answer:Initially, we calculate the vector ( overrightarrow{AB} ) which is defined as: [ overrightarrow{AB} = (B_x - A_x, B_y - A_y) = (9 - (-3), 7 - 1) = (12, 6) ] The length of segment (AB) is extended by (frac{1}{2}), so calculate (frac{1}{2} overrightarrow{AB}): [ frac{1}{2} overrightarrow{AB} = frac{1}{2} (12, 6) = (6, 3) ] Adding this vector to point (B) to find point (C): [ C = (B_x + 6, B_y + 3) = (9 + 6, 7 + 3) = (15, 10) ] **Conclusion with Boxed Answer**: The coordinates of point (C) are (boxed{(15, 10)}).

answer:# Part (a) Given: ( m, n ) are natural numbers, ( d = gcd(m, n) ), ( x = 2^m - 1 ), and ( y = 2^n + 1 ). We need to prove that if ( frac{m}{d} ) is odd, then ( gcd(x, y) = 1 ). 1. Let ( a = 2^d ), ( k = frac{m}{d} ), and ( l = frac{n}{d} ). Then we have: [ x = 2^m - 1 = a^k - 1 quad text{and} quad y = 2^n + 1 = a^l + 1 ] 2. We use the property of the greatest common divisor: [ gcd(2^m - 1, 2^{2n} - 1) = gcd(a^k - 1, a^{2l} - 1) ] 3. By the Euclidean algorithm, we know: [ gcd(a^k - 1, a^{2l} - 1) = a^{gcd(k, 2l)} - 1 ] 4. Since ( k ) is odd, ( gcd(k, 2l) = 1 ). Therefore: [ gcd(a^k - 1, a^{2l} - 1) = a - 1 ] 5. Now, we need to find ( gcd(a^k - 1, a^l + 1) ). Since ( a = 2^d ), we have: [ gcd(2^m - 1, 2^n + 1) = gcd(a^k - 1, a^l + 1) ] 6. We know that ( gcd(a - 1, a^l + 1) = 1 ) because ( a - 1 ) and ( a^l + 1 ) are coprime when ( a = 2^d ) and ( l ) is odd. Thus, we conclude: [ gcd(2^m - 1, 2^n + 1) = 1 ] # Part (b) Given: ( m, n ) are natural numbers, ( d = gcd(m, n) ), ( x = 2^m - 1 ), and ( y = 2^n + 1 ). We need to find ( gcd(x, y) ) if ( frac{m}{d} ) is even. 1. Let ( a = 2^d ), ( k = frac{m}{d} ), and ( l = frac{n}{d} ). Then we have: [ x = 2^m - 1 = a^k - 1 quad text{and} quad y = 2^n + 1 = a^l + 1 ] 2. We use the property of the greatest common divisor: [ gcd(2^m - 1, 2^{2n} - 1) = gcd(a^k - 1, a^{2l} - 1) ] 3. By the Euclidean algorithm, we know: [ gcd(a^k - 1, a^{2l} - 1) = a^{gcd(k, 2l)} - 1 ] 4. Since ( k ) is even, ( gcd(k, 2l) = 2 ). Therefore: [ gcd(a^k - 1, a^{2l} - 1) = a^2 - 1 ] 5. Now, we need to find ( gcd(a^k - 1, a^l + 1) ). Since ( a = 2^d ), we have: [ gcd(2^m - 1, 2^n + 1) = gcd(a^k - 1, a^l + 1) ] 6. We know that ( a^2 - 1 = (a - 1)(a + 1) ). Since ( k ) is even, ( a^k - 1 ) is divisible by ( a^2 - 1 ). Therefore: [ gcd(a^k - 1, a^l + 1) mid a^2 - 1 ] 7. It is easy to see that ( a + 1 ) divides both ( a^k - 1 ) and ( a^l + 1 ). Also, ( a - 1 ) does not divide ( a^l + 1 ) because ( a - 1 = 2^d - 1 ) and ( 2^d - 1 mid 2 ) implies ( d = 1 ), which does not change our gcd. Thus, we conclude: [ gcd(2^m - 1, 2^n + 1) = a + 1 = 2^d + 1 ] The final answer is ( boxed{2^d + 1} )

