PROPOSITION XXX. PROBLEM

304 Upon a given straight line, to describe a segment of a circle which shall contain a given angle.

E

H

DATA. AB is a straight line, and K is a given angle.

REQUIRED. To describe a segment of a circle upon AB which shall contain the angle K.

$ 284

radius,

From A draw a to AT intersecting DH at O.From O as a center, and with OA (= OB) as a describe a circle AEB.

CONCLUSION. The segment AEB is the segment required.

THE ANALYSIS OF PROBLEMS *

305 The analysis of a problem is a course of reasoning by which a solution is discovered.

In attacking a problem, if no obvious solution presents itself, proceed by analysis as follows:

1. Suppose the problem solved and all necessary constructions made. 2. Study this figure carefully to discover some elementary principle upon which the required construction depends. 3. Make this elementary principle the basis of a synthetic solution. 4. If you fail, draw such auxiliary lines and figures as the nature of the problem may suggest, and proceed as before.

Consider the following illustrations:

306 PROBLEM. To construct a triangle, having given the mid-points of its sides.

ANALYSIS. Suppose the problem solved, and that ABC is the ▲ required, D, E, and F being the given mid-points of the sides. Now DF is || to BC, DE is to AC, and EF is to AB (§ 210). This is the elementary principle which we make the basis of the

B

E

F

SOLUTION. Construct the ▲ DEF, and through each vertex draw a line || to the opposite side, intersecting in A, B, and C. ABC is the ▲ required. It remains to prove that E, F, and D are the mid-points of the triangle ABC.

PROOF. BE is to DF, and BD is BEFD is a O. Likewise ECFD is a

to EF by const. .'. ... BE = DF, and

EC = DF. ... BE=EC. .. E is the mid-point of BC. Likewise D and F are the mid-points of AB and AC respectively.

Q. E. D.

* Suggestions on the treatment of exercises, § 219, apply to the solution of problems as well as to the proof of theorems. These suggestions should be carefully observed.

307 PROBLEM.

From a given point in the circumference of a circle to draw a chord which will be bisected by a given chord.

S

ANALYSIS. Let P be the given point in the Oce, AB the given chord. We are required to draw a chord through P which will be bisected by the fixed chord AB. Suppose the problem solved and that PS is the required chord bisected by AB at H.

Now OH is to PS

(§ 245). Hence the problem is reduced to H in AB such that the

this: Find a point

PHO is a rt. Z.

B

Now an angle inscribed in a semicircle is

a right angle (§ 277), and this is the key to the

SOLUTION. On the radius PO as a diameter describe a Oce intersecting AB in H and K. Draw the chords PHS and PKT, and they are the chords required. There are, in general,

two solutions.

PROOF.

PHO and PKO are rt. 4 (§ 277). .. PS and PT are bisected at H and K respectively (§ 245).

When will there be one solution only? When no solution?

308 Another important aid in the solution of many problems is the intersection of loci.

THE LOCUS OF A POINT

309 DEFINITION. The locus of a point is the place where a point may be under a given condition.

This place may be a point, a line (straight or curved), or a group of lines.

The given condition is called the law of the locus. Locus is a Latin word and means place. The plural is loci.

The following illustrations will make the subject clear.

310 PROBLEM. What is the locus of a point in a plane equidistant from the extremities of a given straight line?

A

Let AB be a given straight line. We are required to find where a point may be so that it is always equidistant from A and B. "Where a point may be " is the locus of the point. "So that it is always equidistant from A and B" is the law of the locus.

Let LL' be the perpendicular bisector of AB.

B

Then, from § 186, we know that the point may be anywhere in LL', and from § 187 we know that the point cannot be without LL'.

Therefore the answer to the question asked in the problem is the following

311 THEOREM. The locus of a point equidistant from the extremities of a straight line is the perpendicular bisector of the line.

Let it be clearly observed that § 186 and § 187 are here concisely stated in this one theorem in loci.

312 The solution of problems in loci involves the finding and the proving of a theorem in loci, and this theorem always implies two propositions, namely: a theorem and its converse, or a theorem and its opposite. Thus, in § 311, we must prove

1 THEOREM.

and B.

All points in LL' are equidistant from A

2 CONVERSE. All points equidistant from A and B are in LL'.

Or, we must prove

1 THEOREM.

and B.

All points in LL' are equidistant from A

2 OPPOSITE. All points not in LL' are not equidistant from A and B.

NOTE. Since the converse and the opposite of a theorem are either both true or both false (§ 22), the demands of Logic are fully met by proving the theorem and either the converse or the opposite. The choice we make is largely a matter of convenience, depending upon the nature of the problem under consideration.

313 Propositions in loci may be given in the form of problems or theorems. In § 310 the proposition is given as a problem. It might have been given in the form of the following

THEOREM. The locus of a point in a plane equidistant from the extremities of a straight line is the perpendicular bisector of the line.

The theorem tells what the locus is and requires the student to prove why it is so. The problem requires the student to find out for himself what the locus is, and then prove why it should be so.

NOTE. In this work the author gives most propositions in loci as problems for the reason that the discovery of the locus is the most interesting and profitable part of the exercise.

A few propositions in loci are now given in form.

314 PROBLEM. What is the locus of a point equidistant from the sides of an angle?

B

DATA. BAC is a given angle.

REQUIRED. To find the locus of a point equidistant from AB and AC.