answer:Let the area of each smaller equilateral triangle be denoted as k. Since triangle ABM consists of one of these smaller triangles, its area is k. For triangle ADF, observe that it encompasses the entire hexagon. Since the hexagon is divided into twelve smaller equilateral triangles, the area of triangle ADF which covers the whole hexagon is 12k. Thus, the ratio of the areas is: [ frac{[triangle ABM]}{[triangle ADF]} = frac{k}{12k} = boxed{frac{1}{12}} ]

answer:According to the given information, the line l: x+2y-4=0 intersects the coordinate axes at points (4,0) and (0,2). The circle passing through points O, A, and B is the circumcircle of triangle OAB. Since triangle OAB is a right triangle, its circumcircle has a diameter equal to |AB|, and its center lies on the midpoint of AB. We calculate the length of AB: |AB| = sqrt{(4-0)^2 + (0-2)^2} = 2sqrt{5}. Thus, the radius of the circle is r = sqrt{5}, and the center's coordinates are (2,1). Therefore, the required circle's equation is: boxed{(x-2)^2 + (y-1)^2 = 5}.

answer:Consider a right triangle with sides (a), (b), and (c) where (a) and (b) are the legs and (c) is the hypotenuse. Given that (a), (b), and (c) are natural numbers and pairwise coprime. We start with the Pythagorean theorem for a right triangle: [ a^2 + b^2 = c^2 ] 1. **Examine the parity of (a) and (b):** - Suppose both (a) and (b) are odd. Then we can write: [ a = 2m + 1 ] [ b = 2n + 1 ] for some integers (m) and (n). 2. **Calculate their squares:** [ a^2 = (2m + 1)^2 = 4m^2 + 4m + 1 ] [ b^2 = (2n + 1)^2 = 4n^2 + 4n + 1 ] 3. **Sum these squares:** [ a^2 + b^2 = 4m^2 + 4m + 1 + 4n^2 + 4n + 1 = 4(m^2 + m + n^2 + n) + 2 ] 4. **Analyze the resulting expression:** - Notice that (a^2 + b^2) will be of the form (4k + 2) for some integer (k). - However, (c^2) must be a perfect square. Examining modulo 4, a perfect square (c^2) must satisfy: [ c^2 equiv 0, 1 pmod{4} ] - It is clear that (c^2 equiv 2 pmod{4}) is impossible because squares of integers modulo 4 can only be 0 or 1. Hence our assumption that both (a) and (b) are odd leads to a contradiction. 5. **Conclusion about parity:** - Therefore, (a) and (b) cannot both be odd. Since (a) and (b) are coprime pairs (natural numbers), one must be even, and the other must be odd. 6. **Conclusion about (c):** - We know from the Pythagorean theorem that (a) and (b) having different parities (one even, one odd) ensures (c) must be an odd number because it results from the sum of an even number and an odd number. Thus, the hypotenuse (c) of the right-angled triangle is an odd number, and the legs (a) and (b) have different parities (one is even, and the other is odd). [ boxed{c text{ is odd, and } a text{ and } b text{ have different parities}} ]

answer:By the Fundamental Theorem of Calculus, we have: - a= int_{0}^{2}x^{2}dx= frac{1}{3}x^{3} |_{0}^{2}= frac{8}{3}, - b= int_{0}^{2}x^{3}dx= frac{1}{4}x^{4} |_{0}^{2}= frac{16}{4}=4, - c= int_{0}^{2}sin xdx=-cos x |_{0}^{2}=1-cos 2 < 2, Therefore, c < a < b. Hence, the answer is: boxed{c < a < b}. By applying the Fundamental Theorem of Calculus, we can calculate the values of a, b, and c respectively. This question mainly tests the basic application of the Fundamental Theorem of Calculus, requiring proficiency in the integral formulas of common functions.