The Art and Craft of Problem Solving

66 CHAPTER 3 TACTICS FOR SOLVING PROBLEMS

Thus

Consequently, the value that we seek is

The next problem, from the 1995 IMO, is harder than the others, but only in a

"technical" way. In order to solve it, you need to be familiar with Ptolemy's Theorem,

which states

Let ABCD be a cyclic quadrilateral, i.e., a quadrilateral whose vertices

lie on a circle. Then

AB·CD +AD· BC =AC·BD.

c

See Problems 4.2.41, 8.4.30, and 8.5.49 for different ideas for a proof. A key feature

of cyclic quadrilaterals is the easily verified fact (see page 266) that

A quadrilateral is cyclic if and only if its opposite angles are supple­

mentary (add up to 180 degrees).

Example 3.1. 8 Let ABCDEF be a convex hexagon with AB = BC = CD and DE =

EF =FA, such that L.BCD = L.EFA = 1C/3. Suppose G andH are points in the interior

of the hexagon such that LAGB = L.DHE = 21C/3. Prove that

AG+GB+GH +DH +HE 2: CF.

Solution: First, as for all geometry problems, draw an accurate diagram, using

pencil, compass and ruler. Look for symmetry. Note that BCD and EFA are equilateral

triangles, so that BD = BA and DE = AE. By the symmetry of the figure, it is seems

profitable to reflect about BE. Let C' and F' be the reflections of C and F respectively.