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Solution by E. B. Seitz; A. MARTIN; and others. The roots of the equation are real, if r <

92.

4p When p <fa and q<(4ap)*, R may be taken anywhere within the circle, radius

92

and concentric with the given circle, radius a; and when pcfa and q>(4ap)", R may be taken anywhere within the given circle. When p>ā, from q=0 to q=a, R may be taken anywhere within the circle, radius & Hence the chance that the roots are real

4p
1
q”

παρ. , dq ( 2πη άρ
4p

(4ap) 1

9°

24 pdp. 2trqda

a

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rd

poldp = tá + iz loge 2.

5588. (By T. MITCHESON, B.A., L.C.P.)—If on each side of a triangle ABC triangles similar thereto are drawn, so that the angles adjacent to A are each equal to C, those adjacent to B each equal to A, and those adjacent to C each equal to B: prove that three circles drawn round the outer triangles, and the three lines that join their vertical angles with the opposite angles of ABC, all intersect at a point 0, such that the angles OBC, OCA, OAB are equal to each other; and that cot OBC = cot A + cot B + cot C.

5654. (By T. MITCHESON, B.A., L.C.P.)—If within a triangle a point o be taken such that LACO = BAO = CBO = 0, prove that

cot A +cot B+cot C.

cot 0 =

en

Solution by Professor MOREL; Professor COCHEZ; and others. 1. Les cercles circonscrits aux triangles c' ABC et ACB', se coupent au point 0, et l'on a

AOB + AC'B AOC + ABC 180°. On déduit facilement d'après les données du problème,

B0C + BAC : : 180o. Donc le cercle circonscrit au triangle BA'C passe par le point 0.

2. L'angle AOB' est égal à l'angle ACB' et par suite égal à l'angle B; de même l'angle B’OC est égal à l'angle C, et l'angle COA' est égal à l'angle A. Donc la ligne AOA est une ligne droite, puisque la somme des trois angles est égale à 180°. On verrait de même que BOB' est une ligne droite, et que COC est une ligne droite.

3. Les angles B’AC, ACB étant égaux, les droites AB' et BC sont parallèles, donc l'angle AB'O est égal à l'angle OBC; mais le quadrilatère inscrit ABʻCO nous montre que l'angle AB'O est aussi égal à l'angle ACO. Donc ACO OBC. On verrait de même que CBO BAO. 4. Les triangles OAB, OAC donnent csin (B-0) b sin

b sin B + c sin A cos B ОА

donc cot 0 = sin B sin A'

csin A sin B Remplaçons b par la valeur égale a cos C+c cos A, puis, dans l'equation ainsi obtenue, au numérateur et au dénominateur, les cotés a et c par les quantités proportionnelles sin A et sin C, il vient, après réduction,

cot a cot A +cot B + cot C.

5613. (By R. A. ROBERTS, M.A.)—Prove that the locus of the centroid of a triangle inscribed in a conic and circumscribed to a parabola is a straight line.

Solution by R. F. Davis, B.A.; J. O’REGAN; and others. Let a triangle be inscribed in a circle and circumscribed to a parabola. The centre of the circumscribing circle being fixed and the orthocentre lying on the directrix, the centroid of the triangle which divides the straight line joining these two points in the ratio 2 : 1 will consequently describe a straight line. Projecting orthogonally, the required property is obtained.

137. ON THE SIGN OF ANY TERM OF A DETERMINANT. By G. R. Dick, M.A.

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an, bno

In and suppose we require the sign of the term aa bs h.

Let the determinant be expanded in terms of the constituents of the first column and the corresponding minors, so that A = Eac · Ag, where A, is the minor formed by omitting the first column and the ath row. In the first place, the sign of a, is (-1)--1; to find now the sign of bs in the minor Aa, we remark that, if the suffix B > a, its sign will be changed, but if B <a, will be unaffected; hence its sign will always be the same as that of (-1):-? (2-B). Proceeding in this way, we see that the sign of cy, in its containing minor, will be that of (-1)x-1(a“) (B-y), and so on, till finally we get for the sign of aabe. la Sign of (-1).-1. (-1)2-1(a-B).(-1)-1(a)(-) ......

(-1)^-1 (a-^) (B-1) ... (x-1) = (-1)e+B+. -n (a-B) (a-) (a-1)

(B-) (B-^)

(k-1). But

a + B + 1+2+... in (n + 1); hence we get the following result :—The sign of any term aa, bp, ... lx of the given determinant is the same as the sign of the algebraical product,

(-1)*(n="(a-B) (a-)... (a-1) (B-v)... (6-1) ... (-1). It is perhaps worth noting that the coefficient (+1) of the term an.bs ... l, is the determinant formed by putting au, be, ... la each = 1, and all the remaining constituents each = 0.

