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a circle BHF with radius equal to the side of the given square, and cutting GE produced in B and F. Then BE, EF are the sides of the rectangle required.

3. If two chords AB, AC be drawn from any point A of a circle, and be produced to D and E, so that the rectangle AC, AE is equal to the rectangle AB, AD, then if O be the centre of the circle, AO is perpendicular to DE.

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Since AB. AD AC. AE, a circle may be described about BCED. Therefore the angle BDE=BCA. Hence if A and B be fixed while C moves round the circle, the angle ADE will be constant and the locus of E will be a straight line. Take AC to pass through O and cut the circle in C' and DE in P. Then as before the angle APD = ABC' = a right

angle.

iv. Describe an isosceles triangle having each of the angles at the base double of the third angle.

If A be the vertex, and BD the base of the constructed triangle, D being one of the points of intersection of the two circles employed in the construction, and E the other, and AE be drawn meeting BD produced in F, prove that FAB is another isosceles triangle of the same kind.

For ADE is an isosceles triangle, and the angle AED at the base is the supplement of the angle ACD in the opposite segment of the circle. Hence AED = BCD and therefore by Euclid = ABD, and also the angles ADE, ADB are equal, therefore the third angle DAE= the third angle BAD. Hence the whole angle BAE is double the angle BAD, and therefore equal to ABD. Hence the triangle FAB is isosceles, and each of the angles at the base is equal to the angles at the base of ABD. Therefore, &c.

V. Prove that the straight lines bisecting one angle of a triangle internally and the other two externally pass through the same point.

Let the exterior angles A and C of the triangle ABC be bisected by AD, CO, meeting each other in 0; then BO will bisect the angle ABC. Because AD bisects the exterior

angle A, BA: BD :: AC: CD. And because CO bisects the angle ACD, therefore AC: CD :: AO: OD, therefore BA BD AO: OD, and therefore BO bisects the angle ABD. See fig. 1.

vi. If three straight lines, which do not all lie in one plane, be cut in the same ratio by three planes, two of which are parallel, shew that the third will be parallel to the other two, if its intersections with the three straight lines are not all in one straight line.

This may be easily proved by a "reductio ad absurdum."

vii. Define a parabola: and prove from the definition that it cannot be cut by a straight line in more than two points.

For if possible let a straight line cut the parabola in three points P, Q, R, and let it cut the directrix in T. Draw Pp, Qq, Rr perpendiculars to the directrix, and let S be the focus. Then since SP= Pp, SQ = Qq, it follows that SP : SQ :: PT QT, and therefore ST bisects the exterior angle to PSQ. Similarly ST also bisects the exterior angle to PSR. Which is absurd.

viii. P, Q are points in two confocal ellipses, at which the line joining the common foci subtends equal angles; prove that the tangents at P, Q are inclined at an angle which is equal to the angle subtended by PQ at either focus.

Let the normals at P and Q meet in G, join QP and produce it to any point R. Then the angle between the tangents is equal to the angle PGQ which is

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= RPG − RQG = (RPS — RQS) + (SPG − SQG).

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Now SPG SQG, being the halves of equal angles, and the difference RPS-RQS PSQ. Similarly the angle PGQ may be proved = PHQ.

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ix. If a circle, passing through Y and Z, touch the major axis in Q, and that diameter of the circle, which passes through Q, meet the tangent in P, then PQ = BC.

Let the tangent YZ cut the major axis in T. Then by similar triangles

PQ SY PQ HZ
= and

QTYT QTZT

PQ SY. HZ

=

QT2 TY.TZ

But TY. TZ= TQ2 by Euclid, III. 36, and SY.HZ=BC2;

.. PQ=BC.

11. In an hyperbola, supposing the two asymptotes and one point of the curve to be given in position, shew how to construct the curve; and find the position of the foci.

Let OX, OY be the two asymptotes, and P the given point. Draw PN parallel to OY cutting OX in N. Measure OD = OE along the asymptotes, such that OD=4. ON. NP. Bisect the angle DOE by OA cutting DE in A. Then OA, AD are equal to the axes; and the remainder of the construction is obvious.

12. Given a right cone and a point within it, there are but two sections which have this point for focus; and the planes of these sections make equal angles with the straight line joining the given point and the vertex of the cone.

Let V be the vertex, VCO the axis of the given cone, and P the given point. Then, if two spheres be inscribed in the cone and passing through P, the tangent planes to these spheres will evidently be the only two sections whose foci are at P. Let C and O be the centres of the two spheres, then VC: VO CL: OM :: CP: PO; therefore VP bisects the angle exterior to CPO in the triangle CPO. But the radii CP, PO are perpendicular to the sections AB, DE, therefore VP bisects the angle between these sections. See fig. 2.

TUESDAY, Jan. 3. 1 to 4.

SENIOR MODERATOR. Arabic numbers.

JUNIOR EXAMINER. Roman numbers.

4. (3) SOLVE the equations,

x2-yz = a2, y2 - zx= b2, z2-xy = c2.

Multiplying the second and third of these equations together, and subtracting the square of the first, we get x (3xyz — x3 — y3 — z3) = b2c2 — aa ;

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viii. Trace the changes in sign of sin (π cos 0)

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lies between 0 and

and π.

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π

π

lies between 0 and 6

negative when ✪ lies between and and positive when ✪

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and find sin A and sin B from the equations,

(1)

a sin2 A +b sin2 B = C,

a sin 24 - b sin 2B=0,

sin 343 in A - 4 sin3 A,

cos 34 = 4 cos3 A − 3 cos A ;

.. sin 34- cos 3A

But

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= (sin A + cos A) {3 — 4 (sin2 A + cos2 A − sin A cos A)},

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= 8 sin 45 cos (A — 45) sin (A — 15) cos (A + 15).

sin 3A – cos 3A = sin 3A — sin (90 – 3A)

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