Solutions of Exercises

Annals of Mathematics

Solutions of ExercisesSource: Annals of Mathematics, Vol. 2, No. 3 (Apr., 1886), pp. 63-71Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1967363 .

Accessed: 19/05/2014 05:50

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

Annals of Mathematics is collaborating with JSTOR to digitize, preserve and extend access to Annals ofMathematics.

http://www.jstor.org

This content downloaded from 193.104.110.105 on Mon, 19 May 2014 05:50:31 AMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=annals

http://www.jstor.org/stable/1967363?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


SOLUTIONS OF EXERCISES. 63

x2 log ,u Diff. x2 log IL Diff. X2 log ,u Diff. x 2 log u Diff.

o.o o.ooooo 1.0 0,01084 2.0 0.02I66 3.0 0.03245 io8 0. I 0.0009 I09 I-I O.OI I93 2. I 0 82274 3 I 0 03353 I07 0.2 0.002I7 Iio8 .2 O.0I30I

io8 2.2 0.02382 io8 32 -0346 io8

0.3 0.00326 I09 I.3 O.OI409 I08 2.3 0.02490 io8 3.3 0-o3568 io8 io8 io8 io8 0036 io 0.4 000434 9

I 4 0?015I7 Io 2.4 0.02598 io8 3

04 0-03676 I07 0.5 0.00543 I09 I.5 O.OI626 909 2.5 0 02706 3.5 0.03783 io8 io8 i.6 o.oI734 io8 2.6 0.028I4

i0 3.6 0.389 I o.6 ~oo5 I

io8 io8 io8 I07 0.7 0.00759 I .7 o0oI842 8 2.7 0.02922 3.7 0.03998 io8 o.8 o.oo868 109 i.8 O.OI950 io8 2.8 0.3029 IO7 3.8 0.04I06 I07 0.9 0.00976 I 18 I.9 0.02058 2.9 0.03I37 3-9 0.042I3 108 I.0 0.0184 2.0 0 .042 I 66 I 3.0 00.3245 4.0 0.0432 I

SOLUTIONS OF EXERCISES.

22

THE dome of the rotunda of the University of Virginia is spherical. The length of its meridian section is 85'.2 and the girth of its base is 2I4g. From these data it is required to compute the radius and the surface of the dome.

[William M. Thornton.]

SOLUTION.

We have for the chord of the meridian arc

214 = 68. I.

If therefore x be the half central angle of the arc,

sinx 68i x 852 ?799

Passing to logarithms and expressing x in minutes by putting

ZX io8oo'

we get y = log sin x - log x + 3.6336 = o.

By trial we find, if

x 64?, Y + 0.0030; X -6 , 6 '- 0.0002;



64 SOLUTIONS OF EXERCISES.

therefore x = 64o56' approximately and

R 68.1 376 2 sin x

For the height of the dome we have

H= R(i - CoSX)= 2I.6,

and for the surface

S 2= 7RH= 5 i0oo8.

[W. M. Thornton. 80

INVESTIGATE formulh for the logarithms of 13, I9, and 73 in terms of the logarithms of primes less than 13 and of (N + i)/N where N = 132495 262143; 274625. [F. H. Loud.]

SOLUTION.

Put x log I32496 ) -log 262144 z log 274626. I32495 262I43 274625,

then 3logI3=I9log2-+-lOg II-10g3-310g5-10g7- Y-y5.7y 10 g73 4 log2log 3 -log 5 + 2 10g 7 -2 log I I + 2 log I3x, 10g I9 3 l0g 5 + 3 10g I3 -log 2 -2 10g 3 -log I I -10g 73 + z.

[F. H. Loud.]

38

SHOW that among any three roots a, b, c of a biquadratic, as x4 + px2 + b - o, this relation holds:

I+ I , I I ~+ - a h c a + b + c'

SOLUTION.-

Since the terms containing x and x3 are wanting in the biquadratic given, we have

abc + abd + acd + bcd o, a + b + c + d o,

where d is the remaining root;

I I I I I * a d nb c a+[bA+BcB

[Ashzer B, Evans.]




