answer:1. **Initial Information and Setup:** We are given the circumradius ( r ), the distance ( KS ) from the circumcenter ( K ) to the centroid ( S ), and the distance ( d ) from the centroid ( S ) to the altitude. Specifically, we have: [ r = 7 , text{cm}, quad KS = 3 , text{cm}, quad d = 1 , text{cm} ] 2. **Using Euler's Line Property:** On Euler's line, we know the relationship among the circumcenter ( K ), centroid ( S ), and orthocenter ( M ). The centroid ( S ) divides the line segment ( KM ) in the ratio ( 2:1 ). Consequently: [ SM = 2 cdot KS = 2 cdot 3 , text{cm} = 6 , text{cm} ] 3. **Characterizing the Altitude Line:** The distance ( d = 1 , text{cm} ) from ( S ) represents the shortest distance from ( S ) to the altitude line segment ( m_a ) which can be taken as the tangent to a certain circle centered at ( S ) with a radius ( d ). Hence, the altitude from ( S ) to ( m_a ) forms a circle ( omega_S ) centered at ( S ) with radius ( d ). [ omega_S : , (S, d) ] 4. **Identifying the Tangents and Intersection Points:** The altitude ( m_a ) is thus a tangent to this circle ( omega_S ). Being a tangent implies it intersects at one point on the circle. ( m_a ) will intersect the circumcircle of ( K ) at ( A ). As this tangent is symmetric about Euler's line, two potential intersections form ( A_1 ) and ( A_2 ): [ A_1 , text{and} , A_2 ] 5. **Midpoint and Constructing Other Vertices:** Next, we find the midpoint ( D ) of side ( a ) of triangle ( ABC ). The centroid lies one-third the distance from each vertex to the midpoint of the opposite side: [ SD = frac{AS}{2} ] The line through ( D ) perpendicular to ( m_a ) will be one side of the triangle, thus determining the intersection points, the other vertices ( B ) and ( C ). 6. **Feasibility Conditions:** - The point ( M ) must be able to form a tangent to the circle: [ d leq SM = 6 , text{cm} ] This requirement is satisfied since ( d = 1 , text{cm} leq 6 , text{cm} ). - The circumcircle must intersect the tangent ( m_a ): [ r geq frac{3}{2} d = frac{3}{2} times 1 , text{cm} = 1.5 , text{cm} ] This is also satisfied since ( r = 7 , text{cm} geq 1.5 , text{cm} ). 7. **Distance Considerations Related to ( D ):** - ( D ) lies within the circumcircle ( K ): We need: [ KD < r ] First find: [ KD = frac{AM}{2} ] Compute: [ AM = |A K' pm M K'| ] Where: [ A K' = sqrt{K A^2 - K K'^2} = sqrt{r^2 - left( frac{3d}{2} right)^2} = sqrt{7^2 - frac{9}{4}} = sqrt{49 - 2.25} = sqrt{46.75} ] [ M K' = sqrt{M K^2 - K K'^2} = sqrt{(3KS)^2 - left(frac{3d}{2}right)^2} = sqrt{(3 times 3)^2 - left(frac{3 times 1}{2}right)^2} = sqrt{81 - 2.25} = sqrt{78.75} ] So we compute: [ KD = frac{| sqrt{46.75} pm sqrt{78.75}|}{2} ] Ensuring: [ left| sqrt{4r^2 - 9d^2} - 3sqrt{4KS^2 - d^2} right| < 4r ] Thus ( boxed{7} , text{cm} leq KD ) ensures one solution, confirming the problem's stated constructs. Different configurations may yield other valid triangles typically two or one, considering consistency with constraints ( (4) ) and ( (3) ).

answer:1. **Define the Problem and Graph Representation:** - Let ( n ) be a positive integer. - Consider an ( n times n ) table where each cell is colored in one of ( k ) colors. - Each color is used at least once. - Two different colors ( A ) and ( B ) are said to touch each other if there exists a cell colored in ( A ) sharing a side with a cell colored in ( B ). - The table is colored such that each color touches at most 2 other colors. - We need to find the maximal value of ( k ) in terms of ( n ). 2. **Graph Construction:** - Consider a graph ( G ) where the vertices represent the ( k ) colors. - There is an edge between two vertices if the corresponding colors touch each other. - Since every color is used at least once and the table is connected, ( G ) is connected. - Each vertex in ( G ) has a degree of at most 2, implying ( G ) is either a path or a cycle. 3. **Case Analysis:** - **Case 1: ( G ) is a path.** - If ( G ) is a path, the maximum number of vertices (colors) is ( k ). - **Case 2: ( G ) is a cycle.** - If ( G ) is a cycle, the maximum number of vertices (colors) is ( k ). 4. **Bounding ( k ) for Odd ( n ):** - For odd ( n ), every cell in ( T ) can be reached from the center cell (left(frac{n-1}{2}, frac{n-1}{2}right)) in at most ( n-1 ) steps. - Therefore, ( G ) can have at most ( 2n-1 ) vertices. - Hence, ( k leq 2n-1 ). 5. **Bounding ( k ) for Even ( n ):** - For even ( n ), every cell in ( T ) can be reached from the center cell (left(frac{n}{2}, frac{n}{2}right)) in at most ( n ) steps. - Therefore, ( G ) can have at most ( 2n ) vertices. - Hence, ( k leq 2n ). 6. **Special Considerations for ( n geq 4 ):** - For ( n geq 4 ), consider the cells ((0,0)), ((1,0)), ((0,1)), ((n,0)), and ((0,n)). - If ((0,0)) has a unique color ( C ) with distance ( n ) in ( G ) from the color ( D ) of (left(frac{n}{2}, frac{n}{2}right)), it is the only cell with color ( C ). - If ((1,0)) and ((0,1)) have different colors, ( G ) must be a cycle of length ( 2n ), which contradicts the uniqueness of ( C ). - If ((1,0)) and ((0,1)) have the same color, the colors of ((n,0)) and ((0,n)) are reachable from ((1,0)) and ((0,1)) respectively in ( n-1 ) steps in ( G ) with a path not containing ( C ). - Thus, ( k leq 2n-1 ) for even ( n geq 4 ). 7. **Achieving the Bounds:** - These bounds can be achieved by coloring ((i,j)) with the color ( i+j ) for ( n neq 2 ). - For ( n = 2 ), color each cell with a unique color. The final answer is ( boxed{ k leq 2n-1 } ) for odd ( n ) and ( k leq 2n-1 ) for even ( n geq 4 ). For ( n = 2 ), ( k = 4 ).

