Exercises of Chapter 1 (Part 5/6), Computational Geometry

Exercise 1.45. Find another dissection of the Greek cross, something quite different from that of Figure 1.19, that rearranges to form a square.


Solution. Omitted. □

Exercise 1.46. Find a dissection of two Greek crosses whose combined pieces form a square.


Solution. Omitted. □

Exercise 1.47. Show that any triangle can be dissected using at most three cuts and reassembled to form its mirror image. As usual, rotation and translation of the pieces are permitted, but not reflection.


Solution. (It still took a few minutes for me to figure this out) the solution is quite beautiful and it's simply based on the feasibility of dividing the triangle into three isosceles triangles by cutting along the three straight lines between its circumcenter and its three vertices. □

Exercise 1.48. Assume no three vertices of a polygon P are collinear. Prove that out of all possible dissections of P into triangles, a triangulation of P will always result in the fewest number of triangles.


Proof. We'll use induction as well as reduction to absurdity for the incremental step in the iteration. It's obviously true for triangle. Assume the theorem is true for all polygons with fewer than n vertices. If the theorem doesn't hold for polygon with n+1 vertices, which means the polygon can be dissected into less than n-1 triangles, then reducing the polygon to one with n vertices by contracting any one of the edges will lose at least one triangle (the proof of which also utilises the fact that the polygon doesn't have three vertices collinear), which indicates that the theorem doesn't hold for n. Then we see the statement is evident. □


Exercise 1.51. Let polygon P₁ be scissors congruent to polygon P₂, and let polygon P₂ be scissors congruent to polygon P₃. Show that polygon P₁ is also scissors congruent to P₃. In other words, show that scissors congruence is transitive.


Proof. As P2  can be formed by the pieces dissected from P1 , suppose the dissection is D12, movement (translation and rotation) is M12; And as P3 can be formed by the pieces dissected from  P2, suppose the dissection is D23 and movement adhesion is M23 and adhesion is A23. So we can use the following formula to denote the process of these two transformations. Note that the movement transformation is a set of different transformations applied to the dissected pieces individually, we now clarify the definition of movement transformation to be with respect to the points (positions) in the rigid original shape. Similar definition can be made on the dissection operation (made respect to the lines as set of points in the original shape). With this clarification makes transformations commutable without changing their behaviour.

P2 = M12 D12  P1
P3 = M23 D23  P2

It's not hard to see that

P3 = M23 D23 M12 D12  P1 = M23 M12 D23 D12 P1

So we have M13 = M23 M12, D13 = D23 D12


Then the theorem follows.□

Exercise 1.52. Dissect a 2 × 1 rectangle into three pieces and rearrange them to form a ³√4  × ³√2  rectangle

Solution. Omitted. □

Exercise 1.55. Following the proof of the Bolyai-Gerwein theorem, what is the actual number of pieces that results from transforming the Greek cross into a square? Assume the total area of the square is 5/2 and use Figures 1.24 and 1.25 for guidance.

Solution. Tedious and trivial. Leaving it to the computer to solve is even more interesting. Solution omitted. □

Exercise 1.56. Show that a square and a circle are not scissors congruent, even permitting curved cuts.

Proof.  It's not easy to put it down in detail and in strict mathematical manner, however a sketchy explanation is enough, which reveals the fact that each cut for the arc of the circle will just create one more arc of the same size for it to match, and consequently it will never form a shape without curved boundary. □

References


[1] Math in HTML and CSS, http://www.cs.tut.fi/~jkorpela/math/