprising and natural at the same time.
How do we show that three lines concur? Ceva's theorem comes to mind, but that
requires a triangle. The picture is already crowded, so the prospect of adding auxiliary
constructions seems daunting.
Instead, we will emulate our proof of the centroid theorem on page 27 8. We
showed that the three medians concurred by first inspecting the intersection of two
medians, and characterizing this point.Let's do the same with the problem at hand. Temporarily remove chord EF, and
let X be the intersection point of AB and CD. These are both chords in COJ, so it is
natural to apply POP, and we get
AX ·XB =CX ·XD.
This suggests a strategy: suppose the third chord didn't concur with the first two. We
would still have intersection points (phantoms) to work with; perhaps we can POP
them?
There are two fundamentally different ways of drawing the third chord so that itfails to concur. One way is to join E and F with a "line" that misses X, and hence meets
AB and CD in two different points. The problem with this approach is subtle. We can
generate two new POP equalities, but it is hard to find an algebraic contradiction. Try
this yourself!