answer:To find the domain of the function f(x)=frac{1}{ln(x+1)}+sqrt{9-x^2}, we need to ensure that both the denominator of the fraction and the expression under the square root are valid. This leads to two conditions: 1. The argument of the logarithm must be positive, and the logarithm itself cannot be zero. This translates to: - x+1 > 0 to ensure the argument of the logarithm is positive. - x+1 neq 1 to avoid the logarithm of 1, which is 0 (since ln(1) = 0 and we cannot divide by zero). 2. The expression under the square root must be non-negative to ensure we are dealing with real numbers: - 9-x^2 geq 0. Let's solve these inequalities step by step: For the logarithm condition: - x+1 > 0 simplifies to x > -1. - x+1 neq 1 simplifies to x neq 0. For the square root condition: - 9-x^2 geq 0 can be rewritten as -(x^2-9) geq 0, which factors into -(x-3)(x+3) geq 0. Solving this inequality, we find -3 leq x leq 3. Combining these conditions, we have: - From the logarithm condition, x > -1 and x neq 0. - From the square root condition, -3 leq x leq 3. Overlaying these conditions, we find that the domain of f(x) is (-1,0) union (0,3]. Therefore, the correct answer is: boxed{B}.

answer:1. Let ( f(x) = x^n + a_1 x^{n-1} + ldots + a_n ) be a polynomial with real roots ( b_1, b_2, ldots, b_n ). We need to prove that for all ( x geq max{b_1, b_2, ldots, b_n} ), the inequality [ f(x+1) geq frac{2n^2}{frac{1}{x-b_1} + frac{1}{x-b_2} + ldots + frac{1}{x-b_n}} ] holds. 2. Let ( c_i = x - b_i ). Then ( c_i geq 0 ) for all ( i ). We need to show that [ (c_1 + 1)(c_2 + 1) cdots (c_n + 1) left( frac{1}{c_1} + frac{1}{c_2} + ldots + frac{1}{c_n} right) geq 2n^2. ] 3. By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have: [ frac{1}{c_1} + frac{1}{c_2} + ldots + frac{1}{c_n} geq n sqrt[n]{frac{1}{c_1 c_2 cdots c_n}}. ] 4. For each ( c_i ), we can write: [ c_i + 1 = c_i + (n-1) frac{1}{n-1} geq n sqrt[n]{c_i left( frac{1}{n-1} right)^{n-1}}, ] using the AM-GM inequality again. 5. Therefore, the product of these terms is: [ (c_1 + 1)(c_2 + 1) cdots (c_n + 1) geq n^n left( frac{1}{(n-1)^{n-1}} right)^n = frac{n^n}{(n-1)^{n(n-1)}}. ] 6. Combining these results, we get: [ (c_1 + 1)(c_2 + 1) cdots (c_n + 1) left( frac{1}{c_1} + frac{1}{c_2} + ldots + frac{1}{c_n} right) geq frac{n^n}{(n-1)^{n-1}} cdot n sqrt[n]{frac{1}{c_1 c_2 cdots c_n}}. ] 7. Simplifying further, we need to show: [ frac{n^{n+1}}{(n-1)^{n-1}} geq 2n^2. ] 8. This reduces to: [ left( frac{n}{n-1} right)^{n-1} geq 2. ] 9. Let ( m = n-1 ). Then we need to show: [ left( 1 + frac{1}{m} right)^m geq 2. ] 10. By the binomial expansion, we have: [ left( 1 + frac{1}{m} right)^m = sum_{i=0}^m binom{m}{i} left( frac{1}{m} right)^i geq 1 + m cdot frac{1}{m} = 2. ] 11. Therefore, the inequality holds, and we have proven the required result. (blacksquare)