In some of the earlier chapters we had sections on construction problems. In this chapter, we expand further and describe some useful techniques.
The object is to draw geometric figures in the plane using two simple tools:
The constructions that can be accomplished using these two basic tools are called Euclidean constructions.
There are certain famous construction problems that have been shown to be impossible in the sense that they cannot be accomplished with a straightedge and compass. These are:
Given a segment of length 1 unit, what other lengths can we construct? The construction must use only a straightedge and a compass and must be accomplished in a finite number of steps. This section describes what lengths we can construct. A complete discussion would prove that these are the only lengths that we can construct, but that is well beyond the scope of this book.
It is clear that, given a segment of length 1, we can construct a segment of length n where n is any positive integer.
Given segments of length p and q, we can construct segments of length p + q and p − q.
What about constructing segments of length p/q or pq?
Example 6.2.1. Given segments of lengths 1, p, and q, construct a segment of length p/q.
Solution.
The diagram above reveals how to use similar triangles to accomplish the task. The actual construction is as follows:
The proof that DE has length p/q follows from the fact that ΔABC ~ ΔADE.
Exercise 6.2.2. Given segments of lengths 1, p, and q, construct a segment of length pq.
Example 6.2.3. Given segments of length a and b, show how to construct a segment of length .
Solution. The key to this is to somehow use Pythagoras’ Theorem. It is actually used in conjunction with Thales’ Theorem, which tells us that the angle inscribed in a semicircle is 90°. Here is the analysis figure:
Since ΔADC ~ ΔCDB, we have
so that
and it follows that c = .
Thus the construction is as follows:
Note. Using the power of a point also yields the same construction:
Since CE is perpendicular to the diameter AB, we must have CD = DE. By the power of the point D, we have CD · DE = AD · DB, that is, CD = .
The ancient Greeks described a number x as being constructible if, starting with a segment of length 1, you could construct a segment of length x. For example, 3/5 is a constructible number: starting with a segment of length 1, construct segments of length 3 and 5, and then using Example 6.2.1, construct a segment of length 3/5.
Combining Examples 6.2.1, 6.2.2, and 6.2.3 we have the following theorem:
Theorem 6.2.4. (Constructible Numbers)
If the nonnegative numbers a and b are constructible, then so are the following numbers:
We can build many constructible numbers by taking a succession of these operations. For example:
Example 6.2.5. Show that is constructible.
Solution. The numbers 1 and are constructible, so is constructible, thus is also constructible and then so is . Since
we are finished.
Starting with the number 1, and by taking a finite succession of additions, subtractions, ratios, products, and square roots, with repetitions allowed, we can obtain all of the constructible numbers. For example, the number
can be obtained in the way described, so it is constructible.
These are the only types of numbers that are constructible. This can be rather tricky, because there are many ways of expressing the same number. For example, at first glance
may not appear to be constructible, but it is constructible because
Some numbers that are known not to be constructible are
This explains why the Three Problems of Antiquity cannot be solved.
Remark. To reiterate, a number a is constructible if and only if, given a segment of unit length, it is possible to construct a segment of length |a| using only a straightedge and compass. It can be shown using Galois theory that the only numbers that are constructible are the following:
Recall that a simple polygon is called a regular polygon if all of its sides are congruent and all of its vertex angles are congruent. This section deals with the problem of how to construct some of the regular polygons.
There is little difficulty in constructing the regular n-gon for n = 3 (an equilateral triangle), for n = 4 (a square), and for n = 6 (a regular hexagon).
In an earlier problem, we saw that all of the regular polygons have their vertices on a circle, so construction of the regular n-gon amounts to finding n equally spaced points on a circle. The general problem then becomes:
Given a circle of radius 1, find n equally spaced points on that circle.
The situation for the pentagon is shown on the right.
The central angle α is given by
while the vertex angle β is given by
since α + 2 · (β/2) = 180.
It should be noted that nonsimple polygons can also have all of their sides congruent and all of their vertex angles congruent. The vertices for these polygons also lie on a circle. Such polygons are known as regular star polygons.
The regular star polygon can be obtained from the regular n-gon by joining every d th point.
The regular star polygon is shown on the right.
