Solutions of Exercises

Annals of Mathematics

Solutions of ExercisesSource: Annals of Mathematics, Vol. 3, No. 3 (Jun., 1887), pp. 89-95Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1967368 .

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SOLUTIONS OF EXERCISES. 89

SOLUTIONS OF EXERCISES

32 A TRIANGLE PQR is inscribed in the triangle ABC. Determine the ratios in

which P, Q, R divide the sides BC, CA, AB in order that AQR, BRP, CPQ may be respectively 3, X, + of ABC. [W. M. Thornton.]

SOLUTION.

Put BP = ax, CQ by, AR = cz. Then the hypothesis gives

(I -Y) z-X

(I ZX X (I - X)y =}

Elimination gives 35X2 - 38x + 9 = o.

Whence we find x, and by substitution, y and z. The results are

I9 + j//(46) i6 + V(46) Z 26 + 1/(46) 35 ) Y 42 - 45

in which either value of v/(46) may be taken. [H. N. Drazughon.]

82

THE major axes of two similar and equal concentric ellipses intersect at right angles, and the area common to the two curves is half that of either ellipse. Find the eccentricity. [Ormond Stone.]

SOLUTION.

Let A CB, A' CB' be quadrants of the two ellipses, the curves intersecting in D. Join CD, and draw DMN parallel to AC, intersecting the circle drawn about C with radius CB _ CB' in M, and CB in N, and join CM. The circular sector CBM is formed from the elliptic sector CBD, by reducing the breadths parallel to CA in the ratio b / a. Hence the area of CBYJ is the same part of the whole circle that CBD is of the whole ellipse. But by hypothesis, CBD is E of the whole ellipse. Hence CBA/I is -1 of the whole circle, and the angle MCB is therefore equal to Aw. Or since by hypothesis DN= CAT,

AMN: DN= b: a; b/a =tan 1r;

e2 I - tan2 1= 2V2 - 2. [O. L. Mlathiot.]

107

GIVEN one vertex of a rectangle and the ratio of its sides, construct the rect-



90 SOLUTIONS OF EXERCISES.

angle so that the extremities of the diagonal opposite to the given vertex shall lie

I. On two given parallel straight lines, 2. On two given intersecting straight lines, 3. One on a given line and the other on a given circle, 4. On two given circles. [R. D. Bolannan.]

SOLUTION.

Let C be the given vertex; AB the required diagonal; and a, j the lines on which A, B are to lie. When A describes a, A', the extremity of CA' _ n. CA, will describe a figure a', similar and similarly placed to a and n times as large. B, the extremity of CB = CA', will describe an equal figure ji', located by turn- ing a' through a right angle about C. The points of intersection of j' with jB

furnish solutions of the problem. The application of the general method to case (4) of the problem is easy. In (4) A' is a circle cutting j in two points, each of which gives a solution. [ W M. Thornton.]

[Mr. Henry Heaton sends substantially the same solution. Mr. E. L. Stab- ler suggests the generalized form applicable to other curves.]

108 CONSTRUCT an equilateral triangle of given size inscribed in a given equilat-

eral hyperbola. [R. H. Graves.] SOLUTION.

Any rectangular hyperbola, passing through the vertices of a triangle, passes through its orthocentre. Also the nine-point circle is the locus of the centres of rectangular hyperbolas passing through the vertices of a triangle. Therefore the centre of the required triangle lies on the given hyperbola, and its inscribed circle passes through the centre of the curve. Hence the following construction-

Construct a diameter equal to two-thirds of the altitude of the triangle, with one end of the diameter as a centre and a radius equal to this diameter describe a circle; three of the points of intersection of hyperbola and circle are the vertices of the required triangle. (Omit point of intersection at end of diameter, for otherwise the diameter would bisect a chord perpendicularly, which is, in general, impossible.) If two-thirds of the altitude is less than the transverse axis, the construction is obviously impossible. [R. H Graves.]

110 SOLVE the triangle ABC; given a, A and the product dy of the bisectrices

drawn from B, C to CA, AB. SOLUTION.

