21.

"ABCD is a quadrilateral; O, the intersection of the diagonals; P, 2, points in BD, AC, such that 2A=0 C and P B = OD. Prove that the centre of gravity of the quadrilateral coincides with that of the triangle O P Q:"

SOLUTION BY GEO. M. DAY, LOCKPORT, N. Y.

MS= MA and draw S S" parallel to A C; join M 2, ther S, S', S" are the centres of gravity of A B D, OP 2, DBC respectively. We have A B D: B DC::AO: O C, and S' S" : S S':: A 0:0 C. .. ABD:B DC::S' S": S S', hence S' is the centre of gravity of the quadrilateral A B C D.

22.

"Show that the distance from a vertex of any plane triangle to the point where the opposite escribed circle touches the sides meeting at the vertex is constant and equal to half the sum of the sides of the triangle."

SOLUTION BY E. B. SEITZ, GREENVILLE, O.

=

Let A B C be any triangle, D, E the points at which the escribed circle opposite A touches A B, A C produced. Now BD + CE BC; .. AD+AE= AB+ AC+B C. But ADA E;

· · . A D = 1⁄2 (AB+ AC+B C).

23. "If the brightness of the moon be equal to the brightness of the clouds by day, show that the light of an overcast day is to that of a full moon-lit night as 8 (360)2: ; the diameter of the moon being 30'."

Let

SOLUTION BY PROF. W. C. ESTY, AMHERST, MASS.

the radius of the cloud surface. Then 2 r2 = the visible cloud area,

or the portion of the cloud surface covered by its disk.

24. "There are m labels, to be distributed by lot among m different articles. Required the probable amount of coincidence in two independent allotments."

SOLUTION BY PROF. PLINY EARLE CHASE, HAVERFORD COLLEGE, PA. Suppose the articles to be similarly arranged in the two allotments. Then, among the possible permutations of the labels, there is only one in which they will all coincide, and none in which m -- 1 will coincide. If all but two coincide, those two must change places. There are, as many such arrangements as we can make selections of 2 out of m (m

m, or

--

2!

If all but three coincide, those three must change places in such a way that neither will occupy its original place. This can be done in two ways with each group of 3; there are, ..., twice as many such arrangements as the number of selections that we can make of 3 out of m (m m, or 2 x

1) (m 3 !

2)

Tabulating and differencing those results, it will be seen that the number of arrangements in which there are n displacements out of m, is

The value of this expression can be easily found, for

There will... be m displacements, or 0 coincidences, in

arrangements, leaving for the probability of one or more coincidences, .63212 m! out of m! possible arrangements,-which is a probability of about.

There will also be m-1 displacements, or 1 coincidence, in .367SS m! arrangements. Deducting this amount from .63212 m ! there remains only a chance of 24 for more than one coincidence. There would ... probably be one accidental coincidence, and no more.

100000

SOLUTION BY PROF. W. C. ESTY, AMHERST, MASS.

There are m ways in which the labels may be arranged. If we take any particular set of labels, r in number, and designate by A, the number of arrangements in which this set of labels all fail to coincide and then multiply this number A, by the number of sets consisting of r labels each that may be formed out of m labels, we shall have the entire number of arrangements in which there are r labels that fail of coincidence. This product

It is evident that the same product gives the number of cases in which there are m-r coincidences. Putting, in the above formula, for r every number in succession from r = 0, up to r = m inclusive and taking the sum of the results we have all the arrangements possible m in number i. e.

This is the equation given on page 336 of Todhunter's History of the Theory of Probabilities where a general formula for computing A, is given, which is not needed for our present problem.

The probability that r labels will fail and that mr labels will not fail to coincide may be designated by P, which can be found by the equation

*The student will understand that the notation m! means the same as | m, and that each represents 1. 2. 3. ... .m.—ED.

Let E,

=

the mathematical expectation of coincidence, or the probability of an arrangement containing any assigned number of coincidences as m―r, multiplied by the number of coincidences. Then will

Now the total expectation of coincidence will be a summation of all the terms E, from r = 0 to r = m 1 inclusive. Call this total expectation, or probable amount of coincidence, C. Then

=

m-1

C = 1,. E‚ = 1,.' (m − r) P‚ = TM,1 (m − r) A‚‡

m--1

E,

m-1

Σ

r

m

PROBLEMS.

30. BY OTIS SHEPARD, GOODLAND, IND.—Prove that

√ a + √ b a + b + 2v a b + 2 √ a b + 2 √ a b3

31. BY PROF. D. KIRKWOOD, BLOOMINGTON, IND.-Solve the equation

express the value of x in a finite number of terms.

32. BY PROF. D. TROWBRIDGE, WATERBURGH, N. Y.-There are two spheres of equal size and of exactly the same appearance, one solid

In order that we may obtain the probable number of coincidences in each independent allotment the C, must be divided by the number of permutations, |m; .'.

silver and galvanized with gold, the other hollow and made of gold. Required some means by which one may be determined from the other.

33. BY PROF. W. W. JOHNSON, ST. JOHN'S COLLEGE, ANNAPOLIS, MD.-A circle is referred to rectangular axes passing through its centre. A tangent and ordinate are drawn from any point of the circumference. A distance n times the abscissa of this point is measured upon the axis of x, and from this point a perpendicular falls upon the tangent. Required the equation of the locus of the intersection of the perpendicular and tangent.

SOLUTION OF A QUESTION IN MECHANICS.

BY DR. J. B. HOLCOMB, NEWPORT, N. Y.

AB and C B are two equal lines perpendicular to each other, and DE equal to either is inserted between them. Two H equal and perfectly elastic balls fall from b A and C towards a force at B, which force E varies as the distance from B. At D and E the balls are deflected by planes c d and a b inclined at an angle of 45° to A B and C B.

What will become of the balls?

Will they ever meet, and if so where?

After such meeting, if it should happen, what will become of them then?

There may be two cases:

SOLUTION.

1. When the balls are deflected at E and D at the same instant. 2. When they begin their motion at A and C at the same instant. For the 1st describe on E B and BD as semi-axes the quadrant of an ellipse DFE. Bisect the area DFE B by the line B F, then the two balls will meet in the elliptic arc at the point F.

For the line DE which joins the extremities of the two axes of the ellipse, being equal to either of the lines A B and C B is a measure of the force which either of the balls would have acquired in falling from