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Since a, bac

, co are in geometrical progression, ,

abc (se)"
(acjó = bac,
(ac)" = b,

1

1

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(acyód

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therefore, aró, bone, cos, are in geometrical progression.

4. If either of the two quantities 1 + 3", 1 + 3 m++, is a multiple of 10, prove that the other is also a multiple of 10, m and r being positive integers.

1+ 39+41 Assuming that

is integral, it is evident that the

10 following quantities also are integral:

3°+41 3a + 10

10
3(3+4V-2 – 1)

10
(10 – 1) (3M+4+2 – 1)

10

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3 + 1 Proceeding in the same way, we see that is integral.

10 The reasoning here given, taken backwards, shews that, if 3" +1

3”+45 + 1 is integral,

also is integral. 10

10

The following is a somewhat different solution of the same problem.

Suppose 3" +1 divisible by 10; then 3" must have 9 for its last digit. Now 3* = 81: hence 34 has 1 for its last digit. Hence 3" x 34r has 9 for its last digit, and therefore 1 + 3" x 34" is divisible by 10.

Suppose 3M+4 + 1 divisible by 10: then 3m+4 has 9 for its last digit: therefore 3” must have 9 for its last digit; for otherwise 3" x 34 would not have 9 for its last digit.

Hence 3" + 1 is divisible by 19.

5. Find the value of tana or tans from the equations

tan (a + b) = tana cotß + cota tanß,
tan (a - b) = tana cotß – cota tan ß.

or

Adding together the two equations, we get
tana + tanß tana - tanß
+

= 2 tan a cotß,
1 - tana tanß 1+ tanq tanß

1 + tanß 2 tana.

= 2 tana cotß, 1 – tan’a tan"ß

tanß (1 + tan*B) = 1 – tan'a tan-ß........ ..(1). By symmetry, tana (1 + tan'a) = 1 - tan*ß tan*a...........

.....(2). From (1) and (2), tana – tanß + tan® – tan®ß = 0,

(tana – tanß) • {1+tan’a + tan a tanß + tan*ß} = 0,

(tana – tanß). {(tana + ftan 3)2 + 1 + &tanoß} = 0, and therefore, since the second factor cannot be zero,

tan 4 = tang.............. ... (3).

From (2) and (3),

tan a (1 + tanʼa) = 1 – tan*a,
tan'a + tan a + = 1,

#15 - 1
tana =

= tanß. 2

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6. If A + B + C = 90°, shew that the least value of tan' A + tan'B + tanoC is 1.

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1-tan Atan B-tan B tan- tanCtan A 0 = cot(A+B+C) =

tan A + tan B+tan C -tan A tan B tan therefore tan A tan B + tan B tan C + tanC tan A

= 1.

But since tan’ A + tan’B = (tan A – tan B)? + 2 tan A tan B,

tan'B + tan*C = (tan B – tan C)2 + 2 tan B tan C,

tanoC + tan*A = (tan C- tan A)+ 2 tan C tan A ; therefore tan*A + tan'B + tanC = 1 + 1{{tan A – tan B)'

+ (tan B – tan C)*

+ (tan C – tan A)?}; therefore tan’ A + tan'B + tan’C is not < 1.

7. Lines, drawn through Y, Z, at right angles to the major axis of an ellipse, cut the circles, of which SP, HP, are diameters, in I, J, respectively. Prove that IS, JH, BC, produced indefinitely, intersect each other in a single point.

Let IY, JZ, fig. (4), produced if necessary, intersect the major axis in Y', Z', respectively: then

L SIY' = supplement of SIY,
= L SPY = L HPZ

= L HJZ';
L SY'I = right angle = 4 HZ J,
and therefore

LISY' = L JHZ';

1

whence the triangle formed by producing IS, HJ, is isosceles, and therefore, CS, CH, being equal, the vertex of the triangle must lie in BC produced.

Since the angles SIY', HJZ', are equal respectively to the angles SPY, HPZ, they can never be zero, and therefore si, HJ, can never be perpendicular to the major axis. Thus the point of intersection of IS, JH, BC, can never move off to an infinite distance from C.

(CV

8. From any point T, (fig.5), two tangents are drawn to a given ellipse, the points of contact being Q, Q: CQ, CQ, QQ, CT, are joined; V is the intersection of QQ, CT. Prove that the area of the rectilinear triangle QCQ varies inversely as

TV3
+
TV

CV)
Draw CK at right angles to QQ'. Then
(area QCQ')* = QV. CK

CD
CP2

52: (CP CV*).CK
AC?. BC
CP. PE (CP" – CV*).CK?.

But

CP - CV° = CT.CV CV2 = CV. TV.

CK
1 CV? CV

1

1 Also CP.PE CP CP2 CT.CV

CT

(CV+ TV)

AC. BC. CV.TV Hence (area of QCQ')

(CV + TV)2

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and therefore area of QCQ

AC.BC
CVE TV1 »

+
TV CV

or area of QCQ varies inversely as

TVI

+ TV)

9. A piece of uniform wire is bent into three sides of a square ABCD, of which the side AD is wanting; shew that, if it be hung up by the two points A and B successively, the angle between the two positions of BC is tan18.

Let EF, fig. (6), be drawn parallel to BA, through E the middle point of BC. Then, if G be the centre of gravity of the piece of wire, EG equals two-thirds of BE.

Draw HG parallel to BC, and join AG, BG.

When the wire is hung up by A, AG will be vertical, and when hung up by B, BG will be vertical; therefore the inclinations of BC to the vertical will be equal to the angles which BC makes with AG and BG. Therefore the angle between the two positions of BC, (supposing it to be kept in the same plane,) will be the angle between AG and BG. Now tan AGB = tan (AGH+HGB)

$+

18;

1-3 therefore the angle between the two positions of BC is tan-18.

10. A weight of given magnitude moves along the circumference of a circle, in which are fixed also two other weights: prove that the locus of the centre of gravity of the three weights is a circle. If the immoveable weights be varied in magnitude, their sum being constant, prove that the corresponding circular loci intercept equal portions of the chord joining the two immoveable weights.

Let R, fig. (7), be the moveable weight, P and Q the stationary ones. Let G be the centre of gravity of P and Q, H that of P, Q, R.

R.GR Then

P + Q +R

GH =

OC GR.

But the locus of R is a circle; hence that of H is a circle, G being a similar point in the two circles, and GR, GH,

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