Mathematical Problems on the First and Second Divisions of the Schedule of Subjects for the Cambridge Mathematical Tripos ExaminationMacmillan and Company, 1878 - 480 sider |
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Side xii
... point of intersection will lie on one of two fixed circles concentric with the given circles . 14. A chord CD is drawn at right angles to a fixed diameter AB of a given circle , and DP is any other chord meeting AB in Q : prove that the ...
... point of intersection will lie on one of two fixed circles concentric with the given circles . 14. A chord CD is drawn at right angles to a fixed diameter AB of a given circle , and DP is any other chord meeting AB in Q : prove that the ...
Side 1
... fixed diameter AB of a given circle is taken a fixed point C from which perpendiculars are let fall on the straight lines joining A , B to any point of the circle : prove that the straight line joining the feet of these perpendiculars ...
... fixed diameter AB of a given circle is taken a fixed point C from which perpendiculars are let fall on the straight lines joining A , B to any point of the circle : prove that the straight line joining the feet of these perpendiculars ...
Side 2
... point on its diameter in two points : prove that these six points lie on a circle . 35. The tangents from a point 0 ... fixed circle . 37. From the point of intersection of the diagonals of a quadrangle inscribed in a circle ...
... point on its diameter in two points : prove that these six points lie on a circle . 35. The tangents from a point 0 ... fixed circle . 37. From the point of intersection of the diagonals of a quadrangle inscribed in a circle ...
Side 3
... fixed circles which do not intersect in P and cuts the other in Q , Q ' : prove that there are two fixed points at ... point of the arc intercepted between two of the points per- pendiculars are let fall on the straight lines joining them to ...
... fixed circles which do not intersect in P and cuts the other in Q , Q ' : prove that there are two fixed points at ... point of the arc intercepted between two of the points per- pendiculars are let fall on the straight lines joining them to ...
Side 11
... fixed points . 111. A point has the same polar with respect to each of two circles : prove that any common tangent will subtend a right angle at that point . 112. Given two points A , B , a straight line PAQ is drawn through A so that ...
... fixed points . 111. A point has the same polar with respect to each of two circles : prove that any common tangent will subtend a right angle at that point . 112. Given two points A , B , a straight line PAQ is drawn through A so that ...
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Vanlige uttrykk og setninger
angular points angular velocity asymptotes ax² axes bisected cardioid centre of perpendiculars chord circumscribed circle co-ordinates common point common tangents confocal conicoid conjugate diameters constant continued fraction cos² curve described diagonals directrix envelope equal excentric angles fixed circle fixed point fixed straight line foci focus four points given circle given conic given ellipse given point given the equations inscribed latus rectum length locus major axis meet minor axis normal parabola parallel particle passes plane point of intersection points of contact polar pole prove radical axis radius of curvature ratio rectangle rectangular hyperbola respectively right angles roots self-conjugate sin² sin³ straight line joining subtends a right tangents drawn tetrahedron triangle ABC velocity vertex vertical
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