answer:# Problem: 如图 1, 在 triangle ABC 中， A_B , A_C 分别为 A 一一内切圆与 AC , AB 的切点， A_1 为 A_BB 与 A_CC 的交点， AA_1 与 triangle ABC 外接圆 Gamma 的第二个交点为 A_2。 类似地， 设 B_A , B_C 分别为 B 一一内切圆与 BC , AB 的切点， B_1 为 B_AA 与 B_CC的交点， BB_1 与圆 Gamma 的第二个交点为 B_2。 I_C 为 triangle ABC 的顶点 C 所对应的旁切圆的圆心。 则 angle I_C A_2B = angle I_C B_2A 1. 为便于叙述， 已给出的字母除再一次说明之外， 在下文中含义不变。 2. 先证明三个引理。 - **引理 1** 如图 2， 设 triangle ABC 的内切圆 odot I 与三边分别切于点 D, E, F, J 为弧 overparen{BAC} 的中点。 则 A_2, D, J 三点共线。 - **引理 1 的证明:** - 设 X 为 A_BA_C 与 AA_2 的交点。 - 由曼海姆定理知， I 为 A_BA_C 的中点。 [ frac{B A_C}{C A_B} = frac{B A_C}{A_C I} cdot frac{A_B I}{C A_B} = frac{sin^2 frac{angle ACB}{2}}{sin^2 frac{angle ABC}{2}} ] - 故对 triangle AA_BA_C - 对 triangle AA_BA_C 和 点 A_1 使用 塞瓦定理， 得 [ frac{B A_2}{C A_2} = frac{sin angle B A A_2}{sin angle C A A_2} = frac{X A_C}{X A_B} = frac{B A_C}{B A} cdot frac{C A}{C A_B} = frac{sin angle ABC}{sin angle ACB} cdot frac{sin^2 frac{angle ACB}{2}}{sin^2 frac{angle ABC}{2}} = frac{tan frac{angle ACB}{2}}{tan frac{angle ABC}{2}} = frac{BD}{CD} ] - 由 角平分线 定理知 [ angle BA_2D = angle CA_2D ] - 因此， A_2, D, J 三点共线。 引理 1 得证。 - **引理 2:** 如图 3，设 I_CA_2 , I_CB_2 分别与 AB 交于点 U , V， 则 A_2, E, F, U, B_2, D, F, V 分别四点共圆。 - **引理 2 的证明:** 由引理 1 得 [ triangle A_2DC sim triangle A_2BJ Rightarrow frac{CA_2}{CE} = frac{CA_2}{CD} = frac{JA_2}{JB} = frac{JA_2}{I_CJ} ] - 又 angle ECA_2 = angle I_CJA_2，故 [ triangle ECA_2 sim triangle I_CJA_2 Rightarrow angle CEA_2 = angle AI_CA_2 ] - Rightarrow A_2, E, A, I_C 四点共圆。 [ Rightarrow angle AUI_C = 180^{circ} - angle AI_CU - angle I_CAU = angle AEA_2 - angle AEF = angle FEA_2 ] - Rightarrow A_2, E, F, U 四点共圆。类似地， B_2, D, F, V 四点共圆。引理 2 得证。 - **引理 3:** A_2 , B_2, V, U 四点共圆。 - **引理 3 的证明** 如图 4，设 I_A, I_B 分别为 triangle ABC 的顶点 A, B 所对应的旁切圆圆心， I_C E 和 I_B F , I_C D 分别交于点 T , S。由引理 2 的证明知 [ angle I_B F A_2 = angle I_BAA_2 , , angle I_C E A_2 = angle I_CAA_2 ] - 故 angle T E A_2 + angle T F A_2 = angle I_CAA_2 + angle I_BAA_2 = 180^{circ} Rightarrow T , E, A_2, F 四点共圆。类似地， S, F, B_2, D 四点共圆。易得 I_C D = I_C E, triangle I_AI_B I_C 和 triangle DEF 位似， 故 [ frac{T I_C}{T E} = frac{I_C I_B}{EF} = frac{I_C I_A}{DF} = frac{S I_C}{DS} Rightarrow I_CT = I_CS ] - 再由引理 2 得 [ I_C U cdot I_C A_2 = I_CT cdot I_C E = I_CS cdot I_CD = I_C V cdot I_C B_2 ] - 由 割线定理，知 A_2, B_2, V, U 四点共圆。引理 3 得证。 3. 如图 1, 设 I_CA_2, I_CB_2 与圆 Gamma 的第二个交点分别为 P , Q。由引理 3 知 [ angle VUI_C = angle I_CB_2A_2 = angle QPI_C Rightarrow PQ//AB ] - Rightarrow angle I_CA_2B = angle I_CB_2A。 # Conclusion boxed{angle I_CA_2B = angle I_CB_2A}.

