Weekly problem papers, with notes. [With] Solutions1885 |
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Side vi
... Geometrical Maxima and Minima , and the other on the geometrical method of inves- tigating the envelope of a line which moves subject to certain conditions . The problems bearing upon these two subjects are extremely interesting ...
... Geometrical Maxima and Minima , and the other on the geometrical method of inves- tigating the envelope of a line which moves subject to certain conditions . The problems bearing upon these two subjects are extremely interesting ...
Side vii
... geometrical problems . Those marked with an asterisk are all due to him , and it is a source of great regret to me that I was not able to avail myself still further of his aid . Owing to the nature of the present work it is impossible ...
... geometrical problems . Those marked with an asterisk are all due to him , and it is a source of great regret to me that I was not able to avail myself still further of his aid . Owing to the nature of the present work it is impossible ...
Side xi
... ' longer . ' " 3 294 , " " 21 99 7 , at the end insert ' For geometrical proof see Besant , Rect . Hyp . Props . 9 , 10. ' XIX . 2. ( 1 ) , for 3 read 6 throughout . SOLUTIONS OF WEEKLY PROBLEM PAPERS . PAPER L 1. The.
... ' longer . ' " 3 294 , " " 21 99 7 , at the end insert ' For geometrical proof see Besant , Rect . Hyp . Props . 9 , 10. ' XIX . 2. ( 1 ) , for 3 read 6 throughout . SOLUTIONS OF WEEKLY PROBLEM PAPERS . PAPER L 1. The.
Side 22
... geometrically as follows : - Let OL , OL ' be the two given straight lines , and let OA bisect the angle between them , so that LOA = L'OA = a . Then if P be any point on 04 , P is the centre of one of the circles which touch OL and OL ...
... geometrically as follows : - Let OL , OL ' be the two given straight lines , and let OA bisect the angle between them , so that LOA = L'OA = a . Then if P be any point on 04 , P is the centre of one of the circles which touch OL and OL ...
Side 54
... geometrically . Let Cr , Cy be the equiconjugate diameters , and let the tangent at P meet them in T , t , and bisect Tt in Q. = Then the angle PCT QCt . ( See solution to XLVIII . No. 7 ) . Draw PN , PM , PU perpendicular to Cx , Cy ...
... geometrically . Let Cr , Cy be the equiconjugate diameters , and let the tangent at P meet them in T , t , and bisect Tt in Q. = Then the angle PCT QCt . ( See solution to XLVIII . No. 7 ) . Draw PN , PM , PU perpendicular to Cx , Cy ...
Vanlige uttrykk og setninger
a+b+c ABCD axis bisects the angle Cambridge centre chord circle round circle will go circum-circle Clifton College coef College conic const coordinates cos² cosec Crown 8vo denote diameter directrix draw Edition ellipse envelope equal equation Fcap fixed point geometrical given expression hyperbola inscribed circle intersect Join Let ABC Let the tangent locus Mathematics meet middle point Monday afternoon morning nine-point circle nth term orthocentre PAPER parabola parallel perpendicular plane Professor quadrilateral radical axis radius right angles shew shewn sides Similarly sin² sin³ straight line tan² tangent Todh triangle Trinity College Tripos velocity vertex vertical
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