answer:to Problem 3.23: Given System of Linear Equations: [ left{ begin{array}{rl} 10x_{1} + 3x_{2} + 4x_{3} + x_{4} + x_{5} &= 0 11x_{2} + 2x_{3} + 2x_{4} + 3x_{5} + x_{6} &= 0 15x_{3} + 4x_{4} + 5x_{5} + 4x_{6} + x_{7} &= 0 2x_{1} + x_{2} - 3x_{3} + 12x_{4} - 3x_{5} + x_{6} + x_{7} &= 0 6x_{1} - 5x_{2} + 3x_{3} - x_{4} + 17x_{5} + x_{6} &= 0 3x_{1} + 2x_{2} - 3x_{3} + 4x_{4} + x_{5} - 16x_{6} + 2x_{7} &= 0 4x_{1} - 8x_{2} + x_{3} + x_{4} - 3x_{5} + 19x_{7} &= 0 end{array} right. ] Step-by-Step Procedure: 1. **Rewrite the system in matrix form:** The system can be represented as (A mathbf{x} = mathbf{0}), where (A) is a (7 times 7) coefficient matrix, (mathbf{x}) is a column vector of unknowns, and (mathbf{0}) is a zero vector. [ A = begin{pmatrix} 10 & 3 & 4 & 1 & 1 & 0 & 0 0 & 11 & 2 & 2 & 3 & 1 & 0 0 & 0 & 15 & 4 & 5 & 4 & 1 2 & 1 & -3 & 12 & -3 & 1 & 1 6 & -5 & 3 & -1 & 17 & 1 & 0 3 & 2 & -3 & 4 & 1 & -16 & 2 4 & -8 & 1 & 1 & -3 & 0 & 19 end{pmatrix}, quad mathbf{x} = begin{pmatrix} x_1 x_2 x_3 x_4 x_5 x_6 x_7 end{pmatrix}, quad mathbf{0} = begin{pmatrix} 0 0 0 0 0 0 0 end{pmatrix} ] 2. **Examine the properties of the coefficient matrix (A):** The matrix (A) satisfies the following diagonal dominance condition for each column (j) (also called the strict diagonal dominance property): [ left|a_{jj}right| > sum_{i neq j} left|a_{ij}right| ] Verifying for each column: - For (j = 1): [ left|10right| > left|2right| + left|6right| + left|3right| + left|4right| implies 10 > 2 + 6 + 3 + 4 implies 10 > 15 , text{(not satisfied)} ] 3. **Conclusion:** Since the condition of strict diagonal dominance is not uniformly satisfied for all columns, the initial assumption that one of (x_1, ldots, x_7) is nonzero leads to a contradiction. Therefore, the only solution to the given system of equations is the trivial solution: [ x_1 = x_2 = ldots = x_7 = 0 ] (boxed{x_1 = x_2 = ldots = x_7 = 0})

answer:Since there were 30 males last year, there are 1.10 times 30 = 33 males this year. Let the number of females last year be y. Thus, there are 1.25y females this year. The total number of participants last year was 30 + y, and this year it is 1.15 times (30 + y). Setting up the equation for this year's participants: [ 33 + 1.25y = 1.15 times (30 + y) ] Expanding and solving for y: [ 33 + 1.25y = 34.5 + 1.15y 1.25y - 1.15y = 34.5 - 33 0.1y = 1.5 y = 15 ] Therefore, there were 15 females last year, so there are 1.25 times 15 = 18.75 females this year. Hence, the fraction of the league's participants that is female this year is: [ frac{18.75}{18.75 + 33} = frac{18.75}{51.75} = boxed{frac{25}{69}} ]

answer:1. **Identify Key Points and Definitions:** - Let ( L ) be the midpoint of the arc ( BAC ) on the circumcircle of ( triangle ABC ). - Let ( O_1 ) be the circumcenter of ( triangle ABC ). - Let ( O_2 ) be the circumcenter of ( triangle ADM ). - ( M ) is the midpoint of ( BC ). - ( D ) is the point where the angle bisector of ( angle BAC ) intersects ( BC ). 2. **Cyclic Quadrilateral ( ADML ):** - Since ( L ) is the midpoint of the arc ( BAC ), ( L ) lies on the circumcircle of ( triangle ABC ). - By the properties of the angle bisector and the circumcircle, ( ADML ) forms a cyclic quadrilateral. This is because ( angle ALD = 90^circ ) (as ( L ) is the midpoint of the arc opposite ( A )) and ( angle AMD = 90^circ ) (since ( M ) is the midpoint of ( BC ) and ( AD ) is the angle bisector). 3. **Thales' Theorem:** - By Thales' theorem, since ( ADML ) is cyclic and ( angle ALD = 90^circ ), ( DL ) is the diameter of the circumcircle of ( triangle ADML ). 4. **Radical Axes and Perpendicularity:** - The radical axis of two circles is perpendicular to the line joining their centers. - Since ( ADML ) is cyclic with diameter ( DL ), the line ( AD ) is perpendicular to ( DL ). - Therefore, ( AD perp DL ). 5. **Conclusion:** - Since ( AD perp DL ) and ( DL ) is the line joining the centers ( O_1 ) and ( O_2 ), it follows that ( AD ) is perpendicular to ( O_1O_2 ). - Hence, the line through the centers of the circumcircles of ( triangle ABC ) and ( triangle ADM ) (i.e., ( O_1O_2 )) is parallel to ( AD ). (blacksquare)

answer:1. **Define the Problem in Graph Theory Terms**: Represent each person as a vertex in a graph. Two vertices (persons) are connected by an edge if and only if the corresponding persons know each other. Hence, we have an undirected graph (G = (V, E)) where each vertex has a degree of exactly 3 (each person knows exactly 3 other persons). Additionally, among any three vertices, there exists at least one pair of vertices that are not connected by an edge (i.e., not all persons among any three know each other). 2. **Explore the Meaning of the Degree and Constraints**: - Each vertex in graph (G) has exactly 3 neighbors (degree of 3). - For any three vertices in (G), there must be at least one pair of vertices that are not adjacent. 3. **Check Lower Bound**: Start with smaller numbers and increase until conditions are met. - **For (n = 3)**: If there are 3 people, all three people must know each other (complete triangle). This configuration does not satisfy the condition that among any three, there is one pair of people that do not know each other. - **For (n = 4)**: Each person can know exactly 3 others here, but forming such a configuration is not possible since we would be unable to connect only 3 people to each without creating a cycle where all know each other. - **For (n = 5)**: Similar reasoning shows we cannot have each person knowing exactly 3 people without forming a configuration where a group of 3 people exists with each knowing the other two. 4. **Evaluate (n = 6)**: Construct a 6-vertex graph where each vertex has exactly 3 edges and any subset of 3 vertices includes at least two that are not connected: [ begin{aligned} &text{Vertices: } A, B, C, D, E, F &text{Edges}: {(A, B), (A, C), (A, D)}, {(B, C), (B, E), (B, F)}, &{(C, D), (C, F), (C, E)}, {(D, E), (D, F)} end{aligned} ] This shows we form pairs where each vertex has 3 others it knows, and this setup can be checked to meet the criteria: - Every vertex has exactly 3 connections. - Among any three vertices, you can find pairs of vertices that are not directly connected. **Counting & Consistency**: verify consistent pairings ensuring proper compliance (all constraints fulfilled). After these steps, verification confirms 6 is the minimal configuration meeting all conditions. # Conclusion: The minimal number of people in the room such that each person knows exactly three others, and among any three selected people, there is at least one pair who do not know each other, is: [ boxed{6} ]