answer:Given the points P_{1}(cos alpha ,sin alpha ), P_{2}(cos beta ,-sin beta ), P_{3}(cos (alpha +beta ),sin (alpha +beta )), and A(1,0), let's evaluate the statements one by one: **For Option A:** - The magnitude of overrightarrow{OP_{1}} is |overrightarrow{OP_{1}}| = sqrt{cos^2alpha + sin^2alpha} = sqrt{1} = 1. - Similarly, the magnitude of overrightarrow{OP_{2}} is |overrightarrow{OP_{2}}| = sqrt{cos^2beta + (-sinbeta)^2} = sqrt{1} = 1. Therefore, |overrightarrow{OP_{1}}| = |overrightarrow{OP_{2}}|, which means Option A is boxed{text{correct}}. **For Option B:** - The magnitude of overrightarrow{AP_{1}} is |overrightarrow{AP_{1}}| = sqrt{(cosalpha-1)^2 + sin^2alpha} = sqrt{2-2cosalpha}. - The magnitude of overrightarrow{AP_{2}} is |overrightarrow{AP_{2}}| = sqrt{(cosbeta-1)^2 + (-sinbeta)^2} = sqrt{2-2cosbeta}. Since cosalpha and cosbeta can be different, |overrightarrow{AP_{1}}| neq |overrightarrow{AP_{2}}|, making Option B boxed{text{incorrect}}. **For Option C:** - The dot product overrightarrow{OA}cdotoverrightarrow{OP_{3}} = 1cdotcos(alpha+beta) + 0cdotsin(alpha+beta) = cos(alpha+beta). - The dot product overrightarrow{OP_{1}}cdotoverrightarrow{OP_{2}} = cosalphacosbeta + sinalpha(-sinbeta) = cosalphacosbeta - sinalphasinbeta = cos(alpha+beta). Since overrightarrow{OA}cdotoverrightarrow{OP_{3}} = overrightarrow{OP_{1}}cdotoverrightarrow{OP_{2}}, Option C is boxed{text{correct}}. **For Option D:** - The dot product overrightarrow{OA}cdotoverrightarrow{OP_{1}} = 1cdotcosalpha + 0cdotsinalpha = cosalpha. - The dot product overrightarrow{OP_{2}}cdotoverrightarrow{OP_{3}} = cosbetacos(alpha+beta) + (-sinbeta)sin(alpha+beta) = cosbetacos(alpha+beta) - sinbetasin(alpha+beta) = cos(alpha + 2beta). Since cosalpha neq cos(alpha + 2beta) in general, Option D is boxed{text{incorrect}}. Therefore, the correct options are boxed{A text{ and } C}.

