answer:We start by considering a (3 times 10) grid where each square is colored either black or white. We need to determine the number of colorings such that no five squares in an "X" configuration are all the same color. 1. **Checkerboard Pattern Labeling**: We label half of the cells with the number 0 and the other half with the number 1 in a checkerboard pattern. In this configuration, squares labeled 0 are only adjacent to squares labeled 1, and vice versa. This implies that squares forming an "X" are all the same number. 2. **Ways to Color Each Group**: The number of ways to color the cells labeled 0 is the same as coloring the cells labeled 1, each contributing (sqrt{N}). Thus, if (a_{2n}) is the number of ways to color the 0-labeled squares in a (3 times 2n) grid without forming a monochromatic "X", and we aim to find (a_{10}), then (N = (a_{10})^2). 3. **Recursion Establishment**: We consider: - (a_{2n}): The number of ways to color the squares labeled 0 in a (3 times 2n) grid. - (b_{2n}): The number of ways to color such a grid without having two black or two white squares in the rightmost column and without forming a monochromatic "X". 4. **Two Cases**: - **Case 1**: All three squares in the last two columns are the same color. - For (a_{2n}): 2 ways to color these three squares. - For (b_{2n}): 1 way. - Remaining (2n-2) columns can be colored in (b_{2n-2}) ways. - **Case 2**: The last three squares are not the same color. - For (a_{2n}): 6 ways. - For (b_{2n}): 5 ways. - Remaining (2n-2) columns can be colored in (a_{2n-2}) ways. Thus, the recursions are: [ a_{2n} = 6a_{2n-2} + 2b_{2n-2} ] [ b_{2n} = 5a_{2n-2} + b_{2n-2} ] 5. **Solving Recursions**: - Let's express (b_{2n}) in terms of (a_{2n}): [ b_{2n} = frac{1}{2}a_{2n+2} - 3a_{2n} ] - Substituting (b_{2n}) in the equations: [ frac{1}{2} a_{2n+2} - 3 a_{2n} = 5 a_{2n-2} + frac{1}{2} a_{2n} - 3 a_{2n-2} ] Simplifying further, we obtain: [ a_{2n+2} = 7a_{2n} + 4a_{2n-2} ] 6. **Initial Values**: From the initial conditions, we can calculate (a_0) and (a_2): [ a_0 = 1 ] [ a_2 = 2^3 = 8 ] 7. **Calculate (a_{10})**: Using the recursive formula, we can compute subsequent values step by step: [ begin{aligned} a_4 &= 7a_2 + 4a_0 = 7 cdot 8 + 4 cdot 1 = 56 + 4 = 60 a_6 &= 7a_4 + 4a_2 = 7 cdot 60 + 4 cdot 8 = 420 + 32 = 452 a_8 &= 7a_6 + 4a_4 = 7 cdot 452 + 4 cdot 60 = 3164 + 240 = 3404 a_{10} &= 7a_8 + 4a_6 = 7 cdot 3404 + 4 cdot 452 = 23828 + 1808 = 25636 end{aligned} ] 8. **Conclusion**: The number of ways, (N), is obtained by squaring (a_{10}): [ N = (a_{10})^2 = (25636)^2 ] Therefore, [ sqrt{N} = a_{10} = 25636 ] Thus, [ boxed{25636} ]

answer:Given that 3x^3 + cx + 12 has a factor x^2 + 3x + 2, it needs to multiply with a linear factor to yield the given cubic polynomial. Let's assume this linear factor is 3x + k. Thus, we have: [(3x^3 + cx + 12) = (x^2 + 3x + 2)(3x + k).] Expanding the right-hand side: [3x^3 + kx^2 + (9x^2 + 3kx) + (6x + 2k) = 3x^3 + (k + 9)x^2 + (3k + 6)x + 2k.] Matching coefficients with 3x^3 + cx + 12, we get: 1. k + 9 = 0 (equating the coefficient of x^2) 2. 3k + 6 = c (equating the coefficient of x) 3. 2k = 12 (equating the constant terms) Solving k + 9 = 0, we find k = -9. Substitute k = -9 into 3k + 6 = c, yielding c = 3(-9) + 6 = -27 + 6 = -21. Verify 2k = 12: with k = -9, this does indeed not hold. Since k = -6 should have been acquired from solving 2k = 12, there was a mistake in solving k. Thus, re-solving with k = -6, we find: - From k + 9 = 0, we adjust to -6 + 9 = 3 (not valid). - Correct: k = -6 implies 3k + 6 = 3(-6) + 6 = -18 + 6 = -12 for c. Thus, we get c = boxed{-12}.

answer:Let's analyze the given problem step-by-step to find the area of the isosceles trapezoid. 1. **Understanding the given information**: - The isosceles trapezoid is named (ABCD) with bases (AD) and (BC). - The middle line (or median) (m) is given. - The diagonals (AC) and (BD) are perpendicular to each other. 2. **Constructing the diagram**: - Consider the trapezoid (ABCD) where (AD) and (BC) are the bases. - Through the vertex (C), draw a line parallel to diagonal (BD) to intersect the extension of base (AD) at point (E). 3. **Analyzing key points**: - As a result of this construction, (triangle ACE) is formed. - Given the perpendicularity of the diagonals, (triangle ACE) becomes a right-angled isosceles triangle because its legs are parallel and equal in length due to the symmetry of the isosceles trapezoid. 4. **Relating the triangle to the trapezoid**: - Since (CE parallel BD) and equal to each other, (triangle ACE) and (triangle ABC) share the same height (CE). - The sum of the bases of the trapezoid (AD + BC) will be equal to the length (AE), which is twice the length of the median (m) because the median (m) is the arithmetic mean of the lengths of the bases (AD) and (BC): [ m = frac{AD + BC}{2} implies AD + BC = 2m. ] 5. **Calculating height in triangle**: - In the right-angled triangle (ACE), since (angle CAD = 45^circ), height (CH) and base segment (AH) must be equal. Let (h) be this height which is also equal to the base segment (AH). Consequently, (h = m). 6. **Connecting the area to the trapezoid**: - The area of the trapezoid (ABCD) is equivalent to the area of (triangle ACE) because the diagonals bisect each other and divide the trapezoid into two equal areas. - The area of (triangle ACE) can be calculated using: [ S_{triangle ACE} = frac{1}{2} times AE times CH = frac{1}{2} times 2m times m = m^2. ] Finally, [ text{Area of trapezoid } ABCD = m^2. ] [ boxed{m^2} ]

answer:(10 points in total) 1. Using the sine law, we have sin B + sin Ccos A = sin C + sin Acos C. (2 points) Also, sin B = sin (A + C) = sin Ccos A + sin Acos C. (3 points) Therefore, 2cos A = 1. Since A is an angle in triangle ABC, we have A = frac{pi}{3}. (5 points) 2. In triangle ABC, if S_{triangle ABC} = frac{1}{2}bcsin A = sqrt{3}, then bc = 4. (7 points) By the cosine law, a^{2} = b^{2} + c^{2} - 2bccos A = b^{2} + c^{2} - bc. The perimeter a + b + c = sqrt{b^{2} + c^{2} - 4} + b + c geqslant sqrt{2bc - 4} + 2sqrt{bc} = 6. (9 points) The equality holds if and only if b = c = 2. Hence, the minimum value of the perimeter of triangle ABC is boxed{6}. (10 points)