48

IN the angle A a point P is given. Construct a triangle ABC whose base BC shall contain P, while the sum of the sides AB and AC equals a given length.

[W. M. Thornton.] SOLUTION.

On AB lay off AD = the given length. The problem requires to draw a line BC through P so as to make BD = AC. This can readily be effected by the tentative process known as the mnthode de fausse position, in the following manner: Through P draw any three lines cutting AD in B1, B2, B3; on AC lay off AC1 = B1D, AC2 = B2D, A C3= B3D. The two concentric pencils, P(BI, B2, B3) and P(C1, C2, C3), being in perspective position with the two ident- ical rows B1, B2, B3 and C1, C2, C3 respectively, are projective. Their two common rays, if real, may easily be constructed by means of an auxiliary circle, in the usual way. Each one of these common rays satisfies the conditions of the problem. The construction may, of course, be simplified by selecting particular lines for the three arbitrary rays; e. g. by taking PD and the two parallels to AP and AC for PB1, PB2, PB3 respectively. If, in addition, the arbitrary circle through P be made to pass through A and D, it will be easy to determine the conditions of the possibility of a real solution. [Alexander Ziwet.]

50

ABC is an equilateral triangle inscribed in a circle. From any point P within the angle BAC, straight lines PA, PB, PC are drawn. Show, geome/ri- ca/ty, that PB + PC - PA is a minimum when P is on the circumference of the circle about ABC. [R. D. Bohannan.]

SOLUTION.

Let AP cut the circumference in P'. Then [Eu. 6D]

P'B ? P'C= PA.

Draw BP, CP' and drop on them perpendiculars PX, PY1 Then, since the angles PP'X, PP' Y are each 60?, P'X and PI Y each equal half PP', whence

PTX-+ P'Y= I"P; BX+ CY=PA.

But BX, CY are less than BIP, CP except when P is on the circumference; hence BP + CP- AP is greater than zero except when P is on the circuinfer- ence. [R. D. Bohannan.]

51

A QUADRILATERAL, both circumscriptible and inscriptible, has a given peri- meter 4s and one given angle a. Show that its area is

2S2: (cosec a + cosec x),




where x is one of the angles adjacent to a. And find x when the area is greatest. [W M. Thornton.]

SOLUTION I.

Let r = radius of inscribed circle.

The sum of two opposite sides = r cots + cot -I+ tan- + tan = 2S;

Area =2rs 4s= 2s2:(coseca+ cosec x). cot - + tan-2 + cot- + tan-

2 2 2 2 7r

The area is greatest when cosec x has its least positive value; i. e. when x -.

[R. H. Graves.]

SOLUTION II.

Let 1, a, m, x, n, w, - a, p, 7r - x be in order the sides and angles of the quadrilateral. Since it is circumscriptible,

/ + n = in r P = 2S,

and we have for the area the equivalent expressions

A- lrn+np nn+/p l/n +mn+np+/p (/(I+n)Qn -Ip) 2 cosec a 2 cosec x 2 (cosec a + cosec x) 2 (cosec a + cosec x)

2S2 cosec a + cosec x

This expression is greatest when cosec x is least; i. e. when x = go'. [R. D. Bohannan.]

52

FIND the height to which the Washington Monument must be built so that a body placed on top of it would have no weight. [A. Hall.]

SOLUTION I.

Let x be the distance from the centre of the earth to the top of the monument,

R be the radius of the earth, M mass of the earth, T time of rotation of the earth,

latitude of the monument, mass of the moon,

a mean distance of the moon, t periodic time of the moon.




