answer:(1) Analyzing the function f(x) = |x - 2| + |x + a|: For f(x) to be always greater than or equal to 3 for all x in mathbb{R}, we find the minimum of f(x). We make use of the triangle inequality: |u + v| geq |u| - |v|. Let u = x - 2 and v = x + a, and apply the triangle inequality in reverse: begin{align*} f(x) & = |x - 2| + |x + a| & geq |(x - 2) - (x + a)| & = |-2 - a| & = |a + 2|. end{align*} For f(x) to be always greater than or equal to 3, |a + 2| should also be greater than or equal to 3: |a + 2| geq 3. Now, solving for a, we get two cases: - If a + 2 geq 0, then a + 2 geq 3 implies a geq 1. - If a + 2 < 0, then -a - 2 geq 3 implies a leq -5. Hence, the range for a is a leq -5 or a geq 1. (2) For the given condition x + 2y + 3z = 1, and since x, y, z > 0, we want to minimize the expression frac {3}{x} + frac {2}{y} + frac {1}{z}. We apply the Cauchy-Schwarz inequality: left(sum a_i b_iright)^2 leq left(sum a_i^2right) left(sum b_i^2right), By choosing a_i = sqrt{x}, sqrt{2y}, sqrt{3z} and b_i = sqrt{frac{3}{x}}, sqrt{frac{2}{y}}, sqrt{frac{1}{z}} respectively, we obtain: begin{align*} left(frac {3}{x} + frac {2}{y} + frac {1}{z}right) & = (x + 2y + 3z)left(frac {3}{x} + frac {2}{y} + frac {1}{z}right) & geq left(sqrt{x}sqrt{frac{3}{x}} + sqrt{2y}sqrt{frac{2}{y}} + sqrt{3z}sqrt{frac{1}{z}}right)^2 & = (sqrt{3} + 2 + sqrt{3})^2 & = (2sqrt{3} + 2)^2 & = 16 + 8sqrt{3}. end{align*} Equality holds if and only if: frac{x}{frac{3}{x}} = frac{2y}{frac{2}{y}} = frac{3z}{frac{1}{z}}, which simplifies to x:y:z = 3:sqrt{3}:1. Therefore, the minimum value of frac {3}{x} + frac {2}{y} + frac {1}{z} is boxed{16 + 8sqrt{3}}.

answer:1. **Understanding the problem**: - We are given two islands with rectangular shapes divided into smaller rectangles called graphstum. - Each graphstum has a diagonal road, forming a path where no point is visited twice. - We need to draw the configuration of the larger island if it has an odd number of graphstums. - Finally, we need to determine the number of graphstums on the larger island. 2. **Small Island Configuration**: - From the problem statement, the small island consists of 6 graphstums. - The given configuration is depicted below:  3. **Solution for the larger island**: - We are tasked with a configuration for the larger island such that it contains an odd number of graphstums. - The solution provided depicts an arrangement with 9 graphstums, which is indeed an odd number.  4. **Conclusion**: - The larger island contains 9 graphstums. [ boxed{9} ]

answer:# Problem: Given two points ( G ) and ( H ) inside ( triangle ABC ), satisfying ( angle ACG = angle BCH ) and ( angle CAG = angle BAH ). Verticals ( GD perp BC ), ( GE perp CA ), and ( GF perp AB ) are drawn through point ( G ), and the feet of these perpendiculars are ( D ), ( E ), and ( F ) respectively. Given that ( angle DEF = 90^circ ), we need to prove that ( H ) is the orthocenter of ( triangle BDF ). 1. Since ( GD perp BC ) and ( GE perp CA ), ( C, D, G, E ) are concyclic points. Therefore, ( angle GDE = angle ACG = angle BCH ). 2. Consequently, we have ( DE perp CH ). 3. Similarly, ( EF perp AH ). 4. Since ( angle DEF = 90^circ ), it follows that ( angle AHC = 90^circ ). 5. Considering the sides, we have: [ frac{CD}{CH} = frac{GC cdot cos angle GCD}{AC cdot cos angle HCA} = frac{GC}{AC} ] 6. Given ( angle HCD = angle ACG ), the triangles ( triangle HCD ) and ( triangle ACG ) are similar: [ triangle HCD sim triangle ACG ] 7. Thus, ( angle CHD = angle CAG = angle EFG ). 8. Since ( CH parallel EF ), it follows that ( DH parallel GF ). 9. Therefore, ( DH perp AB ). 10. Similarly, ( FH perp BC ). 11. Hence, ( H ) is the orthocenter of ( triangle BDF ). [ boxed{H} ]

answer:1. **Identify the face with y**: Start with the face marked 'y'. 2. **Visualize folding the cube**: - Evaluate which faces are directly connected to 'y' and which are not. Those directly connected cannot be opposite to 'y'. - Consider the arrangement and configuration of all faces in relation to 'y'. 3. **Analyze the connections**: - Suppose faces A and B are adjacent to 'y' on the net. Upon folding, A moves to the left side and B to the right side of y. - Face C is adjacent to y at the top, and face E is adjacent at the bottom. 4. **Determine the opposite face**: - The face D, which is not directly connected to 'y' in the net and does not get connected to any adjacent sides upon folding, must be opposite to 'y'. 5. **Conclusion**: - Since face D is neither adjacent nor indirectly connected to 'y' in any described folding step, it must be the face opposite to 'y' in the cube. The face opposite to the face marked 'y' is mathrm{D}. The final answer is boxed{mathrm{D}}