answer:1. Define the setup: - Points A, B, and C lie on a circle. - Line b is tangent to this circle at point B. - From a point P lying on the tangent line b, perpendiculars PA_1 and PC_1 are dropped to the lines AB and BC, respectively (points A_1 and C_1 lie on segments AB and BC). 2. Consider the angles and geometric properties: - The points A_1, B, C_1, and P form a cyclic quadrilateral because the angles angle PA_1B and angle PC_1B are both right angles (by construction of the perpendiculars). Therefore, these points lie on a circle with diameter BP. 3. Work with the inscribed angles: - Inscribed angles subtending the same arc are equal. Specifically, angle A_1C_1P and angle A_1BP subtend the arc A_1C_1. Therefore, we have: [ angle A_1C_1P = angle A_1BP = alpha ] 4. Use the tangent-secant theorem: - Theorem: The angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the opposite segment. - Apply the theorem: angle ACB = angle ABP = alpha. 5. Analyze the relationship between angle A_1C_1C and known angles: - angle A_1C_1C is an external angle of triangle BC_1P and: [ angle A_1C_1C = 180^circ - angle BPC_1 = 180^circ - 90^circ = 90^circ - alpha ] - Thus, angle A_1C_1C = 90^circ - alpha. 6. Conclude the perpendicularity: - Since angle A_1C_1C = 90^circ - alpha, we know: [ angle A_1C_1C + alpha = 90^circ ] Therefore, angle A_1C_1C + angle ACB = 90^circ, which implies angle A_1C_1C = 90^circ, showing A_1C_1 perp AC. Conclusion: [ boxed{A_1C_1 perp AC} ]

answer:1. **Understanding the Ratio**: The ratio of boys to girls in Ms. Lee's chemistry class is 3:4. This means for every 3 boys, there are 4 girls. 2. **Setting Up Variables**: Let the number of boys be 3x and the number of girls be 4x. Here, x is a common multiplier for the ratio. 3. **Total Students Equation**: The total number of students in the class is given as 42. The equation based on the ratio is: [ 3x + 4x = 7x ] [ 7x = 42 ] 4. **Solving for x**: Solving for x gives: [ x = frac{42}{7} = 6 ] 5. **Calculating Number of Boys and Girls**: - Number of boys = 3x = 3 times 6 = 18 - Number of girls = 4x = 4 times 6 = 24 6. **Finding the Difference**: The difference between the number of girls and boys is: [ 4x - 3x = x = 6 ] 7. **Conclusion**: There are 6 more girls than boys in Ms. Lee's chemistry class. The answer is 6. The final answer is boxed{B}