To construct the regular pentagon, we actually first construct the regular decagon and then join alternate vertices. The analysis figure for the regular decagon is as follows:
The circle is of radius 1 and center O, and there are 10 vertices, each at distance x from its immediate neighbours. If B and C are two consecutive vertices, then ∠BOC = 360/10; that is, in the isosceles triangle OBC, the angles are 36°, 72°, and 72°. Thus, if BD bisects the base angle OBC, we have ∠OBD = 36, so triangles OBD and CBD are isoceles. Thus, OD = BD = BC = x.
Since ΔOBC ~ ΔBCD we have
so that
that is,
Solving for x, we get the roots
The positive number
is constructible, so given the radius 1 of the circle, we can construct the segment of length x and therefore we can construct the regular decagon and the regular pentagon.
Exercise 6.3.1. Given a circle with radius OP = 1, construct the segment of length
and complete the construction of the regular pentagon.
Remark. The number
(note the plus sign) is called the golden ratio or golden section or golden mean and is usually denoted by the Greek letter ϕ.
Note that
so that
is the reciprocal of ϕ. Note also that
differ by 1.
We can construct a regular n-gon for n = 3, 4, 5, and it is not very difficult to construct a 2n-gon if we can construct an n-gon. For n ≤ 10, this leaves n = 7 and n = 9. Unfortunately, we cannot construct these regular n-gons with straightedge and compass alone. In fact, there are very few n-gons that are constructible, and this section describes all of them. First, we need the following definitions.
A Fermat number is a number of the form 22n +1, where n ≥ 0.
A prime number is a positive integer p > 1 such that p has exactly two positive divisors, namely 1 and p.
A Fermat prime is a Fermat number that is also a prime number.
Here are the first few Fermat numbers:
Remark. As can be seen from the table, the first five Fermat numbers
are all primes, and this led Fermat to conjecture that every Fermat number Fn is a prime.
In 1732, almost 100 years later, Euler showed that this conjecture was false, and he gave the following counterexample:
Even today we do not know if there are an infinite number of Fermat primes, In fact, the only known Fermat primes are the ones in the table above.
In 1796, Gauss found what is probably the most important aspect of the Fermat numbers, the connection between the Fermat primes and the straightedge and compass construction of regular polygons. His result is as follows:
Theorem 6.3.2. (Gauss’ Theorem)
A regular n-gon is constructible if and only if
where k ≥ 0 and p1, p2, …, ps are distinct Fermat primes.
The early Greeks knew how to construct regular polygons with 2k, 3 · 2k, 5 · 2k, and 15 · 2k = 2k · 3 · 5 sides. They also knew how to construct regular polygons with 3, 4, 5, 6, 8, 10, 12, 15, and 16 sides, but not one with 17 sides. Gauss, however, did this at age 19 and so reportedly decided to devote the rest of his life to mathematics. He also requested that a 17-sided regular polygon be engraved on his tombstone (it wasn’t).
Corollary 6.3.3. The regular 7-gon and the regular 9-gon are not constructible.
Proof. When n = 7, n is a prime, but not a Fermat prime. When n = 9, n is the product of Fermat primes since n = 3 • 3, but it is not the product of distinct Fermat primes.
Example 6.3.4. Is an angle of 3° constructible?
Solution. This is the central angle formed by the edges of a 120-gon, since
The question amounts to asking whether we can construct a 120-gon.
Since
and since 3 and 5 are distinct Fermat primes, the construction is possible.
Now we return to the ideas of concurrency and collinearity. First we note that two circles intersect in zero, one, or two points, or they coincide, as in the figure, and therefore:
Theorem 6.4.1. (Miquel’s Theorem)
Given ΔABC and three menelaus points X, Y, and Z, one on each side (possibly extended) of the triangle, then the circles formed using a vertex and its two adjacent menelaus points are concurrent at a point M.
The point of concurrency is called the Miquel point.
Proof. Let two of the circles have a second point of intersection M. We want to show that the third circle also goes through the point M.
We know that the quadrilateral AXMZ is cyclic and that BY MX is also cyclic, and this implies that the angles are as shown in the figure.
Since
then
and the quadrilateral YCZM is also cyclic. Therefore, the circumcircle of ΔYZC also passes through M, and the three circles are concurrent at the Miquel point M.
