answer:**Step 1:** Understanding that N cannot have three distinct prime divisors. To illustrate why N cannot have three distinct prime divisors: 1. Suppose p, q, r mid N where p, q, and r are distinct primes. 2. Consider the triples (p, q, 1), (p, r, 1), and (q, r, 1). By the multichromatic constraint, these triples must have distinct colors. Therefore, the set {p, q, r, 1} must have four distinct colors. 3. Next, consider the triples (pq, r, 1), (pq, pr, p), and (pq, qr, q). These must be distinctly colored as well, suggesting at least five distinct colors are necessary: {pq, p, q, r, 1}. Since this implies needing more than four colors, which is contradictory, it follows that N cannot have three distinct prime divisors. **Step 2:** Addressing other possibilities involving two primes. 1. Suppose ( N = p^3 q ) for distinct primes ( p ) and ( q ). 2. Considering the necessary triples: - (p, q, 1), (left(p^2, q, 1right)), (left(p^3, q, 1right)), - (left(p^2, pq, pright)), (left(p^3, pq, pright)), (left(p^3, p^2q, p^2right)). In each case, the tuples must have distinct colors, leading to the set ({1, q, p, p^2, p^3}) requiring five distinct colors. This leads to a contradiction. 3. Suppose ( N = p^2 q^2 ) for distinct primes ( p ) and ( q ). 4. Consider the necessary triples: - (p, q, 1), (left(p^2, q^2, 1right)), (left(p^2, q, 1right)), (left(p, q^2, 1right)). The distinct coloring of these requires the set ({1, p, q, p^2, q^2}) to consist of five distinct colors, another contradiction. **Step 3:** Categorizing ( N ) into two remaining cases. **Case 1: ( N = pq )**. Here, the only distinct triple is ((p, q, 1)). 1. Colors for ((p, q, 1)): (4 times 3 times 2 = 24) (since we have to choose 3 distinct colors out of 4 for 3 items). 2. The number of choices for ( pq ) seeing 4 colors is (4). 3. Hence, possible colorings are (4 times 24 = 96). **Case 2: ( N = p^2 q )**. The triples include ((p, q, 1)), ((p^2, q, 1)), and ((p^2, pq, p)). 1. Colors for ((1, p, q, p^2)) are four distinct colors: (4 times 3 times 2 times 1 = 24). 2. The number of ways to color ( pq ) is (2) (as it must be different from (p) and (p^2), leaving 2 choices). 3. The number of ways to color ( p^2 q ) subsequently is ( 4 ). Combining these steps, the total number of colorings is (24 times 2 times 4 = 192). **Conclusion:** The maximum possible number of multichromatic colorings for a positive integer N that is not a power of any prime is: [ boxed{192} ]

answer:# Step-by-Step Solution: 1. Finding the value of cos C Given overrightarrow{CA} cdot overrightarrow{CB} = 88, we can express this in terms of side lengths and the cosine of the angle between them, leading to abcos C = 88. Given A = 2B, we can find cos A using the double angle formula: [ cos A = cos 2B = 2cos^2 B - 1 = 2left(frac{2}{3}right)^2 - 1 = -frac{1}{9}. ] For sin B, we use the Pythagorean identity: [ sin B = sqrt{1 - cos^2 B} = sqrt{1 - left(frac{2}{3}right)^2} = frac{sqrt{5}}{3}. ] Similarly, for sin A: [ sin A = sqrt{1 - cos^2 A} = sqrt{1 - left(-frac{1}{9}right)^2} = frac{4sqrt{5}}{9}. ] To find cos C, we use the fact that C = pi - (A + B), and thus: [ cos C = -cos(A + B) = -(cos A cos B - sin A sin B). ] Substituting the values of cos A, cos B, sin A, and sin B: [ cos C = -left(-frac{1}{9} cdot frac{2}{3} + frac{sqrt{5}}{3} cdot frac{4sqrt{5}}{9}right) = frac{2}{27} + frac{20}{27} = frac{22}{27}. ] Thus, cos C = boxed{frac{22}{27}}. 2. Finding the perimeter of triangle ABC Given abcos C = 88 and cos C = frac{22}{27}, we find ab: [ ab = frac{88}{frac{22}{27}} = 108. ] Using the sine rule frac{a}{sin A} = frac{b}{sin B}, we find the ratio frac{a}{b}: [ frac{a}{sin 2B} = frac{b}{sin B} implies frac{a}{b} = frac{2sin Bcos B}{sin B} = 2cos B = frac{4}{3}. ] Solving the system of equations left{begin{array}{l}ab=1083a=4bend{array}right., we find a=12 and b=9. Using the Law of Cosines for side c: [ c^2 = a^2 + b^2 - 2abcos C = 144 + 81 - 2 cdot 12 cdot 9 cdot frac{22}{27} = 49. ] Thus, c = 7. Therefore, the perimeter of triangle ABC is a + b + c = 12 + 9 + 7 = boxed{28}.

