answer:Given: triangle ABC and point M on segment AC (distinct from A and C). We need to construct points P and Q on the sides AB and BC, respectively, such that PQ parallel AC and angle PMQ = 90^circ. 1. **Identify the Objective**: - We need to ensure that PQ parallel AC - angle PMQ must be a right angle. 2. **Homothety Transformation**: - Let's define homothety with center at B, such that the transformation maps: - P to A - Q to C The homothety ratio is given by: [ frac{AB}{BP} = frac{CB}{BQ} ] With this mapping, the segment (PQ) transforms into (AC). Consequently, point (M) maps to a point (N) on (AC). 3. **Perpendicularity Condition**: - Since (PQ parallel AC), and angle PMQ = 90^circ, this implies: [ angle ANC = 90^circ ] - Hence, this means that the point (N) lies on a semicircle defined above segment (AC). 4. **Construct the Points**: - Draw the semicircle on (AC) such that its diameter is (AC). - Let (N) be the intersection of line BM with this semicircle (on the opposite side of triangle ABC). 5. **Parallel Lines Construction**: - Construct two parallel lines through (M): - One parallel to (AN) - Another parallel to (CN) - The intersections of these lines with the sides (AB) and (BC) respectively will give us the points (P) and (Q). 6. **Validate the Construction**: - Since PM parallel AN and QM parallel CN, quadrilateral ANMC ensures angle PMQ = angle ANC = 90^circ - By similarity of triangles triangle BAN and triangle BPM: [ frac{AB}{PB} = frac{NB}{MB} ] - Similarly for triangles triangle BCN and triangle BQM: [ frac{CB}{QB} = frac{NB}{MB} ] - Thus: [ frac{AB}{PB} = frac{CB}{QB} ] 7. **Conclusion**: - By Thales' Theorem, PQ parallel AC. Therefore, the solution is complete, and we have: [ boxed{text{Points } P text{ and } Q text{ correctly constructed as described.}} ]

answer:Option B Key Points: Judgment of necessary conditions, sufficient conditions, and necessary and sufficient conditions. Topic: Basic logic. Analysis: According to the conditions, find sets A and B, and then solve it based on the definitions of sufficient and necessary conditions. Solution: Set A = {x | 2x > 1} = {x | x > 0.5}, Set B = {x | log x > 0} = {x | x > 1}, Thus, B subset A, which means "x in A" is a necessary but not sufficient condition for "x in B", Therefore, the correct option is: boxed{B} Review: This question mainly examines the application of the relationship between sufficient and necessary conditions, which is quite basic.

answer:We are given a country that uses coins of denominations ( 6n + 1 ), ( 6n + 4 ), ( 6n + 7 ), and ( 6n + 10 ) piastres, where ( n ) is a natural number. We need to determine the largest amount that cannot be given using these coins. Let's analyze the problem step by step: 1. **Definitions and Assumptions**: - Define a natural number ( s ) as textit{achievable} if the amount ( s ) can be formed using the given coin denominations. - Take any ( r ) such that ( r = 1, ldots, 6n ). 2. **Finding Minimal Achievable Amount**: - Let ( m_r ) be the smallest achievable number such that ( m_r equiv r pmod{6n+1} ). - We will prove two statements. 3. **Statement 1**: - If ( s equiv r pmod{6n+1} ) and ( s geq m_r ), then ( s ) can be represented using the given denominations. - If ( s < m_r ), then ( s ) cannot be achieved using the given denominations. **Proof of Statement 1**: - If ( s geq m_r ), we can pay ( s ) by first using ( m_r ) and then adding ( s-m_r ) which will be a multiple of ( 6n+1 ). - If ( s < m_r ), suppose ( s ) can be achieved. Since ( s < m_r ), we can subtract some multiples of ( 6n+1 ) to get to ( less) than ( m_r ), which contradicts the minimality of ( m_r ). 4. **Statement 2**: - Numbers of the form ( 3k pmod{6n+1} ) take all possible values ( {1, ldots, 6n } ) as ( k ) varies from ( 1 ) to ( 6n ). **Proof of Statement 2**: - We need to show that all ( 3k ) values are distinct modulo ( 6n+1 ). - If ( 3k equiv 3k' pmod{6n+1} ), then ( 3(k - k') = m(6n+1) ) for some integer ( m ). - Since ( |k - k'| leq 6n ), it must be that ( k = k' ), proving all values are distinct. 5. **Achievability**: - Clearly, sums that are multiples of ( 6n+1 ) can be paid. 6. **Finding ( m_r )**: - To find the largest amount not achievable, we need to find the maximum of ( { m_1, m_2, ldots, m_{6n} } ). - Notice: [ 6n + 4 = (6n+1) + 3, quad 6n + 7 = (6n+1) + 6, quad 6n + 10 = (6n+1) + 9 ] - Using the previous conclusions, there exists a unique ( k ) such that ( 3k equiv r pmod{6n+1} ). - We need to represent ( 3k ) as a sum of the minimum number of coins available. The general form will be ( m_r = 3k + leftlfloor frac{k+2}{3} rightrfloor cdot (6n+1) ). 7. **Calculation**: - If ( r = 6n ), then ( k = 6n ). - Then, [ m_{6n} = 3 cdot 6n + leftlfloor frac{6n + 2}{3} rightrfloor cdot (6n + 1) = 18n + 2n (6n + 1) = 18n + 12n^2 + 2n = 12n^2 + 20n ] 8. **Conclusion**: - Therefore, the maximum amount the bank cannot pay is ( 12n^2 + 20n - 6n - 1 = 12n^2 + 14n - 1 ). Thus, the largest amount of piastres that cannot be paid using these denominations is: [ boxed{12n^2 + 14n - 1} ]

answer:**Analysis** This question examines the application of the formula for the sum of a sequence. It uses ngeqslant 2 and a_n=S_n-S_{n-1} to solve, which is a basic problem. **Solution** Since the sum of the first n terms of the sequence {a_n} is S_n=5n^2+kn−19 (nin mathbf{N}^*), and a_{10}=100, we have S_{10}-S_{9}=100, which means 500+10k-405-9k=100, solving this gives k=5. Therefore, the answer is boxed{text{A}}.