Chapter I: Geometric Constructions¶
Preliminaries, Analytic Method (Analysis, Construction, Proof, Discussion), Geometric Loci, and Indirect Elements
A. PRELIMINARIES¶
1. Notation. We shall frequently denote by:
\(A, B, C, \dots\) the vertices or the corresponding angles of a polygon;
\(a, b, c, \dots\) 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);
\(2p\) the perimeter of a triangle;
\(h_a, h_b, h_c\) the altitudes and \(m_a, m_b, m_c\) the medians of a triangle \(ABC\) corresponding to the sides \(a, b, c\);
\(t_a, t_b, t_c\) the internal, and \(t_a', t_b', t_c'\) 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.
EXERCISES¶
Construct a triangle, given:
\(a, b, c\).
\(a, b, C\).
\(a, B, C\).
\(a, h_a, B\).
\(a, b, m_a\).
\(a, B, t_b\).
\(A, h_a, t_a\).
Construct a right triangle, with its right angle at \(A\), given:
\(a, B\).
\(b, C\).
\(a, b\).
\(b, c\).
Construct a parallelogram \(ABCD\), given:
\(AB, BC, AC\).
\(AB, AC, B\).
\(AB, BD, \angle ABD\).
Construct a quadrilateral \(ABCD\), given:
\(A, B, C, AB, AD\).
\(AB, BC, CD, B, C\).
\(A, B, C, AD, CD\).
With a given radius to draw a circle tangent at a given point to a given (i) line; (ii) circle.
Through two given points to draw a circle (i) having a given radius; (ii) having its center on a given line.
To a given circle to draw a tangent having a given direction.
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 \(ABC\), \(AB^2 : AC^2 = BD : DC\).)
Construct a right triangle, given the hypotenuse and the ratio of the squares of the legs.
Given the segments \(a, p, q\), construct the segment \(x\) so that \(x^2 : a^2 = p : q\).
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.
ANALYSIS. Let CA'B' be the required line, so that (Fig. 1):
and let O \(= (\)AB, A'B'\()\). The two triangles OAA', OBB' are similar; hence:
CONSTRUCTION. Construct the point O such that:
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.
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;
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:
hence:
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.
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.
CONSTRUCTION. Draw the circle \((B, BA)\) cutting AC again in K. Draw the circle \((K, BA)\) cutting AC in L. The parallel through C to BL meets the line AB in the first required point D, and the parallel through D to BK meets AC in the second required point E.
PROOF. The steps in the proof are the same as those in the analysis, but taken in reverse order.
DISCUSSION. The point K always has one and only one position. When K is constructed we find two positions for L, and we have two solutions, DE and \(D'E'\), for the problem.
If A is a right angle the problem becomes trivial.
ANALYSIS. Let P, Q be the required points on the two given circles (A), (B). Through the center A of (A) (Fig. 4) draw a parallel to PQ and lay off AR = PQ. In the parallelogram APQR we have RQ = AP. But AP is a radius of a given circle; hence the length RQ is known. On the other hand the point R is known, for both the direction and the length of AR are given; hence the point Q may be constructed. The point P is then readily found.
CONSTRUCTION. Through the center A of one of the two given circles, (A), draw a line having the given direction and lay off AR equal to the given length, m. With R as center and radius equal to the radius of (A) draw a circle \((A')\), meeting the second given circle in a point Q. Through Q draw a parallel to AR and lay off QP = AR, so as to form a parallelogram in which AR is a side (and not a diagonal). The points P, Q satisfy the conditions of the problem.
DISCUSSION. It is always possible to draw through A a line having the given direction. The point R may then be marked on either side of A, so that we always have for R two positions, R and R'. The circle with either of these points as center and radius equal to the radius of (A) may cut (B) in two points, or may be tangent to (B), or may not cut (B) at all. Consequently the problem may have four, three, two, one, or no solutions.
Figure 5 illustrates the case when there are four solutions.
ANALYSIS. Let PQRS be a square whose sides PQ, QR, RS, SP pass, respectively, through the given points A, B, C, D (Fig. 6).
