Complex numbers in the realm of Euclidean Geometry

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1 Complex numbers in the realm of Euclidean Geometry Finbarr Holland February 7, Introduction Before discussing the complex forms of lines and circles, we recall some familiar facts about complex numbers. These form a field denoted by C. This consists of numbers of the form z = x + iy, where x, y are real numbers, and i is an abstract quantity satisfying the requirement that i = 1. They can be added, subtracted and multiplied as if they were real numbers with the proviso that whenever i occurs in an algebraic expression it is replaced by 1. We call x the real part of the complex number z = x + iy, y its imaginary part, and write x = Rz, y = Iz. We define its complex conjugate, z, and its modulus, z, by Note that z 0, with equality iff z = 0, z = x iy, z = x + y. and Rz = x = z + z z z, Iz = y =, i z = z, z z = (x + iy)(x iy) = x + y = z. Exercise 1. Suppose z, w C. Prove that z + w = z + w. As an exercise in the manipulation of complex numbers, we prove the following basic results. Example 1. If z, w C, then (a) z w = zw, and (b) zw = z w. Proof. Let z = x + iy, w = u + iv, where x, y, u, v R. (a). By definition, applying the usual laws of associativity, commutativity and distributivity of the real number 1

2 system R, z w = (x iy)(u iv) = xu ixv iyu + i yv = xu i(xv + yu) yv = (xu yv) i(xv + yu) = (xu yv) + i(xv + yu). On the other hand, z w = (x + iy)(u + iv) = xu + ixv + iyu + i yv = xu + i(xv + yu) yv = (xu yv) + i(xv + yu) = z w, as stated. (b) We have zw = (zw)(zw) = = zw z w = z z w w = z w, which is equivalent to part (b) since the modulus of a complex number is nonnegative. In the later sections we mae use of a few simple inequalities which we establish here. Proposition 1. For all z C, Rz z, with equality iff Iz = 0. Proof. This is a simple consequence of the fact that z = (Rz) + (Iz) (Rz), with equality iff Iz = 0. Proposition. For all z, w C, z + w z + w, with equality iff z w = w z. This is called the triangle inequality. Proof. z + w = (z + w)(z + w) = (z + w)( z + w) = z z + z w + w z + w w = z + R(z w) + w z + z w + w = z + z w + w = ( z + w ),

3 whence z + w z + w. The inequality is strict unless for some pair z, w, Rz w = z w. But if the latter holds, then w z z w = z w z w R(z w) + z w = z w z w = 0, i.e., z w w z = 0. The result follows. In the following sections we mae repeated use of the following identity: a b = a R(a b) + b, which is a simple consequence of the fact that a b = (a b)(ā b) = aā a b bā+b b = a (a b+bā)+ b = a R(a b)+ b. Written differently, this identity may put you in mind of the Cosine Rule for a triangle. In fact, it s exactly this rule: If a, b are distinct nonzero complex numbers, then the points a, 0, b are the vertices of the triangle AOB (possibly degenerate), and according to the rule, we can calculate the cosine of the angle AOB via the formula cos( AOB) = AO + OB AB AO OB = a + b a b a b = R(āb) a b. Exercise. Under the same assumptions, prove that Exercise 3. Show that sin( AOB) = I(āb) a b. a b + a + b = ( a + b ), a, b C. This identity is called the parallelogram law. We write [a, b] for the set of points z of the form z = λa+(1 λ)b, where 0 λ 1, and refer to it as the line segment containing a, b as endpoints. We say z is an interior point of [a, b] if 0 < λ < 1. Exercise 4. Given three complex numbers a, b, c show that a b = a c + c b iff c = λa + (1 λ)b, for some real number λ [0, 1]. In other words, iff c [a, b]. 3

