The Art and Craft of Problem Solving

286 CHAPTER 8 GEOMETRY FOR AMERICANS

the second solution, line AA' was key. This was no accident, either: since the problem

was set up "symmetrically" with diameters AC and AB, it makes sense that the objects

of interest would involve A. This type of strategic reasoning requires large doses of

peripheral vision and wishful thinking, and is certainly more of an art than a craft. But

it makes the difference between focused investigation and random constructions that

hopelessly clutter one's diagrams!

Area, Cevians, and Concurrent Lines

A line segment joining a vertex of a triangle with a point on the opposite side (ex­

tended, if necessary) is called a cevian. Many figures involve cevians, which can be

investigated by using the principle that two triangles sharing a vertex with bases on the

same line have areas in the same ratio as the base lengths (8.3.4). We have used this

idea several times, but can still squeeze more out of it. Let's use it to solve a famous

puzzle.^5

Example 8.4.2 Points D, E, and F are the first trisection points of BC, CA, and AB,

respectively (i.e., AF = ABI3, DB = BC/3, CE = CAI3). If [ABC] = 1, find [GHI],

the area of the shaded triangle.

A��--�--------�B

Solution: As we did with the centroid problem (Example 8.3.13), let's begin by

considering just two intersecting cevians. We remove BE, but add the auxiliary line I B

so that we can compare areas. Thus if [F I A] = x, then [FBI] = lx, since AF : F B = 1 : 2.

Likewise, if [DI B] = 2y, then [CID] = 4y (we let the first area be 2y instead of y in order

to avoid fractions later).

c

(^5) For several more solutions, see [10] and [28].