answer:Let's analyze each part of the problem step by step. # Part (a) We need to determine if there exists a natural number ( a ) such that: [ left( (a^2 - 3)^3 + 1 right)^a - 1 ] is a perfect square. 1. Let ( k^2 = left( (a^2 - 3)^3 + 1 right)^a - 1 ). Then we have: [ left( (a^2 - 3)^3 + 1 right)^a = k^2 + 1 ] 2. Notice that ( (a^2 - 3)^3 + 1 ) is always odd because ( (a^2 - 3)^3 ) is odd for any natural number ( a ). Therefore, ( left( (a^2 - 3)^3 + 1 right)^a ) is also odd. 3. Since ( left( (a^2 - 3)^3 + 1 right)^a ) is odd, ( k^2 + 1 ) must be odd. This implies ( k ) must be even. 4. Consider the expression ( (a^2 - 3)^3 + 1 ). We can rewrite it as: [ (a^2 - 3)^3 + 1 = (a^2 - 3)^3 + 1 equiv 3 pmod{4} ] because ( a^2 equiv 1 pmod{4} ) for odd ( a ), and thus ( a^2 - 3 equiv -2 pmod{4} ). 5. By Fermat's Two Squares Theorem, a number of the form ( k^2 + 1 ) cannot be divisible by a prime of the form ( 4k + 3 ). Since ( a^2 - 2 equiv 3 pmod{4} ), it implies that ( k^2 + 1 ) cannot be divisible by 3 modulo 4. 6. Therefore, no such ( a ) exists that satisfies the condition for part (a). # Part (b) We need to determine if there exists a natural number ( a ) such that: [ left( (a^2 - 3)^3 + 1 right)^{a+1} - 1 ] is a perfect square. 1. Let ( k^2 = left( (a^2 - 3)^3 + 1 right)^{a+1} - 1 ). Then we have: [ left( (a^2 - 3)^3 + 1 right)^{a+1} = k^2 + 1 ] 2. Notice that ( (a^2 - 3)^3 + 1 ) is always odd because ( (a^2 - 3)^3 ) is odd for any natural number ( a ). Therefore, ( left( (a^2 - 3)^3 + 1 right)^{a+1} ) is also odd. 3. Since ( left( (a^2 - 3)^3 + 1 right)^{a+1} ) is odd, ( k^2 + 1 ) must be odd. This implies ( k ) must be even. 4. Consider the expression ( (a^2 - 3)^3 + 1 ). We can rewrite it as: [ (a^2 - 3)^3 + 1 = (a^2 - 3)^3 + 1 equiv 3 pmod{4} ] because ( a^2 equiv 1 pmod{4} ) for odd ( a ), and thus ( a^2 - 3 equiv -2 pmod{4} ). 5. By Fermat's Two Squares Theorem, a number of the form ( k^2 + 1 ) cannot be divisible by a prime of the form ( 4k + 3 ). Since ( a^2 - 2 equiv 3 pmod{4} ), it implies that ( k^2 + 1 ) cannot be divisible by 3 modulo 4. 6. Therefore, no such ( a ) exists that satisfies the condition for part (b). (blacksquare)

answer:To solve this problem, we start by calculating the probability that the sum of the numbers on the two dice is 8 or less. Here are the possible outcomes where the sum will be 8 or less: - Sum of 2: (1,1) - Sum of 3: (1,2), (2,1) - Sum of 4: (1,3), (2,2), (3,1) - Sum of 5: (1,4), (2,3), (3,2), (4,1) - Sum of 6: (1,5), (2,4), (3,3), (4,2), (5,1) - Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - Sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) Counting these outcomes, the sum is 2 or 3 for three outcomes, 4 for three outcomes, 5 for four outcomes, 6 for five outcomes, 7 for six outcomes, and 8 for five outcomes. Thus, the number of outcomes where the sum is 8 or less is: 3 + 3 + 4 + 5 + 6 + 5 = 26. With 36 possible outcomes when two dice are rolled (since each die has 6 faces and outcomes are independent), the probability that the sum is 8 or less is frac{26}{36} = frac{13}{18}. Thus, the probability that the sum is greater than 8 is 1 - frac{13}{18} = boxed{frac{5}{18}}.