4 Facts about lines in complex form The usual equation of a line in cartesian coordinates is ax + by + c = 0, where a, b, c R, and a + b > 0. With w = a + ib, z = x + iy, this taes the form R( wz) + c = 0, w 0 c R. The solution set of this contains 0 iff c = 0. (In what follows, whenever we spea of a line with such an equation it is to be understood that w 0 and c is real.) Exercise 5. Describe the solution set of the equation R( wz) = c. The equation of a line through a distinguished point a can be written in the form R( w(z a)) = 0. The equation of the line bc through two distinct points b, c is of the form R(i( c b)(z b)) = 0. Two lines R( w 1 z) + c 1 = 0, and R(w z) + c = 0, are said to be perpendicular to each other if R(w 1 w ) = 0, equivalently, if w = λiw 1 for some non-zero real number λ. Thus the lines R( wz) + c 1 = 0, and R(i wz) + c = 0 are perpendicular. The equation of the line perpendicular to bc that passes through a is given by R(( c b)(z a)) = 0. Example. If a, b are distinct, the equation of the perpendicular bisector of the line segment [a, b] can be expressed as z a = z b. Proof. Let m = (a + b)/ be the midpoint of [a, b]. Since the equation of ab is R(i(ā b)(z b)) = 0, the equation of the line perpendicular to this that passes through m is given by R((ā b)(z m)) = 0 R((ā b)z) = R((ā b)(a + b)) = a b. In other words, a R(āz) = b R( bz) z a = z b, whence the result. Consider next the perpendicular distance from 0 to the line R( wz) + c = 0. If z is on this line, then c = R( wz) wz = w z, and so z c w, with equality iff z = c/ w. Thus, the distance in this case is given by c / w, and c/ w is the foot of the perpendicular. 4

5 In the same way it s easy to see that the perpendicular distance from a to R( wz)+c = 0 is given by c + R( wa), w and the foot of the perpendicular is a c + R( wa) w = a (c + R( wa))w w. First, by definition, the required distance is the minimum of z a as z ranges over the line R( wz) + c = 0. If z is such a point, then and so so that The inequality is strict unless 0 = c + R( wz(z a)) + R( wa), c + R( wa) = R( w(z a) w z a, z a z = a Thus the required distance is equal to c + R( wa). w c + R( wa). w c + R( wa), w and the foot of the perpendicular is equal to a c + R( wa). w If the equation of the line L is given in the form 0 = R(i( c b)(z b)), so that b c, and b, c L, then the required distance is equal to min{ z a : z L} = R(i( c b)(a b)), c b and the foot of the perpendicular on L is equal to a R(i( b c)(a b)). i( b c) Example 3. If a, b, c are the vertices of a triangle ABC, its area is given by (ABC) = 1 R(i( b c)((a b)). 5

6 Proof. Using the formula that the area is one-half the base by the height, the result follows because b c is the length of [b, c], and the perpendicular distance from a to bc is equal to R(i( c b)(a b)). c b Given a point a, and a line L : R( wz) + c = 0, we define the reflection ã L of a in L by the formula (c + R( wa)) ã L = a. w Then a, ã L are equidistant from L, and their midpoint (a + ã L )/ belongs to L. Exercise 6. Show that R(i( c b)(a b)) is symmetric in a, b, c. 3 The equation of a circle in complex form The general equation of a circle in cartesian coordinates x, y is x + y + gx + fy + c = 0, where g, f, c R, and g + f > c. Let u = (g + if) and put r = u c. Then, r > 0, and with z = x + iy, the equation taes the form z u = r. The solution set of this is the circle centered at u of radius r. Exercise 7. Show that the line R( wz) + c = 0 intersects the circle z = r iff c w r. Prove that the line is a tangent to the circle iff c = r w. The radical axis of a pair of circles z u i = r i, i = 1,, with distinct centres, is the set of points p such that p u 1 r 1 = p u r. Proposition 3. The radical axis is a line that is perpendicular to the line u 1 u. Proof. The condition that a point z be on the radical axis is equivalent to the statement that z R(zū 1 )+ u 1 r 1 = z u 1 r 1 = z u r = z R(zū )+ u r, i.e., R(z(ū ū 1 )) + r r 1 + u 1 u = 0, the solution set of which is a line, since u 1 u 0. On the other hand, the equation of the line containing the centres u 1, u is given by R(i(ū ū 1 )(z u 1 )) = 0. Clearly, the two lines are perpendicular to each other. 6

