Tower of Hanoi – Inverted Solution:
If the tower consisted of only 1 disk, the puzzle could be completed in only 1 move. If the tower consisted of 2 disks, the puzzle could be completed in 3 moves. if the tower consisted of 3 disks, the puzzle could be completed in 7 moves. For each of these cases, the number of moves is 
. Thus for the 5 disk model, the puzzle could be completed in 
moves.
The Travelling Salesman Solution:
The distances between the various cities, as measured in centimeters on the model, are shown on the table below.
| Olympia | Salem | Boise | Helena | Sacramento | Carson City | Salt Lake City | Cheyenne | Denver | Santa Fe | Phoenix | |
| Olympia | 14.1 | 29.8 | 42.2 | 51.2 | 48.3 | 60.2 | 85.1 | 90.4 | 105 | 98.3 | |
| Salem | 14.1 | 29.5 | 49.1 | 38.1 | 36.9 | 56.8 | 86.2 | 90.3 | 101.5 | 89.7 | |
| Boise | 29.8 | 29.5 | 24.9 | 41.6 | 34 | 30.5 | 56.7 | 61.1 | 75.1 | 72.2 | |
| Helena | 42.2 | 49.1 | 24.9 | 66.5 | 58.4 | 39.6 | 47.8 | 55 | 76.8 | 85.8 | |
| Sacramento | 51.2 | 38.1 | 41.6 | 66.5 | 9.3 | 46.9 | 82.4 | 82.7 | 82.9 | 59.1 | |
| Carson City | 48.3 | 36.9 | 34 | 58.4 | 9.3 | 37.7 | 73.1 | 73.6 | 75.2 | 54.7 | |
| Salt Lake City | 60.2 | 56.8 | 30.5 | 39.6 | 46.9 | 37.7 | 35.3 | 36.2 | 44.7 | 46.2 | |
| Cheyenne | 85.1 | 86.2 | 56.7 | 47.8 | 82.4 | 73.1 | 35.3 | 8.7 | 36.2 | 64 | |
| Denver | 90.4 | 90.3 | 61.1 | 55 | 82.7 | 73.6 | 36.2 | 8.7 | 27.5 | 57.9 | |
| Santa Fe | 105 | 101.5 | 75.1 | 76.8 | 82.9 | 75.2 | 44.7 | 36.2 | 27.5 | 38.8 | |
| Phoenix | 98.3 | 89.7 | 72.2 | 85.8 | 59.1 | 54.7 | 46.2 | 64 | 57.9 | 38.8 |
Selecting Salt Lake City as the starting city, a Mathematica computer program ran for 2 days to arrive at the path that generated the shortest distance of 276.7 cm. That path is
Salt Lake City – Carson City – Sacramento – Salem – Olympia – Boise – Helena – Cheyenne – Denver – Santa Fe – Phoenix
The path from Salt Lake City with the second shortest length of 281.2 cm is
Salt Lake City – Cheyenne – Denver – Santa Fe – Phoenix – Carson City – Sacramento – Salem – Olympia – Boise – Helena
If the starting city is Boise, the shortest path of 276.6 cm is
Boise – Helena – Olympia – Salem – Sacramento – Carson City – Salt Lake City – Cheyenne – Denver – Santa Fe – Phoenix
The path from Boise with the second shortest length of 284.6 cm is
Boise – Helena – Olympia – Salem – Carson City – Sacramento – Salt Lake City – Cheyenne – Denver – Santa Fe – Phoenix
The Which is Fastest Solution:
The How Far Out Solution:
Please click here for a pdf of the solution.
The Penrose Puzzle Solution:
An interesting property of the blue kite and yellow dart is the measures of their interior angles. It turns out that 144 degrees is the measure of an interior angle of a regular decagon while 36 degrees is the measure of the exterior angle of a regular decagon. The central angle and exterior angle of a regular pentagon measures 72 degrees. These angle properties lead to symmetries in the tessellations you create similar to those found in a regular pentagon or regular decagon. For example there are many places in the Penrose tilings where 5 darts or 5 kites come together because 5 angles of 72 degrees is 360 degrees.
In the figure above, notice the decagons outlined in green and the five red lines of symmetry. As mentioned in the information on the table, the angles of the kite and dart are related to the angles of a regular pentagon and regular decagon which contributes to the appearance of the properties of these polygons in the tessellations.
