2 In the illustration, a regular hexagon and a regular octagon have been tiled with rhombuses. In each case, the sides of the rhombuses are the same length as the sides of the regular polygon.
(a) Tile a regular decagon (10-gon) into rhombuses in this manner.
(b) Tile a regular dodecagon (12-gon) into rhombuses in this manner.
(c) How many rhombuses are in a tiling by rhombuses of a 2002-gon? Justify your answer.

Make your drawings on the sheet of decagons and dodecagons provided. Clearly indicate which drawings you would like to have graded.
Solution: For parts (a) and (b), see the tilings below of a decagon and a dodecagon with rhombuses.
(c) Solution 1: We shall describe an algorithm for tiling any regular $2 n$-gon. First construct a rhombus $A$ using two sides of a regular $2 n$-gon. On the two interior sides construct two rhombuses $B$ using a side adjacent and clockwise to the exterior sides of $A$. On the three interior sides of the $B$ rhombuses construct three rhombuses $C$ using the next side of the $2 n$-gon going clockwise. After the $(n-2)^{n d}$ step an outer layer of $n-2$ rhombuses is in place. To see that the $(n-1)^{\text {st }}$ step completes the $2 n$-gon note that the $(n+1)^{s t}$ side after the first side of $A$ is parallel and equal in length to the first side of $A$ and has the same length as the $n$th side of the $2 n$-gon. Therefore it is a rhombus. In the same way, the remaining $n-1$ sides of the $2 n$-gon form rhombuses with segments parallel to the $n^{\text {th }}$ side. The total number of rhombuses is the sum of the first $n-1$ integers. Then it is the case that a regular 2002-gon can be tiled with $1+2+3+\cdots+999+1000=1000(1001) / 2=500(1001)=500,500$ rhombuses.

Solution 2: We will prove the slightly more general fact that any equilateral $2 n$-gon with opposite sides parallel (we will call such polygons "EP-gons") can be tiled with $1+2+\cdots+n-1$ rhombuses. We proceed by induction on $n$. Certainly the result is true for $n=2$, since a 4 -sided EP-gon is simply a single rhombus.

For the inductive step, suppose we have $(2 n+2)$-sided EP-gon. We will use the leftmost figure above for reference; thus in this case $n=4$. Select one side of this polygon (we will use the "bottom" side in the picture, one of the sides of rhombus $A$ ). Then draw $n-1$ lines parallel to this side with lengths equal to this side, each line starting at a vertex of the polygon. Then draw $n$ lines parallel to the sides of the polygon; when we are done we will have drawn a "shell" of $n$ rhombuses on the left of the polygon (rhombus $A$ and the leftmost rhombuses $B, C$, and $D$ ). We claim that the remainder of the $(2 n+2)$-gon (the non-tiled region) is a $2 n$-sided EP-gon. To see this, note that the untiled region consists of $n$ sides of the right half of the original $(2 n+2)$-gon, plus $n$ new sides, each drawn parallel
(and equal in length) to a side of the left half of the original $(2 n+2)$-gon. It is as if we rigidly slided the left half of the original $(2 n+2)$-gon to the right, so that the top and bottom sides of the original polygon disappeared. Thus in this new region, opposite sides are parallel. By the inductive hypothesis, this region can be tiled with $1+2+\cdots+(n-1)$ rhombuses. Thus the original $(2 n+2)$-gon can be tiled by $1+2+3+\cdots+(n-1)+n$ rhombuses.