7 Exercise 8. Show that the set {z : R( wz) c = 0, z = r} consists of an even number of points zero, two or four. By Example, the set of points z that are equidistant from two distinct points a, b is a line whose equation is z a = z b. What s the nature of the set of points z whose distance from a is a positive multiple ( 1) of its distance from b? This was considered a long time ago by the Gree geometer Apollonius. The answer is a circle. Example 4. Suppose 0 < 1, and a, b are distinct. The set S of points z such that z a = z b is the circle z u = r, where Proof setch. Clearly, z S iff u = b a a b, r = 1 1. z R(zā)+ a = z R(z b)+ b (1 ) z R(z(ā b)) = b a, Thus z S iff Adding z Rz(ā b 1 ) = b a 1. a b (1 ) = u to both sides, completing the square of the expression on the left side, and simplifying the right, we see that z u = r, as claimed. Exercise 9. In the same notation show that a u u b = r. In other words, the points a, b are each other s inverse wrt to the circle z u = r. Exercise 10. The equation of any circle γ : z u = r can be written in the from of Apollonius. In other words, there are two distinct points a, b, and a number 1 so that if z γ, then z a = z b. [Hint: Select any v with v = 1, and define a = u + v, b = u + r v. Then, if z γ, z a = z b, where = 1/r.] For instance, the Apollonius form of the unit circle z = 1 taes the form z a z b = a, where a 0, and b = 1 ā. Example 5. Suppose s, t are distinct points on the unit circle z = 1 and a is an interior point of [s, t]. Let b = 1/ā. Then s t = a [ s b + t b ]. 7

8 Proof. As we ve just remared, z a = a z b, z with z = 1. Letting z in turn be s, t in this we deduce that s a + t a = a [ s b + t b ]. But there is a real number λ (0, 1) so that a = λs + (1 λ)t. Hence a s = (1 λ)(t s), a t = λ(t s), and so s a + t a = (1 λ) t s + λ t s = t s, whence the result follow. A further simple result about the mapping z 1/ z follows. Proposition 4. Suppose a 1 and a, 1/ā belong to the circle γ : z u = r. Then γ is invariant under the action of the map z 1/ z. Proof. By assumption, a u = r, 1 āu = a r. Equivalently a R(āu) + u = r, 1 R(āu) + a u = a r. Hence (1 a ) + (1 a ) u = (1 a )r, and so u = r + 1. Suppose now z γ and w = 1/ z. Then z R(ūz) + u = r z R(ūz) + 1 = 0. It follows that z 0, and so ( 1 R ū z ) + 1 z z = 0 1 R(ūw) + w = 0 w Rūw + u = r. Thus w u = r, i.e., w γ. More generally, the following result can be proved in a similar way. Proposition 5. Suppose is a non-zero real number, a, and a, /ā belong to the circle γ : z u = r. Then γ is invariant under the action of the map z / z. Proof. The hypotheses hold iff u = + r. The result follows from this. 4 Triangles inscribed in the unit circle Suppose the vertices of ABC are given by the distinct complex numbers a, b, c, with a = b = c = 1. Proposition 6. The area of ABC is equal to 1 ( I(āb) + I( bc) + I( ca) ). 8

9 Proof. With O standing for the centre, ABC is the disjoint union of the triangles ABO, BCO, CAO. Using the sine formula for the area of a triangle, we see that (ABC) = (ABO) + (BCO) + (CAO) = sin(aob) + sin(boc) + sin(coa) = I(āb) a b + I( bc) b c + I( ca) c a = 1 ( I(āb) + I( bc) + I( ca) ). 4.1 Some concurrency points in a triangle Proposition 7. The perpendicular bisectors of the sides intersect at the origin. Proof. By Example, the perpendicular bisector of bc has equation z b = z c. Since b = c = 1, this bisector contains 0. Similarly, 0 belongs to the perpendicular bisectors of ca and ab. Hence, all three perpendicular bisectors contain 0. In other words, the circumcentre of ABC is the origin. Proposition 8. The orthocentre H of ABC is represented by the complex number h = a + b + c. Proof. Start with the equation of the line bc, viz., R(i( b c)(z b)) = 0. Hence, the equation of the altitude from A is given by Since R(( b c)(z a)) = 0. R( b c)(b + c) = R( b + bc cb c ) = R(1 + ii( bc) 1) = 0, this altitude contains a + b + c. A similar argument shows that this point belongs to the altitudes from B and C also. The result follows. Proposition 9. Let D, E, F stand for the midpoints of BC, CA, AB, respectively. Then h/ is the centre of the circumcircle of DEF. The radius of this semicircle is 1/. Proof. Since d = 1 (b+c) represents D, we see that the distance from D to the point represented by h/ is equal to h + b + c) (b + c) d = (a = a = 1. Thus d belongs to the circle z h = 1 c+a. Similarly, the points,, a+b belong to this circle. Hence, the result. 9