answer:[Analysis] According to the problem, we can find the value of |overrightarrow{c}| and overrightarrow{a}cdotoverrightarrow{c}. By substituting these values into the formula for the cosine of an angle, we can find the cosine value of the angle, and thus the angle itself. This problem tests the understanding of the angle between vectors in a plane and involves the use of the modulus formula and dot product operations. It is a moderate difficulty problem. [Solution] Since overrightarrow{a}+overrightarrow{b}+overrightarrow{c}=overrightarrow{0}, therefore overrightarrow{c}=-(overrightarrow{a}+overrightarrow{b}), therefore overrightarrow{a}cdotoverrightarrow{c}=-overrightarrow{a}cdot(overrightarrow{a}+overrightarrow{b})=-overrightarrow{a}^{2}-overrightarrow{a}cdotoverrightarrow{b}=dfrac{1}{2}|overrightarrow{c}||overrightarrow{a}|, By the modulus formula, |overrightarrow{c}|=|overrightarrow{a}+overrightarrow{b}|, =sqrt{(overrightarrow{a}+overrightarrow{b})^{2}}=sqrt{overrightarrow{a}^{2}+2overrightarrow{a}cdotoverrightarrow{b}+overrightarrow{b}^{2}}, Substituting this into the previous equation, we get therefore cos theta=dfrac{overrightarrow{a}cdotoverrightarrow{b}}{|overrightarrow{a}||overrightarrow{b}|}=dfrac{-dfrac{3}{2}}{1timessqrt{3}}=-dfrac{sqrt{3}}{2}, therefore tan theta=dfrac{sqrt{3}}{3}, therefore The answer is boxed{text{B: }dfrac{sqrt{3}}{3}}.

answer:Step 1: Understanding the Problem We are asked to prove that the center of the inscribed circle (incircle) in a regular polygon is the point where the angle bisectors of two angles of the polygon intersect. Step 2: Reviewing the Basic Properties of a Regular Polygon - A regular polygon has all sides of equal length and all angles of equal measure. - In a regular polygon, the incircle is tangent to every side of the polygon. Step 3: Drawing the Polygon and the Incircle Let's consider a regular polygon (P) with (n) sides. Draw the incircle by placing the center such that it is equidistant from all the sides of the polygon. Step 4: Drawing the Angle Bisectors - Consider two consecutive vertices (A) and (B) of the polygon. Let (O) be the center of the incircle. - Draw the angle bisectors of (angle A) and (angle B). These bisectors will meet at some point which we denote as (I). Step 5: Prove (I) is the Center of the Incircle 1. **Identifying the Angle Bisectors:** Each angle bisector divides the angle into two equal halves. For example, the bisector of (angle A) divides it into two parts of (frac{pi}{n}) each (since each angle in a regular (n)-sided polygon is ( frac{2pi}{n} )). 2. **Intersection Point of Bisectors:** The point (I) where the angle bisectors of (angle A) and (angle B) intersect must be equidistant from the sides of the polygon by definition of the angle bisectors. 3. **Properties of the Incircle Center:** The center of the incircle of any polygon is the point that is equidistant from all sides of the polygon. Therefore, the point (I) which is equidistant from sides (i.e., the point where bisectors intersect) must be (O), the center of the incircle. 4. **Perpendicular Distances:** To further support, from (I), drop perpendiculars to the sides of the polygon. Since (I) is equidistant from the sides, these perpendiculars will be equal in length, demonstrating that (I) is indeed the center of the incircle. Step 6: Conclusion Thus, point (I) where the bisectors intersect is the center of the incircle. [ boxed{ text{The center of the incircle in a regular polygon is the intersection point of its angle bisectors.} } ]