answer:Given two lines (a) and (b) intersecting at point (O), and an arbitrary point (P). A line (l) passing through point (P) intersects lines (a) and (b) at points (A) and (B) respectively. We need to prove that the magnitude of (left(frac{AO}{BO}right) / left(frac{PA}{PB}right)) does not depend on the choice of line (l). 1. We begin by using the Law of Sines in triangles (triangle OPA) and (triangle OPB): [ frac{OA}{PA} = frac{sin angle OPA}{sin angle POA} ] and [ frac{OB}{PB} = frac{sin angle OPB}{sin angle POB} ] 2. Using these relations, we can express (frac{OA}{OB}) and (frac{PA}{PB}) in terms of the sines of the angles: [ frac{OA}{OB} = frac{frac{sin angle OPA}{sin angle POA}}{frac{sin angle OPB}{sin angle POB}} ] Simplifying, we get: [ frac{OA}{OB} = frac{sin angle OPA cdot sin angle POB}{sin angle POA cdot sin angle OPB} ] 3. Similarly, [ frac{PA}{PB} = frac{sin angle POA cdot OB}{sin angle POB cdot OA} ] 4. Hence, [ left(frac{OA}{OB}right) Bigg/ left(frac{PA}{PB}right) = frac{frac{sin angle OPA cdot sin angle POB}{sin angle POA cdot sin angle OPB}}{frac{sin angle POA cdot OB}{sin angle POB cdot OA}} ] 5. Simplifying the right-hand side, the (sin angle POA) and (sin angle POB) terms cancel each other out giving: [ left(frac{OA}{OB}right) Bigg/ left(frac{PA}{PB}right) = frac{sin angle OPB}{sin angle POA} ] Therefore, the ratio (left(frac{AO}{BO}right) / left(frac{PA}{PB}right) = frac{sin angle OPB}{sin angle POA}), which does not depend on the choice of line (l). [ boxed{frac{sin angle OPB}{sin angle POA}} ]

answer:To show that the sum (sum_{i=0}^k binom{n}{i} (-x)^i) is positive for all (0 leq x < frac{1}{n}) and all (k leq n), we can analyze the behavior of the series term by term. 1. **Expression of the Sum:** Let us rewrite the sum: [ S_k = sum_{i=0}^k binom{n}{i} (-x)^i ] 2. **Structure by Grouping Terms:** Consider pairing terms to facilitate analysis. For even (i), group terms (binom{n}{i} (-x)^i) and (binom{n}{i+1} (-x)^{i+1}): [ T_i = binom{n}{i} (-x)^i + binom{n}{i+1} (-x)^{i+1} ] Note that the terms can be grouped as follows: [ binom{n}{0} + sum_{i=0}^{lfloor k/2 rfloor} T_{2i} + text{(possibly unpaired terms)} ] 3. **Analyzing the Typical Grouped Term:** Consider the typical grouped term (T_i): [ T_i = binom{n}{i} (-x)^i + binom{n}{i+1} (-x)^{i+1} ] We will factor this grouping to illustrate positivity: [ T_i = binom{n}{i} (-x)^i left(1 + frac{binom{n}{i+1} (-x)}{binom{n}{i}}right) ] Using the identity for binomial coefficients, we get: [ binom{n}{i+1} = frac{n-i}{i+1} binom{n}{i} ] Substitute this into the expression: [ T_i = binom{n}{i} (-x)^i left(1 - frac{(n-i)x}{i+1}right) ] 4. **Checking the Positivity Condition:** Since (0 leq x < frac{1}{n}), consider the term inside the parentheses: [ 1 - frac{(n-i)x}{i+1} > 1 - frac{nx}{i+1} ] Given that (x < frac{1}{n}): [ nx < 1 ] Therefore: [ 1 - frac{nx}{i+1} > 0 ] Hence, for each (i): [ left(1 - frac{(n-i)x}{i+1}right) > 0 implies T_i > 0 ] 5. **Conclusion of the Sum:** Since each grouped term (T_i) is positive, and any unpaired terms beyond the complete pairings series still respect the initial constraint (0 leq x < frac{1}{n}), we conclude that the sum is positive. [ sum_{i=0}^k binom{n}{i} (-x)^i > 0 ] Thus, the series (sum_{i=0}^k binom{n}{i} (-x)^i) remains positive for all (0 leq x < frac{1}{n}) and all (k leq n). (boxed{text{Positive}})

answer:1. **Analyze the initial condition**: - There are 1001 steps on the mountain. - Initially, 500 stones are placed on the bottom 500 steps. 2. **Steps and rules for Sisyphus and Hades**: - Sisyphus action: He takes a stone and moves it to the nearest free step above. - Hades action: He slides a stone down one step, provided that the previous step is free. - Sisyphus starts the game. 3. **Objective**: - Prove that Hades can prevent Sisyphus from placing a stone on the topmost step. 4. **Key observation**: - After Sisyphus’s move, suppose he removes a stone from the ( n )-th step and places it on the (( n + 1 ))-th step. - Sisyphus’s goal is to eventually place a stone on the topmost (( 1001 )-th) step. 5. **Determine Hades's Strategy**: - Hades can ensure that: - The lowest step is always occupied. - There are no two consecutive free steps at any time. 6. **Proof**: - Assume Hades always moves a stone such that the stone above the lowest free step is moved down. - This ensures after every move by Hades, the sequence of occupied and free steps maintains the condition, i.e., no two consecutive free steps. 7. **Game Dynamics**: - If Sisyphus places a stone on step ( n ), step ( n+1 ) becomes occupied. - Hades then slides a stone from ( n+1 ) to step ( n ), ensuring step ( n+1 ) is free again and maintaining the non-consecutive free step condition. 8. **Implication for Sisyphus's Failure**: - If there are no two consecutive free steps, Sisyphus cannot move a stone continuously up to cover all steps from ( n ) to ( 1000 ). - Thus, the highest Sisyphus can place a stone is the ( 1000 )-th step, but not the ( 1001 )-th step. # Conclusion: Hence, Hades can always perform actions to prevent Sisyphus from placing a stone on the topmost step. boxed{text{Hades can always prevent Sisyphus from reaching the topmost step}}

answer:To analyze the relationship between "x=y" and "cos x = cos y", we proceed as follows: 1. **From x=y to cos x = cos y:** If x=y, then by the properties of trigonometric functions, specifically the cosine function which is a function of the angle, we have cos x = cos y. This step shows that if x is equal to y, then their cosines are also equal. This is because the cosine of an angle is determined solely by the value of the angle itself. 2. **From cos x = cos y to x=y:** However, the reverse does not hold true in general. That is, if cos x = cos y, it does not necessarily mean that x=y. For instance, consider cosfrac{pi}{2} = cosfrac{3pi}{2}. While both sides of the equation are equal (both equal to 0), the angles themselves, frac{pi}{2} and frac{3pi}{2}, are not equal. This example demonstrates that even if the cosines of two angles are the same, the angles themselves might not be the same due to the periodic nature of the cosine function. Therefore, we conclude that "x=y" is a condition that guarantees "cos x = cos y", making it a sufficient condition. However, since "cos x = cos y" does not guarantee "x=y", "x=y" is not a necessary condition for "cos x = cos y". Hence, the correct choice is: boxed{text{A: Sufficient but not necessary condition}}