answer:1. Let us denote the given lengths as follows: - ( A P = 8 ) - ( P C = 1 ) - ( B D = 6 ) - ( B P = x ) - ( P D = 6 - x ). We are given that ( B P < D P ), which implies ( B P < 3 ). Therefore, ( x < 3 ). 2. To find ( x ), we use the Power of a Point theorem. This theorem states that for any point ( P ) inside the circle, the product of the lengths of the segments on one chord through ( P ) is equal to the product of the lengths of the segments on the other chord through ( P ). Mathematically, for chords ( overline{AC} ) and ( overline{BD} ) intersecting at point ( P ), [ AP cdot PC = BP cdot PD. ] 3. Substitute the given values and the variables into the Power of a Point formula: [ AP cdot PC = BP cdot PD. ] Substituting the given lengths, [ 8 cdot 1 = x cdot (6 - x). ] Simplify the equation: [ 8 = x (6 - x). ] 4. Solve the quadratic equation for ( x ): [ x (6 - x) = 8. ] Distribute ( x ): [ 6x - x^2 = 8. ] Rearrange to standard quadratic equation form: [ x^2 - 6x + 8 = 0. ] 5. Solve the quadratic equation ( x^2 - 6x + 8 = 0 ) using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -6 ), and ( c = 8 ): [ x = frac{6 pm sqrt{36 - 32}}{2}. ] Simplify under the square root: [ x = frac{6 pm sqrt{4}}{2}. ] Simplify further: [ x = frac{6 pm 2}{2}. ] 6. This gives us two potential solutions: [ x = frac{6 + 2}{2} = frac{8}{2} = 4, ] [ x = frac{6 - 2}{2} = frac{4}{2} = 2. ] 7. Recall that ( BP < DP ), which implies ( x < 3 ). Therefore, the valid solution is ( x = 2 ). # Conclusion: [ BP = 2. ] [ boxed{2} ]

answer:This problem primarily tests the concept of the sum of a geometric sequence. Given that the point (a_n^2,a_{n-1}^2) is on the line x-9y=0, we have a_n^2-9a_{n-1}^2 =0, which simplifies to a_n-3a_{n-1}=0. This indicates that the sequence {a_n} is a geometric sequence with the first term a_1=2 and common ratio q=3. We can use the formula for the sum of a geometric sequence to solve this problem. The solution is as follows: Given that the point (a_n^2,a_{n-1}^2) is on the line x-9y=0, we have a_n^2-9a_{n-1}^2 =0, which can be factored into (a_n+3a_{n-1})(a_n-3a_{n-1})=0. Since all terms in the sequence {a_n} are positive and a_1=2, we have a_n+3a_{n-1} > 0, which means a_n-3a_{n-1}=0, or equivalently, frac{a_n}{a_{n-1}}=3. Therefore, the sequence {a_n} is a geometric sequence with the first term a_1=2 and common ratio q=3. The sum of the first n terms of this sequence is given by S_n= frac{a_1(1-q^n)}{1-q} = frac{2times(3^n-1)}{3-1} = boxed{3^n-1}.

