these times, or their product if they are prime to each other, will give the time which will elapse between successive conjunctions of the three bodies.

This is essentially the solution given by several of the competitors.

PRIZE SOLUTION OF PROBLEM III.

By STOCKWELL BETTES, Boston, Mass.

The diameter of a circle inscribed in the quadrant of a second circle is equal to the side of the regular octagon circumscribed about the second circle.

Bisect the quadrant by the line AK, and draw the tangent CB at K. Next, bisect the angles CAK and KA B, and IL will be the side of the circumscribed octagon. The centre of the circle inscribed in the triangle CAB is the same as that inscribed in the quadrant, and it is at the intersection, H, of the lines bisecting the angles of the triangle. The triangles AKL and BKH are equal, and therefore KH KL. But KI-KL; therefore IL-2 HK.

SECOND SOLUTION.-LK and LE, tangents at K and E, are half sides of the octagon. Draw LN parallel to AB; then angle KHL=KAB= KLH, . KH=KL=LE=HM=HN, therefore H is the centre of the inscribed circle.-JOHN R. EMERY, Princeton College, New Jersey.

PRIZE SOLUTION OF PROBLEM IV.

By Cadet ARTHUR H. DUTTON, West Point, N. Y., and HIRAM L. GEAR, Marietta College, Ohio. Required the locus of the centres of the circles inscribed within all the right-angled triangles which can be inscribed in a given semicircle.

Let P be the centre of a circle inscribed in any right-angled triangle, AB C, which can be inscribed in the semicircle AEB. Draw the diameter EF perpendicular to A C, and join BF, PC,

and FC. Because B F bisects the angle AB C, it passes through the point P. But FP C-PBC+BCP ACF+PCA

P=

PCF. Therefore the triangle FP C is isosceles, and the point P is in the circumference of a circle of which F is the centre, and radius FC=O C√2.

SECOND SOLUTION. The centre of the inscribed circle is at the intersection of the lines bisecting the angles at the base, and as the sum of these angles is constant, the half sum is also constant ; and hence the vertical angle of this sec

ond triangle is constant, and since the base is constant, the locus of these points must be in a circle.-JOHN A. WINEBRENER, Princeton College, N. Y.

THIRD SOLUTION.-Let (xy) be the co-ordinates of the point P, O being the origin, and r the radius of circle AB C. But tan (PA C +PCA) tan 45°-1, and since tan PAC=" tan PC'A=

r+x2

y

we have by the usual formula, 22 + y2+2ry = r2; or changing the origin to F, x2+y2=22, therefore the locus of the point P is a circle whose centre is F.

All the analytical solutions are essentially the same.

PRIZE SOLUTION OF PROBLEM V.

By W. F. OSBORNE, Wesleyan University, Middletown, Ct.

From a box containing a very large number of white and black balls, of each an equal number, three balls are taken at random and placed in a bag without being seenA takes a ball at random from the bag, observes its color, and replaces it four times in succession. The ball was white on each of the four drawings. What are the respective probabilities that the bag contains 1, 2, or 3 white balls?

The balls as drawn from the box and placed in the bag will be, either no white, bbb; one white, wbb, bwb, bbw; two white, ww b,

wbw, bww; or three white, www. Hence, before drawing from the bag, the chances that the balls are one white, two white, three white are respectively,, .

16

Now, if there is only one white ball in the bag, the chances that it will be drawn four successive times are (); if two of the balls are white, the chances are (); if three of the balls are white the chances are ()*. Combining, we get (3) Combining, we get (1)=16, (3)* = 21%, (3), as the a priori probabilities that a white ball was drawn 27 = 2169 from the bag four times in succession. But since a white ball is drawn four successive times, the respective probabilities are

2169

2169

1. Notes on the Inclined Plane and the Wedge. Let A B C represent any inclined plane; W the weight, placed at 0; P the

is constant and may be denoted by Oa, the radius of a circle; and R the reaction in the line OD perpendicular to the plane. The condition of equilibrium will be represented by the three sides of a triangle. R must always act in the line OD, W must be perpendicular to the horizon, and P may vary in direction, and will give different values for R and W in different positions.

When Pacts in the line Oa,

power, which

=

I must be in the same line, and therefore P W and R = 0. When P acts in the line Og, R must act in the same line, and therefore PR and W= 0. All the positions including the limits are then represented as follows:

In general W increases from an equality with P to its maximum value when P acts parallel to the plane, and afterwards decreases again to its minimum value, or zero, when P acts perpendicular to the plane. R increases from its minimum value, or zero, to its maximum, when P acts parallel to the base of the plane, then decreases again to equality with P, when P acts perpendicularly to the plane. When Pacts in directions beyond the limits Oa and Og, negative results for W and R are obtained, of an interesting character.

=

and beyond O take OX VT, and 0 Y= RM. Next, complete the

parallelogram OXPY; then draw the diagonal OP, and lastly, resolve OP into OP' and P P. MN represents the effective force

of RR, and VS that of TT'. Therefore when there is an equilibrium, MN + VS PP.

+VS

=

From Y draw YK perpendicular to P P', and complete the parallelogram YKLO. The triangles RMN and LKP are equal, since the sides are respectively parallel, and L K=0 Y= RM by construction; therefore P'K= MN. The triangles PKY and TVS are equal, for a similar reason, and PK VS. Consequently MN+VS=PK+KP=PP'. This method of resolution also gives the place E, where the power PP' must be applied to produce equilibrium. When the prolongation of RM and TV meet in a point beyond B C, either to the right or left, negative results are obtained.-Prof. JOHN L. CAMPBELL, Wabash College, Crawfordsville, Ind.

2. Law of Gravity. Solution of the problem on page 204, Vol. II. - Two similar systems of bodies have all corresponding linear dimensions in the same ratio to each other; that is, the linear dimensions and the distances apart of the bodies are in the one system in a fixed ratio to those of the other.

In order that two similar systems may remain similar for successive instants of time, it is necessary that the motions of corresponding bodies be in the same relative directions inter se, and that the velocities of corresponding motions be proportional to the linear dimensions of the two systems.

If, now, these motions be continually modified by the action of central fixed forces, or by forces dependent directly upon the masses of the bodies, and upon some function of their distances apart, then, since the changes in the motions must also be proportional to the dimensions of the systems and to the motions themselves, the values of these central forces will be proportional to the same dimensions. But the forces, so far as they are dependent upon the masses, are proportional to the cubes of the linear dimensions; hence, so far as