LABORATORY METHOD OF MAKING A SELF-STARTING SIPHON.

BY WALLER BONNER,

High School, Montclair, N. J.

This siphon is constructed as shown in the drawing. It con

sists of a long bent tube I, which runs through the center of a glass cylinder H, to which it is attached by rubber stoppers K and L.

Two holes are made in the tube I and are lettered A-A in the drawing. Two other holes made through the lower rubber stopper are marked B-B.

In operation the siphon is thrust into the liquid until the water rises in

tube I to the level F. Water also passes in through openings BB, compressing the air and forcing the air out through A-A into I. This air pushes the water above A-A over the bend in the tube I. When the air in the cylinder has been forced.

out through the opening in I the entire siphon fills with water and operates as a simple bent tube siphon.

By following this method of construction, a siphon can be made from stock materials carried in every laboratory.

MINERAL RESOURCES OF THE KANTISHNA REGION. A preliminary report on the mineral resources of the Kantishna region, Alaska, has just been published by the United States Geological Survey as Bulletin 662-E. This paper gives a rather complete account of the mining activities of this interesting area during the last twelve years and descriptions of the many small producing placer mines and of the promising, but as yet unproductive gold, silver, and antimony prospects. It may be obtained free on application to the Director of the United States Geological Survey, Department of the Interior.

PROBLEM DEPARTMENT,

Conducted by J. O. Hassler.

Crane Technical High School and Junior College, Chicago.

This department aims to provide problems of varying degrees of difficulty which will interest anyone engaged in the study of mathematics. Besides those that are interesting per se, some are practical, some are useful to teachers in class work, and there are occasionally some whose solutions introduce modern mathematical theories and, we hope, encourage further investigation in these directions.

We desire also to help those who have problems they cannot solve. Such problems should be so indicated when sent to the Editor, and they will receive immediate attention. Remember that it takes several months for a problem to go through this department to a published solution.

All readers are invited to propose problems and solve problems here proposed. Problems and solutions will be credited to their authors. Each solution, or proposed problem, sent to the Editor should have the author's name introducing the problem or solution as on the following pages. In selecting problems for solution we consider accuracy, completeness, and brevity as essential.

The Editor of this department desires to serve its readers by making it interesting and helpful to them. If you have any suggestion to make, mail it to the Editor. Address all communications to J. O. Hassler, 2301 W. 110th Place, Chicago.

SOLUTION OF PROBLEMS.

530. Proposed by N. P. Pandya, Sojitra, India. (This is repeated- -no solutions received.-Ed.) If sin10+sec20+tan30

=

cos20+csc20, find sin20+cos20.

Solution by Nelson L. Roray, Metuchen, N. J.

551. Proposed by Murray J. Leventhal, New York City.

Show that the sum of the squares of the reciprocals of the natural numbers is 1/6π2.

Remark by Nelson L. Roray, Metuchen, N. J.

This is a standard series and is proved by considering the following two expansions:

It is evident the above expansions enable one to find the value of S2m.

In Hobb's Trigonometry is found the solution given above.-N. L. R. MR. F. J. McMACKIN sent a solution by means of calculus. lack of space to publish. One incorrect solution also received.-Ed.'

Geometry.

552. Proposed by Norman Anning, France.

We regret

Given a regular n-gon, to construct a regular 2n-gon having the same center and the same length of side.

I. Solution by W. W. Gorsline, Crane Technical High School and Junior College, Chicago.

=

=

A central angle of an n-gon 360/n and that of a 2n-gon 360/2n. Let AB be a side of a regular n-gon in a circle of center O and AC, onehalf of one side. ZCOA will subtend a side of the regular 2n-gon. Extend CO to circumference at D. Let EF be a side of the regular 2n-gon in a concentric circle of center O. A's DOA and EFO are similar and their altitudes or apothems of the 2n- and n-gons are in proportion to EF and AD. The apothem of the regular 2n-gon is a 4th proportional to EF, AD, and the apothem of the n-gon and can, therefore, be constructed and then the 2n-gon.