Corollary 6.4.2. Given ΔABC and three menelaus points X, Y, and Z, one on each side (possibly extended) of the triangle, if X, Y, and Z are collinear, then the circumcircle of ΔABC passes through the Miquel point M.
Proof. We recall that the Miquel point M lies on each of the circumcircles of ΔAXZ, ΔBXY, and ΔCYZ, and therefore:
Therefore, if we let ∠XMC = w, then
that is,
so that
and thus ABCM is cyclic.
The converse is also true.
Corollary 6.4.3. Given ΔABC and three menelaus points X, Y, and Z, one on each side (possibly extended) of the triangle, if the circumcircle of ΔABC also goes through the Miquel point M, then the three menelaus points X, Y, and Z must be collinear.
Proof. In the figure we let ∠XMC = w. Now:
From (2), using Thales’ Theorem, we have
and from (3), using Thales’ Theorem, we have
and to show that X, Y, and Z are collinear, it is enough to show that s = t.
However, from (1), since ABCM is cyclic, the opposite angles are supplementary, so that
While from (4), since BXMY is cyclic, the opposite angles are supplementary, so that
Therefore s = t. and we are done.
Example 6.4.4. Given ΔABC and three menelaus points X, Y, and Z, one on each side, where M is the Miquel point, as in the figure, show that if P, Q, and R are the centers of the circumcircles of ΔAXZ, ΔBXY, and ΔCYZ, respectively, then ΔPQR is similar to ΔABC.
Solution. First, observe that quadrilateral XPMQ is a kite. Let ω be the circumcircle of triangle BXY and draw the common chords , , and . Let the side meet ω at S and the side meet ω at T, as in the figure on the following page.
Since the line joining the centers of two circles is the perpendicular bisector of their common chord, is the perpendicular bisector of and therefore also bisects ∠XQM, so that
Similarly, bisects ∠MQY, so that
Since
and from Thales’ Theorem we have
then
Similarly,
and therefore ΔPQR ~ ΔABC.
Example 6.4.5. Prove Johnson’s Theorem (1916).
Given three circles concurrent at O, all with the same radius r, as in the figure below, then the circumcircle of the other three intersection points A, B, and C has radius r also.
Solution. Let O1, O2, and O3 be the centers of the three circles. Then the quadrilaterals
are rhombii, since all the sides have length r, as in the figure below.
Let AO1BO′ be the completed parallelogram of triangle AO1B, which is in fact a rhombus. Since
then BO′CO3 is also a rhombus, and therefore the circumcircle of triangle ABC is C(O′, r).
Note. In general, the center O′ of the circumcircle of ABC is different from O.
The following result was discovered by Frank Morley in about 1900. He mentioned it to friends in Cambridge and published it about 20 years later in Japan.
Morley’s Theorem states that the points of intersection of the adjacent trisectors of the angles of any triangle are the vertices of an equilateral triangle, as in the figure below.
Before proving this theorem we need a lemma, which is yet another characterization of the incenter of a triangle.
Lemma 6.5.1. (Another Characterization of the Incenter)
The incenter of a triangle ΔABC is the unique point I interior to the triangle which satisfies the following two properties:
Proof. The incenter has property (1) by definition, since it is the intersection of the internal angle bisectors of the triangle.
Also, from the External Angle Theorem, we have
and property (2) holds also.
To prove uniqueness, suppose that the point I′ ≠ I lies on the angle bisector of ∠A, and suppose that I′ also subtends an angle 90 + ∠A with the side BC.
Note that in the figure on the previous page, where I′ is between A and I, the External Angle Inequality implies that
so that
which is a contradiction. Similarly, if I is between A and I′, we again get a contradiction. Therefore, the point I satisfying (1) and (2) is unique.
Theorem 6.5.2. (Morley’s Theorem)
The points of intersection of adjacent trisectors of the angle of any triangle form an equilateral triangle.
Proof. Let ΔABC be a fixed triangle, so that the angles at the vertices A, B, and C are fixed angles. It is enough to prove Morley’s Theorem for a triangle ΔA′B′C′ that is similar to the given triangle, since scaling the sides by a proportionality factor k does not change the angles, and so an equilateral triangle remains equilateral.
In the proof we start with an equilateral triangle ΔPQR and then construct a triangle ΔA′B′C′ that is similar to ΔABC and has the property that the adjacent trisectors form the given equilateral triangle ΔPQR.
