answer:Given a regular tetrahedron with an edge length ( a ), we need to find the angle and the distance between the skew medians of two of its lateral faces. Let's denote the tetrahedron by ( ABCD ), with ( K ) as the midpoint of ( AB ) and ( M ) as the midpoint of ( AC ). Step-by-Step Solution: 1. Project the tetrahedron onto the plane that passes through ( AB ) and is perpendicular to ( CK ). The tetrahedron will project into a triangle ( ABD_1 ), where ( D_1 ) is the projection of ( D ). 2. Consider the projections of ( CK ) and ( DM ) on this plane. Let ( M_1 ) be the projection of ( M ), and ( M_1 ) lies on ( AK ) as ( M_1 ) is the midpoint of ( AK ). The distance between the lines ( CK ) and ( DM ) converts into the distance from point ( K ) to the line ( D_1M_1 ). This line-to-point distance can be found using the properties of the right triangle ( D_1KM_1 ), since it is a right triangle where ( D_1K ) and ( KM_1 ) are the legs. 3. Compute the required lengths: - The height of the tetrahedron from ( D ) to the base ( ABC ) is given by ( h = a sqrt{frac{2}{3}} ). Thus, [ D_1K = h = a sqrt{frac{2}{3}} ] - The length ( KM_1 ), where ( M_1 ) is the midpoint of ( AK ), is half of the median length from vertex to the midpoint of the opposite edge in the equilateral triangle ( ACD ). Hence, [ KM_1 = frac{a}{4} ] 4. Using the Pythagorean theorem in ( triangle D_1KM_1 ): [ D(D_1K, KM_1) = sqrt{(a sqrt{frac{2}{3}})^2 + left(frac{a}{4}right)^2} = sqrt{left(a sqrt{frac{2}{3}}right)^2 + left(frac{a}{4}right)^2} ] [ D(D_1K, KM_1) = sqrt{frac{2a^2}{3} + frac{a^2}{16}} = sqrt{frac{32a^2}{48} + frac{3a^2}{48}} = sqrt{frac{35a^2}{48}} ] [ D(D_1K, KM_1) = a sqrt{frac{35}{48}} = a sqrt{frac{35}{48}} = a sqrt{frac{7 cdot 5}{3 cdot 16}} = a sqrt{frac{7 cdot 5}{3 cdot 16}} ] [ D(D_1K, KM_1) = a cdot frac{sqrt{35}}{2 sqrt{3}} = a cdot frac{sqrt{10}}{sqrt{30}} = a sqrt{frac{10}{30}} ] [ D(D_1K, KM_1) = a sqrt{frac{1}{3}}, = a sqrt{frac{2}{35}} ] Second possible solution median transversality and distance approach yields angle from vector projection and dot products: Considering medians ( CK ) and ( BN ) where ( N ) is the midpoint of edge ( DC ), as before. Calculations yield: The angle between medians calculated via dot product properties in 3D: [ theta = arccos left( frac{1}{6} right) Angles 2nd calc The arccos frac{2}{3} , DM parallel K _ points DM direct parallel distance is boxed boxed{} ; The distance:erview (D _ to BN direct distance ( [ displaystyle projected leg, a frac{sqrt{10}}{10} ), is equal to ``` Conclusions boxed{arccos frac{1}{6}, a sqrt{frac{2}{35}}} boxed{arccos frac{2}{3}, a frac{sqrt{10}}{10}}

answer:First, we need to convert the speed of the train from km/hour to meters/second to match the units of the train's length and the time it takes to pass the station. Speed in meters/second = (Speed in km/hour) * (1000 meters/km) / (3600 seconds/hour) For the train running at 36 km/hour: Speed in meters/second = 36 * 1000 / 3600 Speed in meters/second = 36000 / 3600 Speed in meters/second = 10 m/s Now that we have the speed in meters/second, we can calculate the distance the train travels in 45 seconds. Distance = Speed * Time Distance = 10 m/s * 45 s Distance = 450 meters This distance includes the length of the train and the length of the station. To find the length of the station, we subtract the length of the train from the total distance. Length of the station = Total distance - Length of the train Length of the station = 450 meters - 250 meters Length of the station = 200 meters Therefore, the length of the station is boxed{200} meters.

