Annals of Mathematics

Solutions of ExercisesSource: Annals of Mathematics, Vol. 6, No. 3 (Oct., 1891), pp. 69-75Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1967289 .

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SOLUTIONS OF EXERCISES.

ACKNOWLEDGMENTS.

W. W. Beman 312; W. H. Echols 316; A. Hall 316; J. E. Hendricks 319; William Hoover 315; Jas. M. Ingalls 162; W. W. Johnson 283, 313; F. H. Loud 310; Artemas Martin 311; E. H. Moore 310; H. B. Newson 310, 313, 316; Ormond Stone 283, 288, 296; T. U. Taylor 318; M. E. Rice 312, 318; W. B. Richards 290, 317.

283 Let four circles, A, B, B, )D, pass through a common point, and three

straight lines through the same point meet them again in the points A,, B1, C1, DI1; A2, B2, C2, D2; A3, B3, C3, 1D3.; then, distinguishing lines in opposite directions by opposite signs, will

Al-B, AIC, AID, A2B2 A2C2 A2D 1 O'

A3B3 AC3 A3D3 [. [K JT oltn8ou.]

SOLUTION.

The equations of the circles may be written x2 + y2 + axc - ay 0, x2 + y2 +-b + b'y 0, etc.;

and those of the lines, Y MIX, Y = 112X,, Y M3x,

whence

A-B , - [a - + m21 (a' b')], etc.

Putting a - _b , a' b' j3', a -c = , etc., and omitting the factors i//l1/-l-in2, l/l 1 in92, 1//'1 Mn32, the given determinant becomes

+ +nz' r + " mtr J + m18'

i3 + rn2/ 0r 3110 + M 0.

fi3 msl r +113/ o 2 O'3

[Orrnond Stone.] 286

The squares of the sides of a plane triangle added to the squares of the radii of the four circles respectively touching these sides is equal to four times the square of the diameter of the circumscribing circle. [ Yale.]

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70 SOLUTIONS.

SOLUTION.

Let r _ radius of inscribed circle, ra,, r, -radii of escribed circles,

a) c - the sides of the triangle, R radius of circumscribed circle, / _ the area of the triangle.

Then ,2 + 24 2, + 742_ (s a) (s -b) (s c) ?8 (s -6) (s c)

aibic 8 ~~~~~8 - a

s_(_-_a) (_ - c) s(8 -a) (s b)

8- b 8 C Adding a2 + b2 + c2 to each side, and substituting for 8 its value i (a + b + c), we may readily obtain for the required sum

abEc2 _ 161a2 [T. U. Taylor.]

287 Factor the expression

abc (x3 + y3 + z3) + (a~b + bMc + c2a) (X2y + y2z + ZIX)

+ (a2c + c26 + b6a) (X2z + z2y + y2x) + (a3 + 63 + c3) xyz + 3abexyz. [Yale.]

SOLUTION.

Expanding we get 27 terms, 9 of which can be factored by xc, 9 by yb, and 9 by zb. Taking the first series we see that 3 can be factored by bx, 3 by cy, and 3 by az. The first series then becomes

cx [bx (ax + by + cz) + cy (ax + by + cz) + (z (ax + by + cz)1, or cx (ax + by + cz) (bx + cy + az). The whole becomes

(ax + by J cz) (bx cy + az) (cx + ay + bz). [ T. U. Taylor.]

290 Let A, B, C be any three points in space; find the locus of a point whose

distances to these three points are in the ratio of three given lines. [Yale.]

SOLUTION.

Let A be (0, 0, 0); B, (a, 0, 0); C, (6, c, 0). Then the condition gives x + y2 + Z2 (x -a)2 + y2 + ,z2 (X b)2 + (y C)2 + z2

- 2 ?f- 2 - 2

2ax-a-2 2bx + 2cy -b2 c2 whence 2--in_ - _ 2

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SOLUTIONS. 71

which is of the form Ax + By + C 0, and represents a plane perpendicular to the plane of ABC!. [ If B. Richards.]

296 (1) A complete plane quadrangle ABCD has six sides AB, AC, . .

which intersect an arbitrary line 8 in six points PAB, PAC, Show (a) that a quadrangle may be constructed which shall in this way correspond to any five collinear points to which such a notation has been at random but perma- nently affixed, and (b) that all such quadrangles lead to the same sixth point on the line.