answer:Let's consider a convex polygon inscribed in a circle, where the number of sides is odd, and the length of each side is a rational number. Our goal is to show that the length of each segment created by the tangential points on these sides can also be expressed as a rational number. 1. **Labeling the Sides and Segments**: Let ( a_1, a_2, ldots, a_n ) be the lengths of the consecutive sides of the polygon, where ( n ) is the number of sides and ( n ) is odd. 2. **Defining the Tangent Segment**: We denote ( x ) as the length of the segment of side ( a_1 ) that lies between the first vertex and the point of tangency with the inscribed circle. 3. **Expressing Subsequent Segments in Terms of ( x )**: Since the polygon is inscribed, the points of tangency divide each side in a symmetric manner in terms of the lengths. Therefore, we can express ( x ) in terms of the given lengths of the sides. 4. **Using Symmetry and Algebraic Manipulation**: Let's consider the sum of the lengths of the segments between adjacent vertices and points of tangency. The last segment on side ( a_n ) is equal to the first segment on side ( a_1 ), leading to the equation: [ x = a_n - (a_{n-1} - (a_{n-2} - (ldots + (a_1 - x)))) ] We can simplify this to: [ x = frac{1}{2} left( a_n - a_{n-1} + a_{n-2} - ldots + a_1 right) ] 5. **Ensuring Rationality**: Given that the lengths ( a_1, a_2, ldots, a_n ) are all rational numbers, their linear combination (with rational coefficients) would also be a rational number. Thus, ( x ) is rational. 6. **Conclusion**: Since the reasoning above holds true for the first segment and applies analogously to each subsequent segment on all sides of the polygon, the conclusion follows that the length of each segment between a vertex and a point of tangency is rational. Hence, we have demonstrated that if a convex polygon, inscribed in a circle, has an odd number of sides, each with a length that is a rational number, then the length of each segment formed by the points of tangency is also a rational number. (blacksquare)

answer:Given the functions ( f(x) = log_a (2 + ax) ) and ( g(x) = log_{frac{1}{a}} (a + 2x) ), it is stated that their graphs are symmetric with respect to the line ( y = b ) (where ( b ) is a constant), which implies that: [ f(x) + g(x) = 2b ] First, we can rewrite ( g(x) ) using the change of base formula for logarithms: [ g(x) = log_{frac{1}{a}} (a + 2x) = frac{log_a (a + 2x)}{log_a (frac{1}{a})} = frac{log_a (a + 2x)}{-1} = -log_a (a + 2x) ] Therefore, we obtain: [ f(x) + g(x) = log_a (2 + ax) - log_a (a + 2x) ] Given that ( f(x) + g(x) = 2b ), we have: [ log_a (2 + ax) - log_a (a + 2x) = 2b ] Using the properties of logarithms, this can be simplified to: [ log_a left( frac{2 + ax}{a + 2x} right) = 2b ] Now, raise both sides as powers of ( a ): [ frac{2 + ax}{a + 2x} = a^{2b} ] Since this relationship must hold for all ( x ) in the domain, the fraction ( frac{2 + ax}{a + 2x} ) must be a constant. Let this constant be ( t ): [ frac{2 + ax}{a + 2x} = t ] Solving for ( t ): [ 2 + ax = t(a + 2x) ] [ 2 + ax = ta + 2tx ] Setting coefficients equal for the varying terms, we get two separate equations: 1. ( coefficient of x: quad a = 2t ) 2. ( constant term: quad 2 = ta ) From the second equation: [ 2 = ta ] [ t = frac{2}{a} ] Substitute ( t ) back into the first equation: [ a = 2 left( frac{2}{a} right) ] [ a = frac{4}{a} ] [ a^2 = 4 ] [ a = 2 text{ or } a = -2 ] Since ( a > 0 ), we accept ( a = 2 ). With ( a = 2 ), substitute ( a ) into one of the original function relationships: From ( 2 = ta ): [ 2 = t cdot 2 ] [ t = 1 ] Then, substitute ( t = a^{2b} ): [ a^{2b} = 1 ] [ 2^{2b} = 1 ] Which implies: [ 2b = 0 ] [ b = 0 ] Finally: [ a + b = 2 + 0 = 2 ] Therefore, the correct answer is: [ boxed{2} ]