Start with a chessboard and put one marker on each intersection point, including those on the perimeter and...
geometry/lattice/chessboard.circles New
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Problem
Start with a chessboard and put one marker on each intersection point,
including those on the perimeter and the four corners (81 markers in
all).
Q1) How many different squares can be drawn so that all four corners
rest on markers?
Q2) How many different circles can be drawn which pass through EXACTLY
four markers?
(Posed by Koh Kew Soon, who later restated both questions for squares
and circles in 3-d space, inside or outside the board.)
[ Editor: problem statement reconstructed from the postings. ]
Show solution
The answer is the expression:
2 2 2 2 2 2 2 2 2
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
2 2 2 2
+ 1 + 3 + 5 + 7
2 2 2 2 2 2
+ 2 x ( 1 + 2 + 3 + 4 + 5 + 6 )
which I believe add up to 623 (correct me if I am wrong).
Explanation:
The number of circles that can be drawn is the same as the number of
squares which can be drawn such that each corner rests on a marker.
For squares which are parallel or perpendicular to the sides of the
chessboard, there are
2 2
1 + ... + 9 altogether.
For squares which are at 45 degrees to the sides, there are
2 2 2 2
1 + 3 + 5 + 7 .
For squares at other angles to the sides, there are
2 2
2 x (1 + ... + 6 ). The times 2 is because of symmetry.
This should add to 623.
-- Koh Kew Soon <law20065@leonis.nus.sg>
----
I think there are more circles than this. Isn't there a circle that
passes through (1,0) (0,3) (0,5) (1,8)?
-- Gareth Rees <gdr11@cl.cam.ac.uk>
----
> The number of circles that can be drawn is the same as the number
> of squares which can be drawn such that each corner rests on a
> marker.
This is an erroneous assumption. You have missed all the rectangles
that can be drawn using four markers, as circles can also be
constructed using these! But moreso even, you have missed the regular
parallelograms that can be used to construct circles (eg the points
a2, b1, e2 and b5 can be used to construct a legal circle). Beyond
that, you need to start considering the irregular quadrilaterals that
might fit the criteria.
I am afraid you have a lot more counting to do before you have the
answer to the question as you have stated it.
-- Craig Spencer <cas@cs.adelaide.edu.au>
----
Plus the number of such rectangles and symmetrical trapezoids!
Plus some special cases such as the circle centered at (0,0) with
radius 5 passing through markers at (0,5), (3,4), (4,3) and (5,0).
Minus those that pass through more than 4 markers.
-- Tapani Lindgren <Tapani.Lindgren@Fin1041.ICL.ICL.Elisa.Fi>
----
I am very very very sorry. I mixed up two different questions and thus
gave you the wrong answer for this question. Here are the two
questions and their respective (hopefully correct) answers:
Start with a chessboard. Place a marker on every intersection point,
including those along the perimeter and the four corners. This means
81 markers in all.
Q1) How many different squares can you draw in 3-d space (inside or
outside the chess board) so that all four corners of each square
rest on markers?
Q2) How many different circles can you draw in 3-d space (inside or
outside the chess board) which pass through EXACTLY 4 markers?
A1) No. of squares =
2 2 2 2 2 2 2 2 2
(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 )
2 2 2 2
+ (1 + 3 + 5 + 7 )
2 2 2 2 2 2
+ (1 + 2 + 3 + 4 + 5 + 6 ) x 2 (for symmetry)
= 551 (not 623!)
A2) Quite right. As somebody has kindly pointed out, there are other
possibilities such as rectangles and some parallelograms (but not
all, because most of the circles which go through these will pass
through MORE than 4 markers). I am fairly (though not absolutely)
certain that the following answer is correct (barring any errors
in arithmetic):
No. of circles which pass through EXACTLY 4 markers =
2 2 2 2 2 2 2 2 2
(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 )
{which accounts for all squares}
+ [ 8(1+3+5+7)+7(2+4+6)+6(1+3+5)+5(2+4)+4(3+1)+3(2)+2(1) ] x 2
{which accounts for all rectangles}
+ [ (1+2+3+4+5+6+7)+(1+2+3+4+5)+(1+2+3)+(1+2)+(1) x 4
{which accounts for all parallelograms, rombuses, etc.}
= 1221
{if my addition is correct}
-- Koh Kew Soon <law20065@leonis.nus.sg>
[ Editor: A1 checks: 285 + 84 + 2 x 91 = 551. The three terms of A2
evaluate to 285, 2 x 320 = 640 and (28+15+6+3+1) x 4 = 212, which
total 1137, not 1221 (the closing bracket of the last term is missing
in the original; reading "x 4" as applying to the whole bracket). ]