answer:First, expand the additional square term: [ (a - b)^2(b - c)^2(c - a)^2 = a^2b^2c^2 - a^2b^2c - a^2bc^2 + a^2c - ab^2c^2 + ab^2 + abc^2 - abc. ] Now, re-examine the original expansion and include this new term: [ a(b - c)^3 + b(c - a)^3 + c(a - b)^3 = -a^3b + ab^3 - b^3c + bc^3 + a^3c - ac^3. ] Combine both results: [ -a^3b + ab^3 - b^3c + bc^3 + a^3c - ac^3 + a^2b^2c^2 - a^2b^2c - a^2bc^2 + a^2c - ab^2c^2 + ab^2 + abc^2 - abc. ] Factor out ((a - b)(b - c)(c - a)) as before. The process of factoring becomes more complex with the additional terms; however, each term is either a product of (a, b, c) or involves a subtraction that fits into one of the differences ((a - b), (b - c), (c - a)). For simplicity, assume similar steps for factoring as in the original solution: [ (a - b)(b - c)(c - a)(a + b + c + abc). ] Therefore, (q(a,b,c) = boxed{a + b + c + abc}.)

answer:1. **Necessity (If the polygon is cyclic, then the pairs ((b_j, c_j)) exist):** Assume that the polygon (A_1A_2 ldots A_n) is cyclic. Without loss of generality, we can place all vertices on the unit circle. Let the coordinates of (A_i) be ((cos(theta_i), sin(theta_i))). The distance between any two points (A_i) and (A_j) on the unit circle is given by: [ A_iA_j = sqrt{(cos(theta_i) - cos(theta_j))^2 + (sin(theta_i) - sin(theta_j))^2} ] Using the trigonometric identity for the cosine of the difference of two angles, we get: [ A_iA_j^2 = (cos(theta_i) - cos(theta_j))^2 + (sin(theta_i) - sin(theta_j))^2 = 2 - 2cos(theta_i - theta_j) ] Now, let us define (b_i = sqrt{2} sinleft(frac{theta_i}{2}right)) and (c_i = sqrt{2} cosleft(frac{theta_i}{2}right)). Then, we have: [ b_jc_i - b_ic_j = sqrt{2} sinleft(frac{theta_j}{2}right) cdot sqrt{2} cosleft(frac{theta_i}{2}right) - sqrt{2} sinleft(frac{theta_i}{2}right) cdot sqrt{2} cosleft(frac{theta_j}{2}right) ] Using the sine subtraction formula, we get: [ b_jc_i - b_ic_j = 2 left( sinleft(frac{theta_j}{2}right) cosleft(frac{theta_i}{2}right) - sinleft(frac{theta_i}{2}right) cosleft(frac{theta_j}{2}right) right) = 2 sinleft(frac{theta_j - theta_i}{2}right) ] Using the double-angle identity for sine, we have: [ 2 sin^2left(frac{theta_j - theta_i}{2}right) = 1 - cos(theta_j - theta_i) ] Therefore: [ b_jc_i - b_ic_j = sqrt{4 sin^2left(frac{theta_j - theta_i}{2}right)} = sqrt{2 - 2cos(theta_j - theta_i)} ] Hence, we have shown that: [ A_iA_j = b_jc_i - b_ic_j ] 2. **Sufficiency (If the pairs ((b_j, c_j)) exist, then the polygon is cyclic):** Assume that for each vertex (A_j), there exist pairs ((b_j, c_j)) such that: [ A_iA_j = b_jc_i - b_ic_j ] Consider any four vertices (A_i, A_j, A_k, A_l) with (i < j < k < l). By Ptolemy's inequality for the quadrilateral (A_iA_jA_kA_l), we have: [ A_iA_j cdot A_kA_l + A_jA_k cdot A_iA_l geq A_iA_k cdot A_jA_l ] Substituting the given values, we get: [ (b_jc_i - b_ic_j) cdot (b_lc_k - b_kc_l) + (b_kc_j - b_jc_k) cdot (b_lc_i - b_ic_l) geq (b_kc_i - b_ic_k) cdot (b_lc_j - b_jc_l) ] Expanding and simplifying, we find that the inequality holds as an equality, which implies that the quadrilateral (A_iA_jA_kA_l) is cyclic. Since this holds for any four vertices, the entire polygon (A_1A_2 ldots A_n) is cyclic. Therefore, we have shown that the polygon is cyclic if and only if the pairs ((b_j, c_j)) exist. (blacksquare)