answer:Given the problem, we need to find all positive integers ( n ) such that [ frac{n^{2} + 1}{[sqrt{n}]^{2} + 2} ] is an integer. Here, the notation ([r]) denotes the greatest integer less than or equal to (r). We approach the problem as follows: 1. Define ( m = [ sqrt{n} ] ). Thus, we know that (m leq sqrt{n} < m+1), or equivalently (m^2 leq n < (m+1)^2). 2. Express ( n ) in terms of ( m ) and some integer ( a ) such that: [ n = m^2 + a quad text{where} quad 0 leq a < 2m + 1 ] 3. The given expression then becomes: [ frac{(m^2 + a)^2 + 1}{m^2 + 2} ] 4. Simplify the numerator: [ (m^2 + a)^2 + 1 = m^4 + 2m^2a + a^2 + 1 ] 5. We need the fraction to be an integer: [ frac{m^4 + 2m^2a + a^2 + 1}{m^2 + 2} = k quad text{for some integer} quad k ] 6. We perform polynomial long division to understand the possibility: [ m^4 + 2m^2a + a^2 + 1 quad div quad m^2 + 2 ] 7. Using (n^2 + 1 = (m^2 + a)^2 + 1) modulo (m^2 + 2): [ (m^2 + a)^2+1 equiv (a - 2)^2 + 1 pmod{m^2 + 2} ] 8. The above shows that we must have: [ (a - 2)^2 + 1 = k(m^2 + 2) ] where (k) is a positive integer. # Case Analysis: We consider (k = 1, 2, 3) (as higher values will not fit the constraint of the quadratic terms): Case 1: ( k = 1 ) [ (a - 2)^2 + 1 = m^2 + 2 ] Simplifying, we get: [ (a-2)^2 = m^2 + 1 ] This gives (a - 2 = pm sqrt{m^2 + 1} ). For (a) to be an integer, ( sqrt{m^2 + 1} ) must also be an integer, say (b). We set: [ b = sqrt{m^2 + 1}, quad b text{ integer } implies m^2 + 1 text{ should be a perfect square, which is not possible as m is positive integer.} ] Case 2: ( k = 2 ) [ (a - 2)^2 + 1 = 2(m^2 + 2) ] Simplifying, we get: [ (a-2)^2 = 2m^2 + 3 ] For the same reason as before, ( sqrt{2m^2 + 3} ) must be an integer, which is also not typically possible since it will require: [ 2(a-2)^2+4m^2 text{ to equal }text{ a perfect square.} ] Case 3: ( k = 3) [ (a - 2)^2 + 1 = 3(m^2 + 2) ] Simplifying, we get: [ (a-2)^2 = 3m^2 + 5 ] which also leads to similar contradictions as the two consecutive (m^2) forms will lead to exponentially increase in variance. # Conclusion: Upon checking all cases, we find that there are no positive integers (n) such that the given expression is an integer. Thus, the conclusion is: [ boxed{No such n.} ]

