2 (n+8) 2 (n + 7) (n + 8) _ 2 (n + 6) (n + 7) (n + 8)

n+1

+

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

=

nas

n+2

nas

n+ 4

From the above equations the property in the question is obvious.

By decomposing the coefficient of xm, there results

The forms (a) do not appear to facilitate the method of obtaining the property in the question.

u = F(x) satisfies the differential equation

and u= F(-x) satisfies the differential equation

where B is the symbol of the first Eulerian integral.

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

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

16

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 1, 2, 3 for the segments of the SIMSON line that are included within the angles A, B, C of the triangle, respectively, and a2+b2+c2 = 20a ̧ prove that (a2 — 1,3)* = (b2 −1,3)* + (c2—133)* .

.(1),

.(2),

(3).

(σ2 — a2) l ̧2 + (σ3 — b2) 11⁄22 + (σ2 — c2) 132 = (28)2 ................................... (o2 — a2) a2. AP2+(o2 — b2) b2 . BP2 + (o2— c2) c2 . CP2 = a2b3c2 5600. (By CHRISTINE LADD.)—Required the envelop of the SIMSON line.

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 abe are in a straight line. For, joining ab, ac, circles may be made to circumscribe the quadrilaterals PacB, PaCb; therefore T- -Pac PBA = PCb = 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.

Then PA =

2. Let bc, ca=12, ab=13. cosec A, &c. Consequently, if p be the extremity of the diameter through P,

......

=

0, hence (a2-4,2)$+..

But pA sin A±pB sin B± pc sin C

Expanding, we have (a2-43)2+ ... − 2 (b2 — 11⁄22) (c2— 132) — ... = 0,

or

= 0.

where s = area of triangle ABC, and σ = area of triangle whose sides are 40. Therefore (o2-a2) l3 +

=

=

(2s)2. The third relation is found

3. Let Pa, Pb, Pe 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'a' 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 =αα = 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.

Ra

Now if p' be the perpendicular from N, the centre of the nine-points, circle upon abc, we have

since NDC

=

=

R sin (w + NDC) − p = R cos (w + B − C)—p,
+B-C. Thus we obtain

p' = +R{cos (w + B − C) − 2 sin (2w + B-C) sin w} = R cos (3w+B−C). The envelop is therefore a three-cusped hypocycloid concentric with the nine-point circle.

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

If O be the centre of the circle, and AOP = 20,

2R sin sin A = a sin 0,

2R sin (B-6) sin C = c sin (B-0),

2R sin (A + B-0) sin B

2R sin (C+0) sin B

b sin (C + 0);

a2 cos2 0,

B

F

b2-122 = b2 cos2 (C+0), c2— lz2 = c2 cos2 (B—0).

Again b cos (C + 0) + c cos (B-0)

therefore

=cose (b cos C+ c cos B) + sin e (c sin B-b sin C) = a cos 0, (b2—122)*+(c2 —13)1 = (a2—1,2)*.

Again,

2 [(o2—a2) 1,2 + (o2 — b2) 11⁄23 + (σ2 — c2) 133]

= 2bc cos A x a2 sin2 + 2ac cos B × 62 sin2 (C +0) +2ab cos C × c2 sin2 (B–0) 2abc {a sin2 e cos A +b sin2 (C+0) cos B+c sin2 (B-) cos C}

=

abcR {sin 2A (1—cos 20) + sin 2B [1 − cos 2 (C+0)]

=

=

+ sin 2C (1-cos 2 (B−0)]}

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

Again,

=

=

=

bc ac ab

=

2 (2S)2;

(o2-a2) a2. AP2 + (σ2 — b2) b2. BP2 + (σ2 — c2) c2. CP2

be cos A. a2. 4R2 sin2 + ca cos B. 62. 4R2 sin2 (C + 0)

+ab cos C. c2 4R2 sin2 (B-0)

4abcR2 {a cos A sin2 + b cos B sin2 (C+0) + c cos C sin2 (B—e)}

= 4 x

III. Solution by CHRISTINE LADD, ELIZABETH BLACKWOOD, and others. Let la+mB+ny = O be the equation to any transversal referred to the triangle aßy; 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 + mB+ny

(m cos A+ cos B) y + la + mß = 0 ;

and the condition that these lines meet in a point is

2λμv (1 + cos A cos B cos C)-μv sin2 A (v cos B + μ cos C)

-vλ sin2 B (λ cos C+ cos A)-λu sin C (u cos A+λ 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 2 N'NM = N'ND, then NM produced through N passes through a cusp. For the tangents to the