White Circle in a Chessboard

This solution revolves around the fact that three non-collinear points determine a unique circle. A circle cannot be made from 3 collinear points.

Since this circle lies entirely on white squares, it cannot pass through the edge of any square except at the vertices. So, if one white square has part of the circumference of the circle, one square diagonally adjacent to it must contain the circumference of the circle as well (The exception to this is a circle completely inside one white square, which has a radius of at most $0.5$ . However, we can find bigger circles!).

Now, consider two white squares diagonally adjacent to each other that contain the circumference of the circle. There are 7 vertices between the two squares (1 is shared by both) and we must choose 3 non-collinear points from these 3 such that the radius of the circle that the 3 points define is the greatest. One of these points is clearly the shared vertex. For the other 2 points, they must lie in different squares. There are 2 valid (non-collinear) possibilities:

Case 1: Both of the points are adjacent to the first point (such that the three points are not collinear)

Drawing the circle from this, it's clear that the radius of this circle is $\frac{\sqrt{2}}{2}$ . A little bigger than the simple $0.5$ radius circle we found in the beginning of the problem!

Case 2: One point is adjacent to the first point and the other is diagonally opposite from the first point.

This arc ends up being $1/4$ of the circumference, and by symmetry, the rest of the circle will pass only on white squares as well. Drawing a line segment from the center to one of the corners the circumference hits and using the Pythagorean Theorem, we find that the radius of this circle is $\frac{\sqrt{10}}{2}$ , which is biggest we have found, and the answer is $13$ (Remember that $a=1$ , not $0$ .

First remark that $\frac{1 \sqrt{10}}{2}$ is obtainable by having a cross with $5$ squares and then making a circle go through all the "outer vertices" of the cross, i.e. the vertices of the convex hull of the cross.

Now to prove this is maximal. Remark that our circle must go through the corners of the squares on the chessboard whenever it reaches an edge if it is always to lie on white squares. This means whenever the circle enters a square, there are 3 possibilities where it leaves the square. Now, it is clear that if the circle makes 2 consecutive diagonal movements (i.e. moves from one corner to the opposite one) it is not a circle as then there are 3 collinear points on the circle, contradiction. Furthermore, at most 4 times can the circle make the other type of move (move from one corner to an adjacent one) because every time we do this we define either the x-coordinate or the y-coordinate of the circle and clearly we cannot define either one of those more than twice because a circle centered at $a + 0.5, b + 0.5$ for integers $a,b$ only passes through the lines $x = a, x = a+1$ twice and same for $y = b, y = b+1$ . Thus the circle can go through at most $4 + 4(1) = 8$ squares. From here it is easy to check the remaining cases to deduce indeed $\frac{1 \sqrt{10}}{2}$ is the maximal radius.

Notice that for a curve to pass though only white squares, it either must be in a white $1\times1$ square or pass through corners of white squares in order to the next white square closest to it. We have two cases:

Case 1: The circle doesn't pass through any corners (i.e., it is contained in a single $1\times1$ white square).

The maximum radius of such a circle is $\frac12$ .

Case 2: The circle passed through corners.

Let one of these corner points be point $A$ . Since 3 points defines a circle, we have to choose 2 more corner points that the circle passes through. Let the three other vertices of a white square with point $A$ as a vertex be $B$ , $B'$ , and $B''$ , as shown in the diagram. Similarly, let the three vertices of a white square on the other side of point $A$ be $C$ , $C'$ , and $C''$ . Note that the two vertices we choose cannot be from the same side of $A$ because then the circle will cross into black squares.

Subcase 2.1: The circle passes through $A$ , $B'$ , and $C'$ .

$A$ , $B'$ , and $C'$ are collinear, so there is no circle passing through these points.

Subcase 2.2: The circle passes through $A$ , $B'$ , and $C$ .

This circle has radius $\sqrt{\left(\frac12\right)^2+\left(1+\frac12\right)^2}=\frac{\sqrt{10}}2$ .

Subcase 2.3: The circle passes through $A$ , $B'$ , and $C''$ .

This circle is congruent to the one in Subcase 2.2.

Subcase 2.4: The circle passes through $A$ , $B$ , and $C$ .

This circle has radius $\sqrt{\left(\frac12\right)^2+\left(\frac12\right)^2}=\frac{\sqrt{2}}2$ .

Subcase 2.5: The circle passes through $A$ , $B$ , and $C'$ .

This circle is congruent to the ones in Subcases 2.2 and 2.3.

Subcase 2.6: The circle passes through $A$ , $B$ , and $C''$ .

$A$ , $B$ , and $C''$ are collinear, so there is no circle passing through these points.

Subcase 2.7: The circle passes through $A$ , $B''$ , and $C$ .

$A$ , $B''$ , and $C$ are collinear, so there is no circle passing through these points.

Subcase 2.8: The circle passes through $A$ , $B''$ , and $C'$ .

This circle is congruent to the ones in Subcases 2.2, 2.3 and 2.5.

Subcase 2.9: The circle passes through $A$ , $B''$ , and $C''$ .

This circle is congruent to the ones in Subcases 2.4.

Therefore, the largest radius is $\frac{\sqrt{10}}2$ , so the answer is $1+10+2=\boxed{13}$ .

(Examples of both types of circles are in the figure)

No matter how big the chessboard is, the largest circle that can be drawn completely on white squares will only contain 4 black squares and 1 white square inside it. Here is an illustration: http://i107.photobucket.com/albums/m285/sagisttarius/puzzle/chessboard-circum.jpg

Moreover, we can find the diameter of the circle by getting the diagonal of the 3x1 rectangle inside the circle.

