THE

MATHEMATICAL MONTHLY.

Vol. II.... JULY, 1860.... No. X.

PRIZE PROBLEMS FOR STUDENTS.

I. Solve the equations

by quadratics.

Communicated by E. A. HOPKINS, Cleveland, Ohio.

II. Given the lengths of the three perpendiculars dropped from any point in the plane of an equilateral triangle upon the sides; to find the segments of the sides.- Communicated by F. E. TOWER, Amherst College.

III. If e denote the edge of any regular dodecahedron, and ß= 36°, prove that

Also obtain similar formulas for the solidity of the icosahedron. Communicated by Prof. D. W. HOYT.

IV. A given cylindrical vessel, filled with water, is placed with its base upon a horizontal plane. It is required to determine the angle of inclination to which the plane must be raised before the vessel will fall, the water being at liberty to overflow its top. The base is supposed to be fixed so as to prevent it from sliding, but not from tilting when the plane is inclined. - Communicated by Professor KIRKWOOD.

V. Bisect the attraction which a sphere of varying density exerts

upon an exterior point; that is, divide the sphere so that the two parts shall exert the same attractive force in the same direction.

Solutions of these problems must be received by September 1,

1860.

REPORT OF THE JUDGES UPON THE SOLUTIONS OF THE PRIZE PROBLEMS IN No. VII., Vol. II.

THE first Prize is awarded to JOHN A. WINEBRENER, Princeton College, N. J.

The second Prize is awarded to GEORGE C. ROUND, Wesleyan University, Middletown, Ct.

The third Prize is awarded to LEWIS FOOTE, O. C. Seminary, Cazenovia, N. Y.

PRIZE SOLUTION OF PROBLEM I.

By Miss HARRIET S. HAZELTINE, Worcester, Mass.

Prove that an arithmetic mean is greater than a geometric.

Let xy and xy denote the extremes; then x is the arithmetic and √(22-72) the geometric mean, and it is evident that x = √ x2 > √(x2 — y2).

SECOND SOLUTION.-Let ab; then a-b>0; aa2-2ab+b2> 0; a2 + 2 a b + b2 > 4 ab; a + b > 2 √ ab; :. 1 (a + b) > √ ab. - G. S. MORISON, Harvard College.

PRIZE SOLUTION OF PROBLEM II.

Let three bodies with velocities V, V, V", move uniformly in the same direction, in the circumference of a circle. Required the time of their conjunction, supposing them to quit a given point at the same time.

Let C denote the circumference of the circle; then, since V — V' and V — V′′ are respectively the gains of V upon V' and V" in the same unit of time, will denote the times which

C

and

will elapse between the instant of starting and the conjunctions of V, V and V, V" respectively. And the least common multiple of

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 A K, and draw the tangent CB at K. Next, bisect the angles CAK and KAB, 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 FPC-PBC+BCP ACF+PCA 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 second 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.

y
r + x2

THIRD SOLUTION.-Let (xy) be the co-ordinates of the point P, O being the origin, and r the radius of circle A B C. But tan (PAC +PCA)=tan 45° 1, and since tan PAC= tan PCA=> we have by the usual formula, a2 + y2+2ry=2; or changing the origin to F, x2 + y2 = 2r2, 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 seen. A 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, w wb,