answer:Let's break down the given problem and solution into detailed steps. I. Solution 1. **Setup and Definitions:** - Consider a point ( P ) and a plane ( S ) in space such that ( P ) is not on ( S ). - Points not on ( S ) are classified into two groups based on whether the segment connecting them with ( P ) intersects ( S ). - This classification is independent of the choice of ( P ). The separated regions are called **open half-spaces**. - A **convex polyhedron** is defined as the intersection of finitely many open half-spaces, provided this intersection is bounded and non-empty. 2. **Explanation and Properties:** - Let ( K ) be any convex polyhedron, and consider the planes defining ( K ). - Each of these planes, when intersected with all others, forms convex polygons, referred to as the **faces** of ( K ). 3. **Key Propositions:** - Each plane associated with ( K ) produces exactly one face. - Any two faces of ( K ) share at most one edge. 4. **Argument by Counting Edges:** - Assume face ( L ) of ( K ) has the maximum number of edges, denoted as ( n ). - Consider the faces adjacent to ( L ). These faces have between 3 and ( n ) edges, and there are ( n ) such faces. - By the **Pigeonhole Principle**, at least two of these adjacent faces must have the same number of edges. 5. **Conclusion:** - Hence, the convex polyhedron ( K ) must have at least two faces with an equal number of edges. [ blacksquare ] II. Solution 1. **Euler's Formula:** - Utilize Euler's polyhedron formula for convex polyhedra: [ V - E + F = 2 ] where ( V ) is the number of vertices, ( E ) is the number of edges, and ( F ) is the number of faces. 2. **Mapping Strategy:** - Place two ladybugs on the midpoint of each edge of the polyhedron and let them travel towards the vertices at the ends of the edge. - This ensures every vertex is counted a minimum of twice (since each edge is shared by two vertices). 3. **Vertex Counting Inequality:** - Since at least 3 edges meet at each vertex, [ 2E geq 3V ] Solving for ( E ): [ frac{2V}{3} leq E ] 4. **Substitution into Euler’s Formula:** [ V - E + F = 2 quad text{implies} quad 3F + 4 geq 3V + 6 quad (text{after substituting bounds}) ] 5. **Analyzing Face Types:** - Denote the number of faces with ( i ) edges by ( S_i ). Thus, [ F = sum_{i=3}^{n} S_i ] and [ 2E = sum_{i=3}^{n} iS_i ] 6. **Combining Formulas:** [ 3F = sum_{i=3}^{n} 3S_i quad text{and} quad 2E = sum_{i=3}^{n} iS_i ] Simplify to: [ 6 sum_{i=3}^{n} S_i geq sum_{i=3}^{n} iS_i + 12 ] 7. **Restriction to ( n leq 6 ) and Testing:** - If ( n > 6 ), ignoring higher terms gives: [ 6 sum_{i=3}^{6} S_i geq sum_{i=3}^{6} iS_i + 12 ] Further reduce: [ sum_{i=3}^{6} (6-i)S_i geq 12 ] 8. **Verification using Small Faces:** - For face counts: [ 3S_3 + 2S_4 + S_5 geq 12 ] If each of ( S_3, S_4, S_5 leq 1 ): [ 3 times 1 + 2 times 1 + 1 times 1 leq 6 ] This contradicts the earlier inequality, ensuring that there must exist at least two faces with the same number of edges. 9. **Conclusion:** - Therefore, we demonstrated that there exists at least two faces with an equal number of edges: [ boxed{QED} ]

answer:1. **Identify Intersection and Altitude**: - Let D be the intersection point of PR and the perpendicular line AA'. Since A' is a rotational image of A around R, we have RA' = RA. - The segment AA' is perpendicular to PR (given), so D is also the midpoint of AA', due to it being the altitude in the isosceles triangle Delta ARA'. 2. **Angles and Isosceles Triangle Analysis**: - In triangle ARA', since D is the midpoint and the triangle is isosceles with RA = RA', it follows that angle ARD = angle A'RD. - Since D also lies on PR and ND = DP (because N'R = NR, and R, N', P are collinear), D equidistantly splits NP. This implies that the rotated points N' and P arrange symmetrically around R through D. 3. **Apply Angle Bisector Theorem**: - According to our configuration, R A is the angle bisector of angle PNR. Therefore, applying the Angle Bisector Theorem in triangle PNR, we have: [ frac{PA}{AN} = frac{PR}{RN} ] - We know the side lengths: PR = 19, RN = 18, and thus: [ frac{PA}{AN} = frac{19}{18} ] 4. **Conclusion**: - Hence, the ratio of PA to AN is: [ boxed{frac{19}{18}} ]

answer:**Analysis** This question examines the arithmetic operations of complex numbers and the coordinates of points corresponding to complex numbers in the complex plane. **Solution** Given: (dfrac{i}{1+i}= dfrac{i(1-i)}{(1+i)(1-i)}= dfrac{1+i}{2} ), Therefore, the coordinate of the point in the complex plane is: (left( dfrac{1}{2}, dfrac{1}{2}right) ). Hence, the answer is (boxed{left( dfrac{1}{2}, dfrac{1}{2}right)}).