107.

SOLUTIONS TO (No. IV.)

1. EUCLID, Prop. 1, Book III. See Potts' Euclid, Page

In Analytical Geometry if the equation to a circle is given, the position of the centre and radius may be determined, and the circle is therefore determined in magnitude and position.

^

2. Euclid, Prop. 18, Book x1.

3. (a) Let ABCD (fig. 30) be a quadrilateral figure; draw the diagonals AC, BD intersecting in E; then if a be the angle between the diagonals, ▲ ABC = AC. BE. sina: Also ▲ ADC = 1 AC. DE. sin a; therefore by addition, the area of the quadrilateral ABCD = (AC. BD. sin a): this will be the greatest when AC, BD and sin a are respectively greatest. Now if the quadrilateral be inscribed in a circle, AC, BD are the greatest possible when they are two diameters; and sin a is greatest, when AC, BD are at right angles. Hence if two diameters AC, BD be drawn at right angles, and AB, BC, CD, DA be joined, the square ABCD will be the greatest possible quadrilateral figure which can be inscribed in the circle.

(B) When a quadrilateral figure is inscribed in a circle, the opposite angles are together equal to two right angles; if this condition be not satisfied the quadrilateral figure cannot be inscribed in a circle.

(7) If a circle can be inscribed in ABCD (fig. 31) touching the sides AB, BC, CD, DA in the points a, b, c, d respectively, then

A a

Ad, Ba Bb, Dc = Dd, Cc = Cb,

.. Aa + Ba + De + Cc

or AB + CD

hence the sum of two opposite

the two remaining sides.

Ad + Dd + Bb + Cb,

BC + DA;

sides is equal to the sum of

4. Let the parallelogram ABCD (fig. 32) be described

equal to the given rectilineal figure; produce AB to E making BE = AB; join AC, CE; then

5. Draw CDN (fig. 33) perpendicular to the base, meeting AB, Rr in D and N respectively; then since Rr is bisected in N, RQ+ Qr2 = 2 (RN2 + QN2) = 2 (CN2 + NQ2)

(since ≤ CRN = 1 a right angle = 4 RCN)

2 CA2 = CA2 + CB2 = AB2.

6. See Appendix 1. Cor. Art. 1.

7.

Let 2 Amp (fig. 34) = 0, ≤ Anp = p, ‹ Bpm = 4,

=

also t Am. Bm, t = Cn. An, t Bp. Cp;

m

... tm. t. ť2 = (Am. Bp. Cn) (An. Bm.Cp) = (Am. Bp. Cn)3,

กา

Ρ

or tm • tn. tp = Am. Bp. Cn
= Am. Bp. Cn An. Bm. Cp.

8. Let the fixed point A be taken for the origin, P any point in the locus; then since AP is constant, it is evident that the locus of P is a sphere round the centre A, and radius AP.

9. Find C (fig. 35) the centre of the circle; draw any radius CA, and AB perpendicular to CA and equal to it; join CB and with centre C and radius CB describe a circle, this will be the circle required.

2

For CB2 CA2 + AB2 = 2CAo, and circles are to one another as the squares of their radii; therefore the circle whose radius is CB: circle whose radius is CA :: CB2 : CẤ2 :: 2 : 1.

2

10. The equation may be reduced to the form

(y + x a) (y2 + No12 — a3) = 0,

which is satisfied by making separately

If CA, CB (fig. 36) be taken for the co-ordinate axes, the former equation represents the straight line AB, where CA CB a, and the latter the circle ABD whose centre is C and radius CA. Hence the proposed equation represents the circle ABD, and the chord of the quadrant AB.

=

11. Let the two sides AB, AD (fig. 37) of the parallelogram ABCD be taken for the co-ordinate axes; join AC, BD, intersecting each other in E, draw EF parallel to AD, and let AB = a, AD = b; then the equations to the diagonals AC, BD 20 Y

are y

bx

α

and + = 1, and if x, y' be the co-ordinates of

a

b

a

2

2x

2 = 1, or a = AF = "; and 2 = 1, or y' - EF-,

a

b

2

and the diagonals of the parallelogram bisects each other.

12. Let the equation to the circle referred to the centre C (fig. 38) be a2 + y2 = a2, and the equation to the chord PQ, y = mx + c; then the equation to CR drawn through C perpendicular to PQ is y x, and if X, Y be the co-ordinates

of the point R,

1

m

and to find the co-ordinates of the points of intersection of PQ with the circle, we have

x2 + (mx + c)2 = a2, or (1 + m2) x2 + 2 mc x + (c2 − a2) = 0 ; therefore if x, x" be the two values of a which are the abscissæ of the points P, Q

13.

Let ABC (fig. 39) be the triangle, Ax, Ay the coordinate axes; < BAx=0', CA x = 0", AB = p, AC = p"; then the area of the triangle ABC

AB. AC. sin 4 BAC=1 p'p" sin (0′′ – 0′)

= (p" sine". p'cos 'p" cos 0".
1⁄2

0′ p′ sin 0′)

= 1⁄2) (y′′x' — x′′y').

ST JOHN'S COLLEGE. DEC. 1834. (No. V.)

1. WHAT objections have been urged against the doctrine of parallel straight lines as it is laid down by Euclid? Where does the difficulty originate, and what has been suggested to remove it?

2. Magnitudes have the same ratio to one another which their equimultiples have. When is the first of four magnitudes said to have to the second the same ratio which the third has to the fourth; and when a greater ratio?

Do the definitions and theorems of Book v. include incommensurable magnitudes?

3. If a solid angle be contained by three plane angles, any two are together greater than the third.

Define the inclination of a plane to a plane, and shew that it is equal to the inclination of their normals.

4. If there be two concentric circles, and any chord of the greater circle cut the less in any point, this point will divide the chord into two segments whose rectangle is invariable.

5. Divide algebraically a given line (a) into two parts such that the rectangle contained by the whole and one part may be equal to the square of the other. Deduce Euclid's construction from one solution, and explain the other.

6. Find a straight line, which shall have to a given straight line the ratio of 1: 5.

7. ACB is a triangle whose base AB is divided in E and produced to F, so that AE: EB and also AF: FB as AC CB. Join CE, CF and shew that ECF is a right angle.

8.

The point C is the centre of a given circle, and E is any point in the radius; find that point in the circumference where CE subtends the greatest angle.