... n =

5611. (By Prof. WOLSTENHOLME, M.A.)—Having given that
e (yoza +1) + (y2 + 2*) _ (*.x2 + 1) + (z2 + x2) e (x*y+1) + (x + y)
YZ

xy prove that k = (2-1, and that y2 + 2x + xy + (92) - 1 + (2x) -1 + (xy)-1.

= k,

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e =

...

Solution by D. EDWARDES ; J. McDoWELL, M.A.; and others

z2 xy
From the two first equations we have e = ; therefore

xyz2–1
2- 23 - YZ
22 - 2 + y (:-*) x + y +%

.. (A, B), xyz2_1 yzx2 – 1

xyz (2-2)

XYZ 22 - xy a

-2 2x+ yz-?. xyz2-1 yzx2 - 1 Again, taking the two first equations, we have, by (B) and (A), k

(ez* + 1) (y + 2) – (ez2+1) (exy-1) ez2 (y2 — «2) + y2 — 2?

z (y—~)

e-xyzo +e (xy 22) — 1 = e-xyz2 + e? (1 – xyz2)-1 = 42–1. [Professor WOLSTENHOLME remarks that the question possesses an interest for him as being the algebraical representation of the system e cos (B+8) + cos (B-~) = e cos (9+ a) + cos (y-a)

= e cos (a + b) + cos (a–B), the first system of poristic equations he happened to come across, which he deduced from the geometrical theorem that if an ellipse be inscribed in a triangle with a focus at the centre of the circumscribed circle, its major axis is equal to the radius (R) of the circumscribed circle. The expression of this in polar coordinates (with focus for pole) gives the system. It is very easy to make any number of such poristic systems ; for example, the three equations (a-y) (a-) (a ---) (a-2) (a – x) (a-y)

^(1 + 4a) {1+(y + 2)2}+{1+ (x + x)2}+ {1+ (x + y)2}} are poristic, and can only be satisfied if 1? 1, being then equivalent to

yz + 2x + xy - 2a (x+y+z) 1+a?, xyz + (1 + 3a2) (x + y +z) + 2a3 = 0, deducible from a triangle inscribed in a parabola, and circumscribed to a circle whose centre lies on the parabola.]

5609. (By A. W. PANTON, M.A.)—The four common tangents to two circles being supposed such that the two opposite pairs, external and internal, are at right angles to each other; show that their eight points of contact with the circles lie on two straight lines, every point on each of which subtends the circles in an harmonic system of tangents.

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For any two circles, the polars of A and A'intersect at a limiting point; but on account of the right angles at C, C' these polars are at once seen to be inclined at an angle of 45° to the line of centres; so the polars of A with respect to X and Y coincide with the polars of A' with respect to Y and X, thereby proving the first part of the question.

Let P be a point on one of these polars, AB its polar with respect to X (B being taken on the other polar of A). Then BĀ' is the polar of P with respect to Y (on account of the harmonic pencils at A, A'), and B is the pole of PA with respect to the same circle. Let 00' be the two intersections of opposite lines of the quadrilateral abcd which lie on the line BA. These are conjugate points with respect to X; and PO, POʻ are tangents to Y; for, AB are opposite vertices of a quadrilateral circumscribed to X at abcd, therefore 0, O are the common harmonic conjugates to aß and AB; and from the circle Y, the tangents from Pare harmonic at once with PL, PA and PA, PB, and accordingly coincide with PO, PO'; therefore the tangents from P to Y are conjugate lines to X.

It is worth remarking that an exactly similar proof applies to the corresponding theorem for orthogonal circles_(A, A being then centres of similitude), which is otherwise proved in Townsend's Modern Geometry, Vol. I., p. 287. Both of course are only particular cases of the co-variant F of two conics breaking up into straight lines, and the above method is general and proves geometrically for two conics so related that the tangents from any point on these lines are harmonic.

5557. (By W. H. H. Hudson, M.A.)-Find the shape of a uniform wire such that the moment of inertia of any portion of it bounded by two radii vectores about an axis through the pole perpendicular to its plane, may vary as the angle between them.

Solution by J. L. MACKENZIE, B.A.; G. S. CARR; and others. Since the moment of inertia increases in the same ratio as the angle 0, we have, by differentiating with respect to 0,

ds

= a constant as;
do
dr

do
do

dr (26 — gG!

23 and integrating, 0 = } sin-1

2

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+C.

a3

8

If we choose the prime radius so that C = 0, we have finally

g3 a3 sin 30. The form of the curve is given in the

do figure. If we give the double sign to the radical in the value of

dr' we get in addition the three dotted loops.

The curve 93 = að sin 30 is the inverse of the cubic a'y (3x2 - y2) = k , which consists of three simple hyperbolic branches, the asymptotes being evidently three lines through the origin inclined to each other at angles

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