of snow had the ribbed surface except one high up on Mount Fairchild and it was therefore surmised that the ridges were produced by the motion of the glacier.

The mountains of the Estes Park region consist chiefly of gneiss and mica schist. Even the so-called Pillars of Hercules in Thompson Canyon are but the edges of layers of fine textured mica schist. The boulders of all the moraines have not been moved more than fifteen miles, most of them not more than five or six. They are therefore still angular like the disintegration boulders of Hagues Peak and have not been smoothed by sliding. It may be of interest to the mineralogist to learn that many of the boulders on the slopes of the ridge connecting Hagues Peak and Mummy Mountain contain twin crystals of feldspar.

Several vacations would be required for a complete study of all these morainic ridges in the vicinity of Estes Park, but such studies would take one into a region replete with gorges, snow fields and rugged mountain peaks and would fill the hearts of the fisherman, mountain climber, teacher of physical geography, geologist and artist with the deepest pleasure.

CAROLINA TIN DEPOSITS.

The United States Geological Survey, Department of the Interior, has recently issued a report on the tin resources of the Kings Mountain District, North Carolina and South Carolina. The presence of cassiterite, oxide of tin, at many places in the district has led to much prospecting and to attempts at mining. In at least one place-the Ross mine, near Gaffney-placer mining was temporarily profitable, Practically all the work on the lodes, which are pegmatite dikes carrying cassiterite, has been done at a loss but the results of this work have not been sufficiently conclusive to prove or disprove the value of some of the deposits. The report can be had free on application to the Director, U. S. Geological Survey, Washington, D. C., asking for Bulletin 660-D.

THE WAY TO PEACE.

"Any body of free men that compounds with the present German Government is compounding for its own destruction. Any man

in America or anywhere else that supposes that the free industry and enterprise of the world can continue if the Pan-German plan is achieved and German power fastened upon the world is as fatuous as the dreamers in Russia. What I am opposed to is not the feeling of the pacifists but their stupidity.

"If we are true friends of freedom of our own or anybody else's we will see that the power of this country and the productivity of this country are raised to their absolute maximum, and that absolutely nobody is allowed to stand in the way of it.

"Our duty is to stand together night and day until the work is finished." From President Wilson's Address to the American Federation of Labor.

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, however, 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.

Correction (Problem 527).

The following communication from Prof. B. F. Yanney, Wooster, Ohio, explains itself. The Editor confesses to the mistake, practically the same as pointed out by him in the issue of June, 1917, under the discussion of Problem 511.

"Permit me also to call attention to what I believe to be an erroneous statement in your solution of Problem 527. You develop your formula for n on the theory of remainders after monthly payments of $50 have been made. Now the presumption is, and your result clearly shows, that there is not an exact number of these monthly payments of $50. Evidently, after the 101st payment there is still a residual, not zero. So that your explanation leading up to the formula is hardly the correct one. The formula, however, is all right. One way of explaining is to let n equal the time during which the compound amount of the unpaid balance of $4,000 at the nominal rate of 6% (also the effective rate here) becomes equal to the final value of the $50 monthly payment (annuity) at the effective rate in this case of (1.06)1/12-1, and so on.

Fallacy.

541. Proposed by the Editor. (See December, 1917, number. Problem 529 is restated below.)

Find the error in the published solution of 529. A correct solution would also be acceptable.

I. Solution by B. F. Yanney, College of Wooster, Wooster, Ohio.
The error in question is in the statement,

"/P equals 1/2 half circle HER less are HK," which is tacitly assuming that diameter HP of circle S is tangent to circle T.

The theorem is essentially that of Exercise 615 in the 1882 edition of Todhunter's Elements of Euclid, and its proof depends upon Exercise 614 of the same book, which latter says:

"The opposite sides of a quadrilateral inscribed in a circle when produced meet at P and Q. Show that the square on PQ is equal to the sum of the squares on the tangents from P and Q to the circle.'

We may then prove 529 as follows:

Let O be the center and r the radius of the given circle S. Let C be

the center and R the radius of the circle described on PQ as diameter. Let t1 and to be the tangents from P and Q, respectively, to circle S. Now, as is well known, OP2+OQ2 = 20C2+2R2.

=

But Op2
Therefore, by substitution and reduction, we obtain

r2+t21, OQ2 = r2+t2 2, and 4R2

=

PQ2

= t21+122.

OC2 = r2+R3,

which easily leads to the desired result.

Exercise 614 of Todhunter's Euclid can be proved by aid of Exercise 309 of the same work, and this in turn by means of well-known principles of elementary geometry.

The error in 529 was also pointed out by R. M. MATHEWS. Ed.

Geometry.

529. Proposed by R. T. McGregor, Nord, Cal.

The circle whose diameter is the third diagonal of a quadrilateral inscribed in another circle cuts the latter orthogonally.

Solution by E. L. Brown, Denver, Col.

Let ABCD be the cyclic quadrilateral, EF its third diagonal. Draw FN, EL, tangents to circle O. Let M be mid-point of EF, hence center of circle on EF as a diameter.

