answer:Let's denote the cost price of one pencil as ( C ) and the selling price of one pencil in the initial scenario as ( S_1 ). We are given that the man sells 16 pencils for a rupee, so ( S_1 = frac{1}{16} ) rupees per pencil. In the initial scenario, the man experiences a loss, which means that ( S_1 < C ). The loss percentage can be calculated as: [ text{Loss percentage} = left( frac{C - S_1}{C} right) times 100 ] In the second scenario, to gain 20%, the selling price of one pencil should be ( S_2 ), where ( S_2 ) is 20% more than the cost price ( C ). This can be expressed as: [ S_2 = C + 0.20C = 1.20C ] We are given that approximately 12 pencils should be sold for a rupee to gain 20%, so ( S_2 = frac{1}{12} ) rupees per pencil. Now we can equate ( S_2 ) to ( 1.20C ) to find the cost price ( C ): [ 1.20C = frac{1}{12} ] [ C = frac{1}{12 times 1.20} ] [ C = frac{1}{14.4} ] Now we can calculate the loss percentage using the initial selling price ( S_1 ) and the cost price ( C ): [ text{Loss percentage} = left( frac{C - S_1}{C} right) times 100 ] [ text{Loss percentage} = left( frac{frac{1}{14.4} - frac{1}{16}}{frac{1}{14.4}} right) times 100 ] [ text{Loss percentage} = left( frac{16 - 14.4}{14.4} right) times 100 ] [ text{Loss percentage} = left( frac{1.6}{14.4} right) times 100 ] [ text{Loss percentage} = left( frac{1}{9} right) times 100 ] [ text{Loss percentage} approx 11.11% ] Therefore, the percentage of loss in the initial scenario is approximately boxed{11.11%} .

answer:(1) Solution: According to the problem, the moving point P(x, y) satisfies sqrt{(x - sqrt{2})^{2} + y^{2}} + sqrt{(x + sqrt{2})^{2} + y^{2}} = 4. Since |F_{1}F_{2}| = 2sqrt{2} < 4, the trajectory of the moving point P(x, y) is an ellipse with foci on the x-axis, and begin{cases} 2a = 4 2c = 2sqrt{2} end{cases} Rightarrow b = sqrt{2}. Therefore, the equation of the trajectory of curve C is boxed{frac{x^{2}}{4} + frac{y^{2}}{2} = 1}. (2) Proof: Let the distance from the origin to the line AB be d, and A, B are two moving points on curve C satisfying OA perp OB. 1^st If point A is on the coordinate axis, then point B is also on the coordinate axis, and we have frac{1}{2}|OA||OB| = frac{1}{2}|AB| cdot d, which means d = frac{ab}{sqrt{a^{2} + b^{2}}} = frac{2sqrt{3}}{3}. 2^nd If point A(x_{A}, y_{A}) is not on the coordinate axis, we can set OA: y = kx, OB: y = -frac{1}{k}x. From begin{cases} frac{x^{2}}{4} + frac{y^{2}}{2} = 1 y = kx end{cases}, we get begin{cases} x_{A}^{2} = frac{4}{1 + 2k^{2}} y_{A}^{2} = frac{4k^{2}}{1 + 2k^{2}} end{cases}. Let point B(x_{B}, y_{B}), similarly, we get begin{cases} x_{B}^{2} = frac{4k^{2}}{2 + k^{2}} y_{B}^{2} = frac{4}{2 + k^{2}} end{cases}. Thus, |OA| = 2sqrt{frac{1 + k^{2}}{1 + 2k^{2}}}, |OB| = 2sqrt{frac{1 + k^{2}}{2 + k^{2}}}, |AB| = sqrt{OA^{2} + OB^{2}} = frac{2sqrt{3}(1 + k^{2})}{sqrt{(2 + k^{2})(1 + 2k^{2})}}. Using frac{1}{2}|OA||OB| = frac{1}{2}|AB| cdot d, we get d = frac{2sqrt{3}}{3}. Combining 1^st and 2^nd, we always have d = frac{2sqrt{3}}{3}, that is, the distance from the origin O to the line AB is a constant boxed{frac{2sqrt{3}}{3}}.

answer:To find the next occurrence of Andrea's birthday on a Tuesday after 2015, we note the leap years and count the days accordingly: 1. **2015**: December 18 was a Friday. 2015 is a non-leap year. 2. **2016** (leap year): Adds 2 days, so December 18, 2016, is a Sunday. 3. **2017** (non-leap year): Adds 1 day, so December 18, 2017, is a Monday. 4. **2018** (non-leap year): Adds 1 day, so December 18, 2018, is a Tuesday. 5. **2019** (non-leap year): Adds 1 day, so December 18, 2019, is a Wednesday. After progressing through each year while correctly accounting for normal and leap years, we find that the next time Andrea's birthday falls on a Tuesday is in the year **2018**. Thus, the correct answer is 2018. Conclusion: The calculation is consistent with the number of day increments per year-type (leap vs. non-leap), and the conclusion logically follows from the stated day advancements. The final answer is The correct choice given the problem statement and calculation above is boxed{mathrm{(C)} 2018}.

answer:The problem requires us to determine whether it is possible to place crosses ((X)) and noughts ((O)) on graph paper in such a way that no three identical symbols (either (X) or (O)) appear in a row, either horizontally, vertically, or diagonally. Let's walk through an example configuration that satisfies these conditions: 1. Consider a 5x5 grid of graph paper. 2. Populate each cell with (X) or (O) such that no three identical symbols align horizontally, vertically, or diagonally. Here's an example configuration that meets these requirements: [ begin{array}{|c|c|c|c|c|} hline X & O & X & O & X hline O & X & O & X & O hline X & O & X & O & X hline O & X & O & X & O hline X & O & X & O & X hline end{array} ] In this 5x5 grid: - Horizontally, there are no three consecutive (X) or (O). - Vertically, there are no three consecutive (X) or (O). - Diagonally, there are no three consecutive (X) or (O). Each row and column alternates between (X) and (O), ensuring that three identical symbols do not appear consecutively in any direction. # Verification: 1. **Horizontal Check:** - Each row alternates between (X) and (O). 2. **Vertical Check:** - Each column alternates between (X) and (O). 3. **Diagonal Check:** - Major diagonals and minor diagonals are reviewed and alternate between (X) and (O), preventing three identical symbols from aligning. Thus, a configuration like this one proves that it is possible to place crosses and noughts on graph paper under the given conditions. [ boxed{text{Possible}} ]