Here is an interesting little puzzle sugested to me by Mr W. T. Whyte. Mark off on a sheet of paper a rectangular space 5 by 3 inches, and then find the greatest number of halfpennies that can be placed within the enclosure under the following conditions. A halfpenny is exactly an inch in diameter. Place your first halfpenny where you like, then place your second coin at exactly the distance of an inch from the first, the third an inch distance from the second, and so on. No halfpenny may touch another halfpenny or cross the boundary. ...some comment on an illustration.... How many can the reader place? (H. E. Dudeney, 1917 no. 429)

(Dudeney's solution was 13 halfpennies.)

(You can find Dudeney's book here: Free eBooks - Project Gutenberg)

7 members in the Club had the solution 14 halfpennies. My solution:

We call the radius of the halfpenny 1 and the corners of the rectangle (0,0), (10,0), (10,6) and (0,6). Then place halfpenny 1 in (7,3+x), 2 in (3,3+x), 3 in (y,5-x), 4 in (7,1), 5 in (3,1), 6 in (z,w), 7 in (1,1+x), 8 in (5,1+x), 9 in (9,1+x), 10 in (2+y,5), 11 in (y-2,5), 12 in (1,3+2x), 13 in (5,3+2x) and 14 in (9,3+2x). The constellations of halfpennies (11,13,8), (3,1,4) and (10,14,9) are similar. Call the distance, between halfpenny 1 and 3, d, we will check for overlap. Then we have:

(7-y)**2+(2-2x)**=d**2

Likewise 3 and 4 give: (7-y)**2+(4-x)**2=16

Those equations give us : 4+3x**2=d**2, than d is greater than 2.

Coins 2 and 3 give: (y-3)**2+(2-2x)**2=16

and coins 3 and 4 give: (7-y)**2+(4-x)=16

Those last two equations give us: (20-3*x**2)**2+256*(1-x)**2=1024

x=0.03 make the left part=1024.7192973 and

x=0.04 make the left part=1019.660823 then

x greater than 0.03 and less then 0.04. This will give us all the coordinates.

Sprauge gave the answer: 2+3+4+6+10+15+20+30=90.

But 16 members had the answer: 2+3+4+6+10+15+21+26=88.

Later I found the same answer in problem C9 from Richard K, Guy's book Unsolved Problem in Number Theory.

A man wishes to construct the twelve pentominoes (a shape that cover five squares) out of plywood. His saw will not cut around corners. What is the smallest plywood rectangle from which he can cut all twelve pentominoes? (The U-shaped pentomino will require special effort. Assume it will be cut as a 2*3 hexomino, a hexomino is a shape that cover six squares, and finished it later.)

The answer in the book was 6*13.

But eight members had the answer: 5*15, Mr Adolf Ahl: 7*(8.5+1.5*sqrt(2)) and Mr Fai Lötberg: 8*9. His solution:

LWOTTTOXO 1. cut U

LWWOTVXXX 2. cut X

LOWWTVOXO 3. cut W+L+P+N+F

LLFOIVVVZ and free them in that order

NFFFIYZZZ 4. cut I+T+V+Y+Z

NFOPIYZUU and free them in that order

NNPPIYYUO 5. O is waste

ONPPIYOUU 6. observe the diagonal cut in 3

A 5*15 from my hand:

Place the twelve pentominoes so there are as many single holes as possible.

a) No holes connected in their corners.

b) Corners connection allowed.

The best solver was Mr Christer Lindstedt, 13 and 18 holes :

Cut an 8x8 square into the twelve pentominoes and a 2x2 square. Use all the 13 pieces as a fence around an

The 13 pieces as a square:

The best solution, with 143 fenced squares, was from Mr Ole Björkqvist and Mr Sune Backan.

**Made by Sven Egevad **

*Epost: sven.egevad@telia.com*