The Teaching of Geometry Part 19

PROBLEM. _To let fall a perpendicular upon a given line from a given external point._

This is the first problem that a student meets in most American geometries. The reason for treating the problems by themselves instead of mingling them with the theorems has already been discussed.[69] The student now has a sufficient body of theorems, by which he can prove that his constructions are correct, and the advantage of treating these constructions together is greater than that of following Euclid's plan of introducing them whenever needed.

Proclus tells us that "this problem was first investigated by Oenopides,[70] who thought it useful for astronomy." Proclus speaks of such a line as a gnomon, a common name for the perpendicular on a sundial, which casts the shadow by which the time of day is known. He also speaks of two kinds of perpendiculars, the plane and solid, the former being a line perpendicular to a line, and the latter a line perpendicular to a plane.

It is interesting to notice that the solution tacitly a.s.sumes that a certain arc is going to cut the given line in two points, and only two.

Strictly speaking, why may it not cut it in only one point, or even in three points? We really a.s.sume that if a straight line is drawn through a point within a circle, this line must get out of the circle on each of two sides of the given point, and in getting out it must cut the circle twice. Proclus noticed this a.s.sumption and endeavored to prove it. It is better, however, not to raise the question with beginners, since it seems to them like hair-splitting to no purpose.

The problem is of much value in surveying, and teachers would do well to ask a cla.s.s to let fall a perpendicular to the edge of a sidewalk from a point 20 feet from the walk, using an ordinary 66-foot or 50-foot tape.

Practically, the best plan is to swing 30 feet of the tape about the point and mark the two points of intersection with the edge of the walk.

Then measure the distance between the points and take half of this distance, thus fixing the foot of the perpendicular.

PROBLEM. _At a given point in a line, to erect a perpendicular to that line._

This might be postponed until after the problem to bisect an angle, since it merely requires the bisection of a straight angle; but considering the immaturity of the average pupil, it is better given independently. The usual case considers the point not at the extremity of the line, and the solution is essentially that of Euclid. In practice, however, as for example in surveying, the point may be at the extremity, and it may not be convenient to produce the line.

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Surveyors sometimes measure _PB_ = 3 ft., and then take 9 ft.

of tape, the ends being held at _B_ and _P_, and the tape being stretched to _A_, so that _PA_ = 4 ft. and _AB_ = 5 ft. Then _P_ is a right angle by the Pythagorean Theorem. This theorem not having yet been proved, it cannot be used at this time.

A solution for the problem of erecting a perpendicular from the extremity of a line that cannot be produced, depending, however, on the problem of bisecting an angle, and therefore to be given after that problem, is attributed by Al-Nair[=i]z[=i] (tenth century A.D.) to Heron of Alexandria. It is also given by Proclus.

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Required to draw from _P_ a perpendicular to _AP_. Take _X_ anywhere on the line and erect _XY_ [perp] to _AP_ in the usual manner. Bisect [L]_PXY_ by the line _XM_. On _XY_ take _XN_ = _XP_, and draw _NM_ [perp] to _XY_. Then draw _PM_. The proof is evident.

These may at the proper time be given as interesting variants of the usual solution.

PROBLEM. _To bisect a given line._

Euclid said "finite straight line," but this wording is not commonly followed, because it will be inferred that the line is finite if it is to be bisected, and we use "line" alone to mean a straight line.

Euclid's plan was to construct an equilateral triangle (by his Proposition 1 of Book I) on the line as a base, and then to bisect the vertical angle. Proclus tells us that Apollonius of Perga, who wrote the first great work on conic sections, used a plan which is substantially that which is commonly found in textbooks to-day,--constructing two isosceles triangles upon the line as a common base, and connecting their vertices.

PROBLEM. _To bisect a given angle._

It should be noticed that in the usual solution two arcs intersect, and the point thus determined is connected with the vertex. Now these two arcs intersect twice, and since one of the points of intersection may be the vertex itself, the other point of intersection must be taken. It is not, however, worth while to make much of this matter with pupils.

Proclus calls attention to the possible suggestion that the point of intersection may be imagined to lie outside the angle, and he proceeds to show the absurdity; but here, again, the subject is not one of value to beginners. He also contributes to the history of the trisection of an angle. Any angle is easily trisected by means of certain higher curves, such as the conchoid of Nicomedes (_ca._ 180 B.C.), the quadratrix of Hippias of Elis (_ca._ 420 B.C.), or the spiral of Archimedes (_ca._ 250 B.C.). But since this problem, stated algebraically, requires the solution of a cubic equation, and this involves, geometrically, finding three points, we cannot solve the problem by means of straight lines and circles alone. In other words, the trisection of _any_ angle, by the use of the straightedge and compa.s.ses alone, is impossible. Special angles may however be trisected. Thus, to trisect an angle of 90 we need only to construct an angle of 60, and this can be done by constructing an equilateral triangle. But while we cannot trisect the angle, we may easily approximate trisection. For since, in the infinite geometric series 1/2 + 1/8 + 1/32 + 1/128 + ..., _s_ = _a_ (1 - _r_), we have _s_ = 1/2 3/4 = 2/3. In other words, if we add 1/2 of the angle, 1/8 of the angle, 1/32 of the angle, and so on, we approach as a limit 2/3 of the angle; but all of these fractions can be obtained by repeated bisections, and hence by bisections we may approximate the trisection.