Since the attractive force of the earth must balance the vertical component of the centrifugal force on a body at the top of the monument,

M 472 2 -X= -

xrcos 2;

X3 M. 22 sec2 n

But d1=(+rni) 2 472

* 3 x M (T 2 ai -M+mn 111 sec 1c. Putting M= 81, T I, t= 27.39647, SD = 38?53'I8", a 238800m, we get

X - 30927m,

and the height required is 30927m - 3958n = 26969m. [A. Hall.] SOLUTION II.

With the same notation we have for the weight of unit mass at height x above the centre of the earth

21 2 x2 4Thxcos,

where W is its weight on a still earth at the surface. Hence at the surface J2 g = W - R cos2 (p,

and at the top of the monument

()3 7-2 W 2 UR

_ J ;T-see (D.

Eliminating W we get J = 3 1 + 2 . sec2.

Putting T 86408, g 32.I5f, R= 2o889ooof, c= 38O533I8 V;

we get x = 7.837R,

and the height required is 6.837R = 27050m. [Chas. Puryear.] [A solution nearly the same, but not quite accurate in method, was sent

by S. T. Moreland.-Ed.] 53

SHOW that when x - '3

(COS X + COS 5X) (COS 2x + cos 3z) (cos 4x + cos 6x) - .

[R. H. Graves.]




SOLUTION I.

Converting the sums of cosines into products, the equation to be proved is

COS 3X COS 2X COS 2Ax COS co 1

2~~~~~27 or denoting the product by X and putting (wo -x 26

BY COs (0 COs 2W( COs 4(0 COS 5(O cos 6(o cos I o (- -

Noticing that 3(0, 7(0, 8(w, 9(o, I i(w, and 12 (w are the supplements of the six angles which occur in X, we see that

X2 COS (0 COS 2(w COS 3(W . . . . COS 12(w;

and since 14(0, 15(0, . . 23(o have the same cosines as the angles which occur in X2 we have, since cos o =I and cos 13(0 - I,

X4 COS 0 COS (0 COS 2(0 . . . . COs 23(0.

I have given in the Analyst, Vol. VI, p. io5, the equation whose roots are the

sines of the X angles a., al,. . . -I included in the expression a, + k -27

In - o, I,. . . - I]. From this equation it appears that the product of these sines is, when n is even,

it

( 1)2 sin2 1 na.

Putting a = and v 26, this product becomes the value of -AX4 above; thus 2

-Xt=--4; whence X -+ 2242

or since XY contains but one negative factor, X A, which was to be proved. [The equation referred to above is that which expresses that cos na has the

same value for each of the angles in question, for when n is even the series for cos na in powers of sin a terminates with sin" a. When n is odd the correspond- ing equation is derived from the value of sin na. The article in the Analyst contains a discussion of the symmetrical functions of the sines of the angles in question.] [ Win. Woolsey Johnson.]

SOLUTION II.

In Burnside and Panton's Theory of Equations, p. 99, the example occurs: "Form the cubic whose roots are a + a8 +- a'2 +- a, a' ? a3 + a"l + a1" + -jfa

+ a' + a7, where a is an imaginary root of x'3 - I o." The required equation is found by the usual process to be

X3+-X2 4X+ I4-0. (I)




The given roots may be written ,,+ v-' +' + 5 -$a-5, a2 + a-2 + t3 -28 (3 4 + a-4 + (I -

{X{

or 2 (Cos 6 + Cos 56), 2 (COS 26 + COS 36), 2 (cos 46 + COS 61), 27W

where - I 3

therefore by (I)

8 (CoS 6 + -OS 50C ) (CoS 2J + COS 3A) (cos 4J + cos6J) 6 - I Other relations may evidently be deduced. [R. H. Graves.]

SOLUTION III.

Multiplying together the factors of the left hand member, reducing each product of cosines to sums of cosines, and substituting 13 - n for n (any co- efficient of x) whenever n is greater than 6, we have

n==6 =1+&! 1 O ~(e + - ' cos n 1 + 5 sin 3x os 3x (See Snowball's Trigonometry, p. 84)

sin 3x Cos 3x sin 6x -

sink Ix sin Y since r= 6yx. [Ormnond Stone.]