With the customary notations we get from the triangles BCE, BCF

_ sin C Y sin B a sin (A + B)' a sin (A + IC)'



SOLUTIONS OF EXERCISES 91

and therefore Ar sin B sin C a2 sin (A + MOB) . sin (A + 1C)

Pr Cos2 1(B- C)-cos2' (B + C) 2a coS (B -C) + sinEA

Conversely from this relation 1 (B - C) can be computed, and as 1 (B -- C) is also known, B and C, and from them the other parts of the triangle, are deter- mined. [Henry Heaton; T. U. Taylor.]

112 GENERALIZATION.

To inscribe in the given triangle ABC a triangle PQR whose sides QR, RP, PQ make given angles with BC, CA, AB.

SOLUTION I.

The conditions of the problem give the directions of the sides of PQR. Draw a parallel to RP, intersecting AB in P' and CA in R'. Through P' and R' draw parallels to QR and PQ, intersecting at Q'. Through Q' draw a parallel to BC, intersecting CA at C'. Through C draw a parallel to C'P'. This parallel will in tersect AB at P. The remainder of the solution follows immediately.

[A. B. Evans; E. L. Stabler; Orinond Stone.] SOLUTION II.

Draw the triangle P'IQ'R' whose sides Q'R', R'P', P'Q' pass through A, B, C and are parallel to QR, RP, PQ, and another A'B 'C' whose sides B 'C', C'AI', A 'B ' pass through P', Q ', R' and are parallel to BC, CA, AB. Divide the sides of ABC at P, Q, R in the same ratios in which the corresponding sides of A'B'C' are divided by P', Q', R'. PQR is the triangle sought. [Sarah Szold.]

113 THE transversal MN meets the sides AB, A C of the fixed triangle ABC, and

makes AM31. A N BM . CN.

Find its envelope. [O. Root, Jr.] SOLUTION I.

According to the given condition BM= ANJ the point M divides the

segment AB in the same ratio in which N divides the segment CA; in other words, the variable points AI and N form similar projective ranges on the fixed lines AB, CA. Hence, the variable line MIN which connects corresponding points of these ranges envelopes a parabola. (See, for instance, L. Cremona's Projecyive Geometry, translated by Leudesdorf, Oxford, i885, P. I28.) The lines AB and AC are tangent to this parabola in B and C respectively, these being the




points corresponding to the point of intersection A of the two ranges. [Alexander Ziwet.]

SOLUTION II.

Referring all parts of the figure to AB, AC as co-ordinate axes, and putting

AM=rn, AA=n, AB=b, AC=c,

we have for the equation to MN

- +I - = I, m n

and the parameters in, n are connected by the relation,

m n _ +-- I.

Differentiation and elimination give

,tn2 n2 Ix Jx cy Xb+ c

whence for the equation to the enveloping parabola we have

nab + BY I. [W. M. Thornton.]

114

A CIRCLE intersects a conic in four points, P., P2, P3, P4. Show that if the x-axis be parallel to the axis of the conic the area of their quadrilateral is

(X2 - X4) (Y1 - Y3) [R. H. Graves.] SOLUTION.

Let the x-axis be parallel to, or coincident with, the axis of the conic, and let the co-ordinates of the vertices taken in order be (x,,yl), (x2,y2), (x3,y3), (X4,Y4).

Let d be the acute angle that each diagonal makes with the x-axis.* Then the area of the quadrilateral =- product of its diagonals into sin 20

_ + 1 X2-4 Yl 8-!3Sitl 20 =+(r2 - X4) (Yj - 3). - 2 cosd0 sinO - 0 4(1y)

The same formula may be proved otherwise for the parabola as follows: Let y2 = 4ax be the equation to the parabola. Then

o I 0 I 0 I 0 I

I 0 I 0 I I 0 I 0 Area C- =

Y1 Y2 Y3 Y4 8a Y2 Y3 Y4

X] X2 X3 X4 Y2 Y2 Y2 Y2

* Salmon's Conic Sections, 244.-[ W. M. 7.]



