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FP=cosFPP.P。P=cos- + P1P<cosx P2P<cosx(x−x ̧).

x+x。 2

0

..cosx.(x-x ̧)<sinx-sinx, <cosx, (x-x), which is (10).
The outer part of Fig. 5 can also be used to prove that

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By JOHN JAMES QUINN, Ph. B., Head of the Department of Mathematics and Manual Training, Warren High School, Warren, Pa.

The linkage is a material embodiment of the facts and conditions set forth in the following

THEOREM: If one vertex of a movably pivoted rhombus is constrained to move in the circumference of a directing circle, while the opposite vertex is fixed in the diameter (or diameter produced), the locus of the intersection of the diagonal (produced) through the other two vertices with the radius of the directing circle is a conic. Let BIDF be the rhombus with the vertex B moving in the circumference

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Now suppose that the radius of the directing circle becomes infinite. Then the circumference becomes the line XY, perpendicular to the diameter (the directrix); BF becomes the line B'F'', parallel to the diameter, and the diagonal D'I intersects it in K'.

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COROLLARY 1. The diagonal of the rhombus is a tangent to the curve in every position; and the second diagonal is parallel to the normal.

COROLLARY 2.
COROLLARY 3.

When the diagonal is parallel to the radius it is an asymptote. If a point in one side of a movably pivoted rhombus, at a given distance from the vertex, is constrained to move in the circumference of a directing circle, and a point in the adjacent side equidistant from the same vertex is fixed in the diameter (or diameter produced), the locus of the intersection of the diagonal (produced if necessary) through the other two vertices with the radius (or radius produced) of the directing circle is a conic.

COROLLARY 4. A conic is the locus of a point which moves so that the ratio of its distances from a fixed circle and a fixed point in its diameter (or diameter produced) is equal to unity.

The linkage is very easily constructed by using thin strips of wood or metal for each line in the above figure except the diameter of the directing circle, which should be strong enough to support the rest of the linkage without bending. The links representing the diagonal and the radius should be slotted, then there will be an opening at their intersection in which a pencil can be inserted to describe the curve.

The points F and F represent the foci, one of which, the point F, should be made to slide along the diameter. Then a change in the curve due to a change in the relative position of the foci is made evident.

If the diagonal DI passes through the point F", and an extra link DE is attached to the vertex D with the point E in the diameter, but so situated that DE=EF, then the vertex I will describe a straight line.

For we have FD>FI=constant. Consequently, if the vertex D describes

a circle, the point I must describe its inverse.

The extreme simplicity of this linkage reduces the geometry of the conics to that of the rhombus and the circle.

'DEPARTMENTS.

SOLUTIONS OF PROBLEMS.

ALGEBRA.

Problem 185 was also solved by A. H. Holmes, Brunswick, Maine.

187. Proposed by L. E. DICKSON, Ph. D., Assistant Professor of Mathematics, The University of Chicago. Express by radicals the roots of x2 +рx3 +2 p2x2+p3x+r=0.

Solution by G. B. M. ZERR, A. M., Ph. D., Parsons, W. Va.

Let x=y+z. Then the equation becomes

y2+z2+(7yz+p)[y5 +25 +(3yz+‡p)(y3+z3)

+(5y2z2+5pyz+'p2)(y+z)}+r=0.

Now x may be decomposed into two parts, y and z, in an infinite number of ways; we may, therefore, suppose y and z are such as to satisfy the condition 7yz+p=0. ...yz=-4p, y2+z7=-r.

Let y2=a, 27 =b._.'.a+b=-r, ab=—(4p)7.

... a and b are the roots of the equation u2+ru- (7p) ^ —-0.
... a=−jr+√/[‡r2+(†p)*], b——±r−√/[‡r2 + (7p)2].

Let be an imaginary seventh root of unity, so that

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Va+Vb, @Va+w® 7/b, w2 Va+w5 7b, w3 Va+w+ 7/b,

w° Ya+wVh, w5 Va+w2 Vb, w1 Vi+w3 7/b.

188. Proposed by GUY SCHUYLER.

xy+ab=2ax, x2y2+a2b2=2b2y?.

Solution by O. W. ANTHONY, Head of Mathematical Department, DeWitt Clinton High School, New York City. By squaring the first equation and subtracting from the second we get y=xa/b and y=-2xa/b. Whence easily x=b, 1(−1±1/3)b. y=a(11/3)b/a.

Also solved by G. W. Greenwood, B. A., Professor of Mathematics and Astronomy, McKendree College, Lebanon, Ill.; G. W. Drake, Fayetteville, Ark.; Charles E. Barrett, Louisville, Ky.; H. F. MacNeish, A. B., Instructor in Mathematics in the University High School, Chicago, Ill.; L. E. Newcomb, Los Gatos, Cal.; G. B. M. Zerr, A. M., Ph. D., Parsons, W. Va., and J. Scheffer, Kee Mar College, Hagerstown, Md.

GEOMETRY.

Problem 203 was also solved by Henry A. Converse, Ph. D., Instructor in Mathematics, Johns Hopkins University, Baltimore, Md.

