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from the laws to the meaning of the operations and from the laws and operations to the applications. Let me quote from Chrystal: "It may be well to insist once more upon the exact position which they (the fundamental laws) hold in the science. To speak as is sometimes done of the proof of these laws in all their generality is an abuse of terms. They are simply laid down as the canons of the science. The best evidence that this is their real position is the fact that algebras are in use whose fundamental laws differ from those of ordinary algebra." In support of this he gives as an example the fact that in Quaternious ab-ba.

This confusion results from the following facts, the fundamental operations are very difficult of definition. And the definitions of discrete arithmetic are not satisfactory when we come to extending the number system. 4X5 is a matter of definition so far as the use of the word times is concerned, but if the number system is agreed upon and the word times is understood then problem becomes a matter of calculation. So if we start with certain notions as to how we shall apply the four fundamental operations in algebra, then the results are matters of calculation and demonstration not matters of definition.

There is always a chance that the order of the historical development of a subject is the pedagogical one. Diaphantes knew that "minus into minus gives plus" two thousand years before the process was reversed and "minus into minus gives plus" was made a part of the definition of multiplication. But what shall we say of the case cited by Chrystal Quaternious?

Again an historical incident helps us out. Sir William Hamilton did not start with the law that abba and develop the subject but he started with certain conceptions and interpretations and after long struggles found that as he was representing and interpreting quantities and operations ab ba as he had up to that time supposed.

Therefore I believe that with beginners we are right when we explain how we interpret negative number, and how we apply the fundamental operations and then make the laws of signs. matters of demonstration. If the writers for advanced students can be more logical by reversing the process we surely have no objection.

PROBLEM DEPARTMENT.

IRA M. DELONG,

University of Colorado, Boulder, Colo.

Readers of the Magazine are invited to send solutions of the problems in this department and also to propose problems in which they are interested. Solutions and problems will be duly credited to the author. Address all communications to Ira M. DeLong, Boulder, Colo.

40. Proposed by O. R. Sheldon, Chicago, Ill.

A square garden, sides 12 rods, is planted with trees, no two of which are less than one rod apart, and no tree less than one-half rod from the fence. How many trees can be planted?

[NOTE.-Mr. Sheldon offered a prize of $3.00 for a correct solution which shows that more than 152 trees are possible; but no such solution has been received.-EDITOR.]

Solution by J. W. Ellison, Alcott, Colo.

Planted in squares one rod apart, no tree less than a half rod from the fence, there would be 12 rows of 12 trees each, 144 trees in all. Planted in regular quincunx order (12 trees in the odd rows, 11 trees in the even rows), there would be 7 rows of 12 trees each and 6 rows of 11 trees each, 150 trees in all.

If now the quincunx order be used for the first 9 rows and the square order for the remaining 4 rows, there would be 9 (the last 4, and 5 out of the first 9) rows of 12 each and 4 (the even ones in the first 9) rows of 11 each, 152 trees in all.

43. Proposed by H. C. Whitaker, Ph.D., Philadelphia, Pa.

A heavy sphere of radius R is placed in a glass of water in the shape of a cone, radius R and height h. How much water runs over? Note by H. E. Trefethen, Kent's Hill, Me.

Mr. I. L. Winckler's solution on page 139 of the February issue R2+R2 assumes, as he states, that the sphere sinks in the water, i. e., r< h The limiting case, when D coincides with A, is interesting, the volume of water displaced being readily found from the relation r: R= Vh2+r2: h.

But if r>

RV 2+R2
h

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the sphere is not then tangent to the side of the cone but rests upon the rim of the glass and its sides if produced would cut the sphere. In this case the formula given for finding EF fails and EF may be found from the equation - R2 (r — EF )' or EF=r-vr-R', using only the minus sign before the radical since EF <r.

ALGEBRA.

48. Proposed by I. L. Winckler, Cleveland, O.

A and B run around a course, starting from the same point, in opposite directions. A reaches the starting point 4 minutes, and B 9 minutes, after they have met on the road. If they continue to run

at the same rates, in how many minutes will they meet at the starting point? (From Wells' Algebra.)

Solution by R. P. Harker, Parker, Ind.

Let a distance traveled by A before they meet, y = the distance

y 4

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traveled by B, let t = time. Then =A's rate, = B's rate, and

9

t = x +
5

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Hence x + y = Zy=whole distance. Since A travels y in 4 minutes,

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he travels 24 in 10 minutes. Similarly, B travels the whole distance in 15 minutes. Therefore A and B meet at the starting point again in 30 minutes, the L. C. M. of 10 and 15.

