The Art and Craft of Problem Solving

8.5 TRANSFORMATIONS 297

And of course, what makes this particular symmetry useful is the fact that the sums of

each column are identical!

Here is a geometric problem with a similar flavor.

Example 8.5.1 Hexagon ABCDEF is inscribed in a circle, with AB = BC = CD = 2

and DE = EF = FA = I. Find the radius of the circle.

Solution: The length-l and length-2 segments are segregated on opposite sides

of the circle, making a shape with only modest bilateral symmetry (figure on the left

below). However, each of these segments is the base of an isosceles triangle (whose

vertex is the center of the circle). The entire hexagon is the union of three triangles of

one type and three of the other type, and the side lengths of these isosceles triangles

are the same (i.e., the radius of the circle).

So, why not rearrange these six triangles in a more symmetrical way? If we al­

ternate the length-l and length-2 segments, we get a much more symmetrical hexagon

(figure on the right below).

F

Denote the center of the circle by 0, with XY = 1 and YZ = 2 adjacent sides of the
rearranged hexagon. Notice that quadrilateral X ozr is exactly one-third of the entire
hexagon. Let the radius of the circle be r; thus OX = OY = OZ = r. Since X OZY is
exactly one third of the entire hexagon, we know that LXOZ = 360/3 = 120°. Like­
wise, by the symmetry of the rearranged hexagon, all of its interior angles are equal, so
LXYZ = 720/6 = 120°. (We could have also deduced this with angle chasing, using
the factXOY and YOZ are isosceles.)
Now the problem has been reduced to routine trigonometry. Verify that two ap­
plications of the law of cosines (which you proved in Problem 8.3.32, right?) yield
XZ = .J7 and then r = J773. •

We easily solved this relatively simple problem by imposing a single symmetri­

cal structure on it, making it easier to understand. More complicated problems may

have several different structures, each with competing symmetries. If you can find one

symmetry and apply a transformation that leaves one structure invariant, while chang­

ing the other structures, you may learn something new.^6 That, in a nutshell is why

(^6) See Example 3.1.4 on page 63 for a simple but striking illustration of this.