54 SHOW that if in a plane triangle ABC

COS A + COS B + CO$ C-=,/2,

the centre of the circumscribed circle lies on the circumference of the inscribed circle. [R. D. Bohannan.]

SOLUTION.

Let R = radius of circumscribed circle, and r = radius of inscribed circle. Then (Chauvenet's Tr-zgonoinetry, pp. 78 and 79)

r - cos A + Cos B +cos C -I. (I) R- D2 R2 2Rr,

D being the distance between the centres of the two circles. By substituting the value of R found from (i) in this last equation, since cosA + cos B + cos C - p72, there results after reduction D = r; hence the proposition is proved.

[P. Richardson.]

WHAT is the locus of the vertex of a variable angle whose sides intercept a fixed line and whose bisectrix passes through a given point in that line.

[L. G. Carpenter.]




SOLUTION 1.

Let a, na be the segments of the fixed line, x, y co-ordinates of the vertex of the moving angle.

Let the origin be taken at the point of division of the fixed line which is the- x-axis. The sides bear to each other the ratio I: n. Hence

(x- na)2 + y2 = n2 [(x + a)2 + y2];

. 2 + -2 -ax K2 Y ax.

The locus therefore is a circle described on the segment 6f the fixed line ob- tained by dividing it harmonically in the ratio I: n. [S. J. Cunningham.]

SOLUTION II.

The line joining the given point with its harmonic conjugate with respect to- the extremities of the fixed line always subtends a right angle at the vertex of the variable angle. Therefore the locus is a circle. [R. H. Graves.]

53

CONSTRUCT a triangle; given its circumscribed circle and the radii which bi-- sect its sides.

SOLUTION.

Let OD, OE, OF be the radii; produce DO to meet the circumference in D! and make FA equal to ED'; from A draw the chords AB, AC perpendicular to, OF, OE, and join BC; ABC will be the required triangle.

[R. D. Bohannan.] 58

DIVIDE an isosceles triangle by perpendiculars from a point within it to the sides, into three equivalent quadrilaterals.

SOLUTION.

Let ABC be the given triangle, AD its altitude, P the required point which must lie in AD, and PM the perpendicular on AC; then the triangles APM, A CD are by construction similar and their areas are by hypothesis as I to 3;

AD=A. APJ 3. Accordingly, to locate P we construct an equilateral triangle with AD as al-

titude; its base will be equal to AP. [Charles Puryear.] 59

SHOW that the height of a cone bears to the height H of an equal spherical, dome of diameter D on the same base the ratio

31)- 2H: 2D- 2H.



EXERCISES 71

SOLUTION I.

Let D denote the diameter of the sphere, H altitude of the dome. r radius of its base, /1 altitude of the cone;

then r2= DH -H2,

and the volume of the dome, being the difference of the volumes of the sector arid the cone whose vertex is at the centre of the sphere, is

jirHD2- irr1(D -2H)= 17rH2 (3D- 2H). The volume of the required cone is, again,

jnThr2 = jmff (D -H). The two volumes are equal by hypothesis;

h 3D- 2H H - 2D - 2H [Charles Puryear.]

SOLUTION II.

Let D1 be the diameter of the base of the dome; the condition gives

- DI2 -DI2H?+ H3; 12 ' 8 1 6

A 2H2

* H D D2

But DI2=4H(D H); . Hi - f 3D-2H HA?+ 2D 2H 2D -2H

[R D. Bohmnizan.]

EXERCISES.

70

A RIGHT ANGLE moves so that a given point in one side, distant c from the vertex, lies in a fixed axis, while the other side passes through a fixed point, distant c from this axis. Find the locus of the instantaneous centres or centrodes of the motion. [R. H. Graves.]



Date post:	08-Jan-2017
Category:	Documents
Upload:	ngoduong
View:	214 times
Download:	0 times

Solutions of Exercises

Documents