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2 (n + 4), (n+3) (n + 4) A. = 24 ( 2–

2

n+1 (n + 1) (n + 2) A6

204 02 _az?
205")
= 66

ав 06 06

2 (n +6), 2 (n + 5) (n +6)_(n+4) (n + 5) (n + 6)

2
= 16
n +1 (n + 1) (n + 2) (n + 1) (n + 2)(n+3)

az? Ag

ag ag as as
2 (n + 8), 2 (n + 7) (n + 8)_2 (n + 6) (n + 7) (n + 8)

+
(n+1) (n+2) (n + 1) (n + 2) (n+3)

(n +5) (n + 6) (n+7) (n +8)

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nao

n +3'

as (2

= dg

n +1

паѕ

+

(n +1)4+2)(n+3) (n +33)

n +4

nazm

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Similarly A2m = + according as (m) is odd or even.

n + m
From the above equations the property in the question is obvious.
By decomposing the coefficient of xm, there results
F(x)-1 1

22

24
Xe*

n+2' (n + 3) 2! (n + 4) 3!' (n + 5) 4!
na
n3

04
Be

+Rg. +B4. 2! 3!

4!

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+

+

n+1

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therefore

(a).

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+1, and F(-x) = 1 -**e* dx

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The forms (a) do not appear to facilitate the method of obtaining the property in the question. u = F (2) satisfies the differential equation đều

du +

+ doc?

dx and u'= F(-X) satisfies the differential equation

dru

du' +1

(1 – u) = 0. d.x2

dx

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5374. (By R. TUCKER, M.A.)-Prove that 1

B + 1, n) 22n-1 where B is the symbol of the first Eulerian integral.

Solution by J. HAMMOND, M.A.; the PROPOSER; and others.

(sin 6)2n-1 do = 1 | (sin )2n=do ;

X, da

therefore, putting on the right-hand side sina jo

sin 10 cos jo do,

dx xn-(1x)"of

2 (1-2)

22n-2 B(n,n).

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= 22n - 2

0

22n - 2

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This agrees with the result in the question, for
B (n+1, n) =

r(n+1) r(n) nr() r(n)

r (2n + 1) 2n (2n)

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II. Solution by R. TUCKER, M.A.

r de Let x = tan 10, then u = 支 1

log (1 + cos e).

cos

0

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(See WILLIAMSON’s Integral Calculus, Art. 113, 2nd ed.)

5599. (By the EDITOR.)—Calling the straight line that passes through the feet of the perpendiculars on the sides of a triangle from a point (P) on its circumscribed circle the Simson line of the point P, (from the name of the geometer who seems to have been the first to mention it), and putting 11, la, lz for the segments of the Simson line that arc included within the angles A, B, C of the triangle, respectively, and a + b2 + 02 = 20%, (a? — 1,9)* = (62 1,2)* + (c2—132)*

(1), (-a)/2 + (3-5) 12 + (09-c%) 12 = (28)

.(2), (02–22) a2. APP+(2—62) 62. BPP + (o? —co) cm. CP a25%

. (3). 5600. (By CHRISTINE LADD.)-Required the envelop of the Simson line.

prove that

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B

I. Solution by R. F. Davis, M.A. 1. If P be a point on the circle circumscribing the triangle ABC, and Pa, Pb, Pc be drawn perpendicular to the sides, then abc are in a straight line. For, joining ab, ac, circles may be made to circumscribe the quadrilaterals PacB, PaC6; therefore 1-Pac = PBA = PC6 = Pab.

This theorem is generally attributed to Dr. ROBERT SIMSON, the celebrated translator of Euclid's Elements.

The line abc may be termed the Simson (or pedal) line corresponding to P. 2. Let bc=h, ca=

= la, ab=lz. Then PA = ly cosec A, &c. Consequently, if p be the extremity of the diameter through P, PA2 = 4R?-12 cosec? A (a: -12) cosec? A ;

PA sin A (a® –12). But på sin A +pB sin B £ pc sin C 0, hence (a2—42) + ... = 0. Expanding, we have (a2-49+...-2 (02-1) (-1)-... = 0,

16sé – 4 (0%-aa) 42-... +160o = 0, where s = area of triangle ABC, and o = area of triangle whose sides are lilalz = 0. Therefore (os - a) 23 + : (2s). The third relation is found by putting ly = PA sin A, &c.

R

or

or

= aa =

3. Let Pa, Pb, Pc be produced to meet the circumference again in aßy. Then PaA = PBA = caa, and Aa is parallel to ac. Thus, Aa, BB, Cy are each parallel to the Simson line abc.

4. If Pa, P'á be parallel chords each perpendicular to BC, and P, P' extremities of a diameter, then also will a, a' be extremities of a diameter, and Aa, Aa' at right angles. The Simson lines, therefore, corresponding to the extremities of any diameter are at right angles to each other.

