answer:# Part (a) 1. **Given**: The function ( f: mathbb{R}^2 to mathbb{R}^2 ) preserves distances between any two points in (mathbb{R}^2). This means for any points ( A ) and ( B ), the distance ( d(A, B) = d(f(A), f(B)) ). 2. **To Show**: If ( X ) is a point on the line ( l(C, D) ), then ( f(X) ) is a point on the line ( l(f(C), f(D)) ). 3. **Proof**: - Consider the line ( l(C, D) ) which passes through points ( C ) and ( D ). - Let ( X ) be any point on the line ( l(C, D) ). This means ( X ) can be expressed as ( X = C + t(D - C) ) for some scalar ( t ). - Since ( f ) preserves distances, the distances ( overline{CX} ) and ( overline{DX} ) must be preserved under ( f ). Therefore, ( d(C, X) = d(f(C), f(X)) ) and ( d(D, X) = d(f(D), f(X)) ). - This implies that ( f(X) ) must lie on the line ( l(f(C), f(D)) ) because the distances from ( f(X) ) to ( f(C) ) and ( f(D) ) are the same as the distances from ( X ) to ( C ) and ( D ), respectively. - Hence, ( f(X) ) is a point on the line ( l(f(C), f(D)) ). (blacksquare) # Part (b) 1. **Given**: Lines ( l(E, F) ) and ( l(C, D) ) intersect at an angle ( alpha ). 2. **To Show**: The lines ( l(f(C), f(D)) ) and ( l(f(E), f(F)) ) intersect at the same angle ( alpha ). 3. **Proof**: - Consider the quadrilateral formed by points ( E, C, F, D ). - Since ( f ) preserves distances, the quadrilateral ( ECFD ) is transformed into a new quadrilateral ( f(E)f(C)f(F)f(D) ) with the same side lengths and angles. - The angle between ( l(E, F) ) and ( l(C, D) ) is ( alpha ). Since ( f ) preserves distances and hence the shape of the quadrilateral, the angle between ( l(f(E), f(F)) ) and ( l(f(C), f(D)) ) must also be ( alpha ). - Therefore, the angle ( alpha ) is preserved under the transformation ( f ). 4. **Extension**: - If the lines ( l(C, D) ) and ( l(E, F) ) do not intersect, they are either parallel or skew. - If they are parallel, the distance between any point on ( l(C, D) ) and any point on ( l(E, F) ) remains constant under ( f ). Thus, the lines ( l(f(C), f(D)) ) and ( l(f(E), f(F)) ) will also be parallel. - If they are skew, the property of preserving distances ensures that the relative orientation and distance between the lines are preserved, maintaining the same geometric relationship. (blacksquare)

answer:First, express x^2 + 6xy + 9y^2 in a perfect square form: [x^2 + 6xy + 9y^2 = (x + 3y)^2.] Using AM-GM, we have: [x + 3y geq 2sqrt{3xy},] so, [(x + 3y)^2 geq 12xy.] Hence, [x^2 + 6xy + 9y^2 + 4z^2 geq 12xy + 4z^2.] Apply AM-GM to the expression 12xy + 4z^2: [ 12xy + 4z^2 = 4xy + 4xy + 4xy + 4z^2 geq 4 sqrt[4]{(4xy)(4xy)(4xy)(4z^2)} = 4 cdot sqrt[4]{64x^3y^3z^2}. ] The condition given is xyz = 48, along with: [ 4 cdot sqrt[4]{64x^3y^3z^2} = 4 cdot sqrt[4]{64 cdot 48^2} = 4 cdot 32 = 128. ] Equality holds when x = 3y and 4xy = 4z^2 and xyz = 48. Solving, x = 6, y = 2, z = 4 meet all conditions. Checking, [x^2 + 6xy + 9y^2 + 4z^2 = 6^2 + 6 cdot 6 cdot 2 + 9 cdot 2^2 + 4 cdot 4^2 = 36 + 72 + 36 + 64 = 208.] Thus, the minimum value is boxed{128}.

answer:(I) We can rewrite the function using trigonometric identities as: begin{align*} f(x) &= cosleft(x - frac{pi}{3}right) - sinleft(frac{pi}{2} - xright) &= cosleft(x - frac{pi}{3}right) - cosleft(xright) &= frac{1}{2}cosleft(xright) + frac{sqrt{3}}{2}sinleft(xright) - cosleft(xright) &= frac{sqrt{3}}{2}sinleft(xright) - frac{1}{2}cosleft(xright) &= sinleft(x - frac{pi}{6}right). end{align*} Therefore, the smallest positive period of f(x) is 2pi. So we have: boxed{T = 2pi}. (II) Since we know from (I) that f(x) = sinleft(x - frac{pi}{6}right), begin{align*} fleft(alpha + frac{pi}{6}right) &= sinleft(alpha + frac{pi}{6} - frac{pi}{6}right) &= sin(alpha) &= frac{3}{5}. end{align*} Given that alpha in left(0, frac{pi}{2}right), using the Pythagorean identity, we find cos(alpha): cos(alpha) = sqrt{1 - sin^2(alpha)} = sqrt{1 - left(frac{3}{5}right)^2} = frac{4}{5}. Now we calculate sin(2alpha) and cos(2alpha) using double-angle formulas: begin{align*} sin(2alpha) &= 2sin(alpha)cos(alpha) &= 2 times frac{3}{5} times frac{4}{5} &= frac{24}{25}, end{align*} and begin{align*} cos(2alpha) &= 2cos^2(alpha) - 1 &= 2 times left(frac{4}{5}right)^2 - 1 &= frac{7}{25}. end{align*} Finally, we calculate f(2alpha) using the same identity for f(x): begin{align*} f(2alpha) &= sinleft(2alpha - frac{pi}{6}right) &= frac{sqrt{3}}{2}sinleft(2alpharight) - frac{1}{2}cosleft(2alpharight) &= frac{sqrt{3}}{2} times frac{24}{25} - frac{1}{2} times frac{7}{25} &= frac{24sqrt{3}}{50} - frac{7}{50} &= frac{24sqrt{3} - 7}{50}. end{align*} Thus, the value of f(2alpha) is: boxed{f(2alpha) = frac{24sqrt{3} - 7}{50}}.

answer:From the sum-to-product formulas, we have: [ sin 150^circ - sin x^circ = 2 sin frac{150^circ - x^circ}{2} cos frac{150^circ + x^circ}{2} ] and [ cos 150^circ - cos x^circ = -2 sin frac{150^circ + x^circ}{2} sin frac{150^circ - x^circ}{2}. ] So, substituting these into the equation, we get: [ tan (150^circ - x^circ) = frac{2 sin frac{150^circ - x^circ}{2} cos frac{150^circ + x^circ}{2}}{-2 sin frac{150^circ + x^circ}{2} sin frac{150^circ - x^circ}{2}} = -frac{cos frac{150^circ + x^circ}{2}}{sin frac{150^circ + x^circ}{2}} = -cot left( frac{150^circ + x^circ}{2} right). ] Moreover, [ -cot left( frac{150^circ + x^circ}{2} right) = -tan left(90^circ - frac{150^circ + x^circ}{2} right) = -tan left(frac{30^circ - x^circ}{2} right) = tan left(frac{x^circ - 30^circ}{2} right). ] Thus, solving [ 150^circ - x^circ = 90^circ + frac{x^circ - 30^circ}{2}, ] we find [ 150^circ - x^circ = 90^circ + frac{x^circ - 30^circ}{2} Rightarrow 240^circ = 3x^circ/2 Rightarrow x = 160^circ. ] Therefore, the solution is: (x = boxed{160^circ}).