answer:1. **Identify the problem and its components:** - We are given a regular triangle with side length ( a ). - We need to calculate the fraction of the area of the circumcircle that is covered by three smaller circles of radius ( frac{a}{2} ) drawn around each vertex of the triangle. 2. **Find the radius of the circumcircle:** - The radius ( R ) of the circumcircle of a regular triangle is given by the formula: [ R = frac{a sqrt{3}}{3} ] 3. **Angles and segments involved:** - Denote by ( beta ) the angle subtended at the center of the circumcircle by one of the circles with radius ( frac{a}{2} ). - Since the circumcircle is centered at ( O ) and goes through the vertices of the equilateral triangle, each of these angles corresponds to ( 120^{circ} ). 4. **Calculate (beta):** - By symmetry, the central angle (beta) for one of these circles can be found using the cosine rule. Let’s denote one side of the triangle as (A), and the side after rotating (120^{circ}) around the circle as ( B ). [ cos beta = frac{left(frac{a sqrt{3}}{3}right)^2 + left(frac{a sqrt{3}}{3}right)^2 - left(frac{a}{2}right)^2}{ 2 cdot frac{a sqrt{3}}{3} cdot frac{a sqrt{3}}{3}} ] - Simplify the equation: [ cos beta = frac{frac{a^2 sqrt{3}}{3} + frac{a^2 sqrt{3}}{3} - left( frac{a}{2} right)^2}{2 cdot frac{a sqrt{3}}{3} cdot frac{a sqrt{3}}{3}} = frac{sqrt{3}}{4} ] - Solving for (beta): [ beta = cos^{-1} left( frac{sqrt{3}}{4} right) approx 64^{circ} 20^{prime} ] - In radians: [ beta = frac{64.34 cdot pi}{180} approx 1.123 text{ radians} ] 5. **Calculate the area of each segment and sector:** - The area of one segment in terms of (beta): [ A_{text{segment}} = frac{1}{2} left( frac{a}{2} right)^2 left( beta - sin beta right) ] - The area of one sector can be calculated directly: [ A_{text{sector}} = frac{1}{2} left(frac{a}{2}right)^2 cdot beta = frac{1}{2} cdot frac{a^2}{4} cdot 1.123 = a^2 cdot frac{1.123}{8} ] 6. **Sum up the areas:** - Total area of the three sectors: [ 3 cdot a^2 cdot frac{1.123}{8} = a^2 cdot frac{3 cdot 1.123}{8} ] - Calculate the sectors area: [ 3 A_{text{sector}} = frac{3}{8} a^2 times 1.123 ] 7. **Area of the segments:** - Sum the segment contributions: [ 3 times T_{text{segment}} = 3 times a^2 times frac{1}{8} times 1.123 approx a^2 cdot frac{3 cdot 1.123}{8} ] 8. **Circumcircle area:** [ A_{text{circumcircle}} = pi left(frac{a sqrt{3}}{3}right)^2 = frac{pi a^2}{3} ] 9. **Fraction of the area covered:** [ frac{text{Sum of area of circles}}{text{Circumcircle area}} = frac{3 cdot a^2 cdot frac{1.123}{8}}{frac{pi a^2}{3}} = frac{3 cdot 1.123}{8 pi} approx 0.914 ] Conclusion: [ boxed{0.914} ]

answer:1. **Claim**: All primes of the form (3k+1) can be written in the form (a^2 + ab + b^2) for some positive integers (a) and (b). It is well known that there exist infinitely many primes of this form. 2. **Lemma**: Given any prime (p) of the form (3k+1), there exists a solution to the equation (x^2 + x + 1 equiv 0 pmod{p}). **Proof**: This is equivalent to solving the quadratic congruence ((2x+1)^2 equiv -3 pmod{p}). Using Legendre notation, we need to show that (left(frac{-3}{p}right) = 1). We have: [ left(frac{-3}{p}right) = left(frac{3}{p}right) left(frac{-1}{p}right) ] It is known that (left(frac{-1}{p}right) = (-1)^{frac{p-1}{2}}). By quadratic reciprocity, we have: [ left(frac{3}{p}right) left(frac{p}{3}right) = (-1)^{frac{p-1}{2} cdot frac{3-1}{2}} = (-1)^{frac{p-1}{2}} ] Since (p = 3k+1), we have (left(frac{p}{3}right) = 1) and thus: [ left(frac{3}{p}right) = (-1)^{frac{p-1}{2}} ] Therefore: [ left(frac{-3}{p}right) = (-1)^{p-1} = 1 ] So by our lemma, for any prime (p) of the form (3k+1), there exists an (x) such that (x^2 + x + 1 equiv 0 pmod{p}). 3. **Existence of (a) and (b)**: Given the solution (x) to (x^2 + x + 1 equiv 0 pmod{p}), we can find integers (a) and (b) such that (a^2 + ab + b^2 = mp) for some integer (m). Note that since (x leq p-1), (m < p) and thus (m) is coprime to (p). 4. **Reduction to (m=1)**: Suppose (m) is strictly greater than 1. Let (a = a' pmod{m}) and (b = b' pmod{m}) where (-frac{m}{2} leq a', b' leq frac{m}{2}). Consider the following: [ frac{(a^2 + ab + b^2)(b'^2 + a'b' + a'^2)}{m^2} = left(frac{ab' - ba'}{m}right)^2 + left(frac{ab' - ba'}{m}right)left(frac{aa' + bb' + ba'}{m}right) + left(frac{aa' + bb' + ba'}{m}right)^2 ] From our definitions of (a') and (b'), and from the knowledge that (a^2 + ab + b^2 equiv 0 pmod{m}), it is clear that our new solution consists of integers. Also, the result is still a multiple of (p), as (m) and (p) are coprime. Not only that, but: [ a'^2 + a'b' + b'^2 leq frac{3}{4}m^2 ] Thus, if we were to write this new solution as (np) for some integer (n), it is clear that (n < m). Hence, we repeat this process until (m = 1) and hence we are done. (blacksquare)

