But we have (CHAUVENET'S Trig., p. 251, and Eq. 319),

Therefore, by comparing (1) and (2) we get

Plane Triangle. Since in a plane triangle (A + B + C) = { π,

Taking the sum of (1) and (2), and also of (3) and (4), we obtain

Adding (5) and (6), developing and reducing, we have

4 cos A cos B cos C sin A+ sin B+ sin C.

1⁄2 =

MR. WHEELER'S solution for the spherical triangle is essentially

the same as the one already given.

SIMON NEWCOMB.

W. P. G. Bartlett.

TRUMAN HENRY SAFFORD.

ANOTHER SOLUTION OF PRIZE PROBLEM I., No. IX. By GEORGE EASTWOOD, Saxonville, Mass.

THERE are two ways of solving a geometrical problem geometrically. One way is to solve it by synthesis, the other way is to solve it by analysis. In solving a problem by analysis, we assume that the work is done; and then proceed, as the word implies, to take it to pieces, and to examine and develop the properties of each part and its relations to all the other parts. To solve the same problem by synthesis, we proceed to construct it upon elementary principles, from the given data and such combinations of them as the exigencies of the case may require. We have a beautiful illustration of the synthetic method in Mr. EVANS's solution of Prize Problem I. of No. IX., published in the last Monthly. The following solution of the same problem is offered to young students as a simple specimen of the analytic method.*

Analysis. Suppose the thing done, that A CB is the required triangle, and that AFBE, GHI, are its circumscribing and inscribed circles. Join 0, 0, the centres of the circles, and draw the radi O G, OH, O I, to the points of contact G, H, I. Draw the diameter EF perpendicular to the hypothenuse AB, and CD perpendicular to EF

By the question, we have given,

AC BC (AGGC)-(BH+HC)

AI-BI=2012 MN suppose.

* Strictly speaking, the two ways of solving a geometrical problem, above indicated, are not two distinct and independent methods of doing the same thing, but rather the different parts of one full and perfect method. The one process is the reverse of the other, and the use of each is indispensable to a complete solution.

Also, we have given,

A B — B C = (A I + I B) — (IB +1 0')= A I — 1Ơ, which suppose equal to 2 L K. Take 2 MN from 4 L K, then AI+BI-2 10 = 4 LK — 2 MN;

.. A0=2LK-MN+IO.

By Prop. VI., Liverpool Student, EO. FD= 0 12 MN2; so that when E 0 A 0 is found, FD will be given.

By Prop. 47, Euc. I.

0 12 + 0 r2 = 0′ 02 = A 0 (A 0 — 2 0 I)

= (2 L K — M N )2 — 0 12 — (2 L K — 0 1)2 — 0 12,

=

... 0 12 = 2 LK (LK — MN).

I) —

As L K and MN are both given lines, therefore I is given. Hence this

Construction. Find (Euc. VI. 13) Ø I a mean proportional between 2 L K and L K- MN, and with centre 0 and radius 0 A= 2 LK MNOI, describe a circle. Draw the horizontal and vertical diameters AB, FE, and upon FE apply (Euc. VI. 11) FD a third proportional to E 0, 0 I. Draw DC parallel to AB to meet the circle in C, and join A C, B C; then A CB is the required triangle. The demonstration is evident from the analysis.

NOTES AND QUERIES.

1. Note on Decimals. - Decimals should be taught in written arithmetic in connection with whole numbers. Let them be treated in the same manner as whole numbers, and not as common fractions, there being no necessity for confusing the mind of the pupil by writing the denominator. In 31.3 there is no more need of indicating that the three standing in tenths' place is divided by ten, than

that the three standing in ten's place is multiplied by ten. The pupil can readily be made to understand these relations without the divisor or multiplier. There should be no separate divisions in arithmetic for Decimal Fractions and Federal Money. The four simple rules should embrace these, and Percentage and Interest should be treated immediately after as a development of the decimal notation. The tables and Compound Numbers should immediately precede Vulgar Fractions, -the former increasing in an irregular ratio, and the latter decreasing in an irregular ratio. If authors in writing arithmetics would observe this order, it would result in great advantage to the learner.- SAMUEL P. BATES, Superintendent of Public Instruction, Crawford County, Pa.

After the pupil

2. Reduction of Fractions to a Common Denominator. has thoroughly learned that multiplying or dividing both terms of a fraction by the same number does not change its value, he is then prepared to learn how to reduce fractions to a common denominator. The following process is very simple, and the pupil should be required to repeat it until it is thoroughly understood.

9

Let it be required, for example, to reduce 14, 15, 18, 18, 1 to a common denominator. First, resolve the terms of all the fractions

tors. Let us first compare them with reference to the factor 7. This factor is found in all the denominators but one; and in order

23.5 2.2.7.5

19

5.7 2.2.19

2.2.5 7.5 2.2.2.2.5

to make them common with reference to 7, it must either be introduced into the denominator of the 2d fraction, or removed from the denominators of all the other fractions. But to remove a factor from the denominator of a fraction without changing its value, it must be removed from the numerator also. Now 7 is a factor in only one of the numerators, and the denominators cannot be made common by removing the 7. It must therefore be introduced into the denominator of the second fraction to make the denominators common; and also into the numerator, in order not to change the value of the fraction. We thus get the second row. Next compare the denominators with reference to the factor 3. It is found in only two of them, from which it may be removed, since it is also found in the corresponding numerators. Thus we get the third row. (It is plain, that instead of removing the factor 3 from the 2d and 5th fractions, the denominators might be made common with reference to 3 by introducing it into the terms of all the other fractions. should not, however, in this way reduce the fractions to their least common denominator.) Next introduce the factor 5 and get the fourth row; and lastly, introduce the factors 2. 2, which makes all the denominators common. - TEACHER.

3. Note on the superior limit of the Roots of an Equation.

We

PROP. I. The greatest negative coefficient of an equation, plus unity, is a superior limit of its roots.

PROOF. Let us assume the general equation,

-1 ±Bxn-2 + Nx3-+1± V = 0.

A superior limit of the roots of an equation must produce a positive result when substituted for ; that is, the sum of all the positive terms must exceed the sum of all the negative ones by some quantity R. The most unfavorable case evidently is that in which all the terms after the first are negative and have equal coefficients. Under this assumption our equation becomes