PREFACE. THIS book of the Solutions of the Questions in the Mathematical Tripos of 1875 has been undertaken at the request of several persons who are interested in the teaching of mathematics at Cambridge. The solution of each question has, in general, been furnished by the maker of the question, or is a solution that has been given by a candidato in the Examination. By the new regulations for the Examination several new subjects have been introduced, which have not been treated of in preceding collections of Senate-House Solutions. Complete solutions of the questions on the higher subjects have been given in all cases where it was felt that a reference to the Text-books in ordinary use was not sufficient. An apology must be given for the unavoidable delay in the appearance of the book. ERRATUM. Page 95, in lines 8 and 9, read "If the first three planes meet the fourth plane in BC, CA, AB, &c." SOLUTIONS OF SENATE-HOUSE PROBLEMS AND RIDERS FOR THE YEAR EIGHTEEN HUNDRED AND SEVENTY-FIVE. MONDAY, Jan. 4, 1875. 9 to 12. MR. GREENHILL. Arabic numbers. MR. WRIGHT. Roman numbers. 1. PARALLELOGRAMS and triangles upon the same base and between the same parallels are 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 the triangle ABC and that be passes through A, ca through B, ab through C. For (fig. 1) Δ ▲ ABa+Aac = A Ba + Aa C=ABC=ABB' = ^ ABc ; therefore Bac is a straight line. Similarly it may be proved that Cab is a straight line. And ▲ Bbc therefore = abcaBC= £^ ABC. Δ Also Aab+ Aac = AaB+▲ AaC = ABC=abc; therefore bAc is a straight line. B 2. The angles in the same segment of a circle are 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 to the sides of the quadrilateral lie on a fixed circle, the centre of which is at the middle point of OP. Let Q (fig. 2) be the middle point of OP; a, b, c, d the feet of the perpendiculars from P on AB, BC, CD, DA, and a', b', c', d' the middle points of AB, BC, CD, DA respectively. Then a'b'c'd' is a rectangle. Also LAPa = LABP-L PCD=LCPc'; therefore aPc' is a straight line, and similarly it may be proved that bPd', cPa', and dPb' are straight lines. Therefore Pa'Oc' is a parallelogram, and Q, which is the centre of the parallelogram, is the centre of the rectangle a'b'c'd'. Therefore a circle can be described with centre Q passing through a, a'; b, b'; c, c'; d, d'. Also 2 (rad.)+20Q2 = Oa" + a'P2 = Oa2 + a' A2 = 02A, and therefore the radius of the circle is constant. This circle is analogous to the nine-pointic circle of a triangle. 3. Upon a given straight line describe a segment of a circle which shall contain an angle equal to a given rectilineal angle. Through a fixed point O any straight line OPQ is drawn cutting a fixed circle in P and Q, and upon OP and OQ as chords are described circles touching the fixed circle at P and Q. Prove that the two circles so described will intersect on another fixed circle. If C (fig. 3) be the centre of the fixed circle and OM, ON be drawn parallel to CQ and CP, forming the parallelogram OMCN, then M and N are the centres of the segments. |