208. Proposed by W. J. GREENSTREET, A. M., Editor of The Mathematical Gazette, Stroud, England. Tangents drawn to two confocal parabolae from the point on the common tangent intersect at the same angle as the axes of the parabolae.

I. Solution by G. W. GREENWOOD, A. M. (Oxon), Professor of Mathematics and Astronomy in McKendree College, Lebanon, Ill. Draw PA, PA' paral

PQ is the common tangent; S, the common focus. lel to the axes; PT, PT are the other tangents from P; T, T being the points of tangency. Join PS.

Since a tangent from a point to a parabola makes the same angle with the line joining the point to the focus as the remaining tangent from the point does with the axis, we have TPA / QPS=/TPA'. Hence TPT=APA'.

II. Solution by G. B. M. ZERR, A. M., Ph. D., Parsons, W. Va.

Let y2-4a(x+α)......(1),

y2cos20+2ysin@(xcos0+2b)=-4bxcos0+4b2-x2sino 0....(2),

be the confocal parabolae with axes inclined at an angle 0.
Let the common tangent be y=mx+c.....(3).
If (3) is tangent to (1), c=a(m2+1)/m.

If (3) is tangent to (2), c=b(m2+1)/(mcose+sino).

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From (4), m1=m=asin0/(b-acose). The other values of m, and m, are

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209. Proposed by W. J. GREENSTREET, M. A., Editor of The Mathematical Gazette, Stroud, England. Find by a geometrical method the maximum value of sine cos◊ cos20.

Solution by G. W. DRAKE, Fayetteville, Ark.

In the circle, radius 1 and center 0, let / AOB=0, and /AOC-20. From O drop a perpendicular to AO or AO produced, meeting AO in E. Then CE= sin20, and OE=cos20. sino cos◊ cos20=sin20 cos20 the area of triangle OEC. But triangle OEC is a maximum when OE=CE. Hence sin◊ cos◊ cos20 is a maximum when sin20=cos20, i. e. when 20—90–20, or when 0=221°.

Also solved by L. E. Newcomb, Los Gatos, Cal.; G. B. M. Zerr, A. M., Ph. D., Parsons, West Va. [Dr. Zerr gives the more general result 0=}π(4m+1).]

210. Proposed by L. E. DICKSON, Ph. D., Assistant Professor of Mathematics, The University of Chicago. Let ADC be a triangle with angle C-120°, and let the interior bisector of angle C meet AD in B. Prove that 2. CB is the harmonic mean between CA and CD.

Solution by M. E. GRABER, A. B., Instructor in Mathematics and Physics, Heidelberg University, Tiffin. O., and G. W. DRAKE, Fayetteville, Ark.

On AC produced through C, take a distance CKCD, and join K and D. Since triangle ACB is similar to triangle AKD, .. BC: DK-CA: KA, hence 2.CA.DK 2.CA.DK 2.BC= But because CD=CK, and KCD=60°,

=

KA

CA+KC

··· / CKD=/ KDC=60°, and triangle CKD is equilateral. .. 2.BC= Hence 2.BC is the harmonic mean between CA and CD by definition.

2CA. CD CA+CD

Also solved by R. A. Wells, Bellevue College, Bellevue, Nebr.; G. W. Greenwood, B. A. (Oxon), Professor of Mathematics and Astronomy, McKendree College, Lebanon, Ill.; L. E. Newcomb, Los Gatos, Cal.; G. B. M. Zerr, A. M., Ph. D., Parsons, W. Va.; J. Scheffer, Kee Mar College, Hagerstown, Md.; E. L. Sherwood, Shady Side Academy, Pittsburg, Pa.

211. Proposed by L. E. DICKSON, Ph. D., Assistant Professor of Mathematics, The University of Chicago.

Prove the validity of the following construction of an inscribed regular pentagon and regular decagon: Draw any two perpendicular radii of the given circle with center C. Call E the end of one radius CE and M the middle point of the perpendicular radius CM. Take the point R on CM produced through C such that RCM=EM. Then RC=side of inscribed regular decagon, RE side of inscribed regular pentagon.

Solution by G. W. DRAKE, Fayetteville, Ark., and R. A. WELLS, Bellevue College, Bellevue. Neb.

Join E and M, also E and R. Let r=CE.

(1). ME=r2 +r2==5r2/4. ME=r/5.

...RC=RM-CM-ME-CM=r1/5—dr=r(√/5-1)=a side of a regu

lar decagon inscribed in a circle whose radius is r.

(2). RE2=RC2+CE2=[{r(1/5−1)]2+r2=‡r2 (6−21/5)+r2 =‡r(10— 21/5). ..RE=ry/(10-21/5)=a side of a regular pentagon inscribed in a circle whose radius is r.

Also solved by G. W. Greenwood, A.B. (Oxon), Professor of Mathematics and Astronomy, McKendree College, Lebanon, Ill.; L. E. Newcomb, Los Gatos, Cal.; G. B. M. Zerr, A. M., Ph. D., Parsons, W. Va.; J. Scheffer, Kee Mar College, Hagerstown, Md., G. I. Hopkins, Manchester, N. H.

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