Annals of Mathematics

Solution of a ProblemAuthor(s): Marcus BakerSource: The Analyst, Vol. 10, No. 5 (Sep., 1883), pp. 141-142Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/2635789 .

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?141?

where l is the magnitude of the error which includes the whole number n.

. ? __ ta"'^ = ^^1?) 71

^^W) 2tan"1V2

'

where z -? n is the probability that an error shall not exceed x. It being proved at page 189, Vol. VII, that the constant

Hence

x

"Gwj)^**'2-

= ^ -*??<- *4

And for 3-s-n = J, the probable error is

iV^- A line or surface is in its most probable position when the sum of the

squares of the normals upon it, from the given positions, is a minimum, which normals are the errors.

In problem 239 (Analyst, Vol. VI, p. 49), for A, z-r-n = -fifc, and his

angular error x = tan"1 365QQ = tan""1 t|-q. Hence

A's prob. er. = ;?(V) = tan (fftan"! fr/2) y.tan-*-^ = j^ v B's prob. er. \S(d'12) tan (^ftan"1 Jj/2) tan"1^

SOL UTION OF A PROBLEM.

BY MARCTJS BAKER, DIRECT?R OF TJ. S. MAGNETIC OBS., LOS ANGELES, CAL.

Problem:? In a plane triangle I AB C there is given a+b9 c and m, a perpendicular to BC drawn from B to AC, to solve the triangle.

Solution. From triangle ACN\ we have

b2 = (a + y)2 + c* - f = a2 + 2ay + c2, whence 2a2/ = 62?a2?c2 == (a+6)2?-2a(a + b) ? c2,

or 2/ = _ (a+b)2?c2

2a (a + b) = k

where 2k = (a+6)2?c2 and n = a+b.

a)

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-142?

From the similar triangles CMB and CAN

a _ a+y , or a2c2?aHf = a3m2+2am2;*/+m22/2, m -j/(c2?y2) whence (m2?c2)a2+(m2+a2)y2+2am2y == 0. (2)

Substituting the value of y from (1) we have

(m2?c2)a2+(m2+a2)(-?nY +2am2(-?n\ = 0,

or (TO2_ca)a* + (OTa+a2) (k2?2akn+a2n2) + 2a2m2(k?an) = 0, which by expansion and reduction becomes

(m2+n2?c2)ai?2n(k+m2)a*+[m2(n2+2k)^^ = 0, from which to find a.

Note.? This problem has been known in schools under the following form. A tree of known height n, standing on a side hill, was broken over

by the wind, and while still clinging to the stump its top touched the ground at a distnce c from the foot of the stump. The perpendicular distance from the foot of the tree to the broken over part was measured and found to be m ; required the height of the stump.

Demonstration of the Theorem of Apollonius and its Eecipro- CAL by W. E. Heal.?1. Let ODD'O' be a quadrilateral; A, P' points taken in opposite sides. The diagonals of S^^^SB^S^^^MBIS_^1 the quadrilaterals ODD'O', ODP'A, 0'~ D'P'A intersect in p'ts that lie in a strai't j

line, CRO.

2. In any quadrilateral AB'RS let a p't j B be joined to two opposite vertices, A, R. \ The lines DD', OC, O'C joining the intersection of opposite sides of the

original quadrilateral and of the two derived quadr'ls meet in a point, P'. To prove 1, produce D'D, O'O to meet in P. The lines ABCD, PDP'D'

cutting the pencil AOD give [__UE?GD]* = [PD'P'D~]. The lines PDP'D'

cutting the pencil AO'D' give [PD'P'D~] = [AD'C'B'~]; . ?. \_ABCD~] =

[AD'C'B']. The points A, B, C, D; A, B', C, D' having the same anhar. ratio and one point, A, common, the lines BD', CC', DB' joining the other corr. points meet in a point R; . ?. C, R, C are in the same straight line.

To prove 2, join R, A. The lines ABCD, AB'C'D cutting the pencil ARD give \ABCD~\=IAD'C'B!\ The pencils O'OD, OO'D' having the sameanh. ratio and one ray OO' common, the intersections D,P,D' of the other corr. rays lie in a straight line. That is DD', OC, O'C meet in a point,

*The notation [_1_5C__)] denotes the anharmonic ratio of A, B, G, D.

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