The approximate bisection (or any other division) of an angle may of course be effected by the help of the protractor and a straightedge. The geometric method is, however, usually more accurate, and it is advantageous to have the pupils try both plans, say for bisecting an angle of about 49 1/2.

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Applications of this problem are numerous. It may be desired, for example, to set a lamp-post on a line bisecting the angle formed by two streets that come together a little unsymmetrically, as here shown, in which case the bisecting line can easily be run by the use of a measuring tape, or even of a stout cord.

A more interesting ill.u.s.tration is, however, the following:

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Let the pupils set a stake, say about 5 feet high, at a point _N_ on the school grounds about 9 A.M., and carefully measure the length of the shadow, _NW_, placing a small wooden pin at _W_. Then about 3 P.M. let them watch until the shadow _NE_ is exactly the same length that it was when _W_ was fixed, and then place a small wooden pin at _E_. If the work has been very carefully done, and they take the tape and bisect the line _WE_, thus fixing the line _NS_, they will have a north and south line. If this is marked out for a short distance from _N_, then when the shadow falls on _NS_, it will be noon by sun time (not standard time) at the school.

PROBLEM. _From a given point in a given line, to draw a line making an angle equal to a given angle._

Proclus says that Eudemus attributed to Oenopides the discovery of the solution which Euclid gave, and which is substantially the one now commonly seen in textbooks. The problem was probably solved in some fashion before the time of Oenopides, however. The object of the problem is primarily to enable us to draw a line parallel to a given line.

Practically, the drawing of one line parallel to another is usually effected by means of a parallel ruler (see page 191), or by the use of draftsmen's triangles, as here shown, or even more commonly by the use of a T-square, such as is here seen. This ill.u.s.tration shows two T-squares used for drawing lines parallel to the sides of a board upon which the drawing paper is fastened.[71]

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An ingenious instrument described by Baron Dupin is ill.u.s.trated below.

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To the bar _A_ is fastened the sliding check _B_. A movable check _D_ may be fastened by a screw _C_. A sharp point is fixed in _B_, so that as _D_ slides along the edge of a board, the point marks a line parallel to the edge. Moreover, _F_ and _G_ are two bra.s.s arms of equal length joined by a pointed screw _H_ that marks a line midway between _B_ and _D_.

Furthermore, it is evident that _H_ will draw a line bisecting any irregular board if the checks _B_ and _D_ are kept in contact with the irregular edges.

Book II offers two general lines of application that may be introduced to advantage, preferably as additions to the textbook work. One of these has reference to topographical drawing and related subjects, and the other to geometric design. As long as these can be introduced to the pupil with an air of reality, they serve a good purpose, but if made a part of textbook work, they soon come to have less interest than the exercises of a more abstract character. If a teacher can relate the problems in topographical drawing to the pupil's home town, and can occasionally set some outdoor work of the nature here suggested, the results are usually salutary; but if he reiterates only a half-dozen simple propositions time after time, with only slight changes in the nature of the application, then the results will not lead to a cultivation of power in geometry,--a point which the writers on applied geometry usually fail to recognize.

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One of the simple applications of this book relates to the rounding of corners in laying out streets in some of our modern towns where there is a desire to depart from the conventional square corner. It is also used in laying out park walks and drives.

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The figure in the middle of the page represents two streets, _AP_ and _BQ_, that would, if prolonged, intersect at _C_. It is required to construct an arc so that they shall begin to curve at _P_ and _Q_, where _CP_ = _CQ_, and hence the "center of curvature" _O_ must be found.

The problem is a common one in railroad work, only here _AP_ is usually oblique to _BQ_ if they are produced to meet at _C_, as in the second figure on page 218. It is required to construct an arc so that the tracks shall begin to curve at _P_ and _Q_, where _CP_ = _CQ_.

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The problem becomes a little more complicated, and correspondingly more interesting, when we have to find the center of curvature for a street railway track that must turn a corner in such a way as to allow, say, exactly 5 feet from the point _P_, on account of a sidewalk.

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The problem becomes still more difficult if we have two roads of different widths that we wish to join on a curve. Here the two centers of curvature are not the same, and the one road narrows to the other on the curve. The solutions will be understood from a study of the figures.

The number of problems of this kind that can easily be made is limitless, and it is well to avoid the danger of hobby riding on this or any similar topic. Therefore a single one will suffice to close this group.

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If a road _AB_ on an arc described about _O_, is to be joined to road _CD_, described about _O'_, the arc _BC_ should evidently be internally tangent to _AB_ and externally tangent to _CD_. Hence the center is on _BOX_ and _O'CY_, and is therefore at _P_. The problem becomes more real if we give some width to the roads in making the drawing, and imagine them in a park that is being laid out with drives.

It will be noticed that the above problems require the erecting of perpendiculars, the bisecting of angles, and the application of the propositions on tangents.

A somewhat different line of problems is that relating to the pa.s.sing of a circle through three given points. It is very easy to manufacture problems of this kind that have a semblance of reality.

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