If the perpendicular from D upon AC meets the line QR in F, then DF = AC. Indeed, if we leave the line DF fixed while we revolve the square, and the line AC with it, around its center by an angle of \(90^\circ\) so that the sides PQ, QR, RS, SP shall occupy the present positions of QR, RS, SP, PQ, respectively, the line AC will become parallel to DF; hence AC = DF.
That DF = AC may also be seen in another way. Let the parallels through Q to the lines AC, DF meet RS, SP in the points K, L. In the right triangles PQL, QRK we have:
The equality of these two segments suggests the following
CONSTRUCTION. From the given point D drop a perpendicular DF upon the given line AC and lay off DF = AC. Join F to the fourth given point B, and through D draw a parallel to BF. These two parallel lines and the perpendiculars to them through A and C form the required square PQRS.
PROOF. From the construction it follows immediately that PQRS is a rectangle whose sides pass respectively through the given points A, B, C, D.
DISCUSSION. We obtain a second solution if we make the symmetric F' of F with respect to D play the role of the point F.
Moreover, the perpendicular from D may also be dropped upon either of the other two sides AB, BC of the triangle ABC, and we will obtain two more pairs of solutions. The problem thus has, in general, six different solutions. It is, of course, immaterial to which of the four given points we assign the role of the point D.
Should the point F happen to coincide with B, the direction of the line BF becomes indeterminate, that is, any line through B may be taken as a side of the required square, and the figure completed accordingly. Thus if the segment joining two of the four given points is equal to and perpendicular to the segment joining the remaining two points, the problem has an infinite number of solutions.
Figure 7 presents the case when the problem has six solutions.
10. Suggestions. The figure for the analysis may be drawn freehand and the given elements arranged as conveniently as possible to give a correct idea of the problem and to exhibit the existing interrelations between the elements involved.
The construction should be carried out with ruler and compasses. The given magnitudes, like segments and angles, should be set down before any construction is attempted, and used in the actual construction. If the problem involves points and lines given in position, these data should be marked in the figure before any constructional operations are begun.
In the discussion each step of the construction should be examined for the number of ways it may be carried out and for the number of lines or points of intersection the considered step will yield.
The different aspects which the problem may assume, as indicated by the discussion, should be illustrated by suitable figures. As a general rule a problem or a figure has many more possibilities than is at first apparent. A careful study of a problem will often reveal vistas which in a more casual treatment may readily escape notice.
EXERCISES¶
Through a given point to draw a line making equal angles with the sides of a given angle.
Through a given point to draw a line so that two given parallel lines shall intercept on it a segment of given length.
Through one of the two points of intersection of two equal circles to draw two equal chords, one in each circle, forming a given angle.
Through a given point of a circle to draw a chord which shall be twice as long as the distance of this chord from the center of the circle.
On a produced diameter of a given circle to find a point such that the tangents drawn from it to this circle shall be equal to the radius of the circle.
With a given point as center to describe a circle which shall bisect a given circle, that is, the common chord shall be a diameter of the given circle.
Through two given points to draw a circle so that its common chord with a given circle shall be parallel to a given line.
Construct a parallelogram so that three of its sides shall have for midpoints three given points.
On a given leg of a right triangle to find a point equidistant from the hypotenuse and from the vertex of the right angle.
With two given points as centers to draw equal circles so that one of their common tangents shall (i) pass through a (third) given point; (ii) be tangent to a given circle.
Through a given point to draw a line so that the two chords intercepted on it by two given equal circles shall be equal.
To a given circle, located between two parallel lines, to draw a tangent so that the segment intercepted on it by the given parallels shall have a given length.
Construct a right triangle given the hypotenuse and the distance from the middle point of the hypotenuse to one leg.
Construct a triangle given an altitude and the circumradii (i.e., the radii of the circumscribed circles) of the two triangles into which this altitude divides the required triangle.
C. GEOMETRIC LOCI¶
11. Important Loci. In a great many cases the solution of a geometric problem depends upon the finding of a point which satisfies certain conditions. For instance, in order to draw a circle passing through three given points, it is necessary to find a point, the center of the circle, equidistant from the three given points.
