The Art and Craft of Problem Solving

8.4 THE POWER OF ELEMENTARY GEOMETRY 293

Another way to draw a false "line" is to start at F, intersect X (so as to use a

point we have equations for), and then go on to miss E, intersecting COl and i»2 at

the phantom points E I, E 2 , respectively. This is depicted below with a dashed "line."

(Notice that it is actually two line segments, since a straight line actually does pass

through E. That's the price we pay for accurate computer-drawn diagrams.)

Applying POP to COl yields

AX ·XB =FX ·XEI,

while POP on i»2 produces

ex ·XD = FX ·XE 2.

Combine these with AX. XB = ex. x D, and we get

FX ·XEI =FX ·XE 2 ,

so XEI = XE 2 , which forces EI and E 2 to coincide. But that would mean that this

common point is a point of intersection of COl and i»2. In other words, E I = E 2 = E,

and we are done. _

Problems and Exercises

The problems below use no new techniques other than those introduced in this section, and several
rely only on the "survival geometry" ideas of Sections 8.2-8.3. You may want to first go back to
the problems from this section that you did not solve yet and try them again. You may now find
that you have become more resourceful.

8.4.7 Two circles intersect in points A and B. Let l

be a line that passes through A, intersecting the two

circles at X and Y. What can you say about the ratio

BX /BY?

8.4.8 (Canada 1969) Let ABC be an equilateral trian­

gle, and P an arbitrary point within the triangle. Per-

pendiculars P D, P E, P F are drawn to the three sides of

the triangle. Show that, no matter where P is chosen,

PD +PE+PF I

AB +BC+CA = 2 yS'

8.4.9 (Canada 1969) Let ABC be a triangle with sides

of lengths a, b and c. Let the bisector of the angle C