yard, then the similar triangles will be represented by like parts and equimultiples of the numbers composing each equation, such as ( 3 )2 + 22 = (§)2, and 62 +82 = 102. Rejecting all fractions, and all numbers above 100, the first of the above equations gives 20 similar triangles, the second 7, the third 5, the fourth 4, the fifth 3, and the sixth and seventh 2 each, making in all, 52 right-angled triangles, whose sides may be represented by integers not exceeding 100, when unity represents some fixed linear unit. Those represented by the fifth equation approach nearest to the isosceles rightangled triangle, and that represented by the thirteenth equation has the greatest difference between the sides including the right angle.

When the sides of a right-angled triangle are commensurable, the perpendicular let fall from the vertex of the right angle to the hypothenuse is commensurable with the sides, because the two partial triangles thus formed are similar to the original one. In other words, when the sides of a right-angled triangle can be exactly expressed by numbers, the perpendicular can also be exactly expressed, usually by the aid of fractions, as the greatest common measure of the sides does not measure the perpendicular. In such a triangle, the diameter of the inscribed circle is also commensurable with the sides, as it is equal to the sum of the other two sides minus the hypothenuse. The perpendicular is equal to the product of the other two sides divided by the hypothenuse.-Prof. D. W. HOYT, Fairfax, Vt.

15 (Esin E) 5. The method of developing the expression A= 9E+ sin E employed by Prof. COTTER, in the Mathematical Monthly for March, p. 185, is not the most natural, and does not exhibit the law of progression of the coefficients of the resulting series otherwise than by inspection of the result when obtained. By pursuing the following more direct process, the law of progression is evident at each step.

We have to develop A in terms of T, having T=tan2 E. We take

E=2 tan-1 T2 T (1-1T+ { T2 — & T3 + T — &c.)

sin E

whence

2 T
1+T

= 2 T1 (1+T)-1

=2T (1-T+T2 — T3+ T- &c.);

Ꭲ

15 (E-sin E) = 30 T ( T- T2 + § T3 — § T*+ &c.) =20 T (T- & T2 + & T3 — 12 T + &c.)

T

9 E+ sin E 2 T (10-12 7+ 14 T2 — 10 T3 + &c.)

=

-Prof. W. CHAUVENET, Washington University, St. Louis, Mo.

6. Note on the Theory of Perspective. - The theory of perspective may often be applied to the discovery and demonstration of geometrical truths. I propose to give a few examples of this application. Through the points A, B, C, in the same straight line, let

three sets of parallels be drawn, as in Fig. 1. The points L, M, N are in the same straight line, for

AL: LOAB: BC-PN: NC.

Therefore the line LN passes through the intersection of OC and

AP. Fig. 2 is the projection of Fig. 1, FD being the horizon, if the parallel lines are horizontal. The points L, M, N are still in the same straight line; whence the following theorem of APOLLONIUS.

"If any quadrilateral be divided into two other quadrilaterals by a straight line, the intersections of the diagonals of the three quadrilaterals are on the same straight line."

As a second example, let us take a circle. Draw any two diameters, and at their extremities draw four tangents. If now we connect the intersections of the diameters and tangents by lines, and also draw tangents at the points where these lines cut the circle, a symmetrical figure will be formed. If we take the perspective of this figure, we may assume any ellipse for the projection of the circle, and any point within it for the projection of the centre. Any line drawn through this point may be taken as the projection of one of the diameters; the projection of the other is then determined by the angle which the original diameters make with each other. The original tangents are evidently tangents to the ellipse in the projection of the figure. We have then this general theorem.

I. Any two lines which are parallel in the original figure meet on the horizon in the perspective representation.

II. If any line is bisected in the original figure, this line prolonged to the horizon is divided harmonically.

Problem V. in the Math. Monthly, Vol. I. No. I., is included in the first part of this theorem. - G. B. VOSE, Assistant U. S. Coast Survey, Washington, D. C.

7. Note on the "Rule of False."-This rule admits of a very simple explanation, which has probably occurred to most persons who are in the habit of using the rule.

The explanation suggests the origin of the rule, and cannot be new; still I do not remember having seen it.

In

any equation of one unknown quantity, such as ƒ (x) = 0, let

us for a moment suppose x to be the variable abscissa of a curve, while y, the variable value of f (x), is the ordinate. The equation becomes

f(x)=y.

We wish to find the point where the curve intersects the axis of x; that is, a value of x which shall make y = 0.

Find by trial two values of x, which we may designate by x1, and x2, which shall give two values of y, y1 and y2, nearly equal to zero. The rule of false, or rule of double position, is based on the supposition that the curve near the axis of x is a right line.

If Yı

Fig. 1.

Fig. 2.

and y2 are both positive or both negative, Fig. 1 gives

If y1 be positive and y2 negative, or y2 positive and y1 negative,

These two formulæ, enunciated in words, make the "rule of false." It will be seen that the errors, so called, are true ordinates of the curve, and that the final result will be more or less accurate, according to the nature of the curve and the proximity of y1 and Yı

zero.

Y2

to

The rule as announced in arithmetics does not apply to the case in which the two "positions," x and x2, are unlike.

X1

If we put the known term in the second member f'(x) = p, the solution will give the point or points where the curve intersects a line parallel to the axis of x at the distance above it. Capt. D. P. WOODBURY, U. S. A.

p

9. Solution of Prize Problem III. in Vol. I. No. V.- Problem. "If two sides of a movable right angle are always tangent to a given ellipse, its summit will describe a circle concentric with the ellipse, the radius of which is equal to the chord joining the extremities of the major and minor axes."

E. (See COFFIN's Conic Sections, Prop. XV.) And since MG N is a right angle, HE is a diameter to the circle on GF, and hence it passes through C its centre.

draw CH, CC, CG and CE.

From C, the centre of the ellipse,
Represent the semi-axes as usual by

A and B. Then (LEGENDRE, Prop. XIV. B. IV.)

2 HC220 C2 CH2+ CE2 = 2 A2.

Also, 2 FC + 2 0 02 = F C2 + C G2 = A2 — B2 + C G2

Ᏻ

Whence, since HCFC', 2 A2 A2 - B2 + C C2.

Therefore CG=√(42+ B), which was to be proved.

-M. C. STEVENS, Prof. of Mathematics, Haverford College, Pa.