answer:1. Let ( triangle ABC ) be the given triangle, and ( M ) be the arbitrary point in the plane. 2. Reflect ( M ) over vertex ( A ) to get point ( M' ). By definition of reflection, point ( M' ) is such that ( A ) is the midpoint of the segment connecting ( M ) and ( M' ). 3. Reflect ( M' ) over vertex ( B ) to get point ( M'' ). Similarly, point ( M'' ) is such that point ( B ) is the midpoint of the segment connecting ( M' ) and ( M'' ). 4. Reflect ( M'' ) over vertex ( C ) to get point ( M''' ). Hence, point ( M''' ) is such that point ( C ) is the midpoint of the segment connecting ( M'' ) and ( M''' ). 5. Reflect ( M''' ) over vertex ( A ) again to get point ( M'''' ). Point ( M'''' ) is such that point ( A ) is the midpoint of the segment connecting ( M''' ) and ( M'''' ). 6. Next, reflect ( M'''' ) over vertex ( B ) to obtain point ( M''''' ). This ensures that point ( B ) is the midpoint of the segment connecting ( M'''' ) and ( M''''' ). 7. Finally, reflect ( M''''' ) over vertex ( C ) to obtain point ( M'''''' ). Similarly, ( C ) will be the midpoint of the segment connecting ( M''''' ) and ( M'''''' ). 8. By the properties of reflections and the fact that reflections over vertices of a triangle result in parallel transport (a composition of translations), after reflecting ( M ) two times over each vertex, it will be mapped back to itself. This is because the sum total effect of the reflections is a translation that returns to the original position as sum of translations around the triangle results in a zero vector. # Conclusion: boxed{M = M'''''' }

answer:1. **Objective**: We want to show that (D = d_{1} d_{2} + d_{2} d_{3} + cdots + d_{k-1} d_{k} < n^{2}). 2. **Upper Bound on (d_i)**: For any divisor (d_i) of (n), it holds that: [ d_{k-m} leq frac{n}{m+1} ] This inequality helps us estimate the sum (D). 3. **Sum Representation**: We express (D) in a form easy to estimate: [ D = d_{1} d_{2} + d_{2} d_{3} + cdots + d_{k-1} d_{k} ] 4. **Rewriting the Sum**: Notice that for each term (d_i d_{i+1}), the product of the two consecutive divisors, the terms can be bounded using the inequality: [ D leq sum_{i=1}^{k-1} d_i d_{i+1} leq n^2 left( sum_{i=1}^{k-1} frac{1}{i(i+1)} right) ] 5. **Simplifying the Series**: The series (sum_{i=1}^{k-1} frac{1}{i(i+1)}) is a telescoping series: [ sum_{i=1}^{k-1} frac{1}{i(i+1)} = sum_{i=1}^{k-1} left( frac{1}{i} - frac{1}{i+1} right) ] This telescopes to: [ left( 1 - frac{1}{2} right) + left( frac{1}{2} - frac{1}{3} right) + cdots + left( frac{1}{k-1} - frac{1}{k} right) = 1 - frac{1}{k} ] 6. **Inequality Conclusion**: [ D leq n^2 left( 1 - frac{1}{k} right) < n^2 ] This concludes the proof that (D < n^2): [ boxed{D < n^2} ] --- # Second part: Finding (n) such that (D) is a divisor of (n^2) 1. **Prime (n)**: If (n) is a prime number: [ D = d_1 d_2 = 1 cdot n = n ] Here, (D = n mid n^2). 2. **Composite (n)**: Suppose (n) is composite and (d_2 = p) is the smallest prime divisor of (n). ``` ``` [ D > d_{k-1} d_k = n cdot frac{n}{p} = frac{n^2}{p} ] 3. **Divisibility Condition**: For (D mid n^2) where (D > frac{n^2}{p}), simplify: [ frac{n^2}{p} < D < n^2 ] Since (n^2 / p) is the largest proper divisor of (n^2), (D) cannot be a divisor of (n^2) if (n) is composite. In conclusion: Only when (n) is a prime number will (D) be a divisor of (n^2): [ boxed{text{If } n text{ is a prime number, } D text{ is a divisor of } n^2.} ]

answer:To solve this, since in the Cartesian coordinate system, the slope of the line -x+ sqrt {3}y-6=0 is frac{sqrt {3}}{3}, Let the inclination angle of the line be alpha, so tanalpha= frac{sqrt {3}}{3}, Therefore, alpha=30^circ Since for the linear function x- sqrt {3}y+6=0, b=2 sqrt {3}, Thus, the y-intercept of this function's graph is 2 sqrt {3}. Hence, the answers are: 30^circ, 2 sqrt {3}. By using the equation of the line to find the slope of the line, and then solving for the inclination angle of the line; first determine the value of b based on the analytical expression of the linear function, and then solve it based on the properties of the linear function. This question examines the relationship between the slope of a line and the inclination angle of a line, as well as the method of finding intercepts, testing computational skills. Therefore, the inclination angle of the line is boxed{30^circ}, and its y-intercept is boxed{2 sqrt {3}}.

answer:**Crate A Calculation**: - Each row contains 10 pipes, and there are 25 rows. - The height of each pipe is equal to its diameter, which is 8 cm. - Thus, the total height of the packing in Crate A = number of rows × diameter of a pipe = (25 times 8 = 200) cm. **Crate B Calculation**: - Packed in a honeycomb pattern, the effective height between two consecutive layers of pipes reduces due to the staggered arrangement. - The centers of the pipes in two consecutive layers form an equilateral triangle. - The vertical distance (d) between layers can be calculated as (d = text{height of the equilateral triangle} - r) where (r = 4) cm (radius of a pipe). - The side of the equilateral triangle is the same as the diameter of a pipe, so (d = 4sqrt{3} - 4) cm. - For 250 pipes, assuming 10 pipes per honeycomb row, we have 25 rows. Total height in Crate B = (25 times d = 25 times (4sqrt{3} - 4)) cm. Calculating (d): [ d = 4sqrt{3} - 4 ] [ 25 times (4sqrt{3} - 4) = 100sqrt{3} - 100 ] cm. **Difference in Heights**: [ text{Height of Crate A} - text{Height of Crate B} = 200 - (100sqrt{3} - 100) ] [ = 300 - 100sqrt{3} ] [ = boxed{300 - 100sqrt{3}} , text{cm} ]