IV. SEPTEMBER, 1905.

1. Join AP, and at P make ZAPC=LBAP,..PC=AC. 4. (1) See H. and S., p. 60, Theorem A, of which this is a particular case.

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(2) This is the converse of Mich., 1904, quest. 3. Let BAC be rt/ and D middle point of base. If DA be not = BD, let DE BD, and join BE, CE. Then LEBDL BED and LDCE = L DEC (I. 5)... BEC = Ls EBC, ECB = 90° (I. 32). .. ¿BEC = L BAC, which is impossible (I. 21). .. DA = BD, and the triangles BAD, CAD are isosceles.

(A shorter proof is by using III. 31.) 5. (b) See H. and S., p. 126 (cor.).

6. (b) See Sept., 1902, quest. 10 (p. 150). 7. See Hilary, 1905, quest. 7 (b).

8. The locus the bisectors (internal and external) of the angles between the straight lines. (See H. and S., pp. 90, 190.)

9. (b) Join PQ, and let OX be tangent at X (O being nearer to PX than to QX). ..40XP = ≤ XQP (III. 32) = 180°-4PQS = ≤ PRS (III. 22)... SR is || to OX. 10. Draw OD, PC perp. to XBY, then BC = BY and DB = BX (III. 3). Draw OEF || to DBC. Then OD, QB and PC are ... since OQ = QP, OE = EF (H. and S., p. 60) and .. OB = BC. .. BX = BY.

V. MICHAELMAS TERM, 1905.

1. (b) Since ZPBA = 60° = 4QBC, .. 4 PBC= LABQ, and containing sides PB, BC = sides AB, BQ, .. base PC base AQ (I. 4).

2. ▲ ACB = ▲ ADB (I. 8), .. LDAB = 2 CBA. .. OA OB (I. 6), and OAB is isosceles.

3. (b) Let ABCD be the square. With centre B and radius 3 BC draw a circle and produce AD to meet it in E. Complete parm of which EB, BC are two sides. Then parm sq. ABCD (I. 35) and EB + BC = 3 BC + BC= 4 BC, so that sum of its sides = 8 times side of square.

4. (b) Join CD. .. AC2 CD2-AD2 (I. 47) = CD2 — BD2 = (CX2 + X D2) — (BX2 + DX2) (I. 47)=CX2 — BX2.

6. Join the given point A to B, the centre of the given circle, and draw the diameter CBD at rt/s to AB. Then the circle with AC as radius will pass through D (see H. and S., p. 89).

7. (b) See H. and S., p. 182.

(c) See Hilary T., 1905, quest. 10 (b).

8. (b) Join AB. Then LABX = LAPX, and BAY = LBQY (III. 21). But XPA = 4 YQB (for their containing sides are parallel) * *. .. LABX = LBAY

and BX is to AY.

*If this theorem is not known, it can be proved by I. 29, producing XP and BQ to meet at R. Then XPA == LXRB=LYQB.

VI. HILARY TERM, 1906.

1. (b) BO = CO... LOBC LOCB (I. 5). .. 4 ABC = LACB, and ABC is isosceles (I. 6).

2. On AB, the base, describe an equilateral ▲ DAB, draw AC perp. to AB, and produce BD, AC to meet at C. Then ABC is the required A. For 44

90° and LB = 60°. . 40 = 30°, i. e. A = 3 C and B = 2 C.

3. All equal As on the same base are between the same parallels... the locus of the vertex is a straight line | to the base.

4. Through F draw HF || to AB; through G draw GK || to BC, and produce HF, DC to meet it at L, K. Then AGKD is a parm, and its complements BH, FK are equal (I. 43). But ▲ ABF = 1 BH (I. 34) and ▲ GFC = 1⁄2 FK (Í. 41). .. ABF = GFC.

6. (b) Let AB be the straight line. Draw AC perp. to it, equal to the side of the required square. Through C draw CD to AB. Describe a semicircle on AB, meeting CD at D, and draw DE || to CA, meeting AB at E. Then it can be proved as in II. 14 that

AE.EB DE2 = AC2.

(If AC > AB, the semicircle will not intersect CD.)

8. Let AB be one position of the straight line, and P its middle point. Join OP. Then, since P is the middle point of the hypotenuse of a rt/d ▲, OP = AP = AB = constant. .. the locus is a quarter of a circle, with O

as centre, and half AB as radius (see Sept., 1905, quest. 4 (2)).

9. Let the 3rd circle (centre C) touch externally at D the smaller circle (centre A), and internally at E the larger circle (centre B). Then ADC and BCE are straight lines (III. 11, 12). .. AC+BC = AD+DC+BE-CE = AD+BE = sum of radii of the two first circles. (N.B. The conditions of the problem apparently intend to exclude the case where all three circles touch at the same point.)

10. (a) See III. 36 cor., but it is better proved directly as in H. and S., p. 227.

(b) Draw the tangent OR. Describe a sq. on OR, and draw its diagonal; then draw from 0 the straight line OQP equal to this diagonal, meeting the circle in Q, P. Then OQ.OP = OR2. But by construction OP2 = 2 OR2 (follows from I. 47). .. 20Q.OP = OP2 and 20Q OP, so that OQ = =

QP.

(c) Put for radius of original circle. If O is on a concentric circle whose radius is 3r, then OQP will pass through the centre, for OQ = 2r and QP also = 2r. .. if O is outside the circle just mentioned OQ will be >2r, and therefore QP, which must be < 2r, cannot be equal to it.