Using pythagorean theorem: c^2 = a^2 + b^2 c^2 = 1^2 + 3^2 c^2 = 1 + 9 c^2 = 10 c = \sqrt{10}

Dividing the diameter by two, we will get the radius of the circle. \frac {\sqrt{10}}{2}

writing in the form \frac {a \sqrt {b}}{c} , we get \frac {1 \sqrt{10}}{2} since 1 and 2 are coprime, and 10 is not divisible by the square of any prime, we will get a = 1 b=10 c = 2 Thus, a + b + c = 13

Since every point of the circle is contained in white squares, we may take any particular arc, and it will be completely inside or in the border of a white square.

In particular, if we observe the border line of a white square, we may notices that the circle may not intersect the border since a arc of the circle will be now contained in a black square, which we don't want.

Hence, the circle intersects the white squares in two of their vertex. Now, the circle may pass by two consecutive vertex of the square, or by two diagonally opposite vertex of a white square.

We may also notice that the points where the circle passes are all contained in the same circle, thus, there are only two possible arrangement of the vertex where the circle passes, and they have radius

$\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{10}}{2}$

so the largest radius is $\frac{1\sqrt{10}}{2}$ and thus the answer is $1+2+10=13$.

The corners of the squares are lattice points on a coordinate plane. Call a square $m \Box n$ if its upper left corner is $(m,n)$ . Note that a circle passing through three vertices of a square will circumscribe the square.

Put the positive $y$ extreme of the circle in $0 \Box 0$ . The circle passes through $(0, -1)$ and $(1, -1)$ and is symmetric over $x = \frac {1}{2}$ . Now consider $1 \Box -1$ . The circle either passes through $(1, -2)$ or $(2, -2)$ (if it passed through $(2, -1)$ it would have infinite radius). If it passes through $(1, -2)$ it contains three vertices of $0 \Box -1$ and we're done with a radius of $\frac {1\sqrt2}{2}$ . If it passes through $(2, -2)$ , we must consider $2 \Box -2$ .

In $2 \Box -2$ the circle can only pass through $(2, -3$ . Through $(3, -3)$ it would go through three collinear points, while through $(3, - 2)$ it would circumscribe a non-rectangular parallelogram, which is impossible.

By symmetry, the circle now also passes through $(-1, -2)$ and $(-1, -3)$ . We have three vertices of a square, $(-1, -3)$ , $(0, -1)$ and $(2, -2)$ . Since the circle cannot pass through $(-1, -4)$ nor $(2, -4)$ as it would pass through three collinear points, this is the largest square circumscribed by a visible circle. Its diagonal and thus the diameter of the circle is $\sqrt{(-2 - -3)^2 + (2 - -1)^2} = \sqrt{10}$ and the radius is $\frac {1\sqrt{10}}{2}$ . $1 + 10 + 2 = 13$ .

If the circle is contained within 1 white square, then the radius is $\leq \frac {1}{2}$ . Otherwise, if the circle crosses several white squares, it must cross over in a corner. Mark the corner points on the circle. Consider the points on the right half of the circle. Since the circle has finite radius, it must cross over 2 corners that are directly vertically over each other. Let these points be labeled $A$ and $B$ , with $B$ 1 unit vertically over $A$ . Let $C$ be the corner that is 1 unit vertically above $B$ , $D$ be the corner that is 1 unit to the left of $C$ (diagonally opposite from $B$ ) and $E$ be the corner that is 1 unit to the left of $B$ . The circle must then pass through one of these corners.

The circle cannot pass through $C$ , since $ABC$ is a straight line. If the circle passes through $E$ , the center is the center of the square with vertices $A, B$ and $E$ , so it has a radius of $\frac {\sqrt{2}} {2}$ . If the circle passes through $D$ , it is the circumcircle of triangle $ABD$ . By the extended sine rule , $R = \frac {BD} {2 \sin \angle DAB} = \frac {\sqrt{2} } {2 \frac {1} {\sqrt{5} } } = \frac {\sqrt{10} } {2}$ . This is clearly the largest radius.

Hence, $a + b + c = 1 + 10 + 2 = 13$ .

Denote $A$ as the square that contains the center of the circle. By symmetry, the center of the square is the center of the circle. Let $O$ be this center.

Clearly, there exists a circle that circumscribes $A$ ( $A$ is black in this case) and one that is inscribed in $A$ ( $A$ is white). However, there might be a circle with a larger radius.

Let there be $n$ , $n\ge 1$ squares above this center square inside the circle. By the problem conditions, the topmost square must be black. Then the square to the immediate left of this black square is a white square, and the circle must pass through the white square. In particular, this circle must pass through the point $X$ , the top right vertex of the white square, and the point $Y$ , the bottom left vertex of the white square. We can compute $OX$ and $OY$ in terms of $n$ using the Pythagorean Theorem. But both $OX$ and $OY$ are radii, so they must be equal.

Therefore, $\sqrt{\left(n+\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\left((n-1)+\frac{1}{2}\right)^2+\left(1+\frac{1}{2}\right)^2}$ , or $n^2+n+\frac{1}{4}+\frac{1}{4}=n^2-n+\frac{1}{4}+\frac{9}{4}\implies 2=2n$ , so $n=1$ .

Clearly, this circle is larger than the circumscribed and inscribed circle of $A$ , so when $n=1$ , we get the circle with the maximum radius. This gives us a radius of $\sqrt{\left(1+\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{10}}{2}$ , so $a+b+c=\boxed{13}$ .