bonanova

Related to previous puzzle, which seems mired. There are many ways to arrange 12 unit-length matches to form a polygon whose area is [exactly] an integer. What is the smallest non-zero area you can achieve?

Yes. That's the meaning I intended.

More realistically it's related

EventHorizon points out that in some cases when a polygon whose interior comprises N contiguous unit-square areas, the angle between vertical and horizontal can be adjusted to make the N unit-square areas collectively decrease to N-1, N-2, N-3 ... 1 square units. The idea is that unit squares change into fractional rhombuses [rhombi?] that sum to an integer. The shear can be done in different directions: left, right, upward or downward; and in some cases it can be done locally, leaving the remainder of the polygon unchanged. This increases dramatically the number of 12-match polygons with integral areas. He shows further that when shear is performed independently on separate parts of a polygon -- but this must be done in a way that match count is preserved -- the separate groups need not both [all] have rational area values. That is, unit squares can be sheared in different parts of the polygon to acquire real [not a few rational] area values that sum to an integer. Thus an infinite number of integer-area polygons can be constructed using 12 matches. Bravo. For the purposes of stating the problem in simplest terms, and keeping the solution set finite, let us include shear groups with reflection and rotation symmetries. That is, if there is a class of polygons or classes of classes of polygons whose members have their corresponding vertices connected by edges of the same length, then count only the polygon in that group that has the largest area. How many, now? And what size?

Nice going. When I made mine, I couldn't measure angles other than to make squares. What was the smallest of your 23 "rectilinear" ones.?

Pieces is plural.

Assuming a match is of unit length, it is possible to place 12 matches on a plane in various ways to form polygons with integral areas. I want to use the matches to make as many of these shapes as possible. The entire match length must be used; mirror images and rotations do not count. Examples: A 3x3 square [with interior matches removed] uses 12 matches with an area A = 9 square units. Another shape is a 1x5 rectangle, with A = 5 square units. You can attach two shapes; just remove the common edges, if any. If you attach them only at a corner, there are no common edges. Imagine a 2x2 square [8 matches] glued at a corner to a 1x1 square [4 matches] to give a shape with 5 square units. So far, I have found 30 shapes. Their areas are varioulsy 9, 8, 7, 6, 5, 4 and 3 square units. There may be more. But now I've had a change of heart about gluing shapes together at a point. Those are really shapes that use fewer than 12 matches, that just happen to be touching. That eliminates seven polygons from my collection. Can you find my other 23 shapes -- and possibly others? Hint: Sort shapes by area.

This one ?

Since any 7x7 division includes the center square, you can have only one of them.

Thanks PT, and a tip of the hat rookie! Wolfram has it for "cylinders". Of course I searched on pencils and cigarettes, but forgot to search for matchsticks!!! I think we're agreed now on the pencils and the golf balls. Are we agreed on the coins?

They are in the space directly above the table. They do not all need to contact the table: they may be stacked. Is that what you are asking?

Two friends A and B play poker in a way that eliminates the element of chance. They spread the 52 cards face up on the table so all cards can be see by both players. Play proceeds as follows: A selects any five cards for his hand B does the same A exchanges 0 1 2 3 4 or 5 of his cards for new ones, discarding the old ones, which are then no longer in play B does the same Highest hand wins. Suits have equal value. With best play, can A always win? Since the cards are visible, you can assume both hands are exposed at all times.

A favorite puzzle asks how many people must be in a room for chance to favor there being two with the same birthday We ask a slightly different question: How many people must be in a room for chance to favor every day on the calendar being the birthday of someone in the room? For simplicity you can ignore leap years.

We usually think of things that are scarce as being more valuable. Take that thought to the evaluation of poker hands. Enumerate the card groups that give you a straight flush Enumerate the card groups that give you four of a kind. Which hand should be ranked higher?

1 out of 3. That's a start ...

I have some golf balls, some identical coins, and some pencils. I want to place them on a table in groups so that each object touches all the others in that group. How large can the group be, in each case? Note there are three answers here.

I have a chess board that has 13 squares on a side. I want to break it [along the borders of the individual squares] into smaller square pieces. I could make a 12x12 square and 25 1x1 squares - 26 in all. I could make a 7x7 square, 3 6x6 squares and 12 1x1 squares - 16 in all. What is the fewest I could make?

Cute. Never saw that unit until now.

I think you can see - at 5th grade level - how this works. Read on ...

Hey UR, welcome back. Tell us what you have for part 2?