answer:1. **Consider the tetrahedron**: Let a tetrahedron have vertices ( A, B, C, D ) such that the edge ( AB ) is at most 2 units long. 2. **Assign edge lengths**: Assume ( CD ) is the longest edge. Given ( x leq 2 ), let ( AB = x ) and analyze the volume as a function of ( x ). 3. **Estimate height in triangle ( ABC )**: [ text{In the triangle } ABC, text{ let } T text{ be the foot of the perpendicular from } C text{ to } AB. ] [ text{Assume } T text{ is closer to } A, text{ then } BT geq frac{x}{2}. ] [ text{In triangle } BTC, text{ we have: } ] [ CT = sqrt{BC^2 - BT^2} leq sqrt{4 - left( frac{x}{2} right)^2} ] 4. **Estimate another height ( m_1 )**: [ text{Similarly in triangle } ABD, text{ let } m_1 text{ be the height from } D text{ to } AB. ] [ m_1 leq sqrt{4 - left( frac{x}{2} right)^2} ] 5. **Estimate the height ( m )** of the tetrahedron: [ text{Let } m text{ be the height of the tetrahedron from } C text{ perpendicular to } ABD. ] [ Then } m leq CT leq sqrt{4 - frac{x^2}{4}} ] 6. **Calculate the volume ( V )**: [ V = frac{1}{3} cdot frac{1}{2} cdot AB cdot m_1 cdot m ] [ V leq frac{1}{6} cdot x cdot sqrt{4 - frac{x^2}{4}} cdot sqrt{4 - frac{x^2}{4}} ] [ V leq frac{1}{6} cdot x cdot left(4 - frac{x^2}{4}right) ] [ V leq frac{x}{24} left(16 - x^2right) ] 7. **Show ( frac{x}{24} left(16 - x^2right) leq 1 )** for ( x leq 2 ): We now prove that: [ x (16 - x^2) leq 24 ] 8. **Simplification**: [ 24 - x (16 - x^2) = (x - 2) (x^2 + 2x - 12) geq 0 ] Given ( x in (0, 2] ), [ x - 2 leq 0 text{ and } x^2 + 2x - 12 < 0 ] Both conditions hold, hence, the volume constraint is established. # Conclusion: [ boxed{V leq 1} ]

answer:(1) Using the given functional equation, we compute the following: begin{align*} fleft(frac{1}{4}right) &= fleft(frac{1}{2} cdot frac{1}{2}right) = frac{1}{2}fleft(frac{1}{2}right) + frac{1}{2}fleft(frac{1}{2}right) = 1, fleft(frac{1}{8}right) &= fleft(frac{1}{2} cdot frac{1}{4}right) = frac{1}{2}fleft(frac{1}{4}right) + frac{1}{4}fleft(frac{1}{2}right) = frac{3}{4}, fleft(frac{1}{16}right) &= fleft(frac{1}{2} cdot frac{1}{8}right) = frac{1}{2}fleft(frac{1}{8}right) + frac{1}{8}fleft(frac{1}{2}right) = frac{1}{2}. end{align*} Thus, we have: [ boxed{fleft(frac{1}{4}right) = 1}, quad boxed{fleft(frac{1}{8}right) = frac{3}{4}}, quad boxed{fleft(frac{1}{16}right) = frac{1}{2}}. ] (2) Based on the results above, we can conjecture the formula fleft(2^{-n}right) = n left(frac{1}{2}right)^{n-1}. Now, we use mathematical induction to prove the conjecture: Base case (n=1): fleft(2^{-1}right) = fleft(frac{1}{2}right) = 1, which equals 1 times left(frac{1}{2}right)^{0} = 1. Therefore, the base case holds true. Inductive step: Assume the formula holds for n=k, i.e., fleft(2^{-k}right) = k left(frac{1}{2}right)^{k-1}. We need to prove it holds for n=k+1: begin{align*} fleft(2^{-(k+1)}right) &= fleft(frac{1}{2} cdot 2^{-k}right) = frac{1}{2}fleft(2^{-k}right) + 2^{-k}fleft(frac{1}{2}right) &= frac{1}{2} cdot k cdot left(frac{1}{2}right)^{k-1} + left(frac{1}{2}right)^{k} = k left(frac{1}{2}right)^{k} + left(frac{1}{2}right)^{k} &= (k+1) left(frac{1}{2}right)^{k}. end{align*} Thus, for n=k+1, the induction hypothesis is satisfied. Therefore, for any n in mathbb{N}^*, fleft(2^{-n}right) = n left(frac{1}{2}right)^{n-1} holds true. Finally, boxed{a_n = left(frac{1}{2}right)^{n-1}}.

answer:First, we need to calculate the volume of one box to determine how many boxes the company has in storage. The volume of one box is: Volume = Length × Width × Height Volume = 15 inches × 12 inches × 10 inches Volume = 1800 cubic inches Now, we know that the total volume occupied by the boxes is 1.08 million cubic inches. To find out how many boxes there are, we divide the total volume by the volume of one box: Number of boxes = Total volume / Volume of one box Number of boxes = 1,080,000 cubic inches / 1800 cubic inches Number of boxes = 600 Now that we know there are 600 boxes, we can calculate the total amount the company pays each month for record storage: Total cost per month = Number of boxes × Cost per box per month Total cost per month = 600 boxes × 0.2 per box Total cost per month = 120 Therefore, the company pays boxed{120} each month for record storage.

answer:Part (a): 1. Consider an ( n times n ) chessboard with ( 2n ) pieces placed at the centers of different squares. 2. **Analyze rows**: - Notice that each row can have at most ( n ) leftmost pieces (some rows may not have any pieces at all), hence there are at most ( n ) pieces occupying the leftmost positions in their respective rows. - This implies there must be at least ( 2n - n = n ) pieces that are not the leftmost in their rows. 3. **Record distances**: - Let's denote the distance (in terms of the number of squares) between the leftmost piece and another piece in the same row as a "distance". - There are at least ( n ) such pieces, and consequently, at least ( n ) recorded distances. 4. **Apply Pigeonhole Principle**: - The distances we have recorded range from ( 1 ) to ( n-1 ). Given that there are ( n ) distances recorded but only ( n-1 ) possible values for these distances, by the Pigeonhole Principle, at least two distances must be the same. - This means that there are at least two rows in which two pieces are placed the same distance apart. 5. **Form a parallelogram**: - Let the coordinates of the pieces in one row be ( (i, j_1) ) and ( (i, j_2) ) for ( i )-th row. - Similarly, let the coordinates of the pieces in another row be ( (k, j_1) ) and ( (k, j_2) ) in the ( k )-th row. - These four pieces ({(i, j_1), (i, j_2), (k, j_1), (k, j_2)}) form the vertices of a parallelogram as they have the same horizontal and vertical differences. Conclusion for part (a): blacksquare Part (b): 1. **Construct a configuration**: - Place ( 2n-1 ) pieces in a particular manner on an ( n times n ) chessboard. 2. **Positioning the pieces**: - Place ( n ) pieces in the first column (one piece in each row of the first column). - Place the remaining ( n-1 ) pieces in the first ( n-1 ) squares of the first row. 3. **Verification of no parallelogram formation**: - Observe that with ( n ) pieces in the first column and ( n-1 ) pieces in the first row, it is structurally impossible for these pieces to form a parallelogram. - The pieces in the first column only occupy vertical positions without creating any horizontal pair distances that could match a similar configuration in another row. - Similarly, pieces in the first row are placed in distinct column positions ensuring no matching distances to form another pair in any other row. Conclusion for part (b): blacksquare