5. Let H, R be the orthocentres of the triangles ABC, PBC respectively, and AH meet abc in L. Then Pa.Ra = aB.aC = Pa. aa; hence Ra - AL. But the distance of the orthocentre of any triangle from an angular point is twice the distance of the centre of the circumscribing circle from the opposite side ; so that PR=AH. Therefore Pa-HL; PH is bisected by abc. If a parabola then, whose focus is at P, be inscribed in the triangle ABC, abc is its tangent at the vertex, and consequently the orthocentre H lies on the directrix.

6. Finally, to find the envelop of abc. Let abc be inclined to BC at an angle w. Then, if p be the perpendicular upon it from D, the middle point of BC,

Da sin w.
But Da = R sin (PCB - PBC)

R sin (2w +B-C); therefore

p = R sin (2w + B-C) sin w. Now if p' be the perpendicular from N, the centre of the nine-points, circle upon abc, we have

p' = }R sin (w +NDC)- p = {R cos (w+B-C)-P, since NDC = fm + B-C. Thus we obtain D = {R{cos (w+B-C) - 2 sin (2w+B-C) sin w} IR cos (3w +B-C).

The envelop is therefore a three-cusped hypocycloid concentric with the nine-point circle.

N

Þ

D

II. Solution by W. J. C. SHARP, B.A.; Prof. Evans, M.A.; and others.

If O be the centre of the circle, and ZAOP = 20, we have

AP 2 R sin o,
FE = 2R sin o sin A = a sin ,
ED 2R sin (B-0) sin C = c sin (B-1),
DF 2R sin (A + B-0) sin B

2R sin (C+0) sin B

b sin (C+0); as – 1,2 = acosa 0, 62– 1,2 = 2 cosa (C+0), c- lg = c2 cosa (B-0). Again bcos (C + 0) + c cos (B-0)

= cos 0 (6 cos C+ c cos B) + sin 0 (c sin B-b sin C) = a cos , herefore

(62—132)* +(02—13) * = (a’—1,2;! VOL. XXIX.

B

K

+

Again, 2 [(92-a2) 7,2 + (o2 62) 1,2 + (02—(2) 77*] = 2bc cos A x a2 sina 0 + 2ac cos B x 62 sina (C+0) + 2ab cos Cx co sin? (B-0)

2abc {a sin? cos A +b sin? (C+0) cos B+c sin? (B-0) cos C} abcR {sin 2A (1-сos 20) + sin 2B [1 - cos 2 (C+0)]

+ sin 20 (1-сos 2 (B-1)]} abcR {sin 2A + sin 2B + sin 2C-cos 20 (sin 2A + sin 2B cos 20

+ cos 2B sin 2C) + sin 20 (sin 2B sin 20-sin 2B sin 2C)} = abcR (sin 2A + sin 2B + sin 2C) : 4abcR sin A sin B sin C a2b2c2 28 28 28

= 2 (28)2 ; S bc

ab therefore (q? – a2) 12 + (02—62) 1,2 + (0% — C2) 12 = (28). Again, (q? – a2) a? . AP2 + (62) 62.. BP2 +(02-c%) c2 . CP bc cos A . a? . 4R2 sin? 0 + ca cos B . 62.4R2 sin? (C+0)

+ ab cos C. 4R? sin? (B-0) - 4abcR2 {a cos A sin? @ +6 cos B sin? (C+0)+c cos C sin? (B-C)} abcR3 {sin 2A (1 - cos 2 ) + sin 2B [1- cos 2 (C+0]

+ sin 2C [1 -cos 2 (B-C)]} abcR3 = 4 x

(sin 2A + sin 2B + sin 2C)

Х

Х

Х

ac

+

= 4 x

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m,

n

III. Solution by CHRISTINE LADD, ELIZABETH BLACKWOOD, and others. Let la + + ny = 0 be the equation to any transversal referred to the triangle aby; then the condition that it should be the Simson line is that the perpendiculars to the sides of the triangle at their points of intersection with the line meet in a point. The equations to those perpendiculars are

(n cos B+m cos C)a + + ny = 0,
(l cos C + n cos A) 8+ ng + la 0,

(m cos A +l cos B) 9+ la + = 0;
and the condition that these lines meet in a point is
n cos B +m cos C,
1,
I cos C+n cos A,

0, or 1,

m cos A +1 cos B 21uv (1 + cos A cos B cos C)- uv sina A (v cos B+ cos C) -v1 sinB (a cos C+v cos A)-au sinoC (u cos A +å cos B) = 0, which is the tangential equation to the required envelop, and designates a three-cusped hypocycloid produced by rolling a circle of radius }Ř in a circle of radius R concentric with the nine-point circle.

A cusp of the hypocycloid may be determined in the following manner :-If N' is the foot of the perpendicular from N on BC, D the middle point of BC, M a point on BC such that ZN'NM = ZN'ND, then NM produced through Ñ passes through a cusp. For the tangents to the

n

m,

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