answer:To determine all values of ( n ) for which there exists an ( n times n ) platinum matrix, we need to analyze the conditions given in the problem. 1. **Condition (i):** The ( n^2 ) entries are integers from ( 1 ) to ( n ). 2. **Condition (ii):** Each row, each column, and the main diagonal contains each integer from ( 1 ) to ( n ) exactly once. 3. **Condition (iii):** There exists a collection of ( n ) entries containing each of the numbers from ( 1 ) to ( n ), such that no two entries lie on the same row or column, and none of which lie on the main diagonal of the matrix. To solve this, we need to understand the implications of these conditions. # Step-by-Step Analysis: 1. **Latin Square Property:** - Conditions (i) and (ii) imply that the matrix is a Latin square. A Latin square of order ( n ) is an ( n times n ) array filled with ( n ) different symbols, each occurring exactly once in each row and exactly once in each column. 2. **Diagonal Constraint:** - The main diagonal must also contain each integer from ( 1 ) to ( n ) exactly once. This means the matrix is not just any Latin square but a special type where the main diagonal is a permutation of ( 1, 2, ldots, n ). 3. **Orthogonal Latin Square:** - Condition (iii) requires that there exists a collection of ( n ) entries, each from ( 1 ) to ( n ), such that no two entries lie on the same row or column, and none lie on the main diagonal. This implies the existence of an orthogonal mate to the Latin square, which is another Latin square such that when superimposed, each ordered pair of symbols appears exactly once. # Existence of Orthogonal Latin Squares: - It is known from combinatorial design theory that orthogonal Latin squares exist if and only if ( n ) is not equal to ( 2 ) or ( 6 ). This result is derived from the theory of finite fields and projective planes. # Conclusion: - For ( n = 1 ), the matrix trivially satisfies all conditions. - For ( n = 2 ) and ( n = 6 ), it is impossible to construct orthogonal Latin squares, and hence, a platinum matrix cannot exist. - For all other values of ( n ), orthogonal Latin squares exist, and thus, a platinum matrix can be constructed. The final answer is ( boxed{ n neq 2 } ) and ( n neq 6 ).

answer:Since f(a)=1 and f(x) = begin{cases} 2^{x}-1 & xgeq0 -x^{2}-2x & x<0 end{cases}. So, when ageq0, we have f(a)=2^{a}-1=1 Rightarrow 2^{a}=2 Rightarrow a=1; When a<0, we have f(a)=-a^{2}-2a=1 Rightarrow (a+1)^{2}=0 Rightarrow a=-1. In summary, we get: a=pm1. Therefore, the answer is: boxed{pm1}. Since the function f(x) is a piecewise function, it is necessary to substitute a into the corresponding expressions for the cases ageq0 and a<0 separately to find a, and then synthesize the results. This question tests the simplification and value finding of rational exponentiation as well as the method of finding the value of a piecewise function, and it is a basic question.

answer:If Jack missed 2y - 3 questions out of 8y questions, then the number of correct answers he gave is 8y - (2y - 3) = 8y - 2y + 3 = 6y + 3. The percentage of the questions that Jack answered correctly then is: [ frac{6y + 3}{8y} = frac{3(2y + 1)}{8y} = frac{3(2y+1)}{8y} cdot 100% = frac{300(2y+1)}{8y} = frac{75(2y+1)}{2y} ] Simplify to: [ frac{75(2y+1)}{2y} = 75 left(1 + frac{1}{2y}right) = 75 + frac{75}{2y} ] So, the percent of questions that Jack got correct simplifies to boxed{75 + frac{75}{2y} %}.