answer:Let's start by assigning the breeds that have their own dog houses: 1. German Shepherd: 10 (Since they are 10, they will occupy two dog houses: one with 6 dogs and one with 4 dogs) 2. Bulldog: 6 (They will occupy one dog house with 6 dogs) 3. Beagle: 6 (They will occupy one dog house with 6 dogs) 4. Poodle: 6 (They will occupy one dog house with 6 dogs) 5. Rottweiler: 6 (They will occupy one dog house with 6 dogs) Now we have used up 5 dog houses for the breeds that have their own dog houses. We have 5 dog houses left for the Labs and Golden Retrievers. Since all Labs and Golden Retrievers live together but not in the same dog house, we can distribute them as follows: - Labrador: 8 (They will occupy two dog houses: one with 4 dogs and one with 4 dogs) - Golden Retriever: 8 (They will also occupy two dog houses: one with 4 dogs and one with 4 dogs) Now we have used 4 more dog houses, which brings us to a total of 9 dog houses used. We have 1 dog house left, which must be fully occupied. However, we have already accommodated all the dogs in the available dog houses, and since all dog houses must be fully occupied, we have a problem. We have an extra dog house that we cannot fill with the given distribution of dogs. Therefore, it is not possible to distribute the dogs among the boxed{10} dog houses following the given constraints because we will always end up with one empty dog house.

answer:1. **Assign Variables:** Let AC = b, AB = c, AD = d, and AE = e. 2. **Use the Angle Bisector Theorem:** Since AD is the angle bisector of angle BAC, we have: [ frac{BD}{DC} = frac{AB}{AC}. ] Given BD = 3 and DC = DE + EC = 4 + 3 = 7, we find: [ frac{c}{b} = frac{3}{7}. ] Therefore, b = frac{7c}{3}. 3. **Apply Stewart’s Theorem:** Applying Stewart’s theorem in triangle ABC with cevian AD: [ b^2 cdot 3 + c^2 cdot 7 = AD^2 cdot 10 + 3 cdot 7 cdot c^2. ] We substitute b = frac{7c}{3} into the equation: [ left(frac{7c}{3}right)^2 cdot 3 + c^2 cdot 7 = AD^2 cdot 10 + 21c^2, ] [ frac{49c^2}{3} + 7c^2 = AD^2 cdot 10 + 21c^2, ] [ AD^2 = frac{28c^2 - 21c^2}{10} = frac{7c^2}{10}. ] Simplifying, we find AD = frac{sqrt{70}c}{10}. 4. **Conclusion:** Since c = frac{3b}{7} and b = frac{7c}{3}, comparing these, we find the smallest value is c. Thus, the length of the shortest side of triangle ABC is frac{3b{7}}. The final answer is boxed{textbf{(A)} frac{3b}{7}}

answer:1. **Initialization for ( n = 2 )**: - We start with 4 points: ( {A, B, C, D} ) - We have to place 5 segments. - The maximum number of segments that can connect 4 points is: [ binom{4}{2} = 6 ] - Since we have only 5 segments, exactly one segment must be missing. Let's assume the missing segment is ( [AB] ). - The triangles that can still be formed are ( ACD ) and ( BCD ), and by drawing these segments, we create at least one triangle. This completes the initialization step. 2. **Assume ( n ) points forming a triangle ( forall n geq 2 )**: - Let ( n in mathbb{N} ) and consider the configuration for ( 2(n+1) ) points. - We need ( (n+1)^2 + 1 ) segments to guarantee the formation of at least one triangle. - Choose any segment ( [AB] ), and consider the remaining ( 2n ) points ( {C_1, C_2, ldots, C_{2n}} ). - For each point ( C_i ) (( 1 leq i leq 2n )), if both segments ( [AC_i] ) and ( [BC_i] ) are drawn, then triangle ( ABC_i ) is formed. - If not, we analyze the different possible scenarios: [ forall i, text{ at most one of } [AC_i] text{ or } [BC_i] text{ is drawn} ] - This results in ( 2n ) segments maximum between ( A ) or ( B ) and the remaining points. - Hence, we have initially: [ (n+1)^2 + 1 text{ segments} ] - Remove the segments involving ( A ) and (B). - This leave us with at least: [ (n+1)^2 + 1 - (2n + 1) = n^2 + 1 text{ segments and } 2n text{ points} ] - By the induction hypothesis, this configuration of ( n ) (having ( n^2 + 1 ) segments for ( 2n ) points) ensures that there is at least one triangle formed among these points. 3. **Conclusion**: - By mathematical induction, for any ( n geq 2 ), given ( 2n ) points and ( n^2 + 1 ) segments, there is always at least one triangle formed. [ boxed{text{We have drawn at least one triangle.}} ]

answer:(1) Since tan A+tan B=sqrt{3}(tan Atan B-1), we have tan (A+B)=frac{tan A+tan B}{1-tan Atan B}=-sqrt{3}. Also, tan C=tan [pi-(A+B)]=-tan (A+B), thus tan C=sqrt{3}. Given that 0 < C < pi, we conclude that C=frac{pi}{3}; (2) From the problem, we know that the area of the triangle is given by S_{triangle ABC}=frac{1}{2}absin C=frac{1}{2}absin frac{pi}{3}=frac{sqrt{3}}{4}ab=frac{3sqrt{3}}{2}, thus ab=6. By the cosine rule, we have c^2=a^2+b^2-2abcos C=(a+b)^2-3ab, thus (a+b)^2=3ab+c^2=18+frac{49}{4}=frac{121}{4}, given that a > 0 and b > 0, we conclude that a+b=boxed{frac{11}{2}}.