Step 1. Construct an equilateral triangle ΔPQR.
Step 2. Construct three isosceles triangles on the sides of ΔPQR, with angles as shown,
where
Observations:
Step 3. Extend the sides of the isosceles triangles until they meet as shown to produce a larger triangle ΔA′B′C′.
We claim that ΔA′B′C′ is similar to ΔABC and that the lines
are the angle trisectors of the angles at A′, B′, and C′.
Step 1. α + β + γ + 60 = 180, so that the angles are as shown on the figure.
For example, the angle ∠B′PR′ is a straight angle and ∠QPR = 60 so that α + 60 + β + ∠B′PQ′ = 180, which implies that γ = ∠B′PQ′ and that the vertically opposite angle ∠A′PR′ = γ also.
We should also show that the extensions of the sides of the isosceles triangles actually do intersect at the points A′, B′, and C′.
For example, if we isolate part of the figure, we can show that QR′ and Q′R intersect at a point C′, as shown in the figure on the following page.
If we consider the sum ∠RQR′ + ∠Q′RQ, then
and therefore the parallel postulate says that RQ′ and R′Q intersect on the side of the transversal QR where the sum of the interior angles is less than 180; that is, RQ′ and R′Q intersect on the side opposite P at some point C′, as shown.
Similarly, PQ′ and P′Q intersect at some point A′, while PR′ and P′R intersect at some point B′, as shown.
Step 2. α + β + γ = 120, and therefore
For example, in ΔPB′R, the sum of the interior angles is
so that
and
Similarly,
Step 3. R is the incenter of ΔB′R′C′. Similarly, P is the incenter of ΔA′B′P′, while Q is the incenter of ΔA′C′Q′.
We will show that R is the incenter of ΔB′R′C′ The other two results follow in the same way.
By the characterization theorem for the incenter proven in the lemma, (1) and (2) imply that R is the incenter of ΔB′R′C′.
Therefore,
so that PB′ and RB′ are angle trisectors of ∠B′. Similarly, PA′ and QA′ are angle trisectors of ∠A′, and QC′ and RC′ are angle trisectors of ∠C′.
Therefore, ∠A′ = ∠A, ∠B′ = ∠B, and ∠C′ = ∠C′, and by the AAA similarity theorem, ΔA′B′C′ is similar to ΔABC, and the corresponding segments in ΔABC are the angle trisectors. Therefore, in ΔABC the points of intersection of adjacent trisectors of the angles form an equilateral triangle.
In any triangle A ΔABC, the following nine points all lie on a circle, called the 9-point circle, and they occur naturally in three groups.
Note. In general, the orthocenter H is not the center of the 9-point circle.
Recall the following facts:
Fact 1. In a rectangle,
Conclusion. The center of the circumcircle of a rectangle is the intersection of the diagonals and is also the midpoint of either diagonal.
Fact 2. The Thales’ Locus of points subtending an angle of 90° with a segment is exactly the circle with as a diameter.
Conclusion. If a point P subtends an angle of 90° with any diameter of a circle C, then P must be on the circle.
The 9-point circle theorem was first proved by Feuerbach (1800 – 1834).
Theorem 6.6.1. (Feuerbach’s Theorem)
In any triangle ABC, the following nine points all lie on a circle, called the 9-point circle:
The center of the 9-point circle is denoted by N.
Proof. In the figure on the following page, the following constructions have been performed.
Step 1.
Join M3 to M2, so that M2M3 is parallel to BC, and M2M3 = BC by the Midline Theorem.
Join O2 to O3, so that O2O3 is parallel to BC, and O2O3 = BC by the Midline Theorem.
Step 2.
Join M3 to O2, so that M3O2 is parallel to AH, and M3O2 = AH by the Midline Theorem.
Join M2 to O3, so that M2O3 is parallel to AH, and M2O3 = AH by the Midline Theorem.
Since AH is an altitude and is perpendicular to the side BC, then M3M2O3O2 is a rectangle. Similarly, M3M1O3O1 is a rectangle.
Now note the following:
Note. In the proof, we joined
and
to get rectangles with common diagonals.
The segments O1M1, O2M2, and O3M3 are diameters of the 9-point circle and are also the diagonals of these rectangles. Thus, the point N, the center of the 9-point circle, is the midpoint of each of these segments.
