answer:Consider the triangle (ABC) and its angle bisectors intersecting the circumcircle at points (A', B', C'). We need to show that [ 16 cdot (text{area } A'B'C')^3 geq 27 cdot (text{area } ABC) cdot R^4, ] where (R) is the circumradius of (triangle ABC). 1. **Area Calculation:** The area of (triangle ABC) with circumradius (R) can be expressed using the sine formula: [ text{area} triangle ABC = frac{1}{2} R^2 (sin 2A + sin 2B + sin 2C). ] 2. **Angle Bisectors and Sines:** When the angle bisectors intersect the circumcircle at (A', B', C'), we get the following relationships: [ measuredangle A'C'C = measuredangle A'AC = frac{measuredangle A}{2}, ] [ measuredangle B'C'C = measuredangle B'BC = frac{measuredangle B}{2}. ] Thus, using the angle sum property: [ measuredangle A'C'B' = frac{measuredangle A}{2} + frac{measuredangle B}{2} = frac{A + B}{2}. ] 3. **Area of (triangle A'B'C'):** The circumradius (R) for both ( triangle ABC ) and (triangle A'B'C') remains the same. So, we have: [ text{area} triangle A'B'C' = frac{1}{2} R^2 (sin(A+B) + sin(B+C) + sin(C+A)). ] 4. **Inequality Application:** We need to show: [ 4 (sin(A+B) + sin(B+C) + sin(C+A))^3 geq 27 (sin 2A + sin 2B + sin 2C). ] 5. **Applying AM-GM Inequality:** Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality on (sin(A+B), sin(B+C), sin(C+A)), we have: [ left(frac{sin(A+B) + sin(B+C) + sin(C+A)}{3}right)^3 geq sin(A+B) cdot sin(B+C) cdot sin(C+A). ] Multiplying both sides by (27), we get: [ (sin(A+B) + sin(B+C) + sin(C+A))^3 geq 27 sin(A+B) sin(B+C) sin(C+A). ] 6. **Trigonometric Products and Sums:** Using the product-to-sum identities: [ sin(B+C) sin(C+A) = frac{1}{2} (cos(B-A) - cos(C+180^circ)) = frac{1}{2} (cos(B-A) + cos C). ] This implies: [ sin(A+B) sin(B+C) sin(C+A) = frac{1}{2} sin(A+B) cos(B-A) + frac{1}{2} sin(A+B) cos C. ] Similarly: [ sin 2A + sin 2B = sin((A+B) + (B-A)) + sin((A+B) - (B-A)) = 2 sin(A+B) cos(B-A), ] and [ sin((A+B)+C) + sin((A+B)-C) = 2 sin(A+B) cos C. ] Thus, we get: [ sin(A+B) sin(B+C) sin(C+A) = frac{1}{4} (sin 2A + sin 2B + sin 2C). ] 7. **Combining and Concluding:** Therefore: [ (sin(A+B) + sin(B+C) + sin(C+A))^3 geq frac{27}{4} (sin 2A + sin 2B + sin 2C). ] Rewriting: [ 4 (sin(A+B) + sin(B+C) + sin(C+A))^3 geq 27 (sin 2A + sin 2B + sin 2C). ] Hence, the desired inequality is satisfied, and we can conclude: [ boxed{16 (text{area } A'B'C')^3 geq 27 text{area } ABC cdot R^4}. ]