The problem of drawing a tangent from a given point to a given circle is solved when we find the point of contact, i.e., the point on the circle at which a right angle is subtended by the segment limited by the given point and the center of the given circle.
If one of the conditions which the required point must satisfy be set aside, the problem may have many solutions. However, the point will not become arbitrary, but will move along a certain path, the geometric locus of the point. Now by taking into consideration the discarded condition and setting aside another, we make the required point describe another geometric locus. A point common to the two loci is the point sought.
The problem of drawing a circle passing through three given points may again serve as an illustration. In order to find a point equidistant from three given points, \(A, B, C\), we disregard one of the given points, say \(C\), and try to find a point equidistant from \(A\) and \(B\). Thus stated, the problem has many solutions, the required point being any point of the perpendicular bisector of the segment \(AB\). Now considering the point \(C\) and leaving out the point \(A\), the required point describes the perpendicular bisector of the segment \(BC\). The required point lies at the intersection of the two perpendicular bisectors.
The nature of the loci obtained depends upon the condition omitted. In elementary geometry these conditions must be such that the loci shall consist of straight lines and circles. The neatness and simplicity of a solution depend very largely upon the judicious choice of the geometric loci.
Knowledge about a considerable number of geometric loci may often enable one to discover immediately where the required point is to be located.
The following are the most important and the most frequently useful geometric loci:
For the sake of brevity the expression perpendicular bisector may be replaced conveniently by the term mediator.
Let M (Fig. 8) be a point of the locus so that the angle AMB is equal to the given angle. Pass a circle through the three points A, B, M. At any point M' of the arc AMB the segment AB subtends the same angle as at M; hence every point of the arc AMB belongs to the required locus. On the other hand, any point N not on the arc AMB will lie either inside or outside this arc. In the first case the angle ANB will be larger, and in the second case smaller, than the angle AMB. Hence N does not belong to the locus.
If the condition that the point M is to lie on a given side of the line AB is disregarded, the required locus consists of two arcs of circles congruent to each other and located on opposite sides of AB.
If the given angle is a right angle, the two arcs will be two semicircles of the same radius, and we have: The locus of a point at which a given segment subtends a right angle is the circle having this segment for diameter.
Let P (Fig. 9) be the given point, and M the midpoint of a chord AB of the given circle (O), the chord passing through the given point P.
If the point P lies outside of (O), any point M of the locus must lie on the circle having OP for diameter, but not every point of this circle belongs to the locus. The locus consists of the part of the circle (OP) which lies within the given circle (O).
The chords are all tangent to this circle at their respective midpoints.
Let M (Fig. 10) be a point on the locus. Join M to the midpoint O of the segment AB determined by the two given points A, B and drop the perpendicular MD from M upon AB. From the triangles AOM and BOM we have:
But:
hence, adding, we have:
Now \(AM^2 + BM^2\) is, by assumption, equal to a given constant, say \(s^2\), and \(OA^2 = a^2\), where \(2a\) represents the length of the known segment AB. Hence:
The radius OM may be constructed from the formula (1) as follows. OM is one leg of a right triangle which has for its other leg half the length of the given segment AB, and for hypotenuse the side of a square whose diagonal is equal to \(s\).
Let A, B be the two given points (Fig. 11) and \(p : q\) the given ratio. Let the points C, D divide the segment AB internally and externally in the given ratio, and let M be any point of the locus. We have thus:
CONVERSELY. Let us take any point M on this circle and show that this point belongs to the required locus. Through the point B draw the parallels BE, BF to the lines MC, MD meeting AM in E, F, respectively. We thus have:
But the second ratios in these proportions are equal, by construction, hence:
therefore:
Thus M is the midpoint of the segment EF. But EBF is a right angle, for its sides are parallel to the sides of the right angle CMD, and in the right triangle EBF the line MB is equal to half the hypotenuse EF; hence, replacing in the first proportion of (1) the segment ME by the equal segment MB, we obtain:
which shows that M belongs to the locus.
Where, in this converse proof, has the fact been used that M is a point of the circle?