Although the tiling below is different from the one above, you will still notice many symmetries related to the regular pentagon and regular decagon.
Find the Bears Solution:
The image below displays six bears and another six bears can be found in those same locations by rotating the current bears 180 degrees. Thus the total number of bears is 12. If, however, you do not count the orange bear at the right as being completely within the rectangular boundary then there are only 10.
Find the Fish Solution:
An interesting property of the kite (in our tiling some are blue and some are yellow) and the dart (in our tiling all of the darts are blue) is the measures of their interior angles. It turns out that 144 degrees is the measure of an interior angle of a regular decagon while 36 degrees is the measure of the exterior angle of a regular decagon. The central angle and exterior angle of a regular pentagon measures 72 degrees. These angle properties lead to symmetries in our tiling similar to those found in a regular pentagon or regular decagon. For example, there are many places in our tiling where five darts or five kites come together because five angles of 72 degrees is 360 degrees.
In our tiling, there are five fish similar to the green one in the image below all around that center shape of five kites forming a decagon. You will notice that all polygons immediately adjacent to this decagon are darts. There are no other complete fish on the tiling within the rectangle boundary because there are no other decagons that are completely surrounded by darts. If, however, we extended the boundary, there are several other locations similar to the one in the center where the decagon is surrounded by darts. You can see the possible fish in red in each of these locations. Since none of these red fish are within the rectangle boundary, the total number of fish in our tiling is five!
Tangram:
The solutions for the Tangram table are sown in the figure below.
Polyhedra in a Box:
Since a cube consists of 6 faces in 3 pairs of parallel planes, the key to determining how the polyhedron fits into the cube is to identify the 6 faces of the polyhedron that can map to the faces of the cube.
For example in the polyhedron below, you can see a square on the top that will be in a plane parallel to the plane containing a square at the bottom. The square in the front is in a plane parallel to the plane containing the square in the back. Similarly, there are squares in parallel planes to the left and right. By matching these parallel planes to the faces of the cube allows you to place the polyhedron into the cube.
Julia Robinson Puzzles
A little bit about Julia Bowman Robinson (1919–1985):
Julia Bowman Robinson spent her early years in Ari- zona, near Phoenix. She said that one of her earliest memories was of arranging pebbles in the shadow of a giant saguaro—“I’ve always had a basic liking for the natural numbers.” In 1948, Robinson earned her doctorate in mathematics at Berkeley; she went on to con- tribute to the solution of “Hilbert’s tenth problem.” In 1975 she became the first woman mathematician elected to the prestigious National Academy of Sciences. Robinson also served as president of the American Mathematical Society, the main professional organization for research mathematicians. “Rather than being remembered as the first woman this or that, I would prefer to be remembered simply for the theorems I have proved and the problems I have solved.”
The Travelling Salesman starting at Santa Fe Solution: S
The distances between the various cities, as measured in centimeters on the model, are shown on the table below.
Selecting Santa Fe as the starting city, a Mathematica computer program ran for over a day to arrive at the path that generated the shortest distance of 282 cm. That path is
Santa Fe – Denver – Cheyenne – Salt Lake City – Helena – Boise – Olympia – Salem – Sacramento – Carson City – Phoenix
The second shortest path with a length of 284.1 cm is
Santa Fe – Denver – Cheyenne – Helena – Boise – Olympia – Salem – Sacramento – Carson City – Salt Lake City – Phoenix
The third shortest path with a length of 285.2 cm is
Santa Fe – Denver – Cheyenne – Salt Lake City – Helena – Boise – Olympia – Salem –Carson City – Sacrament – Phoenix
The Einstein Tessellation Solution:
Prior to 2022 the closest solutions to the Einstein question about creating an aperiodic tessellation with a single polygon were aperiodic tessellations that used two different tessellations like those with two Penrose tiles – a kite and a dart. In November of 2022, David Smith (a retired printer systems engineer) of Yorkshire, England found the 13-sided Einstein tiles (shown below) that are commonly thought to have the shape of a hat.
One tessellation with these tiles is seen below.
Smith joined forces with Craig Kaplan (a computer scientist), Joseph Samuel Myers (a software developer) and Chaim Goodman-Strauss (a mathematician) to prove that the “Einstein tile” did, in fact form an aperiodic tessellation of the plane. They submitted a paper with this proof in April of 2023. The results of the peer review of this proof are pending.