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E. J. ROUTH, M. A.
FELLOW AND ASSISTANT TUTOR OF ST PETER'S COLLEGE, CAMBRIDGE
AND EXAMINER IN THE UNIVERSITY OF LONDON.
MACMILLAN AND CO.
The value of a Collection of Solutions depends in great measure on the fact that every Problem is solved by the framer of the question, thus showing the student the manner in which he was expected to proceed in the Senate-House. The Moderators desire therefore to thank the Examiners for the many valuable Solutions of the Problems set by them, by which the book has been made more complete than it would otherwise have been,
The Senior Moderator also acknowledges his obligation to Mr DROOP, Fellow of Trinity College, for much valuable assistance, and particularly for the suggestion and the solution of the three following Problems, viz. No. vi. of Tuesday Morning, Jan. 17, and Nos. 3 and 5 of Wednesday Morning, Jan. 18.
SOLUTIONS OF SENATE-HOUSE
PROBLEMS AND RIDERS
FOR THE YEAR EIGHTEEN HUNDRED AND SIXTY.
TUESDAY, Jan. 3. 9 to 12.
JUNIOR MODERATOR. Arabic numbers.
1. If a straight line DME be drawn through the middle point M of the base of a triangle ABC, so as to cut off equal parts AD, AE from the sides AB, AC, produced if necessary, respectively, then shall BD be equal to CE.
Through C draw CF parallel to AB, and cutting DE in F. Then the two triangles DMB, FMC are clearly equal, and therefore CF= BD. Again, CF being parallel to AB, the angle CFE = the angle ADE, and because AD= AE, the angle ADE = angle AED; whence it easily follows that CH=CE.
2. Shew how to construct a rectangle which shall be equal to a given square; (1) when the sum and (2) when the difference of two adjacent sides is given.
The first case is too obvious to require any solution. In the second case, refer to the figure in Euclid, Book 11. Prop. 14. A little consideration will shewthat GE is twice the difference between the two sides BE, ED. Whence the following construction. Take GE= half' the given difference, describe