answer:Given the problem of finding all possible sets of four real numbers such that the sum of each number with the product of the remaining three equals 2, let us denote these numbers by ( x, y, z, t ). Let ( A = xyzt ). Notice that ( A neq 0 ), because if any of ( x, y, z, ) or ( t ) were zero, it would lead to contradictions in the conditions provided by the problem. 1. **Formulation**: For each number, the equation given can be written as: [ x + yzt = 2 ] By substituting ( yzt ) with ( frac{A}{x} ) (since ( A = xyzt )), we get: [ x + frac{A}{x} = 2 ] This simplifies to: [ x^2 - 2x + A = 0 ] Similarly, we obtain the same form of equations for ( y, z, ) and ( t ): [ y^2 - 2y + A = 0 ] [ z^2 - 2z + A = 0 ] [ t^2 - 2t + A = 0 ] 2. **Solving for the roots**: The quadratic equation ( x^2 - 2x + A = 0 ) can have at most two distinct real roots, given by: [ x = frac{2 pm sqrt{4 - 4A}}{2} = 1 pm sqrt{1 - A} ] From here, we'll consider different scenarios: 1. **Case 1: All numbers are equal**: - If ( x = y = z = t ), replace the equation: [ x + x^3 = 2 ] which simplifies to: [ x^3 + x - 2 = 0 ] Factoring ( x^3 + x - 2 ): [ (x - 1)(x^2 + x + 2) = 0 ] This gives us the roots ( x = 1 ) and the complex roots ( x = frac{-1 pm sqrt{-7}}{2} ). Thus, in real numbers, the only solution is: [ x = y = z = t = 1 ] 2. **Case 2: Three numbers are the same, the fourth is different**: - Let ( x = y = z ) and ( t ) can be different. The equations become: [ x + x^2 t = 2 ] [ t + x^3 = 2 ] Subtracting these: [ x^3 - x^2 t = x - t ] Factorizing, we get: [ (x - t)(x^2 + x t - 1) = 0 ] This implies either ( x = t ) (which returns us to the previous case) or: - If ( x = 1 ) in ( x(t + x^2) = 2 ): [ 1 + t = 2 implies t = 1 quad (text{already included in previous case}) ] - Otherwise, if ( x = -1 ): [ -1 + t^2 = 2 implies t = 3 quad (text{and vice versa}) ] 3. **Case 3: Two pairs of equal numbers**: - Let ( x = y ) and ( z = t ). The equations become: [ x + x z^2 = 2 ] [ z + x^2 z = 2 ] Subtracting these: [ x - z x (1 - x) = 0 implies (x - z)(1 - xz) = 0 ] This gives either: [ x = z rightarrow (x = 1 text{ already known case}) ] Or: [ xz = 1 quad text{and} quad x + z = 2 ] Solving these,: [ x = 1, z = 1 text{ already known case also covered by previous case} ] # Conclusion: Combining all cases, we find: - The only solutions, including all distinct values and non-zero conditions obtained are: [ boxed{x = y = z = t = 1} quad text{Or} quad boxed{x = y = z = -1 ,, text{and} ,, t = 3} ] Thus, these are the possible sets of numbers satisfying the given problem conditions.

answer:Since (x - y)^2 geq 0, we have xy leq frac{(x + y)^2}{2} - xy = frac{36}{2} - xy, which implies xy leq 9. Therefore, z^2 = xy - 9 leq 0. Given that z^2 geq 0, we conclude that z = 0. Hence, the correct option is boxed{text{B}}.

answer:Given an acute-angled triangle ABC, with points P and Q selected on the extensions of its altitudes BB_1 and CC_1, respectively, such that angle PAQ = 90^circ. Let AF be the altitude of the triangle APQ. We need to prove that angle BFC is a right angle. 1. **Consider the Circumcircle**: - Points B_1 and C_1 lie on the circumcircle constructed with BC as the diameter. 2. **Key Insight**: - To solve the problem, we show that point F also lies on this circumcircle. This means we need to prove that the quadrilateral CB_1FC_1 is cyclic. 3. **Right Angles in the Triangle**: - Since angle AB_1P = angle AFP = 90^circ, points B_1 and F lie on a circle with AP as the diameter. Therefore, angle PFB_1 = angle PAB_1. 4. **Analogous Argument for Point Q**: - Similarly, angle QFC_1 = angle QAC_1 due to angle AC_1Q = 90^circ and angle AFQ = 90^circ. Thus, points C_1 and F also lie on a circle with AQ as the diameter. 5. **Sum of the Angles in Quadrilateral**: - We will now compute angle B_1FC_1: [ angle B_1FC_1 = 180^circ - angle PFB_1 - angle QFC_1 ] - Using the fact that both B_1F and C_1F are circumradius segments: [ angle PFB_1 = angle PAB_1 quad text{and} quad angle QFC_1 = angle QAC_1 ] - Since angle PAQ = 90^circ, we substitute: [ angle B_1FC_1 = 180^circ - left(angle PAB_1 + angle QAC_1right) ] - Next, observe that angle PAB_1 + angle QAC_1 = angle BAC = 90^circ - angle ACB_1: [ angle B_1FC_1 = 180^circ - angle PAQ - angle BAC ] - Notice angle PAQ = 90^circ, thus: [ angle B_1FC_1 = 180^circ - 90^circ - left(angle BAC - angle ACB_1right) = 90^circ + angle B_1AC_1 - angle ACB_1 ] 6. **Cyclic Conclusion**: - Since angle B_1FC_1 = 180^circ - angle B_1CC_1, it follows that CB_1FC_1 is a cyclic quadrilateral. Therefore, the supplementary angle properties of cyclic quadrilaterals ensure that angle BFC is indeed 90^circ, satisfying our requirement. boxed{text{This concludes the proof.}}

answer:B Since the equation has a repeated root, we have D = 4cos^2 q - cot q = 0, Since 0 < q < pi, we get 2sin^2 q = 1, which implies q = frac{pi}{6} or frac{5pi}{6}. Therefore, we choose option B. So, the answer is boxed{B}.