We will first prove EF2 = FN2+EL2.

About CDF describe a circle CDFG, cutting EF in G. Join CG. The angles BAD, BCD are supplementary; also the angles DFG, DCG are supplementary. Therefore the angles BAD, BCD, DFG, DCG together make four right angles. The angles BCD, BCG, DCG also make four right angles. Therefore, BCG ·BAD+DFG, and BCG+ BEG EAF+AFE+AEF. Hence angles BCG and BEG are supplementary, and BCGE is a cyclic quadrilateral. Therefore FE⚫ EG = DE EC EL2, and EF FG But FE EG+

=

=

=

BF FC

=

FN2.

=

E.

Now let OL = ON=r, and EM = MF R. Join O to L, N, F, M,

Hence circles whose centers are O and M are orthogonal.

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

Let A, B, C, D be four points on the circumference of a circle, and let connectors AD and BC intersect at G, the connectors CD and AB intersect at F, the connectors BD and AC at E.

Then EFG is a self conjugate triangle (Lachlan's Modern Pure Geometry, §270).

Hence, G is the pole of EF and F is the pole of GE, and G and F are conjugate points with respect to the given circle.

Therefore, the circle whose diameter is FG cuts the given circle orthogonally. (Lachlan, §260).

NOTE: Problem 529 is one case of the following:

Circles whose diameters are the sides of the central tetrastigen inscribed within a circle cut the given circle orthogonally.

One incorrect solution received.-Ed.

Algebra.

542. Selected from College Entrance Examinations (September, 1917), Advanced Algebra.

How many numbers between 100 and 1,000 may be formed from 0, 1, 2, 3, 4, 5, 6, no digit being repeated in any number? How many of these are even?

I. Solution by L. E. Lunn, Heron Lake, Minn.

As the problem is stated, three place numbers are required. For the first place, i. e., the hundreds digit, we have only six choices, since we cannot use the O in this place. Then for the tens digit there remain six numbers to choose from, and for the units digit five choices.

The total number of numbers is then 6x6x5 or 180.

Since 0 cannot be used in hundreds place, there are three even and three odd digits for use in this place. Then just half, or 90 of the numbers will have an even number in hundreds place.

Of the 90 numbers having an even number in hundreds place there will remain three even numbers for units place, and 1, 3, and 5 odd numbers in units place. This will give just half of these 90 numbers even and half odd. There are then 45 even numbers in this group.

Of the 90 numbers having an odd number in hundreds place, there remain 0, 2, 4, and 6, even numbers, and but two odd digits for use in units place. Then two-thirds of this group will be even. This gives 60 even numbers out of this group.

There will then be a total of 105 even numbers in the 180 three place numbers.

Writing the numbers verifies this result.

II. Solution by R. M. Mathews, Riverside, Cal., and R. T. McGregor, McArthur, Cal.

Using the notation Pm for permutations of n things m at a time, we have,

Since half the digits are even and half odd, just half the first group will end in an even digit and so be even numbers. All of the last group will be even. Thus 105 of the 180 will be even numbers.

Also solved by C. E. GITHENS.-Ed.

Geometry.

543.

Selected from College Entrance Examinations (September, 1917), Elementary Geometry.

If ABC is a right triangle with the right angle at B, and if D is a point on AC such that AB is the mean proportional between AC and AD, prove that ADB is a right angle.

I. Solution by Geo. H. Olson, Nevis Minn. (Solutions differing from this only in minor details received from Edward S. Mooney, South Dayton, N. Y., W. W. Gorsline, Chicago, Isidore Ginsberg, New York City, and R. M. Mathews, Riverside, Cal.

Join D with A. The triangles ABC and ADB are similar, having an acute angle of the one equal to an acute angle of the other, and the including sides proportional. Similar triangles have their corresponding angles equal, hence angle ADB is equal to angle ABC. Angle ABC is by hypothesis a right angle, hence angle ADB is a right angle.

II. Solution by Grover C. Koffman, Hopkinsville, Ky.

If BD is not perpendicular to AC, draw BM perpendicular to AC.
Then

Therefore, AD coincides with AM and BD is perpendicular to AC.

III. Solution by L. E. Lunn.

Given: Right triangle ABC, and point D on AC, and (AB)2 XAD.

3. Then DB and BC are either parallel or anti-parallel. (Two lines which cut off proportional sects from the sides of an angle are either parallel or anti-parallel, and the triangles formed are co-sensal or non-co-sensal respectively.)

It then follows that DB and CB are anti-parallel and that the triangles are similar non-co-sensal triangles.

4.

5.

Since B is given a right angle, then by 4 it follows that
ZD is also a right angle.

Solutions were also received from NELSON L. RORAY, R. T. MCGREGOR, M. T. NOLAN, and ERNEST H. H. SANGER. A second solution was re

ceived from GEO. H. OLSON.-Ed.

Trigonometry.

Proposed by Clifford N. Mills, Brookings, So. Dak.