The Art and Craft of Problem Solving

8.4 THE POWER OF ELEMENTARY GEOMETRY 295

Our weight assignment appears to be consistent, since

the interior balancing point I receives 7 units, regard­

less of whether we use AD or C F.

Now, looking at the endpoint weights on AD, we

deduce that the length ratio AI : ID must equal 3 : 4 if

the configuration is to balance properly at I. Likewise,

F I : IC = I : 6. This was much easier than comparing

areas!

This wonderful and intuitively plausible proce­

dure seems like magic. Can you prove, rigorously, that

it works?

8.4.22 Can the method of weights be used to prove
Ceva's theorem?

8.4.23 (AIME 1988) Let P be the intersection of ce­
vians AD, BE , CF of triangle ABC. If PD = PE =
PF = 3 and AP + BP +CP =43, find AP ·BP·CP.

8.4.24 Example 8.4.6 on page 292 can be general­
ized to include cases where the circles do not intersect.
Given a point P and a circle with center 0 and radius
r, define the power of P with respect to the circle to be
(OP +r)(OP - r). Then, given two circles 0) 1 and W,z,
define their radical axis to be the set of points P such
that the power of P with respect to 0) 1 is the same as
the power of P with respect to W,z.
(a) Explain what this definition of power has to do
with the Power of a Point theorem. What does
it mean when the power is positive, negative, or
zero?
(b) Prove that the radical axis is a line that is per­
pendicular to the line joining the centers of the
two circles.
(c) Following Example 8.4.6, prove that in general,
the radical axes of three circles will either be
concurrent, coincide, or be parallel. Draw pic­
tures to illustrate these different cases.
8.4.25 Let AB and CD be two different given diame­
ters of a given circle. Let P be a point chosen on this

circle, and drop perpendiculars from P to the two di­

ameters, meeting them in points X and Y. Prove that

XY is invariant; i.e., its value does not change no mat­

ter where P is placed on the circle.

8.4.26 A sister theorem to Ceva is Menelaus's theo­

rem, which you should now have no trouble proving

on your own. Let ABC be a triangle, and let line m in­

tersect the sides (extended, if necessary) at D, E, and

F. Then

AD CF BE

= 1.

DC FB EA

Hint: Draw a line through C that is parallel to AB, and

hunt for similar triangles.

8.4.27 Formulate and prove the converse of

Menelaus's theorem, which is a useful tool for proving

that points are collinear.

8.4.28 (USAMO 1994) A convex hexagon ABCDEF

is inscribed in a circle such that AB = CD = EF and

diagonals AD,BE,CF are concurrent. Let P be the in­

tersection of AD and CE. ProveCP/PE = (AC/CE)^2.

8.4.29 (Bulgaria 2001) Given a convex quadrilateral

ABCD such that OA = OB · OD/(oe + OD), where 0

is the intersection of the diagonals. The circumcircle

of /::,ABC intersects the line BD at the point Q. Prove

that CQ is the bisector of LDCA.