(2) A complete quadrilateral k1k2k3k, has six vertices k1k2, kjk3, . . . and three diagonals 712.3 - (klk2) (k3k), . A transversal 8 cuts the sides in the points -.K, K2, K3, K4, and the diagonals in the, say, diagonal points D12.3, D13.42, D14-2 Show (by a double application of (1, b) in each case) that these seven collinear points with this notation are determined

(c) by the four points X and any one of the diagonal points D, (d) by any three of the four points X and any two of the diagonal points

D; i. e. (ca) a quadrilateral may be constructed which shall in the way defined correspond to any five collinear points to which the notation (c) has been at random (but, during the consideration of case (c), permanently) affixed, and (cb) all such quadrilaterals lead to the same two remaining points; and simi- larly for case (d). [ Yale.]

SOLUTION.

(1) Let PCD be the sixth point. Through any point A draw the trans- versals AB, AC(, AD intersecting s in PUB, PAC, PAD. Let B be any point in A B. Through B draw BC, BD intersecting s in Pc, PBD. C will be the intersection of AC with BC; similarly, D will be the intersection of AD with BD.

Let Ebe the intersection of BC and AD. Let (BCEPBc), etc. represent the anharmonic ratio of the points designated by the letters within the parentheses. From transversals through BC and s, taking A as the centre of the system, we have (BCEPBC) (PABPACPADPBC), a constant. Taking D as the centre (BCEPBC)- (PBDPCDPADPBc), also necessarily a constant; whence PGD is fixed.

(2) Call the three diagonals d1, d2, d/3, and their intersections with 8 D1, D,9 D3. Through any point draw kj, k2, d, intersecting s in the fixed points

K , K2, D1, Through the fixed point A3 draw at random k3. (c) Through K, and k3d1 draw k4. Through k1k3 and k~k2 draw d2. Or, (d) through 112 and k1k3 draw d2. Through kod2 and k3d, draw Kit. In either case, through hAjk and k2k3 draw d3.

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72 SOLUTIONS.

From the quadrilateral thus formed, three quadrangles may be formed by omitting, in turn, each of the diagonals and its corresponding vertices. If any five of the points El, K2, 14, K+, D,, 1)2 be fixed, (1, b) applied to the corre- sponding quadrangle fixes the remaining point. If 1)3 be substituted for D, or Do and (1, b) again applied, the seventh point 1) is fixed.

[Ormond Stone.] 310

Solve for xl: x, : x3 the continued equality (2 + 1r-3) (, X) (X.3 XI) (X3 x )2 (XX 4 2) (x1 -2)2e

[E. If. Moore.] SOLUTION.

The geometrical construction of the equations, while yielding the required roots with very little labor of computation, indicates as well the purely alge- braic process by which they may be obtained.

Using an equilateral triangle as the triangle of reference, I designate its centre as the point 0, and I further name as A, B, and C the vertices of a new triangle formed by drawing through the vertices of the triangle of refer- ence lines parallel to its sides; in other words, these points are the mutual intersections of the lines x2 +- .1, x3 + X1, and , + X,

The locus of the equation ('xr2 + X3) (X2 .j -3)2 ( -I1)(x3 (1)

apparently a cubic, is shown to break up into lower curves by expanding each member of the equation and transposing one of them, when we have

(x3 x3) x3 (X?2 X12) - X3 (2 X1)- 0,

or (xl d) (X12 + X?-j x22 - x1x3 X2X3 X:32) - 0

and, as the factor in the last parenthesis may be written (X1 + x2 + X3) (X1 + x2 X3) (X2T3 + X3X1 - X1YI2),

it is apparent that the locus consists of the right line 2' x2 and a circle. But in its original form, equation (1) shows that the locus has one of its

double points at the intersection of X2 x3 with 3 - x, that is, at the point O ; likewise that X3 + x, , or A C, is a tangent at the point where it meets X'2 -3

i. e. at A; and similarly that BC is a tangent at B. From these data the locus is readily constructed, and found to consist of the line OC and the circle AOB. In like manner, the locus of the equation

(X3 + X1) (X3 - X)2 (XI 1 am) (x, x9)2 (2) is the line OA and the circle BOC.

The nine intersections of these loci are obviously made up as follows: that of OC with OA, viz. 0; those of OC with BOC, viz. 0 and C; those of OA with A OB, viz. 0 and A; and those of A OB with BOC, which are 0 and B; together with the imaginary circular points I and PJ. Hence, as roots of

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SOLUTIONS. 73

the two equations, we have, counting as four equal roots the co-ordinates of 0, viz. 1: 1: 1; and, each counting singly, those of A, B, and C, viz. -1 :1 :1, 1: 1: 1., and 1: 1:- 1; and finally those of I and J, or 1:j(-1 -/3) : i (-1 + i 3)and 1: (- 1 +/ -3): i (-1 -1__3)-

The number of distinct points of intersection is, of course, not increased by adding for symmetry the locus of the third equation,

(2?1~~ +Xt)(tl 2) t2 + X3) ('X2 _3 * (3 (XI +X2) (X 1-)-(X_ -x)(- )2_ (3) [F H. HLoud.]