10 4. The Euler line and the radical axis of two specific circles The Euler line of a triangle is the line that contains its orthocentre and circumcircle. In the present case, its equation is R(i hz) = 0. This line contains the centroid G of ABC, which is given by (a + b + c)/3 = 1 h. By the previous proposition, the Euler 3 line also contains the centre of the circumcircle of DEF. Exercise 11. Prove that the midpoints of AH, BH and CH belong to the circumcircle of DEF. Proposition 10. The equation of the radical axis of the circumcircles of ABC and DEF is given by R( hz) = 3 + h. 4 Proof. The unit circle z = 1 is the circumcircle of ABC and z h/ = 1/ is the circumcircle of DEF. Hence, applying Proposition 3 with u 1 = 0, r 1 = 1, u = h/, r = 1/, we see that the equation of the radical axis of these circles is equivalently, as claimed. R(ū z) + (1/) h/ = 0, R( hz) = 3 + h, 4 Proposition 11. The radical axis of the circumcircles of ABC and DEF meets the Euler line at the point z = 3+ h = 3+ h h. 4 h 4 h Proof. It is plain that the given point is common to the Euler line R(i hz) = 0, and the radical axis. (clearly, these two lines are perpendicular.) 4.3 The altitudes of ABC Proposition 1. Let X denote the foot of the altitude from A. Then { R( b+ c)(a+c), if H A, b+c HX = Ri( b c)(a b), if H = A. b c Proof. If H doesn t coincide with A, then b + c 0, in which case the equation of bc is given by R( b + c)(z b) = 0. According to one of the formulae given in Section, the distance from h to bc is therefore equal to R( b + c)(h b) b + c = R( b + c)(a + c). b + c On the other hand, if H = A, the equation of bc is given by Ri( b c)(z b) = 0, and the distance formula tells us that the distance from A to bc is equal to HX = Ri( b c)(a b). b c 10

11 Corollary 1. AH HX = R( b + c)(a + c). Proof. If H A, then this is so by the first formula, because AH = a h = b+c. On the other hand, if H = A, both sides are zero, because then b + c = 0. Noting that (a, b, c) = R( b + c)(a + c) = R( bc + ca + āb) + 1 is symmetric, we can deduce the following Proposition 13. Denoting by Y, Z the feet of the altitudes from B and C, then AH HX = BH HY = CH HY =. Exercise 1. Prove that = h 1. Exercise 13. Prove that the points X, Y, Z lie on the circumcircle of DEF. Coupling this with the fact that the midpoints of AH, BH, CH also lie on this circle, we have identified nine distinct points on the same circle. For this reason, the circumcircle of DEF is often referred to as the Nine-point circle of ABC. From geometric considerations it s to be expected that the line AX cuts the circle z = 1 at a point other than A. Similarly for BY, CF. This is proved next. Proposition 14. The line AX contains the point bc/a, whose modulus is 1. Proof. Since the equation of AX is of the form R(i(ā x)(z x)) = 0, where x represents X, and x = a R(i( b c)(a b)) i( b c) a K i( b c), we have to show that R(i(ā x)(w x)) = 0 when w = bc/a. But and so, for any z, i( x ā) = K b c, R(i(ā x)(z x) = KR{ z x b c } = KR{ z a + K i( b c) } b c = K[R{ z a b c + K i b c }] = R{ z a b c }. 11

12 So, our tas is to show that R{ w a b c } = 0. To accomplish this we ll exploit the facts that ā = 1/a, b = 1/b, c = 1/c, w = 1/w. In fact, R{ w a b c } = w a b c + w ā b c = w a a w b c + aw c b bc = w a b c + bc aw = w a b c = 0, ( 1 + bc aw a w c b ) as desired. This completes the proof. Of course, by symmetry, the altitudes BY, CZ meet the unit circle at ca/b, ab/c, respectively. Exercise 14. Prove that bc/a is the reflection of h in the line bc. 4.4 Properties of an inverse mapping Note that if ABC is not right-angled, then the complex form of the feet of the altitudes can be expressed as: x = h b + c = h h ā, y = h Thus, defining the mapping w w by b + ā = h w = h h w h b, z = h ā + b = h h c. on the plane less h, we see that w = w, and a = x, b = y, c = z. We refer to w as the inverse of w wrt h, Proposition 15. Suppose H A. Then any circle that contains a and a is invariant under the map w w. Proof. This follows from Proposition 5 by using translation. For suppose a u = r = a u, then also x v = r = y v, where x = a h, y = a h =. Hence, x if z u = r, then w v = r where w = h u, and so v = r, equivalently, w z u = r. Theorem 1. Suppose ABC is not right-angled. Suppose P H. Then the circumcircles of AP D, BP E, CP F intersect at P and at its inverse P wrt h. 1

