[Guest]

A chessboard has squares that are two inches on the side. What is the radius of the largest circle that can be drawn on the board in such a way that the circle's circumference is entirely on black squares?

[Guest]

A chessboard has squares that are two inches on the side. What is the radius of the largest circle that can be drawn on the board in such a way that the circle's circumference is entirely on black squares?

if the circle were all on a single black square, that would count for 1 inch radius.

Or it could lie on 4 black squares, touching at the corners... that's a radius of somewhere around 1.4 inches i guess.

or that long diagonal of 8 black squares which would make a radius of about 11.31, because if the circle touched those two corners, it'd touch the white ones as well.

The 4x4 black square looks too big, i'm guessing it'd bulge out to white on that one, with a minimum radius to hit the corner blacks at ~5inches.

I'm too stupid today to calculate it, but I'm guessing it'd be in a section where it could be inscribed on a square of black squares, 3blackx3black (or 6"x6"). which would make the radius somewhere just over 3 inches? Just a guess, of course, I don't even have the back of an envelope available to do the math.

[Guest]

Actually, I can think of two ways to do it ...

One is to place the circle inside one black square ... pretty easy to figure out the radius on that one.

The other one is to place the circle with a white square centered within the circle so that the circumference passes between the tangent points of the black squares.

For the first option, the radius is obviously 1 inch.

For the second option, the radius would be 2.8284 ... Thank you, Pythagoras!

[Guest]

1 inch

Edited July 1, 2009 by nikkie

[Guest]

zero. A circle cannot be dawn only touching either the black or white squares

I suppose that depends on how the board is drawn. It does sound reasonable that if the intersection between 2 black squares (at the corners) counts as all black, then that same intersection of 2 white squares' corners should count as all white. But it can't be both.

On my chessboard, the black ink bled a little, so it's all black for that tiny space. And, now I can go beyond 1 square!

[Guest]

5 inches

[Guest]

While centered on a black square, the radius can be just over 3.16 (divide the square root of 40 by 2; this circle intersects corners that are on opposite corners of a three square line. This means, by pythagoras, we sum the squares of 6 and 2; 36+4=40. The square root is the diameter, half that is the radius). This creates intersections at 8 corners and keeps the circle on the 8 black squares which surround the center square. Any larger circle passes into a white (or red if you use on a checkers board) square.

[Guest]

maximum radius is equal to square root of 10 ~ 3.16"

A chessboard has squares that are two inches on the side. What is the radius of the largest circle that can be drawn on the board in such a way that the circle's circumference is entirely on black squares?

[Guest]

I believe the largest radius would be the square root of 5 which is 2.23606...

[Guest]

It's the square root of 2, not five - 1.414...

I believe the largest radius would be the square root of 5 which is 2.23606...

[Guest]

I'd go for root 10 if it can pass through the corners, which I got from pythag root(1^2 + 3^2). I don't see how it could be any bigger. I can't be bothered to draw & upload a diagram though.

[Guest]

i admit i havent much thinking on this,

logically if we are accepting that black corners are continuous or whatever then it still only happens at an exact point meaning logically the only possibilities for crossing between squares it to be going straight up, side ways, or at an exact 45 degree angle. The straight up is discounted because then the circle would curve the same direction above and below that point making it go into a white square in one of them. Same logic for horizontal.

So the only possibility is 45 degrees which would be a circle centered on one white square and going around with radius exactly to the corner. so that diaonal would be

2^2+2^2=8

tkae the root

2square root 2

half the distance so square root of two.

[Guest]

i admit i havent much thinking on this,

logically if we are accepting that black corners are continuous or whatever then it still only happens at an exact point meaning logically the only possibilities for crossing between squares it to be going straight up, side ways, or at an exact 45 degree angle. The straight up is discounted because then the circle would curve the same direction above and below that point making it go into a white square in one of them. Same logic for horizontal.

So the only possibility is 45 degrees which would be a circle centered on one white square and going around with radius exactly to the corner. so that diaonal would be

2^2+2^2=8

tkae the root

2square root 2

half the distance so square root of two.

In order for the answer to be anything other than 1" radius we have to acknowledge that intersections of squares can be used as continuations of the black squares. The intersections do not have to occur at any partivular angle as long as the point of intersections lie on the circle. This leads me to the conclusion that I posted earlier.

square root of 10 = approximately 3.16"