50. Anniversary Contest

To our call for a nice match statement we received many responses.
Many thanks for all the great problems we have got.
Therefore, we thought we let the solver choose.
We hope that we have not forgotten anyone. Else send an email.
This is probably our last competition.
We start with the problem of Kate Jones because we got the following mail "Attached is a "50" puzzle I designed some time ago, one copy of which we will be happy to give as a prize for your winner. It is a $45 value and contains all the polyominoes 1 through 5. It has many solutions and is based on our Poly-5 puzzle, www.gamepuzzles.com/polycub2.htm#P5"

We thank Kate and Kadon Enterprises for the prize.

Problem of Kate Jones.

Fit all the polyominoes from 1 through 5 into the grid within the number 50, so that no two of the tetrominoes, trominoes, the domino and monomino touch each other, not even at the corners. It's not difficult but very pretty!
Kate:"I would like to know whether it was solved by a computer or by a human!"
 
To help you with solving we made an excel file and also a geogebra applet.
If you have questions about them, just ask
o.d.m@fulladsl.be

Flat Poly calculated that there are many different solutions.
You can download FlatPoly at
http://home.wxs.nl/~avdw3b/aad.html

The idea of our 34th competition is also by Kate Jones.

Problem of Bob Harris
(A)  Choose a pentomino and make 13 copies of it, colored yellow, and choose a tetromino, and make 4 copies of it, colored black.
(B) Place the pieces so that a 7x7 square is covered to a "depth" of 1. The yellow pieces are "positives" and count as +1; the black pieces are "negatives" and count as -1. Pieces may be placed on top of each other and wherever this occurs a black cell will cancel a yellow one. For example, a cell covered by a single yellow piece has a depth of 1, as will have a cell covered by two yellows and one black piece.
Pieces needn't stay within the 7x7 square, but any cell outside the square must have a depth of zero (which usually means that it is covered by a single yellow and a single black piece).

Bob's motivation for this problem: "I've been investigating polycube puzzles made from several copies of two (or three) pieces.  One such puzzle has been on my website for years:
http://www.bumblebeagle.org/polycubes/7andF.solution.gif
Now, if I want to have a heptacube and a pentacube fill a 4x4x4, how many copies of each would I need?  This is solved by 7a + 5b = 64, looking for positive integer solutions.  Integer solutions are a=7-5k b=3+7k (for some integer k), and restricting this to positive integers the only solutions are a,b=7,3 or a,b=2,10.
Looking at the same sort of thing in 2D, suppose I want to pack a 7x7 with pentominoes and tetrominoes.  Then I have 5a+4b = 49 with integer solutions (a,b)=(9-4k,1+5k).  Solutions in positive integers are (9,1), (5,6), (1,11).  But there are also solutions with negative integers. In this example is (a,b) = (13,-4), or 13 pentominoes and negative 4 tetrominoes.
We choose the  L-pentomino and the L-tetromino.
Use the mouse pointer to see the example

Place the 12 pentominoes in a 12x12 square so that no two pentominoes
touch each other, not even corner to corner.
Count how many edges of the square (12x4=48 in total) are touched by the pentominoes. What are the maximum and minimum values? In the example below this value is 32 (1+1+4+3 = 9 units at the top edge, and so on.) Notice that a pentomino cell which covers a corner of the square counts for 2 edges.


Problem of Peter Jeuken
Peter created this pentomino puzzle especially for us.
12 sets of 12 pentominoes and 12 sets of 5 tetrominoes together form a 32x32 square where the four corners (4x4 squares) are missing.
If you prefer to see it larger, then click on the picture.

Cut the figure into 12 pieces so that each piece contains exactly 12 different pentominoes and 5 different tetrominoes.
Only cuts along the grid lines are allowed. Pentominoes and tetrominoes must not be cut in two.
If you find more than one solution, then submit the solution with the smallest total perimeter of the 12 pieces.

Problem of Aad Thoen : Pentomino-islands
Place all pentominoes as islands in a sea such that the area of this sea should be as small as possible. The pentominoes must not touch each other, not even at the corners.

In our example the sea area is 122 units.

A variant of this problem is us potpourriproblem 11 (Dutch) where the sea has to be a rectangle.

Another problem is that we allow the pentominoes to be peninsulas.

Problem of Aad van de Wetering
We would like to thank Aad van de Wetering very much for doing a great job in establishing our pentomino site.
1 + 1 = 3
This problem was created by our friend Pieter Torbijn to whom we dedicate the whole competition.
Enclose a set of pentominoes with two other sets. Make the area of the circumscribed rectangle as small as possible.



In our example the rectangle is 16 x 14.
You can find another solution of Michael Keller on http://www.solitairelaboratory.com/pentmino.html.
 This solution is a rectangle of 18 x 13. One can do much better.

Pento-Product-Xudoku
On request, because we know we have many visitors who love sudoku, ​​Aad van de Wetering made for us this great hint free xudoku.

Put the numbers from 1 to 8 in each row, in each column and in the two diagonals.
The products of the numbers in each pentomino are as follows:

L:420, X:96, U:1080, V:11760, Z:576, Y:336, F:1440, W:84, P:720, I:1120, T:384 , N:1680
Would you like a hint? Mail to o.d.m@fulladsl.be

'50. Fifty-Fifty' Competition by Edo Timmermans
We asked Edo to create a competition for our 50th Jubilee because we consider him to be one of our most creative problem creators. (see the Pentomino postcard competition and the Isle of count Penteviticole)

His competition consists of 2 parts.

Take a red 10x10-square with a diagonal green square inside having 50% of the surface of the 10x10-square. The pentominoes are to be placed within the big square in such a way that they are all partially red and partially green.


In the first 50% of the competition, you are invited to search for solutions where 10 pentiminoes are 50% red and 50% green. Also, all pentominoes together must be 50% red and 50% green.
The examples show that this is not very difficult when the remaining 2 pentominoes are allowed to be 10% red or 10 % green. That is why these 2 pentominoes must minimally be 20% red as well as minimally 20% green, yet the closer to 50% the better, of course.

In the second 50% of the competition you are invited to look for a symmetrical solution. The demand that 10 pentominoes must be half red and half green is dropped, but each pentomino must still be partially red and partially green.

 The example is not a valid solution, only because the W-pentomino is entirely green.
Computer programmers are invited to look for symmetrical solutions where at least 50% of the pentominoes are 50% red and 50% green.

Pentorigami-Problem of Bob Henderson
Cut 12 pentominos from stiff paper of different colors.
For your convenience, we put 6 copies of each pentomino on a sheet.
So you get different sets of pentominoes
F I L N P T U V W X Y Z
Cut 1 of the 4 grid lines that meet at the center of P.
There are eight pentominoes which can be folded into an open cube (a cube without top face). They are F,L,N,T,W,X,Y and Z.
Take 6 of these 8 open cubes and join them at the edges of their openings. They will look like 6 cubes attached to the faces of a seventh (hidden) cube.