GEOMETRY.

49. Proposed by Byron E. Toan, Boulder, Mont.

In a circle, radius R, given an arc of 45 degrees. To find the radius of a circle passing through the extremities of the given are and having the area common to the two circles equal to 2/5 of the area of the required circle.

I. Solution by H. E. Trefethen, Kent's Hill, Me.

Let the center of the required circle be exerior to the given circle. Let r be the radius of the circle sought and r the angle between its radii drawn to the extremities of the common chord. Put a = 45°. The area common to the two circles consists of two segments separated by their common chord. The area of each is the difference between a TaR? 1 πX12 360 2 360

sector and a triangle. Therefore,

R2 sin a +

1⁄2 sin = 2/5 π r2, and since R sin 1⁄2 a = chordr sin 1⁄2, R': r2 = sin2 1⁄2 x sin 1⁄2 a. writing (1 cos x) for sin2 2 we have Rr2 π (x 144) : πα

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half the common Simplifying and

= 180 sin x 180 sin a and R2 ; r2 = 1 cos x 2 sin2 1⁄2 a.

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π, a, sin a, and sin' 1⁄2 a their numerical values, and using logarithms in parentheses to represent the coefficients, (1.7219819) sin x + (1.1489865) cos x

(9.9638593) (x

144) 14.092449. Whence

☛ = 147° 1′ 22.8′′, r = 0.3990956 R, π r2 = 0.5003845 R2, and 2/5 π r2 = 0.2001538 R2, the last of which is the same as the area of the space common to the two circles as required.

II. Solution by E. L. Brown, M.A., Denver, Colo.

Let C and O be the centers of the given and required circles, radii R and r respectively.

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50. Proposed by Mary E. Barkley, Sacramento, Cal.

Find the locus of all the points, the sum of the squares of the distances of any one of which from three given points is equivalent to a given square.

Solution by J. F. West, San Diego, Cal.

Let A, B, C be the given points, k the side of the given square; M the centroid of triangle ABC; E the mid-point of CM and D the midpoint of AB. Denote the sides AB, BC, AC by c, a, b respectively.

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The locus of P is therefore a circumference whose center is the centroid of the triangle ABC and whose radius is

1
= √/ 3k2 − ( «2 + b2 + c* ).
3

[NOTE. This problem may be solved in the following neat manner: On page 404 of Volume VI, June, 1906, it was shown that "if P be the point of intersection of the medians of a triangle ABC, and Q any other point in the plane, then QA2 + QB2 + QC2 = PA2 + PB2 + PC2 + 3PQ2.'' In the present problem, PA, PB, PC and QA2 + QB2 + QC2 are given quantities, therefore the required locus is a circle whose center is the point P.-EDITOR.]

51.

TRIGONOMETRY.

Proposed by T. M. Blakslee, Ph.D., Ames, Iowa.

If X + Y + Z = 180°, tan X + tan Y + tan Z = tan X tan Y tan Z.

Use this to find the radius of a circle inscribed in a triangle in terms of the sides.

Solution by Proposer.

If we join the center with the three vertices and contact points, we thus form three pairs of equal angles about this point. Their tangents

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52. Proposed by Charles H. Smith, M.E., Chicago, Ill.

A car trust is to be formed by the different companies now making the separate parts, pooling their interests; the capital is to be one million dollars. The trust is to pay the different firms by issuing 60 notes payable one each month for 60 months. The face value of each of the notes, including interest at six per cent, is to be the same. What is the face value?

Solution by H. E. Trefethen, Kent's Hill, Me.

There are as many solutions of this problem as there are methods of computing interest on notes when payments in part have been made. Whatever the method it should be noted that any payment must exceed the interest on the principal for the preceding interval in order to cancel the debt as required. Of the two methods in general use the United States Supreme Court rule compounds the interest, while the Mercantile rule allows only simple interest on both the principal and the payments.

Put $1,000,000 = k, 60 = n, and let a be the face of each of the equal notes.

1st. By the U. S. Rule, put 1.005 = r, and rk after first payment. In like manner kr2

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sum left

xr

xr

- xr2 X, kra F -- Ꮖ, x = sums left after the second,

third and nth payments, which last must equal zero. Whence =

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values and solving by use of logarithms, x = $19,332.52.

The same result is obtained by the Compound Interest method, compounding the interest every month. For the amounts of the several notes at compound interest from the time each is paid to the end of the sixty months taken together should equal the amount of the capital at the same rate of interest for the sixty months and we have, since the last payment with no interest is x, x (1 + r + p2 + p3 +.............. + r¤−1) = kra.

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