answer:1. Observe the given figure drawn on a grid and recognize the problem: to find a ray starting from point ( A ) that divides the figure into two parts of equal area, pointing out a grid point ( B ) (other than ( A )) through which the ray passes. 2. Notice that the entire figure consists of 9 "large" squares. Each square is a unit of area. Thus, the total area of the figure is: [ text{Total Area} = 9 text{ squares} ] 3. To achieve the division into two equal areas, each part should be ( frac{9}{2} = 4.5 ) squares. 4. Identify the right part of the figure, which includes three whole squares and an additional triangular section ( triangle ABC ) (highlighted in red). To determine if this right part constitutes exactly 4.5 squares, we calculate the area of ( triangle ABC ): - Consider ( triangle ABC ). By visual assumptions or drawing, extend it to form rectangle ( A C B D ). The area of this rectangle is: [ text{Area of } ABCD = 3 text{ squares} ] - Since ( triangle ABC ) is half of the rectangle ( ABCD ), its area is: [ text{Area of } triangle ABC = frac{1}{2} times 3 = 1.5 text{ squares} ] 5. Sum the areas of the parts on the right side: - The three large whole squares: [ 3 text{ squares} ] - The triangular section ( triangle ABC ): [ 1.5 text{ squares} ] 6. Therefore, the total area of the right part is: [ text{Area of the right part} = 3 + 1.5 = 4.5 text{ squares} ] 7. Conclude that the ray, starting from ( A ) and passing through the grid point ( B ), indeed divides the figure into two equal parts: [ text{Each part has area } 4.5 text{ squares.} ] Hence, the required condition of dividing the figure into two equal areas is met. This concludes the proof: [ blacksquare ]

answer:(1) Since a_n > 0 and (a_n+1)^2=4(S_n+1), When n=1, we have (a_1+1)^2=4(a_1+1), which gives a_1=3. For n geq 2, (a_{n-1}+1)^2=4(S_{n-1}+1). We can derive that (a_n+1)^2-(a_{n-1}+1)^2=4a_n, which simplifies to: (a_n+a_{n-1})(a_n-a_{n-1}-2)=0. Hence, a_n-a_{n-1}=2. So, the sequence {a_n} is an arithmetic sequence with a common difference of 2. Therefore, a_n=3+2(n-1)=2n+1. (2) From part (1), we have S_n=frac{n(3+2n+1)}{2}=n^2+2n. Also, b_nS_n-1=(n+1)^2, where n in mathbb{N}^*. Hence, b_n=frac{n^2+2n+2}{n^2+2n}=1+(frac{1}{n}-frac{1}{n+2}). Therefore, the sum of the first n terms for the sequence {b_n}, T_n, is: T_n=n+(1-frac{1}{3})+(frac{1}{2}-frac{1}{4})+(frac{1}{3}-frac{1}{5})+ldots+(frac{1}{n-1}-frac{1}{n+1})+(frac{1}{n}-frac{1}{n+2}) T_n=n+frac{3}{2}-frac{2n+3}{(n+1)(n+3)}. The final answer is T_n=boxed{n+frac{3}{2}-frac{2n+3}{(n+1)(n+3)}}.

answer:1. Start with the given expression and convert the trigonometric functions to their ratio forms: [ frac{left( operatorname{tg} alpha + cos^{-1} alpha right) (cos alpha - operatorname{ctg} alpha)}{(cos alpha + operatorname{ctg} alpha) left( operatorname{tg} alpha - cos^{-1} alpha right)} ] Recall the following trigonometric identities: - (operatorname{tg} alpha = tan alpha = frac{sin alpha}{cos alpha}) - (cos^{-1} alpha = sec alpha = frac{1}{cos alpha}) - (operatorname{ctg} alpha = cot alpha = frac{cos alpha}{sin alpha}) 2. Substitute these identities into the expression: [ frac{left( frac{sin alpha}{cos alpha} + frac{1}{cos alpha} right) left( cos alpha - frac{cos alpha}{sin alpha} right)}{left( cos alpha + frac{cos alpha}{sin alpha} right) left( frac{sin alpha}{cos alpha} - frac{1}{cos alpha} right)} ] 3. Simplify the terms inside the parentheses: - For the numerator: [ left( frac{sin alpha + 1}{cos alpha} right) left( cos alpha - frac{cos alpha}{sin alpha} right) ] - For the denominator: [ left( cos alpha + frac{cos alpha}{sin alpha} right) left( frac{sin alpha - 1}{cos alpha} right) ] 4. Combine the fractions in the numerator and the denominator: - Numerator: [ left( frac{sin alpha + 1}{cos alpha} right) left( cos alpha - frac{cos alpha}{sin alpha} right) = frac{sin alpha + 1}{cos alpha} cdot frac{cos alpha (sin alpha - 1)}{sin alpha} ] - Denominator: [ left( cos alpha + frac{cos alpha}{sin alpha} right) left( frac{sin alpha - 1}{cos alpha} right) = left( frac{cos alpha sin alpha + cos alpha}{sin alpha} right) cdot frac{sin alpha - 1}{cos alpha} ] 5. Simplify the combined fractions: - Numerator: [ frac{(sin alpha + 1)(sin alpha - 1)}{cos alpha sin alpha} ] Notice that ((sin alpha + 1)(sin alpha - 1) = sin^2 alpha - 1) [ frac{sin^2 alpha - 1}{cos alpha sin alpha} ] - Denominator: [ frac{cos alpha (sin alpha + 1)}{sin alpha cos alpha} cdot frac{sin alpha - 1}{cos alpha} ] After cancelling (cos alpha): [ frac{(sin alpha + 1)(sin alpha - 1)}{sin alpha} ] 6. Final simplification: The expressions simplify to (frac{sin^2 alpha - 1}{cos alpha sin alpha}) over (frac{sin^2 alpha - 1}{sin alpha cos alpha}), which both simplify directly: [ frac{sin^2 alpha - 1}{cos alpha sin alpha} div frac{sin^2 alpha - 1}{sin alpha cos alpha} = 1 ] Conclusion: [ boxed{1} ]

answer:To find the maximum value of x^{2}+y^{2} given the equation 5x^{2}-10x+4y^{2}=0, we start by rearranging the equation to express y^{2} in terms of x. This gives us: [ y^{2} = frac{1}{4}(10x-5x^{2}) ] This expression is valid for 0 leqslant x leqslant 2 because the quadratic equation in x must have real roots for y^{2} to be non-negative. Next, we express x^{2}+y^{2} in terms of x: [ x^{2}+y^{2} = x^{2} + frac{1}{4}(10x-5x^{2}) = -frac{1}{4}(x-5)^{2} + frac{25}{4} ] This equation is in the form of a downward-opening parabola, where -frac{1}{4}(x-5)^{2} represents the parabolic term and frac{25}{4} is the maximum value the expression can reach, which occurs when the square term is zero. The axis of symmetry of this parabola is at x=5, but since our domain is restricted to 0 leqslant x leqslant 2, we need to find the maximum value within this domain. At x=2, we substitute into the equation to find the maximum value within the allowed range: [ x^{2}+y^{2} = 2^{2} + frac{1}{4}(10cdot2-5cdot2^{2}) = 4 + frac{1}{4}(20-20) = 4 ] Therefore, the maximum value of x^{2}+y^{2} within the given domain is boxed{4}, which corresponds to option boxed{B}.