answer:1. **Determine the Circumference of Each Track**: - Leah's track radius = 40 meters, so the circumference is C_L = 2pi times 40 = 80pi meters. - Jackson's track radius = 55 meters, so the circumference is C_J = 2pi times 55 = 110pi meters. 2. **Calculate the Speed in Terms of Radians per Minute**: - Leah's speed = 200 m/min, so in terms of radians per minute, her angular speed is omega_L = frac{200}{80pi} times 2pi = 5 radians/min. - Jackson's speed = 280 m/min, so his angular speed is omega_J = frac{280}{110pi} times 2pi approx frac{560}{110} approx 5.09 radians/min. 3. **Relative Angular Speed**: - Since they are running in opposite directions, their relative angular speed is omega_L + omega_J = 5 + 5.09 = 10.09 radians/min. 4. **Time to Meet**: - They meet every time they cover an angle of 2pi radians relative to each other. - Time to meet once, k = frac{2pi}{10.09} approx frac{6.28}{10.09} approx 0.622 minutes. 5. **Total Number of Meetings in 45 Minutes**: - Total meetings = leftlfloor frac{45}{0.622} rightrfloor = leftlfloor 72.35 rightrfloor. 6. **Conclusion**: - They can only meet a whole number of times, so we take the floor of 72.35, which is 72. Thus, Leah and Jackson pass each other 72 times. The final answer is boxed{textbf{(C) } 72}

answer:1. **Intersection Points of Circles**: Let the circles ( w_1 ) and ( w_2 ) intersect at points ( A_1 ) and ( A_2 ). We can denote the centers of ( w_1 ) and ( w_2 ) as ( O_1 ) and ( O_2 ) respectively, with radii ( r_1 ) and ( r_2 ). 2. **Arbitrary Point on Circle ( w_1 )**: Let ( B ) be an arbitrary point on circle ( w_1 ). The line ( BA_2 ) intersects circle ( w_2 ) at point ( C ). 3. **Orthocenter of ( Delta BA_1C )**: We need to find the orthocenter ( H ) of ( Delta BA_1C ). The orthocenter is the point where the altitudes of the triangle intersect. 4. **Properties of the Orthocenter**: To find the orthocenter, we use the fact that the altitudes of a triangle are concurrent. The altitude from ( B ) to ( AC ) is perpendicular to ( AC ), the altitude from ( C ) to ( AB ) is perpendicular to ( AB ), and the altitude from ( A_1 ) to ( BC ) is perpendicular to ( BC ). 5. **Cyclic Nature of Points**: Since ( A_1 ) and ( A_2 ) are the intersection points of ( w_1 ) and ( w_2 ), they lie on both circles. This implies that ( A_1 ), ( A_2 ), ( B ), and ( C ) are concyclic (lie on a common circle). 6. **Fixed Circle**: The key observation is that the orthocenter ( H ) of ( Delta BA_1C ) lies on a fixed circle. This fixed circle is known as the nine-point circle (or Euler circle) of ( Delta BA_1C ). The nine-point circle passes through the midpoint of each side of the triangle, the foot of each altitude, and the midpoint of the segment from each vertex to the orthocenter. 7. **Nine-Point Circle**: The nine-point circle of ( Delta BA_1C ) is fixed because it depends only on the configuration of the points ( A_1 ) and ( A_2 ), which are fixed intersection points of ( w_1 ) and ( w_2 ). Therefore, regardless of the position of ( B ) on ( w_1 ), the orthocenter ( H ) will always lie on this fixed nine-point circle. 8. **Conclusion**: Since the nine-point circle is fixed for the triangle ( Delta BA_1C ), the orthocenter ( H ) of ( Delta BA_1C ) lies on this fixed circle for any arbitrary point ( B ) on ( w_1 ). (blacksquare)

answer:1. We begin by identifying the key elements of the problem: the diagonals ( AC ) and ( BD ) of the cyclic quadrilateral ( ABCD ) intersect at point ( P ). Point ( Q ) is chosen on segment ( BC ) such that ( PQ perp AC ). 2. Our goal is to demonstrate that the line passing through the centers of the circumcircles of triangles ( APD ) and ( BQD ) is parallel to line ( AD ). 3. We select a point ( T ) on line ( QP ) such that ( DT perp DA ). From the given information, we know ( angle APT = 90^circ ) and ( angle ADT = 90^circ ). This implies that points ( A ), ( P ), ( D ), and ( T ) lie on a common circle (cyclic quadrilateral), since opposite angles sum to ( 180^circ ). 4. Therefore, the center of the circumcircle ( omega_1 ) of triangle ( APD ) lies on the perpendicular bisector of segment ( DT ). 5. Since ( ABCD ) is a cyclic quadrilateral, we have ( angle QBD = angle PAD ). Because ( APDT ) is also cyclic, ( angle PAD = angle QTD ). Thus, ( angle QBD = angle QTD ), indicating that points ( B ), ( Q ), ( D ), and ( T ) also lie on a common circle (cyclic quadrilateral). 6. Consequently, the center of the circumcircle ( omega_2 ) of triangle ( BQD ) lies on the perpendicular bisector of segment ( DT ). 7. Therefore, the perpendicular bisector ( ell ) of segment ( DT ) passes through the centers of both circumcircles ( omega_1 ) and ( omega_2 ). 8. Given that ( ell perp DT ) and ( AD perp DT ), we conclude that ( ell parallel AD ). # Conclusion: [ boxed{text{The line passing through the centers of the circumcircles of triangles } APD text{ and } BQD text{ is parallel to } AD.} ]

answer:Solution: (1) Since the distance from one endpoint of the minor axis to the right focus is sqrt{3}, we have a= sqrt{3}. Given e= frac{c}{a}= frac{sqrt{6}}{3}, we get c= sqrt{2}. Therefore, b^2=a^2-c^2=1, so the equation of the ellipse is frac{x^2}{3}+y^2=1; (2) By solving begin{cases} frac{x^2}{3}+y^2=1 y=x+1end{cases} for y, we get 2x^2+3x=0, which gives x_1=0 or x_2=-frac{3}{2}. Thus, y_1=1, y_2=-frac{1}{2}, so the two intersection points are: A(0,1), Bleft(-frac{3}{2}, -frac{1}{2}right), then |AB|= sqrt{left(-frac{3}{2}-0right)^2+left(-frac{1}{2}-1right)^2}= frac{3sqrt{2}}{2}. Therefore, the answers are: (1) The equation of the ellipse is boxed{frac{x^2}{3}+y^2=1}. (2) The distance between points A and B is boxed{frac{3sqrt{2}}{2}}.