Complex Numbers in Geometry

Transcription

1 Complex Numers in Geometry Seastian Jeon Decemer 3, 206 The Complex Plane. Definitions I assume familiarity with most, if not all, of the following definitions. Some knowledge of linear algera is also recommended, ut not required. Susequently, let i e the imaginary unit satisfying i 2. Define the set of complex numers C {z z a + i, a, R} where a is the real part of z and is the imaginary part. The magnitude of a given z a + i C is z a The conjugate of z a + i will e z a i, implying the property z 2 z z. As an exercise, show that z + w z +w and zw zw. Every z C can e expressed as z cos θ + i sin θ for some angle θ [0, 2π, or alternatively z e iθ. This angle θ will e referred to as the argument of z. Lastly, any complex numer satisfying z n 0 will e referred to as an n th root of unity. It can easily e verified that these numers take the form e 2kπi/n for 0 k n..2 Applications to the Complex Plane The complex plane assigns a complex numer to every point in the plane such that the point P with Cartesian coordinates a, is assigned a + i. The counterpart to the xy-axes in the complex plane are the real and imaginary axes, respectively. From this definition, the similarity etween the complex plane and the Cartesian plane should e evident. Consequently, the representation z z cosθ + i sin θ z e iθ corresponds to polar coordinates in the Cartesian plane. Throughout the lecture, the lowercase letter p of a point P will correspond to the complex coordinate, or affix, of P unless otherwise noted. As an exercise, verify that: - z is the reflection of z over the real axis - z z.3 Transformations in the Complex Plane Geometrically, in order to find the coordinates of the sum of two complex numers, one may simply perform a vector head-to-tail addition. Multiplication of complex numers is slightly more interesting. For any complex numers z, w, the map z zw corresponds to a spiral similarity composition of a dilation and rotation aout the origin. Furthermore, the magnitudes and arguments of zw are determined independently in this transformation. More specifically, we have wz w z and argw + argz argwz. Convince yourself that multiplication y a real numer is equivalent to a dilation and multiplication y e iθ is equivalent to a counterclockwise rotation of θ. As a side-note, any transformation z az + cz + d Moius transformation. where a,, c, d C, ac > 0, ad c 0 is a

2 2 Metric Propositions One significant advantage of complex numers over Cartesian coordinates is that every point is assigned a single numer as opposed to an ordered pair, allowing concise algeraic expressions of geometric concepts. From these fundamental propositions, the advantages of complex numers in various configurations should manifest themselves. Proposition 2.. Angle Between Two Lines. Let the angle formed y two lines AB, CD in the clockwise direction from AB to CD e θ AB, CD. Then c d eiθ c d. Proof. Translate the segments such that B, D coincide with the origin, with A, C c d. Then a a c d c d r e iθ /r r 2 e iθ2 /r 2 e iθ /e iθ2 e iθ θ2 e iθ. This result serves as the asis for many useful corollaries. Corollary 2.2. Alternate form of Squaring oth sides of Proposition 2. yields the more applicale form e2iθ c d c d. Corollary 2.3. Collinearity Points A, B, C are collinear iff c c. Proof. It remains to show that the angle etween AB, CB is 0, which follows from letting θ 0 in Corollary 2.2. Note that this can also e rearranged to a a c. c Corollary 2.4. Equation of a Line From Corollary 2.3, letting c e a variale point gives the equation of a line. Corollary 2.5. Perpendicularity Segments AB, CD are perpendicular iff Proof. Let θ π/2 in Corollary 2.2. c d c d. Proposition 2.6. Directly Similar Triangles ABC DEF and ABC, DEF are similarly oriented iff a c d e d f. Sketch. Consider the magnitudes of oth sides of the equation to get AB AC DE. Now dividing DF this with the given condition and applications of Proposition 2., we get BAC EDF. Corollary 2.7 Spiral Similarity The center of spiral similarity taking AB CD is given y Proof. By Proposition 2.5, it suffices to solve p a p p c p d 2 ad c a + d c. for p.

3 Proposition 2.8 Cyclicity Points A, B, C, D are concyclic iff a c d a c d a d c a d c. Sketch. Rewrite ACB ADB with Proposition 2.. Proposition 2.9. The area of triangle ABC is i a a 4 a c a c + c c. c c Sketch. Expand with Shoelace formula and Cartesian coordinates; details are left as an exercise to the reader. Proposition 2.0. Reflection Aout a Line The reflection of P with respect to line AB, denoted y z, satisfies p + a a z. Sketch. The proof of this uses the fact that linear transformations preserve reflections, so taking z z a and noting that AB is mapped to the real axis more specifically, the line segment a containing 0 and reaches the conclusion. Proposition 2.. Circumcenter Formula a aa c cc The circumcenter of ABC, denoted y x, satisfies x. a a c c Sketch. Consider the radius R of the circumcircle, Then x a 2 x 2 x c 2 r 2 so expanding we get a system of equations in x, x, r 2 x 2 which we can solve with Cramer s rule. 3 The Unit Circle One of the fundamental properties of the complex plane that make enormous computations viale is the fact that for any point p on the unit circle, p. Since any circle can e mapped to the unit p circle via a composition of translation and dilation, it is often useful to let a cumersome circle to e the unit circle. Proposition 3. Equation of a Chord If AB is a chord of the unit circle, the equation of line AB is given y z a + az. Proof. From Corollary 2.4, z a z a z a a and upon expanding we z a get the desired result. Remark on the simplicity of this equation, possile only y this property of the unit circle. Corollary 3.2 Chord Intersection Let AB, CD e two chords on the unit circle. If P AB CD then p ac + d cda + a cd 3

