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An Introduction to the Modern Geometry of the Triangle and the Circle
By Nathan Altshiller-Court
Dover Publications, Inc.Copyright © 1980 Arnold Court
All rights reserved.
1. Notation. We shall frequently denote by:
A, B, C, ... the vertices or the corresponding angles of a polygon;
a, b, c, ... the sides of the polygon (in the case of a triangle, the small letter will denote the side opposite the vertex indicated by the same capital letter);
2 p the perimeter of a triangle;
ha, hb, hc the altitudes and ma, mb, mc the medians of a triangle ABC corresponding to the sides a, b, c;
ta, tb, tc the internal, and ta', tb', tc' the external, bisectors of the angles A, B, C;
R, r the radii of the circumscribed and inscribed circles (for the sake of brevity, we shall use the terms circumcircle, circumradius, circumcenter, and incircle, inradius, incenter);
(A, r) the circle having the point A for center and the segment r for radius;
M = (PQ, RS) the point of intersection M of the two lines PQ and RS.
2. Basic Constructions. Frequent use will be made of the following constructions:
To divide a given segment into a given number of equal parts.
To divide a given segment in a given ratio (i) internally; (ii) externally (§ 54).
To construct the fourth proportional to three given segments.
To construct the mean proportional to two given segments.
To construct a square equivalent to a given (i) rectangle; (ii) triangle.
To construct a square equivalent to the sum of two, three, or more given squares.
To construct two segments given their sum and their difference.
To construct the tangents from a given point to a given circle.
To construct the internal and the external common tangents of two given circles.
Construct a triangle, given:
1.a, b, c.
2.a, b, C.
3.a, B, C.
4.a, ha, B.
5.a, b, ma.
6.a, B, tb.
7.A, ha, ta.
Construct a right triangle, with its right angle at A, given:
Construct a parallelogram ABCD, given:
12.AB, BC, AC.
13.AB, AC, B.
14.AB, BD, [angle]ABD.
Construct a quadrilateral ABCD, given:
15.A, B, C, AB, AD.
16.AB, BC, CD, B, C.
17.A, B, C, AD, CD.
18. With a given radius to draw a circle tangent at a given point to a given (i) line; (ii) circle.
19. Through two given points to draw a circle (i) having a given radius; (ii) having its center on a given line.
20. To a given circle to draw a tangent having a given direction.
21. To divide a given segment internally and externally in the ratio of the squares of two given segments p, q. (Hint. If AD is the perpendicular to the hypotenuse BC of the right triangle ABC, AB2:AC2 = BD:DC.)
22. Construct a right triangle, given the hypotenuse and the ratio of the squares of the legs.
23. Given the segments a, p, q, construct the segment x so that x2:a2 = p:q.
24. Construct an equilateral triangle equivalent to a given triangle.
3. Suggestion. Most of the preceding problems are stated in conventional symbols. It is instructive to state them in words. For instance, Exercise 4 may be stated as follows: Construct a triangle given the base, the corresponding altitude, and one of the base angles.
B. GENERAL METHOD OF SOLUTION OF CONSTRUCTION PROBLEMS
4. Analytic Method. Some construction problems are direct applications of known propositions and their solutions are almost immediately apparent. Example: Construct an equilateral triangle.
If the solution of a problem is more involved, but the solution is known, it may be presented by starting with an operation which we know how to perform, followed by a series of operations of this kind, until the goal is reached.
This procedure is called the synthetic method of solution of problems. It is used to present the solutions of problems in textbooks.
However, this method cannot be followed when one is confronted with a problem the solution of which is not apparent, for it offers no clue as to what the first step shall be, and the possible first steps are far too numerous to be tried at random.
On the other hand, we do know definitely what the problem is — we know what figure we want to obtain in the end. It is therefore helpful to start with this very figure, provisionally taken for granted. By a careful and attentive study of this figure a way may be discovered leading to the desired solution. The procedure, which is called the analytic method of solving problems, consists, in outline, of the following steps:
ANALYSIS. Assuming the problem solved, draw a figure approximately satisfying the conditions of the problem and investigate how the given parts and the unknown parts of the figure are related to one another, until you discover a relation that may be used for the construction of the required figure.