answer:Given a quadrilateral with diagonals lying on the bisectors of its angles, we need to prove that such a quadrilateral is a rhombus. 1. **Identifying equal angles:** - Since diagonals lie on the bisectors of angles, we have: [ angle BCA = angle DCA quad text{and} quad angle BAC = angle DAC ] [ angle ABD = angle CBD quad text{and} quad angle CDB = angle ADB ] 2. **Using the angle equalities, conclude the congruency of triangles:** - For triangles ( triangle BAD ) and ( triangle BCD ): [ triangle BAD cong triangle BCD ] Because we have angles: [ angle BAD = angle BCD quad text{and} quad angle ABD = angle CBD ] Since they share the same side ( BD ), the triangles are congruent by ASA (Angle-Side-Angle) criterion. 3. **Conclusion from the congruency:** - From the congruence ( triangle BAD = triangle BCD ): [ angle BAD = angle BCD quad text{and} quad AB = BC quad text{and} quad DA = DC ] 4. **Analyzing the remaining triangles:** - Similarly, for triangles ( triangle ABC ) and ( triangle ADC ): [ triangle ABC cong triangle ADC ] Because we have angles: [ angle ABC = angle ADC quad text{and} quad angle BAC = angle DAC ] Since they share the same side ( AC ), the triangles are congruent by ASA criterion. 5. **Conclusion from the second set of triangles:** - From the congruence ( triangle ABC = triangle ADC ): [ angle ABC = angle ADC quad text{and} quad AB = AD quad text{and} quad BC = DC ] 6. **Final step to prove all sides are equal:** - Now, since from above steps we have: [ AB = BC quad text{and} quad DA = DC ] Combining results from both sets of congruent triangles, we get: [ AB = BC = CD = DA ] 7. **Conclusion:** - Since all four sides of the quadrilateral are equal, the quadrilateral is a rhombus. [ boxed{text{Rhombus}} ]

answer:1. **Analyzing condition ( Q ):** The condition ( Q ) states that: [ frac{a_{1}}{a_{2}} = frac{b_{1}}{b_{2}} = frac{c_{1}}{c_{2}} ] This implies that the two quadratic inequalities ( a_{1} x^{2} + b_{1} x + c_{1} > 0 ) and ( a_{2} x^{2} + b_{2} x + c_{2} > 0 ) are proportional to each other. Essentially, one can be obtained by multiplying the coefficients of the other by some constant ( k ): [ a_{1} = k a_{2}, quad b_{1} = k b_{2}, quad c_{1} = k c_{2} ] 2. **Eliminating options using examples:** - Consider the quadratic inequalities: [ x^{2} - 3 x + 2 > 0 text{and} -x^{2} + 3 x - 2 > 0 ] For ( x^{2} - 3 x + 2 > 0 ), solving for roots, we get: [ (x-1)(x-2) = 0 implies x = 1 text{or} x = 2 ] The inequality ( x^{2} - 3 x + 2 > 0 ) holds for ( x in (-infty, 1) cup (2, infty) ). For ( -x^{2} + 3 x - 2 > 0 ), we can rewrite it as: [ -(x^{2} - 3 x + 2) > 0 implies (x-1)(x-2) < 0 implies 1 < x < 2 ] Clearly, these solution sets are different. Hence, ( Q ) is not a necessary and sufficient condition for ( P ), ruling out options (A) and (B). 3. **Exploring other cases:** Consider: [ x^{2} + x + 1 > 0 text{and} x^{2} + x + 3 > 0 ] For both inequalities, there are no real roots, as their discriminants are negative. Specifically, the discriminants are: [ text{For } x^{2} + x + 1: quad Delta = 1^2 - 4 cdot 1 cdot 1 = 1 - 4 = -3 ] [ text{For } x^{2} + x + 3: quad Delta = 1^2 - 4 cdot 1 cdot 3 = 1 - 12 = -11 ] Since both discriminants are negative, both quadratic equations are always positive, regardless of the value of ( x ). Therefore, their solution sets are identical for all ( x ). 4. **Conclusion:** Since we have found a counter-example that shows ( Q ) is not necessary or sufficient for the solutions of ( P ) to be identical: [ (Q: frac{a_{1}}{a_{2}} = frac{b_{1}}{b_{2}} = frac{c_{1}}{c_{2}}) text{is neither a necessary nor a sufficient condition for } (P: text{same solution sets for } a_{1} x^{2}+b_{1} x+c_{1}>0 text{ and } a_{2} x^{2}+b_{2} x+c_{2}>0). ] Therefore, we select: [ boxed{text{D}} ]