311 To find four biquadrate numbers whose sum is a square number.

[A rtemas Martin.] SOLUTION.

Let x, (x - ay), (x - by), and (x -- cy) represent the numbers required. Then we must have

x 4 + (X -ay)4 + (X- by)4 + (X cy)4 0 Expanding and arranging the terms according to the powers of , we get

4X4 x 4 (a + b + c) t3y + 6 (a2 + b2 + c2) x2y2 - 4 (a3 + b3 + c3) Xy3 + (a4+ b4 + c4) y4 L

Putting this expression - {2x2 (a + b + c) Xy + i L6(a2 + b2 + c2) -(a + b + C)2] y2}2,

expanding, cancelling, and reducing, we find x [6 (a2+ b2 + c2) -(a + b + C)2]2 16 (QZ4 + b4 + C4)

y 8{(a + b + c)[6 (a2 + b2 + C2) (a + 6 + c)2] 8 (a + b3 + [Artemas Martin.]

312 Given

U- X Sinf + Cos X, V sin X - X Cos X;

to compute rfx2d1tX J'x2dx b 2dx

. V2 ' J (au + bv)2 [C. fIermite.]

SOLUTION.

2 1 1 (1 2 _ sec%2 sec X --2 2

1 ~ ~ ~ scv 2 2 1tanx t tan x +_ Itnx + --I L xJ L JLa X

see x - cot 2X

- [tn 2 +

(X + cot X)2

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74 SOLUTIONS.

fz2dx 1 1 -xcos +sin x v

tan x + - x+cotx xsi-F-cosx i

x

(2) Similarly, -f2- x = U 11V

bx2 C bX2 dx C 2dx 1

(3) J (au + bv)2 JT 2 = au + bau L U I

[ JE'. W. Bemnan.] 315

A man runs a race starting with velocity v, the direction of the wind appear- ing to him to make an angle a with his course. It then appears uniformly to veer around through an angle a during the race. The wind blows uniformly throughout and at right angles to his course. Show that if t be the time of the

race, the length of the course is vt tan a log (2 cos a.).

[William 16oover.] SOLUTION.

Let x - the distance run at the end of any time t' from the beginning of the race, k - the constant velocity of the wind, and (o -- the change per unit of time of the direction of the wind. Then a + (Vt--= the apparent direction

of the wind at the time t', and dt = the man's velocity at the same instant.

sin [a + (Vt- --1 dx: (1) _ _ _

_L -_ _ _ _ d t

sin(a+(Vt) - When t' O. dx/dt' _v; .. -cota v/k,

whence k - v tan a, and (1) becomes dx/dt' v tan a. cot (a + t'). (2)

Multiplying by di' and integrating, we get x v tan " log sin (a + (Vt') + C. (3)

When x-O t' O; V tan log sin a, and (3) is

tan al sin (a + (Vt') = -(V log i(4) So i n a. When t' t,)' t' (t a, (o V a/t, and the length of the course

tan elI xi Vt alog (2 cos a.) .

[1 'illiam Jnloover.3

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SOLUTIONS. 75

316 The distance between the heels of a man's feet is 2b, and the length of

each foot is a. As the body sways, the vertical through the centre of gravity must always pass through the area contained by the feet. The toes should therefore be turned out at such an angle that the area contained by the feet is a maximum. Show (1) that a circle can be described about the feet with its centre on the straight line joining the toes, and (2) that the diameter of the circle is b + (2a2 + _2). [E. J. Routth.]

SOLUTION.

Let T - 6 be the angle between the feet. The area contained is 9 _ 2ab sin 6+ Jf12 sin 20.

The condition for maximum is dQ/dO = 2ccb cos 6 + a2 cos 2 -O.

or a see 6 2b + 2a cos 6. Therefore the diagonal of the trapezoid is perpendicular to the sloped side and (1) is true. The diameter of the circle is

(d = 24 + 2a cos 0. Eliminating 6,

d b + V,(2a2 + b2). [TV. 1. Echols.]

EXERCISES.

321 A CIRCLE meets a hypocycloid of class 3 at six finite points. Show that

the tangents to the hypocycloid at these six points touch a conic. [hrank Mlforey.]

322 *

THE arc of a limaqon is shown in works on the Calculus to be equivalent to the arc of a certain ellipse. Show that the double point on the limaqon corresponds with Fagnani's point on the ellipse. [ IF. B. Richards.]

323 ONE Of two casks contains a gallons of wine, and the other & gallons of

water; c gallons are taken from the first and poured into the second cask, and then c gallons are taken from the second and poured into the first. Required the quantity of wine in the second cask after n2 such operations.

LArtem.as fiartin.j * Exercise 317 corrected.

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