must be a right angle. The angle CBB' must therefore be less than half a right angle. Draw AN perpendicular to BB', then the triangles ABN and AB'N are similar; therefore AN2 = BN. B'N, and the locus of A is a rectangular hyperbola with its vertices at B and B'. xx. A hollow acute-angled triangular prism, whose ends are perpendicular to its axis, is capable of reflecting light at its inner surfaces; if it is placed with one face on a horizontal table and a small pencil of light is admitted through a hole in one face immediately opposite an edge so as to be incident upon the bottom face directly under the top edge, prove that the axis of the pencil will emerge at the hole after five reflections at the faces and one at each end of the prism if the direction of first incidence makes with the axis of the prism an angle where ABC is a transverse section and the length of the prism. The course of the light is the same as that of a perfectly elastic ball. The projections of the path on a transverse section ABC perpendicular to the axis of the prism must be the triangle DEF, where D, E, F are the feet of the perpendiculars from A, B, C on BC, CA, AB. The projection of the ball will describe the perimeter of DEF twice while the ball moves once up and down the prism. Therefore the ratio of the velocities perpendicular and parallel to the axis of the prism 2 perimeter of the triangle DEF 2 length of the prism a cos A+ b cos B+ c cos C WEDNESDAY, Jan. 6, 1875. 1 to 4. MR. GREENHILL. Arabic numbers. MR. FREEMAN. Roman numbers. 1. STATE and prove Newton's first lemma. Prove that the quadrilateral of maximum area that can be formed with four straight lines AB, BC, CD, DA of given lengths is such that a circle can be described about it. Hence, prove that the curve of given length which on a given chord encloses a maximum area is an arc of a circle. Let AD, BC (fig. 36) intersect in O. Keeping AB fixed displace C to C and D to D'; O is the instantaneous centre of CD. The area being a maximum ABCD = ABC'D' to the first order and ODC= OD'C' to the same order. and therefore a circle can be described about ABCD. If BC, CD, DA be replaced by any number of straight lines BC, CD,... KA, a successive application of the method will prove that the area enclosed will be a maximum when the angular points of the polygon lie on a circle; and proceeding to the limit when the number of the sides is indefinitely increased, the curve of given length, which encloses the maximum area, will be the arc of a circle. H 2. State and prove the eleventh lemma. If OP, OQ be the tangents to a small arc PQ of continued curvature, prove by Newton's method that the ratio OP+0Q- arc PQ: arc PQ-chord PQ tends to the limit 2: 1 as the arc PQ is indefinitely diminished. Let the involute PB (fig. 37) of the arc PQ be drawn meeting OQ in B, and PN be drawn perpendicular to OQ to meet it in N, and let the tangent PT to the involute meet OQ in T. TP2 Then lt =2, and therefore lt TB.TQ diameter of curvature at P of arc PQ. Since IB varies as TP, therefore lt the arc PC with centre Q, then t and therefore lt = TA 2; therefore lt If the circular arc PA be described with centre 0, and = TP3 TQ2 = = 4 TB2 TP TB TB , or lt BN lt OP+0Q-arc PQ: arc PQ-chord PQ-2: 1. 3. A body moves in a plane curve and the radius drawn to it from a point in the plane which is either fixed or moves uniformly in a straight line describes areas about the point proportional to the time. Prove that the body is acted on by a force tending to the point. If a body moves in a conic section so that the resolved part of the velocity perpendicular to the focal distance is constant, the force tends to the centre of the conic section. Let v be the velocity at the point P (figs. 38, 39). The resolved part of the velocity perpendicular to the focal distance Therefore v.CY is constant, and the force tends to C. 4. Find an expression for the law of force under which a body will describe a given orbit about a centre of force. If S be the centre of force and SY the perpendicular on the tangent at a point P of the orbit, prove that the acceleration at Pis equal to the product of the velocities of Pand Y divided by SY. Let SV represent the velocity at P (fig. 40), SV. SY is constant. Then if T be the time the body takes to move from P to P', the acceleration at P = product of the velocities of P and Y divided by SY. 5. A body describes an orbit under the action of a force tending to and varying as the distance from a fixed point, prove that the orbit is an ellipse, and shew from elementary considerations, that the periodic time is the same for all circumstances of projection. A number of bodies which describe ellipses about the centre of force as centre in the same periodic time, are projected from a given point with a given velocity in different directions in a plane. Prove that their paths will all touch a fixed ellipse with the given point as focus. If C (fig. 41) be the centre of force, P the point of projection, since the velocity at P and the periodic time are constant, therefore CD the diameter conjugate to CP is constant. If the tangent at Q be at right angles to the tangent at P and intersect it in Y, then CY-CP+CD" is constant. QV the normal at Q is an ordinate of the diameter PP', and therefore bisects the angle PQP'. Therefore PQ+QP' = 2CY, and therefore the locus of Q is an ellipse with foci P, P', and all the orbits will touch this ellipse. Compare the Rider to ix, Monday morning, Jan. 4. 6. A body describes an ellipse about a centre of force in the focus, find the law of force. If S be the centre of force, A the nearer apse, P the body, and a small impulse which generates the velocity Tact on the body at right angles to SP, prove that the change in the direction of the apse line is given approximately by where e is the excentricity of the orbit and h twice the rate of description of area about S. Let the normal at P (fig. 42) meet the apse line in G, and let GL be drawn perpendicular to SP. The radial velocity being unaltered by the impulse, if PHg be the new normal and GH be drawn parallel to PĹ to meet PHg in H, then If Sg be the new apse line, and gl be drawn perpendicular to SP, then PL being the old and Pl the new semilatus-rectum neglecting T2; therefore Ll=2 GH. LASP=0, 4 SPG = ↓, h Let e sin The change of direction of the apse line = LGSg=LGSH+L HSg = |