SOLUTIONS OF EXERCISES. 93

0 0 0 I

I 0 0 I 0

8 A -Y3 y2

Y4 Y4= + 8a 'Y3) (A2' 4|

A2 _2 A2 _42 A2 Y2 y1 -y3 y2 -y42 y32 j2 2 I

Ya 4 +Y + Y Y)(YI ThY3)( 2 |Y4)

I I I - -(Y1- A)(Y2 -y4) 4a 0A y2Y4

- 4a (Y2 Y3) (y2- 2) - (X2 - X4) (YI -y3) [R. H. Graves.] 117

SHOW that the angle between the tangent to the ellipse b2X2 + a2y2 = a2b2 and the tangent at the corresponding point of the principal circle is given by the formula,

tanJ= a tan 1~ (I - a) + tan2

2 '

where a = (a - b) / a is the ellipticity, and (n is the eccentric anomaly. SOLUTION.

Let (x', y') be the co-ordinates of the point of tangency with the principal circle; (x',y") those of the corresponding point of the ellipse. The equations of the tangents to the circle and the ellipse are

x' a2 y X ? -

y y

P~rf b2 and Y 2 ,X +- .

Hence the tangents p, q of the angles which the tangents to the circle and the ellipse make with the axis of x, are

Xf b2-r f

,f q = ff = - y79 f vyt

But x' a cos 1t, y' _ a sin (p, y" b sin (p; _ _ b

tan 5 and q atan (p

Then tan a_ qp (I a tan

1n0 I +Pq (I-[a) +i tan 2.

[ William Hoover; 6C. D. Schmitt.]




118 SHOW that the eccentric anomaly which gives the maximum deviation is

tan- 1 /(I - a), and find the co-ordinates of the point of contact T and the equation to the tangent t.

SOLUTION.

The deviation will be a maximum when tan a is a maximum. Equating the first differential coefficient to zero,

a [(I - a) + tan2 (p] sec2 p - a tan Sc . 2 tan SD sec2 (p = 0;

whence o tan-' /(i - a);

sin c? '|f 2aJcos (P [ (2 a)'

and a

y - a (I a)@ v'(2-a) 1' -a

The equation of the tangent t is therefore

a3 (I- a)j ab 22P

V/(2 -a) 1+/(2 - a) [William Hoover.]

119 SHOW that the portion of t intercepted between the axes equals in length

the sum of the semi-axes a, b, and is divided at Tinto these two parts.

SOLUTION.

Let x = o andy = o in order in the equation of the tangent. The correspond-

ing intercepts are = P

1/ (2 -a) X a I (2- (). Then I V(XJ2 + y12) a ' (I - a)i x1a/2a) hn=j/r2y2 _ a + b. If z be the segment adjacent to the axis of x, z: y" I yl, whence z b; and the other is a. [ William Hoover.]

1Q0

SHOW that the circumcircle of try passes through the centre of curvature at T, and that the area of the circle of curvature equals that of the ellipse.

SOLUTION.

If p = the radius of curvature, we have for the ellipse,

- (a2sin2 5p + b2 COS2ab c) V(

by using sin sc, cos (p given in solution of I I8. The radius of the circumcircle of try X (a + b), and the distance of its centre from T 2 1 (a - b). .a. since p, i (a + b), fi (a - b) are the sides of a right triangle, the circumcircle passes through the centre of curvature X We have wp2 = rab. [William Hoover.]



EXERCISES. 95

121 SHOW that when the ellipse varies subject to the condition, a + b constant,

the locus of T is the hypocycloid, x! + yi (a + b)3,

which touches each ellipse at the point T. SOLUTION.

Y2 2 The equation ofthe ellipse is -2 + I, (I)

and we have a + b = c = a constant. (2)

Differentiating (I) and (2), a and b only variable,

X: da ay- d (3)

da db; (4) X2 ay12

* a2 h3 * (5)

(I) is (a + b) I, or b = ciyl, and a cAxs.

These in (2) give xi + yl = (a + b)?, a hypocycloid. From this equation dy Y which for x=x' - 1 _

a Y ,Y

a (I -a) b dx Xi' P (2 -aQ)

" ' =U V(2--a-~)) eoe

-j/(I -a), showing that the ellipse and hypocycloid have a common tangent at T. [William Hoover.]

EXERCISES.

137 INTEGRATE the differential,

sin d cosd

sin4 d + cos4 HIM' [0. Root, Jr.]

138 ABOUT the vertices of an equilateral triangle three spheres are drawn with

radii equal to the side of the triangle. Find the volume common to them all. [U 2V1. Tzornton.]



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