VII. TRINITY TERM, 1906.

2. Let ABC be the isosceles A, and BD perp. to AC. Then A+B+C 180° and BC. A + C = 90°. But 4 DBC+C 90° (I. 32), .. 4 DBC = A.

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=

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1⁄2A.

3. Since AEAHA HDG (I. 4) .. EH = HG and EFGH is equilateral. Also AE = AH, LAEH = LAHE, and A 90°, .. LAHE 45°. Similarly LDHG 45°, .. LEHG 90° and EFGH is rectangular. Again join EG, HF which will be || to the sides, and let them meet at K. Then A EHK = 1 sq. AK, ▲ HKG A =sq. DK, &c., &c. .. EFGH = 1⁄2 ABCD.

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4. Bisect AB in D; draw DE perp. to AB, and draw CE to meet it || to AB. Join EA, EB. Then EA = EB (I. 4) and A EABA CAB (I. 37).

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6. (b) OC2 AC. CB+OB2 (II. 6) = AB2+OB2 = BX2 + OB2 = 0X2, .. 0X = OC.

8. Let D be middle point of BC, and join AD. Then A ADC A ADB (I. 4) and ADB = 90°. .. semicircle described on AB as diameter will pass through D (III. 31). Similarly it will pass through middle point of AC.

10. Let P be one of the points from which a tangent PA of the given length can be drawn to the circle whose centre is C. Join CP, CA. Then CP2 = CA2 + AP2 = constant. .. the locus of P is a circle whose centre

is C.

VIII. SEPTEMBER, 1906.

2. On any base AB describe an equilateral A, and bisect its base-angles by AD and BD, meeting at D. Then each of the base-angles of ▲ ABD = 30°, and.. the vertical 120°.

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3. Since PBA = 60° ≤ QCA and ABC = LACB, .. whole PBC= whole 4 QCB, .. their supplementary angles OBC, OCB are equal (H. and S., p. 11), and OBC is isosceles.

5. Draw XY || to AB. Then XB bisects AY and XC bisects DY (I. 34). .. ▲ XBC= parm. Again AOBCAABC= AABC= parm... figure BXCO .. figure BXCO = parm.

7. Let ABC be the triangle and AD the line drawn to a point in base. Draw AE perp. to base BC, then BC is bisected at E (I. 26). .. BD.DC = BE2 – DE2 (II. 5) (BE2+EA2)-(DÉ2 + EA2) = BA2-DA2.

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9. Let B be point of contact and ABC given straight line, (A being on larger circle, whose centre is D). Join DB and produce it to meet the smaller circle in E; then BE is a diameter (III. 12). Draw DG perp. to AB, then BG = AB (III. 3). In As DGB, BCE the Ls DGB, BCE are rt/s, the vertical 4s GBD, EBC are equal and DB = BE (hyp.). .. GB = BC and AB = 2 BC.

10. Since all the chords are equal, the perps. on them from the centre are equal. Hence a concentric circle, described with one of the perps. for radius, will pass through the feet of the others, and touch all the chords.

B. FURTHER EXAMINATION IN MATHEMATICS.

I. SEPTEMBER, 1902.

9. On any base BC describe segment of circle_with <= 30° (i. e. half the of an equilateral A). Draw DA bisecting BC at rt. angles, to meet circumference

at A. Then ABC is required A. For each of base 2 times vertical 4.

4s (180°-30°) 75° = 1⁄2 =

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10. Let the diagonals of parm ABCD meet at E. Then BE is median of ▲ ABC. Apply H. and S., Theorem 56, p. 223.

Otherwise take B as the acute angle of parm and draw AE and DF perp. to BC and BC produced. Then AC2 = AB2 + BC2-2 BC. BE (II. 13), and BD2 = BƠ2 +CD2+2 BC.CF (II. 12). But As ABE. DCF are equal (I. 26) and BE CF = by addition AC2+BD2 = AB2 + BỞ2 + BC2 + CD2 = sum of sq. on sides of parm. 11. Let common chord AB meet PQ, the straight line joining the centres, at C. Then A APQ = A BPQ (I. 8) and AQP = ≤ BQP. .. ▲ ACQ: .. ▲ ACQ = ▲ BCQ (I. 4), and CB and Zs at Care rt/s.

AC

II. MICHAELMAS TERM, 1902.

2. Let ABC be the A, and D the middle point of BC. Produce AD to E, making DE = AD.

Then (I. 4) ABEC is a parm and its diagonals bisect one another. .. AB, AC AB, BE> AE >2 AD.

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4. Bisect ABC by BD, meeting AC at D; draw DP to CB, meeting AB at P... ¿PBD=¿CBD=2PDB. .. PB = PD.

8. Let 0, P be the centres, then OXP is a straight line (III. 12). Join OC, PD. 48 OXC, CXD, DXP = 180° = 48 OCD, PDC (III. 16). But 20XC-LOCX and 4DXP =LPDX... LCXD=4s XCD, XDC and CXD is a rt2.

III. HILARY TERM, 1903.

7. Let ABC, DBC be As, and let AD cut BC at E. Draw AF, DG perp. to BC. Then ABC, DBC have equal altitudes, i. e. AF DG (I. 39, and H. and S., p. 97). .. ▲ AEF = ▲ DGE (I. 26) and AE = DE.

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