4 if a cd 0. Sketch. From Proposition 3., we know that p is a solution to oth { p a + ap p c + d cdp. Sutracting these two equations and solving for p, we consequently get p after some simplification. Corollary 3.3 Tangent Intersection If the tangents to the unit circle at A, B intersect at P, then p 2a a +. Proof. Note that a tangent is simply a degenerate chord; sustituting AA, BB into Corollary 3.2 we get the desired. Another nice property of unit circles regards fundamental triangle centers of triangles inscried in the unit circle. Proposition 3.4 Orthocenter, Centroid, Nine-Point Center For any triangle ABC inscried in the unit circle, its orthocenter, centroid, and nine-point center is given y a + + c, a + + c, a + + c 3 2 respectively. Proof. Note that from knowledge of Cartesian coordinates, the centroid is a + + c regardless of 3 origin. The rest follows from the Euler Line. Proposition 3.5 Incenter, Excenters, and Midpoints of Arcs Let ABC e a triangle inscried in the unit circle, and let the complex coordinates of A, B, C e a 2, 2, c 2 for complex numers a,, c. Let A, B, C have coordinates c, ca, a, I have coordinate a c ca, and finally I a, I, I c have coordinates ca + a c, a + c ca, c + ca a respectively. Then A, B, C are midpoints of arcs BC, CA, AB, I is the incenter of ABC, and I a, I, I c are the A, B, C-excenters. Proof. Since c c it lies on the unit circle. Furthermore, y Proposition 2. we have e 2i AAB 2 a 2 2 a 2 c a62 c a 2 a2 2 a 2 c c. Similarly e2i AAC c so A is on the A-angle isector, implying that it is indeed the midpoint of arc. We otain similar results for, c. To prove that I has coordinate a c ca, note that I is the orthocenter of A B C and apply Proposition 3.4. The proof for the excenters is left to the reader as an exercise. Proposition 3.6 Regular Polygons and Roots of Unity Let P, P 2,, P n in the complex plane satisfy P k ω k, where ω e 2πi/n. Then P P 2 P n is a regular n-gon. Proof. Evidently p k, so all of the points lie on the unit circle. Furthermore, it can easily e confirmed that p k+ p k is constant for k n, p n+ p and similarly that P k P k+ P k+2 is constant, done. 4

5 4 Some Instructive Examples When ashing, it is helpful to let a prominent circle in the diagram e the unit circle, or otherwise configure the diagram such that the coordinates have nice values. However, it should e noted that when doing so, the diagram should e general enough to encompass all possile cases. Prolem 4. Let ABC e a triangle with circumcenter O. Suppose D and E lie on AB and AC, respectively, such that O lies on DE. Let M and N e the midpoints of CD and BE, respectively. Prove that MON BAC. Iran 2004 / WOOT Proof. WLOG let ω, the circumcircle of ABC, e the unit circle, and further let DE coincide with the real axis. Convince yourself that these assumptions are sufficiently general. Then from the definition of point D, A, B, D are collinear a d. But since D lies on the real a d axis, d d and solving for d yields a d a d d a d a d aa ad a + d aa a ad + d d a + a + a a + a a + a + a a + a a a 2 a aa + + a a +. Hence, the midpoint of CD is simply m d + c ac + c + + a 2 2a + m ac+a++c n Thus e 2i MON 2ac+ n m ac+c++a 2a+ ac+c++a 2ac+ ac+a++c 2a+, and similarly n ac + a + + c 2ac + ac + a + a + ac + ac + a + c + ac + ac c. But e 2i BAC a c a c a2 c ac 2 a c a a 2 a 2 ac a c, so we may conclude. Prolem 4.2 Let ABC e an acute triangle with AB, AC, BC. Denote y O, H the circumcenter and orthocenter of ABC. Suppose that the circumcircle of AHC intersects AB again at M and the circumcircle of AHB intersects AC again at n. Prove that the circumcenter of MNH lies on line OH. APMO 200 Proof. This prolem illustrates the importance of synthetic oservations when complex ashing. Once we assume that ω is the unit circle, it is very difficult to otain coordinates for points M, N without massive computations. Hence, we need some sort of synthetic insight. Let D e the foot of the C-altitude, and P e the circumcenter of MNH. Easily otain the coordinates of D y reflecting C across AB, and taking the midpoint of CC. Note that CMB π AMB π AHB C, so BMC is isosceles, and it thus follows that D is the midpoint of MC. Now d a + + c ac m a + c ac, and similarly 2 we get n a+ ac. Note that O 0, H a++c so a translation of a c will still yield the circumcenter of MNH on line OH. Then m a + c, and similarly n ca +, h 0. c Then y the circumcenter formula, p a+c c a+c a ca+ ca+ a+ ac a+ ac a+c a a+c c 5

