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PRE FAC E.
THE Moderators and Examiners have been induced to
publish the present volume, mainly on the following account.
The value of a problem frequently depends in great measure upon its illustrating clearly some general principle or exemplifying some analytical process; and thus a solution, which is as it were forced out, and which misses the method designed, is worth little in point of the instruction it affords.
It is hardly possible for any but the framers of the questions to produce a complete series of solutions, shewing the method which they wished the student to pursue.
In the present instance the writers have availed themselves of their opportunities of inspecting the answers returned by the candidates for honours, and have appended to their own solutions some of the more striking of those which were submitted to them.
Cambridge, Oct. 1854.
SOLUTIONS OF SENATE-HOUSE PROBLEMS
FOR THE YEAR EIGHTEEN HUNDRED AND FIFTY-FOUR.
THURSDAY, Jan. 5, 1854. 1 to 4.
1. ABD, ACE are two straight lines touching a circle in B and C, and, if DE be joined, DE is equal to BD and CE together; shew that DE touches the circle.
If DE, fig. (1), be not a tangent, from D draw DFG to touch the circle. Then, since (Euc. I. 47) BD is equal to DF, and CG to GF; therefore, BD and CG are together equal to DG. But BD and CE are together equal to DE. Therefore the difference between DG and DE is equal to the difference between CG and CE, which is EG: that is two sides of the triangle DEG are equal to the third, which is impossible; therefore no line except DE can be drawn from D to touch the circle; therefore DE touches it.
Direct Proof.—Let 0, fig. (2), be the centre of the circle. Make DF equal to DE. Join OB, OC, OD, OE, OF, and draw OG perpendicular to DE.
Since DF is equal to DE, therefore DF is equal to DB and EC together; therefore BF is equal to EC; and OB = 0C and LOBF = OCE, therefore OF = OE; therefore, in the triangles DOF, DOE, DO, OF, DF are equal to DO, OE, DE respectively, therefore ZODF = ODE; again, in the triangles ODB, ODG, OD is common, and the angles ODB, OBD, are equal to ODG, OGD, respectively, therefore OB = 0G; therefore the circle passes through G, and, since OG is perpendicular to DE, DE touches it.
2. O, A, B, C, are four points arranged in order in a straight line, so that 0A, OB, OC, form an harmonic progression. Prove that, A and C being stationary, if O move towards A, B will also move towards A.
3. If a, b, c be positive integers, and aś
, ba, có be in geometrical progression, shew that ad, Bover
, com are also in geometrical progression.