Let AB be a side of the given regular n-gon, and O its center. On AB, on the side remote from O, construct a rectangle ABCD, having BC = OB.

Then DC is a side of the required regular 2n-gon. This method yields easiest construction of regular dodecagon.

III. Solution by Nellie F. Henderson, Martins Ferry, O.

Let AB be one side of a regular polygon of n sides, having its center at O.

Draw AD and BC 1 AB at points A and B, respectively.

Draw OA and OB; then draw OE 1 AB from O.

Bisect the Zs AOE and BOE and extend their bisectors until they cut AD and BC at F and G, respectively.

FG will be a side of a regular polygon of 2n sides.

Solutions were also received from A. F. FAIRBANK, GERTRUDE BUCK, T. H. BRIGGS, and PHILOMATHE.-Ed.

Of all possible triangles inscribed in a given segment the isosceles triangle has the maximum perimeter.

I. Solution by Nelson L. Roray, Metuchen, N. J.

Let c be the chord of the given segment, and a and b the variable sides of the triangle.

II. Solution by Philomathe, Montreal, Canada.

Let AACB be isosceles, and ADB any other triangle. Produce AC and AD in E and F respectively, making CE CB, and DF ᎠᏴ ; join EB and FB.

=

=

=

Angles E and F are equal, therefore points A, B, F, E are concyclic. But, because AC CB = CE, angle ABE is right and AE is a diameter. Therefore, AE>AF, or AC+CB>AD+DB; that is, perimeter of A ABC is greater than perimeter of ADB.

Solutions were also received from W. W. GONSLINE, MURRAY J. LEVENthal; an additional one from NELSON L. RORAY, and one incorrect solution.-Ed.

554. Proposed by the Editor.

Prove by a direct method that if two angle bisectors of a triangle are equal the triangle is isosceles.

[This theorem being a familiar one, a recently published solution by Dr. Artemus Martin in the American Mathematical Monthly attracted our attention and it is desired to make a collection of different direct methods. Send also copied methods (with references).-Ed.]

I. Solution by Irvin E. Kline, Atlantic City, N. J., and by W. W. Gorsline, Chicago.

=

=

< BAD, ZADB.

Let ABC be the triangle in which the bisectors AD and BE are equal. Through E draw a line EF, intersecting BC so that BEF and through B draw a line intersecting EF so that 4FBE Draw FG AC, and AHL FB produced. Since 4s BAD AND ADB are two 4s of a A, they are together less than two right Zs and hence EF and BF will meet.

:: AABD

=

AEFB; BD = BF and AB = EF.

K is the point where bisectors AD and BE intersect.

ZKAE+ZAEK

ZKDB +
<FBA;

DBK
. <FEG

=

<FBK+KBA

BD; . AABE
BC.

ZABH; AFGE
AFGA = AAHF.

= ДАНВ.

HB and GF

..GA = HF and AE

.. ZEAB

Note: The above proof is taken mostly from Loomis' Original Investigations. Loomis also gives an indirect method of proof.-I. E. K.

II. Solution by Nelson L. Roray, Metuchen, N. J., Irvin E. Kline, Atlantic City High School, and Murray J. Leventhal, New York City. The formula for the bisector of an angle is

III. Second solution by Nelson L. Roray, Metuchen, N. J.

Let CAB be the given triangle, and the bisector of ZB intersect CA at D, of ZA, CB at E.

IV.

Third solution by Nelson L. Roray, Metuchen, N. J.

Let CAB be the given triangle, and the bisector of ZB intersect CA at D, of ZA, CB at E.

V. Fourth solution by Nelson L. Roray, Metuchen, N. J.

Let the interior bisectors of angles A and B intersect CB and AC at E and F, respectively, and each other at I, and the exterior bisectors of angles B and C at R and R', respectively.

=

Let AE BF = m, IE = a. RE Then we have at once,