Note. The point C divides the segment AB internally so that \(AC : CB = p : q\). But we may also construct the point \(C'\) such that \(BC' : C'A = p : q\). Similarly for the external point of division. Thus the locus actually consists of two Apollonian circles, unless the order in which the two given points are to be taken is specified in the statement of the locus. In actual applications the nature of the problem often indicates the order in which the points are to be considered.
Let A, B (Fig. 12) be the two given points, and let M be a point such that:
where \(d\) is given.
If MC is the perpendicular from M to AB, we have from right triangles:
hence:
or:
and therefore, denoting AC + BC = AB by \(a\):
Note. We will obtain a different line for the locus of M if we consider that \(BM^2 - AM^2 = d^2\). In fact, the locus actually consists of two straight lines, unless the order in which the two given points are to be considered is specified in the statement of the locus.
EXERCISES¶
A variable parallel to the base \(BC\) of a triangle \(ABC\) meets \(AB, AC\) in \(D, E\). Show that the locus of the point \(M = (BE, CD)\) is a straight line.
On the sides \(AB, AC\) of a triangle \(ABC\) are laid off two equal segments \(AB', AC'\) of variable length. The perpendiculars to \(AB, AC\) at \(B', C'\) meet in \(D\). Show that the locus of the point \(D\) is a straight line. Find the locus of the projection of \(D\) upon the line \(B'C'\).
Find the locus of a point at which two consecutive segments \(AB, BC\) of the same straight line subtend equal angles.
The base \(BC\) of a variable triangle \(ABC\) is fixed, and the sum \(AB + AC\) is constant. The line \(DP\) drawn through the midpoint \(D\) of \(BC\) parallel to \(AB\) meets the parallel \(CP\) through \(C\) to the internal bisector of the angle \(A\), in \(P\). Show that the locus of \(P\) is a circle having \(D\) for center.
The following are examples of the use of loci in the solution of problems.
ANALYSIS. Let (O) be the required circle passing through the two given points A, B (Fig. 13). The angle formed by the tangents CT, CT' from the given point C to (O) is given; hence in the right triangle OTC we know the acute angle OCT, i.e., we know the shape of this triangle. Thus the ratio OT \(:\) OC is known, therefore also the ratios:
CONSTRUCTION. On the internal bisector \(PQ\) of the given angle \(P\) take an arbitrary point \(Q\) and drop the perpendicular \(QR\) upon one of the sides. Divide each of the two given segments AC, BC internally and externally in the ratio \(QR : QP\) in the points \(E, F\) and \(G, H\), respectively. A point O common to the two circles having \(EF\) and \(GH\) for diameters is the center of the required circle.
The proof and the discussion are left to the student.
ANALYSIS. Let A, B be the two given points of the circle, center O, and let AC, BD be the two required chords. In the isosceles trapezoid ABDC (Fig. 14) CD = AB, and the length AB is known; hence CD is tangent to a known circle having O for center and touching CD at its midpoint F (§ 11, locus 9).
If E is the midpoint of AB we have:
Now the point E and the length \(AC + BD\) are known; hence we have a second locus for the point F.
CONSTRUCTION. Draw the circle (O, OE). If \(2s\) is the given length, draw the circle (E, s) meeting (O, OE) in F. The tangent to (O, OE) at F meets the given circle (O) in C and D. The lines AC, BD are the required chords.
PROOF. The two chords AB, CD are equal, for they are equidistant from the center O of the given circle (O). Hence ABDC is an isosceles trapezoid, and therefore:
Now \(EF = s\), by construction; hence \(AC + BD\) has the required length.
DISCUSSION. The circle (E, s) will not cut the circle (O, OE), if \(s\) is greater than \(2\,OE\). If \(s < 2\,OE\), we obtain two points of intersection F and F', and therefore two solutions.
The tangent to the circle (O, OE) at F determines the two points C, D on the circle (O). These two points and the given points A, B determine four lines, namely two sides and the two diagonals of the isosceles trapezoid. The figure will show which two of these four lines are those required.
