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22 + 5x + 24 5. (1) (x + 1) (2c2 + x + 1);

149 24202 + 5x+1' 6. (1) (b+c-a+d) (6+c+a-d) (a+d-b+c)

(a+d+b-c);

7. (1) a = 3; (2) æ = p; 8. 15 mendicants. 65 shillings. 10. 4 yards; 5 yards.

(3)

x = 51
y = 3

9. 25, 26.

XVII. MICHAELMAS TERM, 1905.

1. (1) 2. 2. (1) 2x3 + 3 – 2x; (2) 425 + 1224 +9x3 – 4x. 3. (1) 23 +222 – 2x–1; (2) (x2 – 1)? (x2 + 3x + 1). 4. (1) 2x2 – wy+yo; (2) ya(y=2-3); (3) 242 6. 22-16. (1) (6-13) (x – 6); ,

(2) (a-b+c+d) (a-b-c-d);

(3) 9 (a + b) (a? – 3 ab + b2). 7. (1) x = 11;

X = (2) x = 7-1: y = -1 9. ac + bc- ab . 10. 32 sq. yards.

8. 60.

9

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XVIII. HILARY TERM, 1906. 1. 16. 2. (1) 2«— 3 ; (2) xy (24 – 81 y^). X-5

lol X+1 x + 1:

(3)

æ+ 3
4. (1) x2 + x + 3; (2) 124 x + 125 y.

(1) (Cy) (2c2 + xy + y2);
(2) (c + y) (22 - xy + y^) («y) (x2 + xy + y^).

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XIX. TRINITY TERM, 1906. 1. 1453. 2. (1) x2 + x +1; (2) (x − 3)? (x + 4) («—5).

8a22 3. (1) ; (2) me 4. (1) (x – 11) (x +9);"' (2) (a +c) (a b)2.

9 P 6. (1) x = 2117; (2) x = a; 8. £60. 9. 20 francs.

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10. 12.

XX. SEPTEMBER, 1906. 2. (1) 62; (2) – 20. 3. (1) 22 + 2x + 2; (2) (x − 1)2 (a— 6)2 4. 203 + 2% + 30.+ 1.

62

5. (1) +386 – 26

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GEOMETRY

SOLUTIONS OF THE PROBLEMS SET

(The solutions have been made as clear as is consistent with brevity. In order to fully understand them, the figures should be drawn and the references verified. Propositions in Euclid are referred to by number onlye. g. I. 47 means Euclid, Book I, proposition 47. H. and S. refers to Hall and Stevens's School Geometry (edition of 1903). A very slight knowledge is assumed of the measurement of angles by degrees. The abbreviations A, L, ll, are used for triangle, angle, and parallel, and some words, such as right, square, perpendicular, parallelogram, have been shortened.)

A. STATED SUBJECTS.

I. MICHAELMAS TERM, 1904. 1. (b) In proving this, it is best to assume no property of the rhombus, only the def. Let the diagonals of rhombus ABCD meet at E. Then A ABD = A CBD (I. 8) and LABD=LCBD, .. also A ABE = ACBE (I. 4), so that AE = EC and angles at E are rt s.

2. (See H. and S., p. 89, Problem 14.) The point required is the intersection of the fixed 'straight line with the locus of points equidistant from the two fixed points. If the fixed straight line is || to this locus, the construction fails.

3. RtZ BAC = sum of Zs ABC, ACB (I. 32). But Z BAD = LDBA (I. 5) .. LDAC = LDCA, and DC = DA = DB.

4. (6) (See H. and S., p. 104.) The rectangle DBCE = 2 A ABO. Bisect BC at G, and draw GH || to BD. The rectangle DBGH = A ABC.

5. (6) (See H. and S., p. 98.) Join BD. The area of ADAB is obviously half the rectangle. 8. (a) See H. and S., p. 223, Theorem 56.

(b) See Michaelmas Term, 1902, question 2 (p. 150). 7. (a) See H. and S., p. 174.

(6) Find centre of circle, and draw through it diameter at right angles to given line. This will meet circle at points of contact of tangents required.

8. (6) Let AB produced meet tangent at C in Q. (1) If Ċ and P are on the same side of Q, let BP cut circle again at D and join AD. LACB = LADB

>LAPŘ. (2) If C and P are on different sides of Q, the theorem only holds if L AQC is acute. The proof is rather complicated.

9. Let OS be the chord through the centre A; then A is on the locus. Let B be the middle point of another chord OP. Then OBA is a rtZ (III. 3), ... the locus is a circle described on 0 A as diameter (see III. 31).

II. HILARY TERM, 1905. 1. (6) The bisectors of two adjacent angles are at right angles. (See H. and S., p. 13.) 3. (6) In (1) the vertical <= of 2 rtZs. (I. 32.)

„ (2) » » „ = a right angle.

.. ratio = 4:1 = 2:5. 3. (6) See H. and S., p. 62, ex. 12. 4. See Michaelmas Term, 1904, question 4 (6).

7. (6) The straight line perp. to both chords from the centre bisects both chords (III. 3). Hence both the middle points must lie on a diameter.

8. (6) If they touch externally the distance is the sum of the radii (III. 12); if internally, the difference (III. 11).

9. (6) The locus is a circle described on the line joining the two fixed points as diameter (see III. 31). 10. (a) See H. and S., p. 182.

b) Let C, D be the points of contact, and produce BA to meet CD in E. Then

CEP = EA, EB (III. 36) = EDP, .:CE = ED.

III. TRINITY TERM, 1905. 1. (6) Let ABC be the A, and AD, BE, CF the medians.

... AB+ AC>2 AD (Mich., 1902, quest. 2, p. 150). · Similarly AB + BC> 2 BE and AC + BC > 2 CF.!

.: 2 (AB+ AC + BC)>2 (AD+ BC + CF) and perimeter > sum of medians.

2. Let ABC be the rt 2. In BC take any point C, and describe on BC an equilateral A DBC. Bisect Z DBC by EB. Then DB and EB trisect LABC. (For rtZ = 90° and L of equilateral A = 60°.)

3. Let ÅB be the base and CAB the given angle. Make AC = sum of other sides. Join BC and make ZDBC = LACB. Then DC = DB, .. AD, DB = AC, and ADB is required triangle.

5. (6) In figūre ABCD let AB = DC and AD = BC. Join AC. Then A ADC = A ABC (I. 8). .. ZDAC = LACB and AD is || to BC. So also AB is || DC.

6. Since LACB is acute AB2 = AC2+ BC2 – 2 AC.CD (II. 13). But ABP = AC?, .. BC2 = 2 AC.CD.

8. (6) Let the chords AB, CD meet at E. Take 0 the centre, join OE, and draw OF, OG perp. to AB, CD. Then ÅB, CD are bisected at F, Ğ (III. 3). Also OF = OG (III. 14), OE is common, and the Zs OFE, OGE are rt s.. EF = EG (H. and S., p. 51). But AF = CG,.. AE = EC and = DE.

9. (6) From P, a point on the common chord AB, draw tangents PD, PC, one to each circle. Then

PD2 = PA, PB = PC2 (III. 36),.. PD = PC. 10. Let A, B be the fixed points. “Draw CD bisecting AB at rt s.. Then CD is the locus of points equidistant from A and B (H. and S., pp. 89, 144), so that it is the locus of the centres of all circles passing through A, B. .: centre of required circle is intersection of CD with given straight line.

(1) When the given line is | to CD.
(2)

identical with CD.

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