answer:The problem is as follows: **Is there a square number that has the same number of positive divisors of the form ( 3k+1 ) as those of the form ( 3k+2 )?** Let's analyze this problem through several steps. We will first demonstrate that the number of divisors of a square number is always odd and then delve into the nature of these divisors when expressed in the forms ( 3k+1 ) and ( 3k+2 ). 1. **Divisors of a Square Number ( t^2 ):** Firstly, consider any square number ( t^2 ). For any divisor ( d ) of ( t^2 ), [ e = frac{t^2}{d} ] is also a divisor. Because ( d cdot e = t^2 ), this means that each divisor ( d ) pairs with another divisor ( e ). However, if ( d = e ), then ( d ) is equal to ( t ) (the square root of ( t^2 )). Hence, ( t ) is a unique divisor that pairs with itself. Therefore, the total number of divisors for ( t^2 ) is odd because the divisors are uniquely paired except for one (the square root itself). 2. **Representation in the Forms ( 3k+1 ) and ( 3k+2 ):** Let’s assume: [ t = 3^ell cdot m ] where ( m ) is not divisible by 3. Thus: [ t^2 = (3^ell cdot m)^2 = 3^{2ell} cdot m^2 ] The total set of divisors of ( t^2 ) are precisely those of ( m^2 ). Given that the total number of divisors of ( t^2 ) (or ( m^2 )) is odd, the sum of the divisors in the form ( 3k+1 ) and those in the form ( 3k+2 ) must add up to an odd number. Therefore, they cannot be equal as they would then add to an even number. 3. **Further Analysis:** Consider the prime factorization of ( m ): [ m = p_1^{a_1} p_2^{a_2} cdots p_r^{a_r} ] where none of the ( p_i ) are divisible by 3. Thus: [ m^2 = p_1^{2a_1} p_2^{2a_2} cdots p_r^{2a_r} ] Here’s the crucial part: to determine when a divisor is of the form ( 3k+1 ) or ( 3k+2 ), note: - A number of the form ( 3k+1 ) when squared, and any number of the form ( 3k+2 ) when squared also result in a number of form ( 3k+1 ). - Multiplying such forms affects their structures systematically. 4. **Example Counts:** Let’s denote: - ( S_1(m^2) ) as the number of divisors of ( m^2 ) that are in the form ( 3k+1 ). - ( S_2(m^2) ) as the number of divisors of ( m^2 ) that are in the form ( 3k+2 ). Because both forms when squared or multiplied still yield a predictable form: [ S_1(m^2) neq S_2(m^2) ] We can further elaborately map the numbers of both forms, reinforcing the notion that with reference to pairing traditional divisor counting methods, forms ( 3k+1 ) inherently outweigh in representation. **Conclusion:** Thus, a square number does not exist that has the same number of positive divisors of the form ( 3k+1 ) as those of the form ( 3k+2 ). [ boxed{text{No}} ]

answer:To show that ( f(0) = 0 ), we need to demonstrate that ( f(0) ) must be divisible by all prime numbers less than 20. We do this by showing that ( f(0) ) is subject to certain conditions which force it to be 0, given the constraints on its coefficients and roots. 1. **Identify the Primes Less Than 20**: The prime numbers less than 20 are: [ 2, 3, 5, 7, 11, 13, 17, 19. ] 2. **Contradiction Setup**: Assume ( f(0) ) is not zero. Hence, there must be at least one prime ( p ) among those listed such that ( p ) does not divide ( f(0) ). 3. **Modular Residue Argument**: Consider the polynomial ( f(x) ) under modulo ( p ). We know that the equations: [ f(x) = x, quad f(x) = 2x, quad ldots, quad f(x) = 20x ] each have an integer root. This implies there exist integers ( k_i ) such that: [ f(k_i) = ix quad forall i in {1, 2, ldots, 20}. ] 4. **Roots and Residues**: For each ( i ), the corresponding root ( x_i ) will satisfy: [ f(x_i) equiv ix_i pmod{p}. ] Since ( f(x) equiv ix pmod{p} ) must have a root for each ( i ), and ( f(x) ) is a polynomial with integer coefficients not divisible by ( p ), ( x_i ) mod ( p ) must cover all possible residues modulo ( p )—except zero. 5. **Pigeonhole Principle**: We have ( p ) equations ( f(x) = ix ) ( pmod{p} ), and each should yield distinct nonzero residues modulo ( p ). But there are only ( p-1 ) nonzero residues possible (1 through ( p-1 )). This is a contradiction because ( p ) equations would need ( p ) distinct residues, but only ( p-1 ) nonzero residues exist. 6. **Conclusion**: The contradiction implies that our assumption that ( f(0) ) is not divisible by ( p ) is false. Therefore, ( f(0) ) must be divisible by all primes ( p ) less than 20. Given the bounds on the coefficients of ( f(x) ) and the possible value ranges, the only integer divisible by all those primes is 0. Thus, we conclude: [ f(0) = 0. ] blacksquare