Let G (Fig. 15) be the point of contact of the second tangent issued from A to the circle (O, OE). If \(s < EG\), the tangent to (O, OE) at F will cross the chord AB, and in the resulting trapezoid the line AB will be a diagonal. The line EF is equal to one-half the difference of the two bases.
Consider the case when the two chords are required to have a given difference.
Let the lines AM, BM joining the given points A, B on the circle to the required point M meet the given line FPQ in the points P, Q (Fig. 16). If F is the given fixed point and the parallel QL to MA through Q meets the line FA in L, we have angle LQB = AMB, and the latter angle is known, for the chord AB is given.
On the other hand we have:
The proof and discussion are left to the reader.
EXERCISES¶
Construct a triangle, given:
\(a, b, A\).
\(a, c, h_b\).
\(a, h_a, m_a\).
\(a, h_a, b : c\).
\(a, m_a, b : c\).
\(a, t_a, b : c\).
Construct a parallelogram, given:
An altitude and the two diagonals.
The two altitudes and an angle.
The two altitudes and a diagonal.
A side, an angle, and a diagonal.
A side, the corresponding altitude, and the angle between the diagonals.
Construct a quadrilateral \(ABCD\), given:
The diagonal \(AC\) and the angles \(ABC, ADC, BAC, DAC\).
The sides \(AB, BC\), the diagonal \(CA\), and the angles \(ADB, BDC\).
The sides \(AB, AD\), the angle \(DAB\), and the radius of the inscribed circle.
Construct a quadrilateral given three sides and the radius of the circumscribed circle. Give a discussion.
Given three points, to find a fourth point, in the same plane, such that its distances to the given points may have given ratios.
With a given radius to draw a circle so that it shall touch a given circle and have its center on a given line.
With a given radius to draw a circle so that it shall pass through a given point and the tangents drawn to this circle from another given point shall be of given length.
In a given circle to inscribe a right triangle so that each leg shall pass through a given point.
Construct a triangle given the base, the opposite angle, and the point in which the bisector of this angle meets the base.
Construct a triangle given the base and the angles which the median to the base makes with the other two sides.
About a given circle to circumscribe a triangle given a side and one of its adjacent angles, so that the vertex of this angle shall lie on a given line.
Construct a triangle given the base, an adjacent angle, and the angle which the median issued from the vertex of this angle makes with the side opposite the vertex of this angle.
D. INDIRECT ELEMENTS¶
Among the conditions which a figure to be constructed may be required to satisfy, elements may be given which do not occur directly in the figure in question. For instance, it may be required that the sum of two sides of a triangle shall have a given length, or that the difference of the base angles shall have a given magnitude, etc. In order to arrive at a method of solution of such a problem, it is necessary to introduce this “indirect element” in the analysis of the problem.
Let ABC (Fig. 17) be the required triangle. Produce BC on both sides and lay off BE = BA, CF = CA. Thus EF \(= 2p\).
The triangles EAB, FCA are isosceles, hence:
hence:
Note
Editorial Continuation (Folios ``0041``–``0052``, Pages 22–33)
Article § 15 continues onto Folio 0041 (p. 22): “and therefore the angle \(EAF\) is known. The altitude \(AD\) of the triangle \(ABC\) is also the altitude of \(AEF\) … Since \(BA = BE, CA = CF\), the vertices \(B, C\) are the traces on \(EF\) of the mediators of \(AE\) and \(AF\) .”
All 300 DPI archival scans for the remainder of Chapter I: Geometric Constructions (0041.png–0052.png, covering §§ 15–22 and Review Exercises 1–74) are ingested in resources/college_geometry/01_geometric_constructions/ and indexed below for Phase 2 transcription.
Remaining Chapter I Architecture (Folios 0041–0052)¶
Section |
Title |
Folios |
Pages |
Articles |
|---|---|---|---|---|
|
Preliminaries |
|
1–2 |
§§ 1–3 |
|
General Method of Solution of Construction Problems |
|
3–11 |
§§ 4–10 |
|
Geometric Loci |
|
11–21 |
§§ 11–14 |
|
Indirect Elements |
|
21–29 |
§§ 15–22 |
|
Review Exercises (1–74) |
|
29–33 |
— |