answer:(1) Since A and B are on the ellipse, we have |AF_1| + |AF_2| = |BF_1| = |BF_2| = 2a. Given that the perimeter of triangle ABF_1 is 8, we have |AF_1| + |AF_2| + |BF_1| = |BF_2| = 4a = 8, thus a = 2. By the symmetry of the ellipse, triangle AF_1F_2 forms an equilateral triangle if and only if A is at the vertex of the short axis of the ellipse, which implies a = 2c, hence c = 1, and b^2 = a^2 - c^2 = 3. Therefore, the equation of ellipse C is boxed{frac{x^2}{4} + frac{y^2}{3} = 1}. (2) Proof: If the slope of line l does not exist, i.e., l: x = 1, we find |AB| = 3, |MN| = 2sqrt{3}, thus frac{|MN|^2}{|AB|} = 4. If the slope of line l exists, let the equation of line l be y = k(x - 1). Let A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), D(x_4,y_4), substituting into the equation of the ellipse frac{x^2}{4} + frac{y^2}{3} = 1, we get (3 + 4k^2)x^2 - 8k^2x + 4k^2 - 12 = 0, yielding x_1 + x_2 = frac{8k^2}{3 + 4k^2}, x_1x_2 = frac{4k^2 - 12}{3 + 4k^2}, |AB| = sqrt{1 + k^2} cdot sqrt{(x_1 + x_2)^2 - 4x_1x_2} = frac{12(1 + k^2)}{3 + 4k^2}, substituting y = kx into the ellipse equation, we get x = ± frac{2sqrt{3}}{sqrt{3 + 4k^2}}, |MN| = 2sqrt{1 + k^2} cdot frac{2sqrt{3}}{sqrt{3 + 4k^2}} = 4sqrt{frac{3(1 + k^2)}{3 + 4k^2}}, thus frac{|MN|^2}{|AB|} = 4. In conclusion, frac{|MN|^2}{|AB|} is a constant value boxed{4}.

answer:1. **Define the variables:** Let ( m ) represent Mary's age, ( s ) represent Sally's age, and ( d ) represent Danielle's age. 2. **Relationships between the ages:** - Since Sally is 50% younger than Danielle, Sally's age is 50% of Danielle's age: [ s = 0.5d ] - Since Mary is 25% older than Sally, Mary's age is 125% of Sally's age: [ m = 1.25s ] 3. **Express Mary's age in terms of Danielle's age:** - Substitute ( s = 0.5d ) into ( m = 1.25s ): [ m = 1.25(0.5d) = 0.625d ] 4. **Equation for the sum of their ages:** - Given the sum of their ages is 42 years: [ m + s + d = 0.625d + 0.5d + d = 2.125d ] - Solve for ( d ): [ 2.125d = 42 implies d = frac{42}{2.125} = 19.7647 text{ (approximately 20)} ] 5. **Calculate Mary's current age and next birthday age:** - Substitute ( d approx 20 ) back into the equation for ( m ): [ m = 0.625 times 20 = 12.5 ] - Mary's age on her next birthday: [ 13.5 text{ (approximately 14)} ] Conclusion: The problem is valid as it provides a consistent and calculable scenario where the ages can be logically deduced and are in agreement with the initial assumptions. The final answer is boxed{B} (14)

answer:1. **Calculation of the equation of a line passing through point ( A ).** Consider a line ( d ) that passes through point ( A ). Let ( P ) be the intersection of this line with the line segment ( BC ). The baricentric coordinates of point ( P ) can be expressed using weights ( v ) and ( w ) such that ( P ) is the barycenter of ( (B, v) ) and ( (C, w) ). 2. **Barycentric coordinates condition for a line:** A point with barycentric coordinates ( (alpha, beta, gamma) ) lies on the line ( d ) if and only if: [ frac{gamma}{beta} = frac{mathrm{BP}}{mathrm{PC}} = frac{w}{v} ] Therefore, the equation of the line ( d ) can be written as: [ beta w - gamma v = 0 ] 3. **Equation for any line:** For a generic line, consider points ( P ) and ( Q ) on the line with homogeneous barycentric coordinates ( (x_1, y_1, z_1) ) and ( (x_2, y_2, z_2) ), respectively. A point ( M ) with coordinates ( (alpha, beta, gamma) ) lies on the line ( PQ ) if there exist real numbers ( lambda ) and ( mu ) such that ( M ) is the barycenter of ( (P, lambda) ) and ( (Q, mu) ): [ M : (lambda x_1 + mu x_2, lambda y_1 + mu y_2, lambda z_1 + mu z_2) ] 4. **Linear algebra properties for barycentric coordinates:** Using properties of linear algebra, the vector ( (alpha, beta, gamma) ) is a linear combination of the vectors ( (x_1, y_1, z_1) ) and ( (x_2, y_2, z_2) ). This is equivalent to the determinant condition: [ operatorname{det}begin{pmatrix} x_1 & x_2 & alpha y_1 & y_2 & beta z_1 & z_2 & gamma end{pmatrix} = 0 ] This expands to the equation: [ (y_1 z_2 - z_1 y_2) alpha + (z_1 x_2 - x_1 z_2) beta + (x_1 y_2 - y_1 x_2) gamma = 0 ] 5. **General form of a line in barycentric coordinates:** Therefore, the equation of a line in barycentric coordinates takes the form: [ u alpha + v beta + w gamma = 0 ] where ( (u, v, w) ) is a triplet of real numbers unique up to a multiplicative constant. --- 6. **Equation of a circle:** The equation of a circle in barycentric coordinates is generally more complex. It takes the form: [ -a^2 beta gamma - b^2 gamma alpha - c^2 alpha beta + (u alpha + v beta + w gamma)(alpha + beta + gamma) = 0 ] where ( (u, v, w) ) is a triplet of real numbers unique up to a multiplicative constant. # Conclusion: Hence, the equation of a line passing through a point in barycentric coordinates and the equation of a circle can be determined as specified above. [ boxed{} ]

answer:The probability of getting a head in each coin toss is frac{1}{2}. Therefore, the probability of getting exactly two heads in three tosses can be calculated using the combination formula C_{3}^{2} (representing the number of ways to choose 2 successes out of 3 trials) multiplied by the probability of getting a head twice and a tail once, which is (frac{1}{2})^{2}(frac{1}{2}) = frac{1}{8} times frac{1}{2} = frac{1}{16}. However, since there are 3 ways to get exactly two heads (HHT, HTH, THH), we multiply this probability by 3, resulting in boxed{frac{3}{8}}. This solution is derived from the formula for the probability of exactly k successes in n independent trials, which is C_{n}^{k}p^{k}(1-p)^{n-k}, where p is the probability of success in each trial and C_{n}^{k} is the binomial coefficient. In this problem, we are looking for the probability of exactly 2 successes (heads) in 3 independent trials (coin tosses), where the probability of success (getting a head) in each trial is frac{1}{2}.