answer:To determine all pairs ((m, n)) for which there exists a simple polygon (P) with vertices in (S_{m,n}) such that all points in (S_{m,n}) are on the boundary of (P), all interior angles of (P) are either (90^circ) or (270^circ), and all side lengths of (P) are (1) or (3), we proceed as follows: 1. **Initial Observations**: - Since all points in (S_{m,n}) must be on the boundary of (P), (m) and (n) must be even. This is because each side of the grid must have an even number of points to ensure that the polygon can close properly with the given constraints on side lengths and angles. 2. **Special Cases**: - For (m = 2) and (n = 2), we can construct a simple polygon that satisfies the conditions. The polygon is simply the boundary of the (2 times 2) grid. - For (m = 4) and (n = 2), or (m = 2) and (n = 4), we can also construct simple polygons that satisfy the conditions. These constructions can be visualized as follows: - For (m = 4) and (n = 2), the polygon can be constructed by following the boundary of the grid in a zigzag manner. - For (m = 2) and (n = 4), the polygon can be constructed similarly by following the boundary in a zigzag manner. 3. **General Construction**: - When both (m) and (n) are even and at least one of them is divisible by 4, we can construct a simple polygon that satisfies the conditions. The construction involves creating a zigzag pattern along the boundary of the grid, ensuring that all side lengths are either (1) or (3) and all interior angles are (90^circ) or (270^circ). 4. **Checkerboard Argument**: - Consider the grid as a checkerboard with alternating black and white squares. When traveling along the boundary of the polygon, each movement flips the color of the square. This ensures that horizontal movements end on white squares and vertical movements end on black squares. - If two parts of the path are adjacent without any squares in between, they must flow in opposite directions. This prevents the construction of paths that move in the same direction adjacent to each other. 5. **Key Claim**: - If there is a horizontal side of length 3, the horizontal movements must alternate between left and right. This is because any deviation would result in paths moving in the same direction adjacent to each other, which is not allowed. 6. **Contradiction for (m, n equiv 2 pmod{4})**: - If both (m) and (n) are congruent to 2 modulo 4 and at least 6, it is impossible to construct a polygon (P) that satisfies the conditions. This is because any attempt to construct such a polygon would result in paths moving in the same direction adjacent to each other, leading to a contradiction. 7. **Conclusion**: - The pairs ((m, n)) for which there exists a simple polygon (P) that satisfies the conditions are ((2, 2)) and all even pairs for which 4 divides at least one of the two numbers. The final answer is ( boxed{ (2, 2) } ) and all even pairs for which 4 divides at least one of the two numbers.

answer:1. **Identify the given information and setup the problem:** - The diameter of the larger circle is (20), so its radius (R) is (10). - Points (A, B, C, D, E, F) are equally spaced along the circumference of the larger circle. - A second circle is tangent to the lines (AB) and (AF) and internally tangent to the larger circle. - We need to find the diameter of the second circle, which is given as (sqrt{m} + n), and then find (m + n). 2. **Determine the angle subtended by the points:** - Since the points are equally spaced, the angle subtended by each pair of adjacent points at the center (O) of the larger circle is (frac{360^circ}{6} = 60^circ). 3. **Position the points and the second circle:** - Place (A) at ((10, 0)) and (F) at ((5, 5sqrt{3})) in a coordinate system with (O) at the origin. - The second circle is tangent to (AB) and (AF) and internally tangent to the larger circle. 4. **Use geometric properties to find the radius (r) of the second circle:** - The center (O') of the second circle lies along the angle bisector of (angle AOF), which is (30^circ) from the x-axis. - The distance from (O) to (O') is (R - r), where (R = 10). 5. **Set up the equation using the tangent properties:** - The distance from (O') to the line (AF) is (r). - The distance from (O') to the line (AB) is also (r). - Using the coordinates, the distance from (O') to the line (AF) can be calculated using the perpendicular distance formula. 6. **Calculate the radius (r):** - The coordinates of (O') are ((10 - rcos(30^circ), rsin(30^circ))). - Using the distance formula, we get: [ r = frac{10 sin(30^circ)}{1 + cos(30^circ)} = frac{10 cdot frac{1}{2}}{1 + frac{sqrt{3}}{2}} = frac{5}{1 + frac{sqrt{3}}{2}} = frac{5}{frac{2 + sqrt{3}}{2}} = frac{10}{2 + sqrt{3}} ] - Rationalize the denominator: [ r = frac{10}{2 + sqrt{3}} cdot frac{2 - sqrt{3}}{2 - sqrt{3}} = frac{10(2 - sqrt{3})}{4 - 3} = 10(2 - sqrt{3}) = 20 - 10sqrt{3} ] 7. **Find the diameter of the second circle:** - The diameter of the second circle is (2r = 2(20 - 10sqrt{3}) = 40 - 20sqrt{3}). 8. **Express the diameter in the form (sqrt{m} + n):** - Comparing (40 - 20sqrt{3}) with (sqrt{m} + n), we see that (m = 1200) and (n = 40). 9. **Calculate (m + n):** [ m + n = 1200 + 40 = 1240 ] The final answer is (boxed{1240}).