THURSDAY, MAY 8, 1937
UNIVERSITY DAILY KANSAN, LAWRENCE, KANSAS
PAGE THREE
Here on the Hill an account of Mt. Oread Society
-an account of Mt. Oread S
MARY JOHNS, Society Editor
Before 1 p.m. call K.U. 21; after 5, call 2702-K3
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Alpha Kappa Pai, professions commerce fraternity, held formal initiation Sunday morning for Lester Phlepa, b'uncl; Vern Dane, c'40 Neel Dem, c'40; Frank Gaddee, c'40; Paul Henderson, c'40; and Clifford Black, b'38.
The Jay James entertained yesterday afternoon from 3 to 5 with a tea in the Cafeteria. Miss Maule Elliot poured. The decorations consisted of green candles and a centerpiece of roses and snagdraps.
Wayne Clover, c'29, was elected rush captain of the Kappa Sigma fraternity in the election of officers held Monday night, in addition to the new officers announced in yesterday's Kansas.
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Sunday dinner guests at the Alpha Kappa Psa fraternity house were Prof. Jens P. Jensen, Prof. D. Gageilardo, and Don Crane, of Lawrence.
Mrs. F. C. Hensler of Kansas City Mo., was a guest at the Alpha Delta Pi sorority house yesterday.
W. H. Towers, attorney at law, o
Kansas City, was a visitor last night at the Alpha Phi Alumnity fraternity house. Mr. Towers is a member of the Kansas legislature,
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Dr. J. J. Wheeler, and Robert Reed. c'38, were dinner guests at the Sigma Nu fraternity house Tuesday evening.
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Mr. Elmer C. Jackson of Kansas City, was a visitor last evening at the Alpha Phi Alpha fraternity house.
Georgia Whitford, c37, and Pesgy Morgan, c37, were dinner guests at the Gamma Phi Beta house Tuesday.
The K. U. Dames' afternoon bridge group will meet this afternoon at 2 o'clock with Mrs. G. D. Carter, 2245 Vermont street.
Geltch Will Play
For University Women's Club
Waldemar Glethner, professor of virology will play a group of violin selections at a final meeting at Wake Forest University's *teronomy* at 3 o'clock in Myers Hall.
PHONE K.U. 66
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Men's leather half soles 60c
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Three Ways To Trisect an Angle
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dicular HP. Move the angle up and down until the apex is on the line, one leg is on the point B and the other leg of the angle is tangent to
Trisection of any given angle was one of the mathematical problems that ancient mathematicians used to puzzle over, along with squaring a circle, or duplicating a square.
Continued on page four
Ancient mathematicians attempted to accomplish these feats with compass and straight edge, but never succeeded. The trisection of an angle still lures many, although mathematicians have proved by suitable formulae that the problem is impossible of solution.
Angles, of course, can be triseeded, but only with special apparatus, and not by ruler and compass alone. A thesis showing three score true and false triesections of an angle was submitted last commencement by S. Reid Hemphill of Emporia. He received his master's degree in mathematics.
One of the simplest methods of triseasing an angle is by means of a carpenter's square. (See Plate III.) Given any angle, BAC, the A representing the apex of the angle, and AC the base line. With a carpenter's square having the larger leg two inches wide, draw a line DF parallel to AC and two inches from it, and intersecting the line AB. Place the square on this figure with the inside of the larger leg of the square at the intersection point of the square on the line DF, and move it along until a point 4 inches from the point of the square is on the line AB, marking the point G on the line AB. Extend the inside of the main arm of the square to H.
Another simple solution is by the use of three arms of equal length, linked into a hinged form (Plate I). The second arm equals CD, with bards at B and C.
trace the leg CD on the baseline ACD, and the end A on the same baseline. Move the end A along the line until the angle formed by the extension of the side AB (as AF) and the line from B to D is the angle to be trieved. The angle CDB equals one-third of the angle DBE.
**roof.** Triangle AEF equals triangle AFH, since both are right angles having the hypotenuse AF in common, and the side JF equal to side FH (each being 2 inches). In the same way triangle AFH equals triangle AHG. Since all three triangles are equal, their angles are equal, and the three angles at A are each one-third of the given angle BAC.
B. making HB equal HO. At H erect a perpendicular. Place the angle to be trisected (represented by an angle from a card if desired) so that the apx of the angle is on the perpen-
Since the exterior angle of a triangle equals the sum of the two opposite angles, then if CDB equals x, then ACB equals 2x. (Angle CBD equals angle CBD, opposite equalides of an isosceles triangle.)
On a base line AB, with O as a
center, strike a semicircle, cutting the
base line at A and N. Mark the point
The angle CBE, therefore, is 4x,
and since CBD is x, then DBE is 3x,
and the measure of its trisection has
been found.
In the same way, angle BAC equals angle ACD, and each may be designated 2x.
Still another trisection, and again using a special device, is shown in Plate II.
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Advertising discusses realities romantic ones often enough but actual articles you can have for your own,new joys for your family,stepping stones to your happiness.
But your daily paper finds you in a more practical mood. Interested in real people, in the facts of the day. Looking for news of things that you may buy and enjoy—here and now.
You may find in the advertisement just what you want in the way of a better breakfast food, or a new radio, or an improved face powder. If the thing fits into your living, is practical and possible and promisingyou are vitally interested.Because you are reading about yourself!
Today, a great many things that were romantic dreams only ten years ago are common realities. Life has more color, more charm, more adventure And the things that give it all these may be found in the advertising columns.
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The advertisements in the Daily Kansan are written for you. They are real. They are reliable. Take their advice.
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