For an isosceles triangle which is not an equilateral triangle, only eight of the nine points on the 9-point circle are distinct.
For an equilateral triangle, only six of the nine points on the 9-point circle are distinct. The center N of the 9-point circle is H, and the circle is just the incircle.
For a right-angled isosceles triangle, as below, where AB = BC and ∠B is a right angle, only four of the nine points on the 9-point circle are distinct.
Theorem 6.6.2. (Euler Line)
Given a nonequilateral triangle ABC, the circumcenter S, the centroid G, and the orthocenter H are collinear and form the Euler line.
In fact, G is a trisection point of HS; that is, G is between H and S and GH = 2GS.
Note. If ΔABC is equilateral, then S = G = H, and conversely, if S = G = H, then ΔABC is equilateral
Proof. Since ΔABC is nonequilateral, then G ≠ S. Extend SG to SO, with S – G – O and GO = 2GS. If we can show that O = H, then we are done.
Let M be the midpoint of BC. Since S is the circumcenter of ΔABC, then SM is perpendicular to BC.
Now join A and M. Then AM is a median and so passes through G. Next, join A to O, and by construction GO = 2GS.
Also, since G is the centroid of ΔABC and AM is a median, GA = 2GM, and since vertically opposite angles are equal, ∠AGO = ∠MGS.
By the sAs similarity theorem, ΔAGO ~ ΔMGS, with proportionality constant k = 2, so that SG = 2GO and AO = 2SM.
Now ∠SMG = ∠OAG, and the alternate interior angles formed by the transversal AM of SM and AO are equal so that SM is parallel to AO, and if AO is extended it hits BC at 90°. Thus, the altitude from A goes through O. Similarly, the altitudes from B and C also pass through O, and O = H.
We note for future reference that AH = 2SM.
Theorem 6.6.3. The center N of the 9-point circle of ΔABC lies midway between the circumcenter S and the orthocenter H.
Note. This gives us four special points on the Euler line
and we can compute all of these distances.
Proof. Note that if the triangle is equilateral, then S = N = G = H and there is nothing to prove.
Suppose that ΔABC is nonequilateral, so that S ≠/ H. Introduce the midpoint M1 of BC so that SM1 is perpendicular to BC.
Join AH and let O1 be the midpoint of AH, and then join O1M1 intersecting SH at T. Since H is the orthocenter and AH (extended) is perpendicular to BC, then AH is parallel to SM1.
Since SH and O1M1 are criss-crossing transversals of the parallel lines AH and SM1, then
However,
so the proportionality constant is 1, and
Therefore, ST = TH, and T is the midpoint of SH.
Also, TO1 = TM1 and T is also the midpoint of O1M1, but O1M1 is a diameter of the 9-point circle, and therefore T = N, the center of the 9-point circle.
Theorem 6.6.4. The radius of the 9-point circle is one half the radius of the circumcircle.
Proof. In the figure below, NO1 is the radius of the 9-point circle, AS is the radius of the circumcircle, and
by the Midline Theorem.
Theorem 6.6.5. The 9-point circle of ΔABC bisects every segment connecting the orthocenter H to a point P on the circumcircle.
Proof. In the figure below, let M be the midpoint of PH. We will show that the 9-point circle passes through M.
By the Midline Theorem, MN = PS = R, where R is the radius of the circumcircle, but MN is the radius of the 9-point circle by the previous theorem, so M is on the 9-point circle.
We note the following two theorems.
Theorem 6.7.1. If the altitudes of a triangle are congruent, then the triangle is isosceles.
Proof. In the figure on the following page, if BD = EC, then ΔBDC ≡ ΔCEB by the HSR congruence theorem.
Therefore, ΔADB ≡ ΔAEC by the ASA congruence theorem, so that AB = AC.
Therefore, ∠B = ∠C and ΔABC is isosceles.
Theorem 6.7.2. If two medians of a triangle are congruent, then the triangle is isosceles.
Proof. In the figure below, G is the centroid of ΔABC and medians BD and CE are equal.
Since G is the centroid, we have
and
and since ∠EGB = ∠DGC, then by the SAS congruency theorem we have
Therefore, BE = DC and 2BE = 2DC; that is, AB = AC, so ΔABC is isosceles.
