answer:Given the condition ( sin^{2} A + sin^{2} B + sin^{2} C = 1 ), where ( A, B, ) and ( C ) are all acute angles, we need to prove the inequality: [ frac{pi}{2} leq A + B + C leq pi. ] 1. **Express ( sin^{2} A ) in terms of ( sin B ) and ( sin C ):** [ sin^{2} A = 1 - sin^{2} B - sin^{2} C ] Utilizing the identity ( sin left( frac{pi}{2} - B right) = cos B ), we have: [ sin^{2} A = cos^{2} B - sin^{2} C ] 2. **Simplify using the product-to-sum identities:** [ cos^{2} B - sin^{2} C = left( cos B - sin C right) left( cos B + sin C right) ] 3. **Apply angle addition and subtraction formulas:** [ (cos B - sin C)(cos B + sin C) = cos(B + C)cos(B - C) ] 4. **Since ( B ) and ( C ) are acute angles:** [ cos(B - C) > 0 implies cos(B + C) geq 0 ] This implies that ( B + C ), being an acute angle, must satisfy: [ B + C leq frac{pi}{2} ] 4. **Since ( cos(B-C) geq cos(B+C) ):** [ sin^{2} A = cos(B + C) cos(B - C) geq cos^{2}(B + C) = sin^{2} left( frac{pi}{2} - B - C right) ] 5. **Relate angles ( A ) and ( B+C ):** Since both ( A ) and ( B+C ) are acute angles, we have: [ A geq frac{pi}{2} - B - C ] This implies: [ A + B + C geq frac{pi}{2} ] Finally, we combine the inequalities: [ frac{pi}{2} leq A + B + C leq pi. ] Thus, the desired result is: [ boxed{frac{pi}{2} leq A + B + C leq pi} ]

answer:**Answer** (1) f() = sin^2 - sin^2 = 0. (4 points) (2) Since f(x) = sin^2x - sin^2x = f(x) = sin^2x + cos^2x - 1 = sin(2x + pi) - 1. Therefore, when x in [0, pi], -pi leq 2x + pi leq 3pi, so -1 leq sin(2x + pi) leq 1, Thus, the maximum value of f(x) is boxed{0}. (12 points)

answer:1. **Define the problem and notation:** - Let ( triangle ABC ) be a triangle with sides ( AB = c ), ( AC = b ), and ( BC = a ). - Let ( alpha = angle A ), ( beta = angle B ), and ( gamma = angle C ). - The excircle corresponding to ( A ) touches ( BC ) at ( A_1 ), ( CA ) at ( B_1 ), and ( AB ) at ( C_1 ). - Similarly, the excircle corresponding to ( B ) touches ( BC ) at ( A_2 ), ( CA ) at ( B_2 ), and ( AB ) at ( C_2 ). - The excircle corresponding to ( C ) touches ( BC ) at ( A_3 ), ( CA ) at ( B_3 ), and ( AB ) at ( C_3 ). 2. **Sum of the perimeters of the triangles:** - The sum of the perimeters of ( triangle A_1B_1C_1 ), ( triangle A_2B_2C_2 ), and ( triangle A_3B_3C_3 ) can be expressed as: [ 2 sum a cos alpha + (sum a)(sum sin frac{alpha}{2}) ] 3. **Ratio ( k ) of the sum of the perimeters to the circumradius ( R ):** - The ratio ( k ) is given by: [ k = frac{2 sum a cos alpha + (sum a)(sum sin frac{alpha}{2})}{R} ] - Using the Law of Sines, we can rewrite ( k ) as: [ k = 4 sum sin alpha cos frac{alpha}{2} + 2 sum sin alpha sum sin frac{alpha}{2} ] 4. **Applying Jensen's Inequality:** - By Jensen's Inequality, we have: [ 2 sum sin alpha sum sin frac{alpha}{2} leq 2 cdot 3 cdot frac{sqrt{3}}{2} cdot 3 cdot frac{1}{2} = frac{9sqrt{3}}{2} ] - Also, we have: [ 4 sum sin alpha cos frac{alpha}{2} = 2 sum (sin frac{3alpha}{2} + sin frac{alpha}{2}) ] [ 2 sum sin frac{alpha}{2} leq 3 ] [ 2 sum sin frac{3alpha}{2} leq 6 ] [ 4 sum sin alpha cos frac{alpha}{2} leq 9 ] 5. **Combining the inequalities:** - Adding the two parts, we get: [ k = 4 sum sin alpha cos frac{alpha}{2} + 2 sum sin alpha sum sin frac{alpha}{2} leq 9 + frac{9sqrt{3}}{2} ] 6. **Equality condition:** - The equality holds if and only if ( alpha = beta = gamma = frac{pi}{3} ). 7. **Conclusion:** - Hence, the maximum possible value of ( k ) is: [ k_{text{max}} = 9 + frac{9sqrt{3}}{2} ] The final answer is ( boxed{ 9 + frac{9sqrt{3}}{2} } )

answer:The remainder when a number is divided by 10 is the units digit of that number. Thus, we need to find the units digit of the product of 1265 times 4233 times 254 times 1729. Focusing only on the units digits of these numbers: - The units digit of 1265 is 5. - The units digit of 4233 is 3. - The units digit of 254 is 4. - The units digit of 1729 is 9. Calculating sequentially and keeping only the units digit at each stage: 1. 5 times 3 = 15; units digit is 5. 2. 5 times 4 = 20; units digit is 0. 3. 0 times 9 = 0; units digit remains 0. Since the calculations produce a final units digit of boxed{0}, the remainder when this product is divided by 10 is 0.