The Art and Craft of Problem Solving

8.5 TRANSFORMATIONS 303

is transformed and the rest is left alone. Here is a dramatic example of a problem that

is quite thorny without transformations, yet nearly trivial once we focus on the natural

symmetries involved.

Example 8.5.5 Lines f1, f 2 , f3 are parallel. Let a and b be the distance between fl

and f 2 , and f 2 and f 3 , respectively. The point A is given on line fl. Locate points B,C

on f 2 ,f 3 , respectively, so that triangle ABC is equilateral.

Solution: This is certainly doable with cartesian coordinate geometry. There are
just two variables (the coordinates of B, for example; or the length AB and the angle
AB makes with line f 1), and plenty of equations determined by the angles and inter­
sections. But it will be ugly and unilluminating.

In the diagram, we are assuming we have solved the problem (wishful thinking!) so

we have a nice equilateral triangle to ponder. The natural transformations to ponder

are rotations that leave parts of this triangle invariant. Consider clockwise rotation by

60 ° about A. This takes AC to AB. Now imagine letting f 3 go along for the ride as

well. Thus AC plus f 3 will be rotated into a line segment plus a line. Since f 3 intersects

C, the image of f 3 will intersect f 2 at B. So we're done: all we need to do is rotate f 3

by 60 ° clockwise! In the diagram below, f4 is the image of f 3 under this rotation. The

intersection of f4 with f 2 is the point B, and once we know the location of B, we can

easily construct the equilateral triangle ABC. We drew in the perpendiculars from A to

f 3 and f4 to suggest a Euclidean construction.

«

«

A

• Drop a perpendicular from A to f3.


• Rotate this perpendicular segment by 60 °, which should be easy since you know