6 p a + ca + a+ ac a+c a a + ca + c a 2 ac 2 It suffices to show p, o, h collinear, ut this follows from and we may conclude. p o p o ca++c 2 +c+c 2 c+ca+a a 2 +c+c 2 aca + + c a + c + ca ca + + c 2 + c + c 2. h o h o, Prolem 4.3 In ABC, let O e the circumcenter of ABC, A to e the antipode of A WRT ABC, A 2 to e the reflection of O across AB. Let O A e the circumcenter of A A 2 O. Define O B, O C similarly. Prove that O A, O B, O C are collinear. Lemmas in Olympiad Geometry Proof. Evidently O A BC and similar, since the perpendicular isector of OA 2 is BC y definition of reflection. Now it suffices to show that BO A CO B AO C y Menelaus. To do CO A AO B BO C so, we use complex numers. Let ABC e the unit circle so that A a, A 2 + c. Then a a 2 a + c c O A. a a 2 a a 2 a + c a + c Now BO a++c+a+a A CO A 2 +ac a+c a+c a++c+ac+ac 2 +ac a+c a+c a + + c + a + a 2 + ac a + + c + ac + ac 2 + ac a + + c + a + a 2 + a + + c + ac + ac 2 + ac ac ac c a2 + a 2 + ac + a c a 2 c + ac + ac 2 + a 2 + c 3 + c 2 c + c a 2 + a c a 2 + ac + c 2 c a2 + a + 2 a 2 + ac + c 2, at which point it is clear that multiplying cyclically will give the desired result. Prolem 4.4 In ABC with incenter I, the incircle is tangent to CA, AB at E, F. The reflection of E, F across I are G, H. Let Q BC GH, and let M e the midpoint of BC. Prove that IQ IM. Proof. Let the incircle e tangent to BC at D, and let the incircle e the unit circle. This gives nice coordinates from chord intersection formulas, and moreover we get g e and h f. Then A EE F F so a 2ef and similar coordinates can e derived for B and C. Next e + f Q BC GH DD GH d2 g + h ghd + d d 2 gh M 2fd f+d + 2 2de d+e fd2 + 2def + ed 2. d + fd + e By the perpendicularity condition, it suffices to show q q m m But q d m d so we may conclude. de+df+2ef d 2 ef df+2ef+de d+fd+e 5 Practice Prolems d2 + ed + df + ef d 2 ef d2 e + f 2def d 2, ef q m q m d + 2 ed + df + ef d 2 ef. q, m Prolem 5. A quadrilateral ABCD is inscried in unit circle.if P AB CD and Q AD BC, then prove that p q + q p 2. Prolem 5.2 Let ABC e a triangle with incricle Γ and let D, E, F e the tangency points 6

7 of Γ with BC, CA, AB, respectively. Let K e the orthocenter of triangle DEF and let Γ A, Γ B, Γ C e circles centered at A, B, C with radii AD, BE, CF, respectively. Prove that K is the radical center of Γ A, Γ B, Γ C. AMSP Geo 3 Prolem 5.3 Oscar is drawing diagrams with trash can lids and sticks. He draws a triangle ABC and a point D such that DB and DC are tangent to the circumcircle of ABC. Let B e the reflection of B over AC and C e the reflection of C over AB. If O is the circumcenter of DB C, help Oscar prove that AO is perpendicular to BC. ELMO, 206 Prolem 5.4 Quadrilateral AP BQ is inscried in circle ω with P Q 90 and AP AQ < BP. Let X e a variale point on segment P Q. Line AX meets ω again at S other than A. Point T lies on arc AQB of ω such that XT is perpendicular to AX. Let M denote the midpoint of chord ST. As X varies on segment P Q, show that M moves along a circle. USAJMO, 205 Prolem 5.5 Let P e a point in the plane of triangle ABC, and γ a line passing through P. Let A, B, C e the points where the reflections of lines P A, P B, P C with respect to γ intersect lines BC, AC, AB, respectively. Prove that A, B, C are collinear. USAMO, 202 Prolem 5.6 Let ABC e a triangle and let P e a point on its circumcircle. Let X, Y, Z e the reflections of P over lines BC, CA, AB respectively. Then X, Y, Z, H are collinear where H is the orthocenter of ABC. Steiner Line Prolem 5.7 Prove that the incircle of ABC is tangent to the nine-point circle of ABC. Feuerach Prolem 5.8 Let ABCD e a cyclic quadrilateral centered at O and let E AC BD, F AB CD, and G AD BC. Prove that O is the orthocenter of EF G. Brokard 7