answer:# Part 1: Prove that ( K ) always lies on a fixed line. 1. **Reflection and Collinearity**: - Let ( M' ) be the reflection of ( M ) through ( d ). - Since ( d ) is the external bisector of (angle xOy), ( M' ) lies on the circle centered at ( I ) with radius ( IM ). - We have ( angle ONM' equiv angle ONP + angle OPM' equiv angle ONP + angle OMQ equiv 0 pmod{pi} ). - Therefore, ( O ), ( N ), and ( M' ) are collinear. 2. **Angle Bisectors**: - Since ( NP ) is the bisector of (angle ONM) and ( MQ ) is the bisector of (angle OMN), ( K ) is the incenter of (triangle OMN). - Thus, ( OK ) is the internal bisector of (angle xOy). 3. **Conclusion**: - Since ( OK ) is the internal bisector of (angle xOy), ( K ) always lies on the fixed line ( OK ). # Part 2: Prove that ( EN ), ( FM ), and ( OK ) are concurrent. 1. **Intersection and Circle**: - Let ( G ) be the intersection of ( EM ) and ( FN ). - Since ( O ), ( M ), ( G ), and ( N ) lie on a circle and ( GM = GN ), ( G ) is the midpoint of the arc ( MN ) not containing ( O ). 2. **Cyclic Quadrilaterals**: - We have ( angle MPK equiv angle OPK equiv angle OKQ equiv angle NKQ equiv frac{pi}{2} pmod{pi} ). - Hence, ( O ), ( P ), ( M ), ( K ) and ( O ), ( Q ), ( N ), ( K ) lie on circles. 3. **Angle Relationships**: - We have ( angle ONM equiv angle ONK + angle OKM equiv angle QNK + angle PKM equiv angle INM pmod{pi} ). - This implies ( O ), ( M ), ( N ), and ( I ) lie on a circle, and thus ( O ), ( M ), ( G ), ( N ), and ( I ) lie on a circle. 4. **Concurrency**: - Since ( GM = GN ), ( OG ) is the internal bisector of (angle xOy). - Therefore, ( O ), ( K ), and ( G ) are collinear, implying ( EN ), ( FM ), and ( OK ) are concurrent. (blacksquare)

answer:To solve the problem, we start by expressing the given function f(x) in a more simplified form. The function is given as: f(x)=sin^2omega x + sinomega xcosomega x - 1. We can use trigonometric identities to simplify this expression. Recall that sin^2theta = frac{1-cos2theta}{2} and sinthetacostheta = frac{sin2theta}{2}. Applying these identities, we get: begin{align*} f(x) &= frac{1-cos2omega x}{2} + frac{sin2omega x}{2} - 1 &= -frac{1}{2} + frac{sin2omega x}{2} - frac{cos2omega x}{2} &= frac{sqrt{2}}{2}sinleft(2omega x-frac{pi}{4}right) - frac{1}{2}. end{align*} The problem states that f(x) has exactly 4 distinct zeros in the interval left(0, frac{pi}{2}right). Setting f(x) = 0 and solving for x gives us: frac{sqrt{2}}{2}sinleft(2omega x-frac{pi}{4}right) = frac{1}{2}. This equation implies that sinleft(2omega x-frac{pi}{4}right) = frac{sqrt{2}}{2}, which corresponds to angles that are frac{pi}{4} and frac{3pi}{4} (modulo 2pi) away from any integer multiple of pi. Given the interval for x is left(0, frac{pi}{2}right), we consider the range of 2omega x - frac{pi}{4}, which is left(-frac{pi}{4}, omegapi - frac{pi}{4}right). To have exactly 4 distinct zeros, the upper bound of this range must fall between the angles that give us 4 solutions for sinleft(2omega x-frac{pi}{4}right) = frac{sqrt{2}}{2}, which are frac{11pi}{4} and frac{17pi}{4}, inclusively. Therefore, we have: frac{11pi}{4} < omegapi - frac{pi}{4} leq frac{17pi}{4}. Solving this inequality for omega gives us: 3 < omega leq frac{9}{2}. Therefore, the range of the real number omega that satisfies the given conditions is boxed{(text{D}) (3, frac{9}{2}]}.

answer:The left-hand side of the equation, when multiplied out, is a polynomial of degree 7. By Vieta's formulas, the product of the roots of a polynomial is determined by the ratio of its constant term to its leading coefficient, with a sign depending on the degree. - The leading coefficient of the resulting polynomial is the product of the leading coefficients of the individual polynomials: [3 cdot 6 = 18.] - The constant term of the resulting polynomial is the product of the constant terms of the individual polynomials: [(-7) cdot (-4) = 28.] - Applying Vieta's formulas, the product of the roots, with the degree being 7 (odd), gives: [frac{28}{18} = frac{14}{9}.] Therefore, the product of the roots is boxed{frac{14}{9}}.

answer:Let's denote the speed of the second train as ( v ) miles per hour. Since the two trains are traveling towards each other, their relative speed is the sum of their individual speeds. So, the relative speed is ( 40 + v ) miles per hour. We know that 1 hour before they meet, they are 70 miles apart. This means that in the last hour, they cover the remaining 70 miles to meet each other. Since their relative speed is ( 40 + v ), we can write the equation: [ (40 + v) times 1 = 70 ] Solving for ( v ): [ 40 + v = 70 ] [ v = 70 - 40 ] [ v = 30 ] So, the speed of the other train is boxed{30} miles per hour.