The Art and Craft of Problem Solving

8.5 TRANSFORMATIONS 311

After the inversion, we still have four "circles," but the situation is easy to analyze.

Circles c 2 and c; are sandwiched between the parallel lines c� and c 4. Straightforward

angle chasing (Problem 8.2.30) verifies that in this situation, the three points of tan­

gency D', A', B' are collinear. The fourth point of tangency is C' = 00. Again, it all

makes sense. The images of A, B, C, D are points on a line, i.e., a "circle." If we repeat

the inversion, we will restore these back to A, B, C, D. But images of "circles" are

"circles," so A, B, C, D must be either collinear or concyclic. Obviously, they are not

collinear. We're done! _

Problems and Exercises

Many of these problems have multiple solutions. However, all of them can be solved using trans­
formations. Some of the problems are rather challenging, so feel free to consult the hints appendix
(available online at http://www.wiley. com/college/ zei tz).

8.5. 17 Let ABCD be a square with center at X and side

length 8. Let XYZ be a right triangle with right angle

at X with XY = 10, XZ = 24. If XY intersects BC at

E such that CE = 2 and EB = 6, find the area of the

region that is common to the triangle and the square.

8.5.18 Let ABCDEF be a hexagon, with AB =

DE ,BC=EF, CD=FA, and AB II DE, BC II EF, CD II

FA. Prove that AD, BE, CF are concurrent.

8.5.19 For the following, either discover how to per­

form the given task, or explain why it may not be pos­

sible.

(a) Find the vertices of a triangle, given the mid­

points of the sides.

(b) Find the vertices of a pentagon, given the mid-

points of the sides.

(c) Find the vertices of a parallelogram, given the

midpoints of the sides.

(d) Find the vertices of a quadrilateral, given the

midpoints of the sides.

(e) Find the vertices of a hexagon, given the mid­

points of the sides.

(f) Find the vertices of a 17-gon, given the mid-

points of the sides.

8.5.20 Let a triangle have a fixed area and base. Prove

that the perimeter is minimal when the triangle is

isosceles.

8.5.21 Let ABCDE be a pentagon (not necessarily

regular) inscribed in a circle. Let A' be the midpoint