Thus, we see that if two altitudes of a triangle are congruent, or if two medians of a triangle are congruent, then the triangle must be isosceles.
In 1880, Lehmus conjectured that the result was also true for the angle bisectors, and in about 1884 Steiner proved that Lehmus’ conjecture was true. Today almost 60 proofs have been given. We give below one of the simplest proofs.
Theorem 6.7.3. (Steiner-Lehmus Theorem)
If two internal angle bisectors of a triangle are congruent, then the triangle is isosceles.
Proof. In the figure below, let BD and EC be the internal angle bisectors at B and C, respectively, and suppose that BD = CE but that ΔABC is not isosceles, so that x < y.
Transfer x to C, as shown. Then by the Angle-Side Inequality applied to ΔBFC. since 2x < x + y, we have FC < FB.
Now transfer CF to BG as shown, and draw GH, making an angle s = ∠EFC at G. Then by the ASA congruency theorem, we have
so that BH = CE. However, BH < BD, which implies that CE = BE < BD, a contradiction.
Similarly, x > y is impossible. Hence x = y, and ∠B = ∠C, so that ΔABC is isosceles.
Example 6.7.4. Let ABC be a triangle, and let L and M be points on and , respectively, such that AL = AM. Let P be the intersection of and . Prove that PB = PC if and only if AB = AC.
Solution. Suppose that AB = AC, so that ΔABC is isosceles. Then
and by the SAS congruency theorem, ΔLBC ≡ ΔMCB.
Therefore, ∠PCB = ∠PBC and ∠BPC is isosceles, and thus, BP = PC.
Conversely, suppose that AL = AM but AB > AC. Extend the side AC to D so that AB = AD, as in the figure below. We will show that this implies PB > PC.
Let l be the common perpendicular bisector of LM and BD. By symmetry, BM and DL intersect at a point Q on l.
Since C is between M and D, the point P is on the same side of l as D so that PB > PD.
Now,
Hence, PD > PC and the conclusion follows.
We have used the fact that an exterior angle of a triangle is greater than either of the interior angles and the fact that the larger angle is opposite the longer side.
In this section we will prove the Apollonian circle theorem, but first we give a characterization of points P on a line l determined by distinct points A and B in terms of the ratio
We give the line an orientation such that the positive direction corresponds to going from A towards B, and we use the notation A − P − B to mean that P is between A and B.
We have the following result.
Lemma 6.8.1. If A, B, and P are distinct points on the line l, and
then:
Proof.
Theorem 6.8.2. (Circle of Apollonius)
Given two fixed points A and B, with A ≠ B, together with a fixed positive constant γ ≠ 1, then the locus of points P that satisfy
is a circle with center on the line joining A and B, called the circle of Apollonius.
Note. We have avoided γ = 1, since
is the perpendicular bisector of the segment AB.
Proof. Let C be an internal point of the segment AB such that
The previous lemma implies that there is only one such point C.
Let D be an external point to the segment AB such that
Again, the lemma implies that there is only one such point D.
Let C be the circle with CD as diameter. Then we claim that C is the circle of Apollonius, that is,
and C is internal to the segment AB.
Thus, in ΔAPB, PC is the internal bisector of ∠APB by the converse of the internal bisector theorem.
Also,
and D is external to the segment AB.
Thus, in ΔAPB, PD is the external bisector of the external angle at P.
Now, 2x + 2y = 180, so that x + y = 90, and ∠CPD = 90. Therefore, P is on the circle C with diameter CD.
From the lemma, the internal bisector of ∠P divides the side opposite P in the ratio
at a point C1 to the left of C.
Again, from the lemma, the extenal bisector of ∠P divides the side opposite P in the ratio
at a point D1 to the right of D.
However, ∠C1 PD1 = 90 since the internal and external angle bisectors are perpendicular, and ∠CPD = 90 since P is on the circle C with diameter CD. This is a contradiction since ∠CPD < ∠C1 PD1. Thus, it is not true that PA/PB < γ. Similarly, if we assume that PA/PB > γ, we get a contradiction. Therefore, we must have PA/PB = γ.
In the following figure, triangles ABC and ADE are similar, so
If we choose y = 1 we get w = xz. Let one of x or z be p, the other be q, and then w = pq.
Here is a fairly efficient construction.