Solutions of the Cambridge Senate-house Problems and Riders for the Year 1875Sir George Greenhill |
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Page 1
... equal . A ' , B ' , C ' are the middle points of the sides of the triangle ABC , and through A , B , C are drawn three parallel straight lines meeting B'C ' , C'A ' , A'B ' in a , b , c respectively ; prove that the triangle abc is half ...
... equal . A ' , B ' , C ' are the middle points of the sides of the triangle ABC , and through A , B , C are drawn three parallel straight lines meeting B'C ' , C'A ' , A'B ' in a , b , c respectively ; prove that the triangle abc is half ...
Page 2
... equal to one another . If the diagonals AC , BD of the quadrilateral ABCD , inscribed in a circle the centre of which is at 0 , intersect at right angles in a fixed point P , prove that the feet of the perpendiculars drawn from O and P ...
... equal to one another . If the diagonals AC , BD of the quadrilateral ABCD , inscribed in a circle the centre of which is at 0 , intersect at right angles in a fixed point P , prove that the feet of the perpendiculars drawn from O and P ...
Page 3
... equal sides such that a regular pentagon can be inscribed in the circle having these portions as two of its sides ... Equal triangles which have one angle of the one equal to one angle of the other have their sides about the equal angles ...
... equal sides such that a regular pentagon can be inscribed in the circle having these portions as two of its sides ... Equal triangles which have one angle of the one equal to one angle of the other have their sides about the equal angles ...
Page 4
... equal , and therefore FA bisects BD , and its parallel CE . vi . If a straight line stand at right angles to each of two straight lines at the point of their intersection , it shall also be at right angles to the plane which passes ...
... equal , and therefore FA bisects BD , and its parallel CE . vi . If a straight line stand at right angles to each of two straight lines at the point of their intersection , it shall also be at right angles to the plane which passes ...
Page 5
... equal to twice the diameter of the circle which is the locus of the point of intersection of tangents at right ... equal to half of it ; therefore CP = perimeter of parallelogram or perimeter of parallelogram = 4 CP = twice diameter of ...
... equal to twice the diameter of the circle which is the locus of the point of intersection of tangents at right ... equal to half of it ; therefore CP = perimeter of parallelogram or perimeter of parallelogram = 4 CP = twice diameter of ...
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Solutions of the Cambridge Senate-House Problems and Riders for the Year 1875 George Greenhill No preview available - 2016 |
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