answer:First, we need to convert the speed of the trains from km/hr to m/s because the length of the trains is given in meters. To convert km/hr to m/s, we use the conversion factor: 1 km/hr = 1000 m / 3600 s. So, the speed of each train in m/s is: 108 km/hr * (1000 m / 3600 s) = 30 m/s Since the trains are running in opposite directions, their relative speed is the sum of their individual speeds. Therefore, the relative speed is: 30 m/s + 30 m/s = 60 m/s The total distance to be covered when the trains cross each other is the sum of their lengths, which is: 120 meters + 120 meters = 240 meters Now, we can use the formula: Time = Distance / Speed To find the time it takes for the trains to cross each other, we divide the total distance by the relative speed: Time = 240 meters / 60 m/s = 4 seconds Therefore, it takes boxed{4} seconds for the two trains to cross each other.

answer:1. **Define Variables**: Given ( n = 2018 ) circles, let's analyze the situation and ensure our understanding of terms: - **Vertex**: The point where two circles cross. - **Yellow vertex**: A vertex whose color differs on each circle that crosses at that point. - We need to show that if a circle has at least 2061 yellow points, there exists a region where all vertices are yellow. 2. **Lemma 1**: - **Statement**: If two circles intersect at points ( x ) and ( y ), then both ( x ) and ( y ) are yellow or neither. - **Proof**: Since each circle is alternately colored, the parity (odd/even nature) of vertices between intersections ensures consistency. Thus, if one intersection point is yellow, both must be, and vice versa. 3. **Lemma 2**: - **Statement**: For three intersecting circles ( C_1, C_2, C_3 ) at points ( x, y, z ), the set ( {x, y, z} ) contains an odd number of yellow vertices. - **Proof**: Consider ( C_1 cap C_2 = x ), ( C_2 cap C_3 = y ), ( C_3 cap C_1 = z ). Note that each circle crossing the cycle ( overline{xy} cup overline{yz} cup overrightarrow{zx} ) does so at an even number of points. Thus, counting bichromatic pairs and color transition consistency leads us to deduce an odd total of yellow vertices in ( {x,y,z} ). 4. **Bounding the Number of Yellow Vertices**: - If each region has at least one non-yellow vertex, consider a circle ( C ) with ( k ) yellow points. - These ( k ) yellow vertices form ( k/2 ) pairs, each pair resulting from ( k/2 ) intersecting circles (Lemma 1). - Lemma 2 reassures that additional intersecting circles deliver at least ( 2binom{k/2}{2} ) new yellow vertices. - Remaining ( n - k/2 - 1 ) circles intersect in non-yellow vertices adding ( 2binom{n-k/2-1}{2} ) yellow points. This yields: [ k + 2binom{k/2}{2} + 2binom{n - k/2 - 1}{2} = frac{k^2}{2} - (n - 2)k + (n - 2)(n - 1) ] 5. **Graph Analysis**: - ( G ) (the geometric graph of our circles) holds ( n(n-1) / 2 + 1 ) non-yellow vertices (since it composites degree 4 vertices, Euler's formula, and consistent face boundary non-yellow alternation). Thus: [ frac{n(n-1)}{2} - 1 geq frac{k^2}{2} - (n-2)k + (n-2)(n-1) ] Solving above restricts: [ k leq n + lfloor sqrt{n-2} rfloor - 2 ] Plugging actual values: [ 2018 + lfloor sqrt{2018-2} rfloor - 2 = 2018 + 44 - 2 = 2060 ] 6. **Consequently**: If some circle contains at least 2061 yellow points (contradicting the above): - It forces multiple circles' intersections to yield all yellow vertices in some region, hence proving our assertion. # Conclusion: [ boxed{text{The vertices of some region are all yellow.}} ]

answer:Let's denote the price of a pair of pants as P. We know the price of a T-shirt is 5, and the price of a skirt is 6. The refurbished T-shirts cost half the original price, so they cost 5/2 = 2.50 each. Now, let's calculate the total income from the items sold: - Two T-shirts: 2 * 5 = 10 - One pair of pants: P (we're trying to find this) - Four skirts: 4 * 6 = 24 - Six refurbished T-shirts: 6 * 2.50 = 15 The total income is the sum of all these amounts: Total income = 10 (from T-shirts) + P (from pants) + 24 (from skirts) + 15 (from refurbished T-shirts) We know the total income is 53, so we can set up the equation: 10 + P + 24 + 15 = 53 Now, let's solve for P: 10 + 24 + 15 = 49 49 + P = 53 Subtract 49 from both sides to find the price of the pants: P = 53 - 49 P = 4 Therefore, Karl sells a pair of pants for boxed{4} .

answer:1. **Given Information and Initial Setup:** - Let ( triangle ABC ) have circumcenter ( O ) and centroid ( M ). - Lines ( OM ) and ( AM ) are perpendicular. - Let ( AM ) intersect the circumcircle of ( triangle ABC ) again at ( A' ). - Let lines ( BA' ) and ( AC ) intersect at ( D ). - Let lines ( CA' ) and ( AB ) intersect at ( E ). 2. **Properties of the Centroid and Circumcenter:** - The centroid ( M ) divides the median ( AM ) in the ratio ( 2:1 ). - Since ( OM perp AM ), ( O ) lies on the perpendicular bisector of ( AM ). 3. **Symmetry and Homothety:** - Since ( A' ) is the reflection of ( A ) across ( O ), ( overline{AM} = overline{MA'} ). - Let ( N ) be the midpoint of ( AC ). Then ( MN parallel A'C ) because ( M ) is the centroid and ( N ) is the midpoint. 4. **Intersection Points and Equal Segments:** - Since ( MN parallel A'C ), triangles ( MNA ) and ( A'CA ) are similar. - This implies that ( overline{EB} = overline{AB} ) and ( overline{CD} = overline{AC} ). 5. **Homothety and Circumcenter:** - We can obtain ( triangle ADE ) from ( triangle ABC ) by a homothety with center ( A ) and ratio ( 2 ). - The circumcenter of ( triangle ADE ) is the symmetric point of ( A ) with respect to ( O ). 6. **Conclusion:** - Since the symmetric point of ( A ) with respect to ( O ) lies on the circumcircle of ( triangle ABC ), the circumcenter of ( triangle ADE ) also lies on the circumcircle of ( triangle ABC ). (blacksquare)