answer:1. **Given Setup**: Let (ABCD) be a convex quadrilateral. Let ((I)) and ((J)) be the incircles of triangles (triangle DAB) and (triangle BCD), respectively. These incircles are tangent to the diagonal (BD) at points (U) and (V), respectively. The internal homothety center of ((I)) and ((J)) is denoted by (H), which is the intersection of the line of incenters (IJ) with (BD). 2. **Symmetry Condition**: We need to prove that the tangency points (U) and (V) are symmetrical with respect to the midpoint of (BD) if and only if the line of incenters (IJ) passes through the intersection point of the diagonals (AC) and (BD). 3. **Assume (BU = DV)**: - If (BU = DV), then the lengths of the tangents from (B) and (D) to the incircles are equal. This implies: [ AB + BD - DA = CD + DB - BC ] - Rearranging, we get: [ AB + BC = CD + DA ] - This means that the quadrilateral (ABCD) is tangential, i.e., it has an incircle ((E)) that is tangent to all four sides. 4. **Homothety Centers**: - (A) is the external homothety center of ((I)) and ((E)). - (C) is the internal homothety center of ((J)) and ((E)). - Therefore, the internal homothety center (H) of ((I)) and ((J)) lies on (AC), implying (H equiv IJ cap BD cap AC). 5. **Assume (H equiv IJ cap BD cap AC)**: - Let ((E)) be the A-excircle of (triangle DAP). - The internal homothety center of ((J)) and ((E)) lies on their common internal tangent (CD equiv PD). - (A) is the external homothety center of ((I)) and ((E)), and (H in AC) is the internal homothety center of ((I)) and ((J)). - This implies that the internal homothety center of ((J)) and ((E)) is on (AC), and it is identical with (C equiv CD cap AC). - Consequently, the other tangent (BC) of ((J)) through (C) is also a tangent of ((E)), making (ABCD) tangential with the excircle ((E)) in the angle (angle DAB). 6. **Conclusion**: - Therefore, (AB + BC = CD + DA) or equivalently: [ AB + BD - DA = CD + DB - BC ] - This implies (BU = DV). Thus, we have shown that the tangency points (U) and (V) are symmetrical with respect to the midpoint of (BD) if and only if the line of incenters (IJ) passes through the intersection point of the diagonals (AC) and (BD). (blacksquare)

answer:Given the equation ( x^{4} - 7x - 3 = 0 ), it is known that this equation has exactly two real roots ( a ) and ( b ), where ( a > b ). We need to find the value of the expression ( frac{a - b}{a^{4} - b^{4}} ). 1. **Identify the roots:** Since ( a ) and ( b ) are roots of the equation ( x^4 - 7x - 3 = 0 ), we can write: [ a^4 - 7a - 3 = 0 ] [ b^4 - 7b - 3 = 0 ] 2. **Express ( a^4 ) and ( b^4 ):** From the above equations, solve for ( a^4 ) and ( b^4 ): [ a^4 = 7a + 3 ] [ b^4 = 7b + 3 ] 3. **Substitute into the expression:** Substitute ( a^4 ) and ( b^4 ) into ( frac{a - b}{a^4 - b^4} ): [ frac{a - b}{a^4 - b^4} = frac{a - b}{(7a + 3) - (7b + 3)} ] 4. **Simplify the denominator:** Simplify inside the denominator: [ frac{a - b}{(7a + 3) - (7b + 3)} = frac{a - b}{7a + 3 - 7b - 3} = frac{a - b}{7(a - b)} ] 5. **Simplify the fraction:** Note that ( a - b ) is a common factor in both the numerator and the denominator: [ frac{a - b}{7(a - b)} = frac{1}{7} ] # Conclusion: The value of the expression ( frac{a - b}{a^4 - b^4} ) is: [ boxed{frac{1}{7}} ]

answer:To identify the inequalities from the given expressions, let's examine each one: - Expression ① is 3 < 5. This is a basic inequality comparing two numbers. - Expression ② is 4x + 5 > 0. This is an inequality involving a variable x. - Expression ③ is x = 3. This is an equation, not an inequality. - Expression ④ is x^2 + x. This is an expression, not an inequality or equation. - Expression ⑤ is x neq 4. This represents an inequality because it specifies that x cannot equal a certain value. - Expression ⑥ is x + 2 geqslant x + 1. This is an inequality showing that one expression is greater than or equal to another. From the above analysis, expressions ①, ②, ⑤, and ⑥ are inequalities. Therefore, there are a total of boxed{4} inequalities. Hence, the correct answer is boxed{C}.