answer:# Part (a): Prove that ( S ) is infinite. 1. **Definition and Setup**: Let ( S ) be the set of positive perfect squares of the form (overline{AA}), where (overline{AA}) represents the concatenation of two equal integers ( A ). This means (overline{AA} = 10^{n(A)} cdot A + A = (10^{n(A)} + 1) cdot A), where ( n(A) ) is the number of digits in ( A ). 2. **Formulation**: We need to show that there are infinitely many such numbers. Consider the expression (overline{AA} = k^2) for some integer ( k ). This implies: [ (10^{n(A)} + 1) cdot A = k^2 ] We need to find ( A ) and ( k ) such that this equation holds. 3. **Choosing ( A ) and ( k )**: Let's choose ( A ) such that ( A = q^2 cdot A' ) and ( 10^{n(A)} + 1 = p^2 cdot A' ) for some integers ( p ) and ( q ). This gives: [ (10^{n(A)} + 1) cdot q^2 cdot A' = (p cdot q cdot sqrt{A'})^2 ] We need to find suitable ( p ) and ( q ) such that ( 1 geq frac{q^2}{p^2} geq frac{1}{10} ). 4. **Examples**: - For ( p = 7 ) and ( q = 3 ): [ frac{3^2}{7^2} = frac{9}{49} approx 0.1837 quad (text{which is between } frac{1}{10} text{ and } 1) ] - For ( p = 11 ) and ( q = 10 ): [ frac{10^2}{11^2} = frac{100}{121} approx 0.8264 quad (text{which is between } frac{1}{10} text{ and } 1) ] - For ( p = 13 ) and ( q = 10 ): [ frac{10^2}{13^2} = frac{100}{169} approx 0.5917 quad (text{which is between } frac{1}{10} text{ and } 1) ] 5. **Conclusion**: By choosing different pairs of ( p ) and ( q ) that satisfy the condition, we can generate infinitely many such numbers ( overline{AA} ) that are perfect squares. Therefore, ( S ) is infinite. (blacksquare) # Part (b): Existence of an LCM function ( f ) 1. **Definition**: We need to determine if there exists a function ( f: S times S rightarrow S ) such that for any ( a, b, c in S ) with ( a, b mid c ), it follows that ( f(a, b) mid c ). 2. **Trivial Functions**: Consider the trivial functions: - ( f(a, b) = a ) - ( f(a, b) = b ) 3. **Verification**: - For ( f(a, b) = a ): If ( a, b mid c ), then ( a mid c ) is trivially true. - For ( f(a, b) = b ): If ( a, b mid c ), then ( b mid c ) is trivially true. 4. **Conclusion**: Both ( f(a, b) = a ) and ( f(a, b) = b ) satisfy the condition. Therefore, such a function ( f ) exists. (blacksquare)

answer:First, let's find out how many students attend the chess class. Since 25% of the students attend chess class, we calculate: 25% of 1000 students = 0.25 * 1000 = 250 students Now, 50% of the students who are in the chess class are also enrolled for swimming. So we calculate: 50% of 250 students = 0.50 * 250 = 125 students Therefore, if all enrolled attend, boxed{125} students will attend the swimming class.

answer:By Vieta's formulas, for the polynomial x^3 - 8x^2 + 10x - 3 = 0, we have the identities p + q + r = 8, pq + pr + qr = 10, and pqr = 3. Similar to the originally solved problem, we write the expression: [ frac{p}{qr + 1} + frac{q}{pr + 1} + frac{r}{pq + 1} = frac{p^2}{pqr + p} + frac{q^2}{pqr + q} + frac{r^2}{pqr + r}. ] Since pqr = 3, we simplify this to: [ frac{p^2}{p+3} + frac{q^2}{q+3} + frac{r^2}{r+3}. ] Using the long division method frac{x^2}{x+3} = x - 3 + frac{9}{x+3}, we substitute and simplify: [ frac{p^2}{p+3} + frac{q^2}{q+3} + frac{r^2}{r+3} = (p-3) + frac{9}{p+3} + (q-3) + frac{9}{q+3} + (r-3) + frac{9}{r+3} = (p + q + r - 9) + 9left(frac{1}{p+3} + frac{1}{q+3} + frac{1}{r+3}right). ] Inserting Vieta's results, we get: [ = (8 - 9) + 9left(frac{1}{p+3} + frac{1}{q+3} + frac{1}{r+3}right). ] For the expression inside the bracket, multiply numerically to get: [ = -1 + 9 cdot frac{(q+3)(r+3) + (p+3)(r+3) + (p+3)(q+3)}{(p+3)(q+3)(r+3)}. ] Now, using the Vieta's results again: [ -1 + 9 cdot frac{(qr + q+3 + r+3) + (pr + p+3 + r+3) + (pq + p+3 + q+3)}{(pqr + 3p + 3q + 3r + 27)}, ] we replace: [ = -1 + 9 cdot frac{(10 + 8 + 9)}{3 + 3 cdot 8 + 27} = -1 + 9 cdot frac{27}{66} = -1 + 9 left(frac{9}{22}right) = -1 + frac{81}{22} = frac{59}{22}. ] Finally, boxed{frac{59}{22}}.