answer:1. **Define the sum of consecutive terms:** Let [ S_k(i) = a_i + a_{i + 1} + ldots + a_{i + k - 1} ] be the sum of the k consecutive terms starting with a_i. The given condition implies that for every k geq 1 and every m geq 0, there exists n in [m + 1, m + r] such that S_k(m) = S_k(n). 2. **Claim: For each k geq 1, there are at most r distinct values in the list S_k(i) for i = 0, 1, 2, ldots.** - **Proof:** Assume that for some value of k, there are at least r + 1 distinct values. Consider the first r + 1 such values to appear, and take m such that for each of these values x, there is some i < m with S_k(i) = x (so each of these r + 1 values has appeared before m). Then consider [ S_k(m), S_k(m + 1), ldots, S_k(m + r - 1). ] There are at most r values in this list, so there is at least one x of those first r + 1 values which is not in this list. Consider the largest i < m such that S_k(i) = x, which must exist since there is at least one such i. Then there is no j with i + 1 leq j leq m + r - 1 and S_k(j) = S_k(i), which is a contradiction (as i + r leq m + r - 1). So there can be at most r distinct values. 3. **Define n and M:** Let n = r^{2r} + 1. For each 1 leq k leq n, let M_k be the last index where a new value appears in the list S_k(i), meaning the greatest i such that S_k(i) neq S_k(j) for all 0 leq j < i. (This must exist since there can be at most r such indices i.) Let M = max(M_1, ldots, M_n) + 1, so then for all 1 leq k leq n, any value in the list S_k(i) appears at some i < M. 4. **Claim: There exists some k with 1 leq k leq n, and some m geq M, for which S_k(m) = S_k(m + 1) = ldots = S_k(m + r).** - **Proof:** Consider the lists S_k(M), S_k(M + 1), ldots, S_k(M + r) for each 1 leq k leq n. We will show that two of these lists differ by a constant. First, note that for any i and k, [ S_k(i + 1) - S_k(i) = a_{i + 1} + ldots + a_{i + k} - (a_i + ldots + a_{i + k - 1}) = a_{i + k} - a_i. ] Since there are at most r distinct values in a_0, a_1, ldots, there are at most r^2 possible values for this (over all i and k). For each 1 leq k leq M, consider the sequence of differences between consecutive terms, S_k(M + 1) - S_k(M), S_k(M + 2) - S_k(M + 1), ldots, S_k(M + r) - S_k(M + r - 1). There are r terms in this sequence, and each term must be one of the at most r^2 possible values, so then there are at most (r^2)^r = r^{2r} distinct sequences possible. Since we have n = r^{2r} + 1 such sequences, at least two must be the same. Suppose these are at k_1 and k_2, with k_1 > k_2. Then we have S_{k_1}(i + 1) - S_{k_1}(i) = S_{k_2}(i + 1) - S_{k_2}(i) for all M leq i leq M + r - 1. This implies S_{k_1}(i + 1) - S_{k_2}(i + 1) = S_{k_1}(i) - S_{k_2}(i) for all M leq i leq M + r - 1, so then [ S_{k_1}(i) - S_{k_2}(i) = c ] for all M leq i leq M + r and some constant c. However, we have [ S_{k_1}(i) - S_{k_2}(i) = a_i + ldots + a_{i + k_1 - 1} - (a_i + ldots + a_{i + k_2 - 1}) = a_{i + k_2} + ldots + a_{i + k_1 - 1} = S_{k_1 - k_2}(i + k_2). ] So then we have [ S_{k_1 - k_2}(i + k_2) = c ] for all M leq i leq M + r. Taking k = k_1 - k_2 (so 0 < k < n) and m = M + k_2 (so m > M), we have S_k(m) = S_k(m + 1) = ldots = S_k(m + r). 5. **Claim: For the k from the previous claim, S_k(i) is constant (for all i geq 0).** - **Proof:** We know S_k(m) = ldots = S_k(m + r) is some constant c. If there is some i < m with S_k(i) neq c, then we get a contradiction by taking the greatest such i, as then there is no i < j < m or m leq j leq m + r with S_k(i) = S_k(j). So then S_k(i) = c for all i leq m. However, we have m geq M, and all values in the sequence S_k(i) must appear for some i < M, as we chose M to come after the first occurrence of any value in the list. So there can be no values other than c in the list, so S_k(i) = c for all i > m as well. So then we have S_k(i + 1) - S_k(i) = 0 for all i geq 0. However, S_k(i + 1) - S_k(i) = a_{i + k} - a_i, so this means a_{i + k} = a_i for all i geq 0, so the sequence is periodic. blacksquare

answer:To find the total area of the wet surface, we need to calculate the area of the bottom, the two longer sides, and the two shorter sides that are in contact with the water. 1. The area of the bottom (which is fully in contact with water) is: Length × Width = 6 m × 4 m = 24 m² 2. The area of the two longer sides: Since the water is up to a breadth of 1 m 25 cm, we need to convert this to meters to be consistent with the other measurements. 1 m 25 cm = 1 m + 0.25 m = 1.25 m The area of one longer side is: Height (water level) × Length = 1.25 m × 6 m = 7.5 m² Since there are two such sides, we multiply by 2: 7.5 m² × 2 = 15 m² 3. The area of the two shorter sides: The area of one shorter side is: Height (water level) × Width = 1.25 m × 4 m = 5 m² Since there are two such sides, we multiply by 2: 5 m² × 2 = 10 m² Now, we add up all the areas to find the total wet surface area: Total wet surface area = Area of bottom + Area of longer sides + Area of shorter sides Total wet surface area = 24 m² + 15 m² + 10 m² Total wet surface area = 49 m² Therefore, the total area of the wet surface is boxed{49} square meters.