CONSTRUCTION. Utilizing the information obtained in the analysis, carry out the actual construction.
PROOF. Show that the figure thus constructed satisfies all the requirements of the problem.
DISCUSSION. Discuss the problem as to the conditions of its possibility, the number of solutions, etc.
The following examples illustrate the method.
5. Problem.Two points A, B are marked on two given parallel lines x, y. Through a given point C, not on either of these lines, to draw a secant CA'B' meeting x, y in A', B' so that the segments AA', BB' shall be proportional to two given segments p, q.
ANALYSIS. Let CA'B' be the required line, so that (Fig. 1):
AA':BB' = p:q,
and let O = (AB, A'B'). The two triangles OAA', OBB' are similar; hence:
AO:BO = AA':BB'.
But the latter ratio is known; hence the point O divides the given segment AB in the given ratio p:q. Thus we may construct O, and OC is the required line.
CONSTRUCTION. Construct the point 0 such that:
AO:BO = p:q.
The points O and C determine the required line.
PROOF. Left to the student.
DISCUSSION. There are two points, O and O', which divide the given segment AB in the given ratio p:q, one externally and the other internally, and we can always construct these two points; hence the problem has two solutions if neither of the lines CO, CO' is parallel to the lines x, y.
Consider the case when p = q.
6. Problem.Through a given point, outside a given circle, to draw a secant so that the chord intercepted on it by the circle shall subtend at the center an angle equal to the acute angle between the required secant and the diameter, produced, passing through the given point.
ANALYSIS. Let the required secant MBA (Fig. 2) through the given point M cut the given circle, center O, in A and B. The two triangles AOB, AOM have the angle A in common and, by assumption, angle AOB = angle M; hence the two triangles are equiangular. But the triangle AOB is isosceles; hence the triangle AOM is also isosceles, and MA = MO. Now the length MO is known; hence the distance MA of the point A from M is known, so that the point A may be constructed, and the secant MA may be drawn.
CONSTRUCTION. Draw the circle (M, MO). If A is a point common to the two circles, the line MA satisfies the conditions of the problem.
PROOF. Let the line MA meet the given circle again in B. The triangles AOB, AOM are isosceles, for OA = OB, MA = MO, as radii of the same circle, and the angle A is a common base angle in the two triangles; hence the angles AOB and M opposite the respective bases AB and AO in the two triangles are equal. Thus MA is the required line.
DISCUSSION. We can always draw the circle (M, MO) which will cut the given circle in two points A and A'; hence the problem always has two solutions, symmetrical with respect to the line MO.
Could the line MBA be drawn so that the angle AOB would be equal to the obtuse angle between MBA. and MO? If that were possible, we would have:
[angle]AOB + [angle]OMA = 180°
[angle]OMA = [angle]OAB + [angle]OBA.
But in the triangle OBM we have:
We are thus led to a contradiction; hence a line satisfying the imposed condition cannot be drawn.
Consider the problem when the point M is given inside the given circle, or on the given circle.
7. Problem.On the sides AB, AC, produced if necessary, of the triangle ABC to find two points D, E such that the segments AD, DE, and EC shall be equal (Fig. 3).
ANALYSIS. Suppose that the points D, E satisfy the conditions of the problem, and let the parallels through B to the lines DE, DC meet AC in K, L. The two triangles ADE, ABK are similar, and since AD = DE, by assumption, we have AB = BK, and the point K is readily constructed.
Again, the triangles DEC, BKL are similar, and since DE = EC, by assumption, we have BK = KL; hence L is known.
Excerpted from College Geometry by Nathan Altshiller-Court. Copyright © 1980 Arnold Court. Excerpted by permission of Dover Publications, Inc..
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