13 Proof. By the previous proposition, all three circles contain P. This establishes Part I of the following problem communicated to me by Izán Péraz in an . Part II is covered by Proposition 11. Problem (Izán Péraz). Let P a point in the plane of a triangle ABC a triangle with orthocenter H. Let D, E and F be the respective foots of the heights from A, B and C. Prove that the circumcircles of AP D, BP E and CP F intersect (in P and) also in a point which is the intersection of the Euler line of ABC and the radical axe of the circumcircle of ABC and the nine points circle of ABC. There are three parts to this proposition. Part I: The circumcircles of AP D, BP E and CP F intersect at a point Q P. Part II: The Euler line and the radical axis of the circumcircle of ABC and the nine points circle of ABC intersect, at a point R, say. Part III: Q = R. We ve verified the truth of Parts I and II, but there is no obvious reason why the intersection points as described coincide. In fact, Part III is false. We construct a simple counterexample in the next section. 5 A specific example to illustrate some of the preceding NB. In what follows, in contrast to the notation in the problem posed by Péraz, we use D, E, F for the midpoints of the sides, and X, Y, Z for the feet of the altitudes. Let a = i, b = 1, c = (3 4i)/5. These lie on the unit circle. The orthocentre of the triangle ABC with these vertices is h = i 1 + (3 4i)/5 = ( + i)/5. The midpoints D, E, F of the sides BC, CA, AB are given by d = b + c Also, so that = ( + 4i)/10, e = c + a = (3 + i)/5, f = a + b 1 h d = i, 1 h e = 1, 1 h f = 3 4i 10, = ( 1 + i)/. 1 h t = 1, t {d, e, f}. Hence the circumcircle of DEF, aa the nine-point circle of ABC, is z 1h = 1. The centroid is 1h = ( + i)/15. The equation of the Euler line is R(i hz) = 0, i.e, 3 R((1 i)z) = 0. Since h = 1/5 the equation of the radical axis of the circumcircles of ABC and DEF is R( hz) = 4/5, i.e., R(( + i)z) = 4. Thus the Euler line and the radical axis meet at 4( + i)/5 = 4h. Next = R(i( b+ c)(a+c)) = R(i( 1+(3+4i)/5)(i+(3 4i)/5)) = R(i( +4i)(3+i)/5) = 5. 13

14 Since ABC is not right-angled the feet X, Y, Z of the altitudes from A, B, C, respectively, can be calculated using the formulae and x = h z = h h ā = h + 5(( i)/5 + i) = h i = 3 + i 5 ; y = h + h b = h + 3 i = 1 + i ; 5 h c = h i (3 + 4i) = h 5(1 + i) 3 + i =. 5 We tae P to be the origin, and determine the cartesian equations of the circumcircles of AP X, BP Y, CP Z. These are found to be x + y + x y = 0; x + y + x y = 0; 5(x + y) + 17x + 19y = 0. In complex form they are of the form z u i = r i, i = 1,, 3, where u 1 = u = ( 1 + i)/, u 3 = ( i)/10, r 1 = r = 1/, r 3 = 6. These circles are invariant under the mapping z z = h +, z h. 5( h z) In particular, they contain the image of 0, namely 0 = h + 5 h = + i 5 + i = + i (4 i) 5 = 6 + 3i 5 = 3h, something that can be verified directly. However, this point is not the point of intersection of the radical axis for the circumcircles of ABC and DEF, which as we ve seen above is 4h!. In other words, Part III of Péraz s problem is false. 14

Berkeley Math Circle, May

Berkeley Math Circle, May 1-7 2000 COMPLEX NUMBERS IN GEOMETRY ZVEZDELINA STANKOVA FRENKEL, MILLS COLLEGE 1. Let O be a point in the plane of ABC. Points A 1, B 1, C 1 are the images of A, B, C under symmetry