Geometry in the Complex Plane

Transcription

1 Geometry in the Complex Plane Hongyi Chen UNC Awards Banquet 016

2 All Geometry is Algebra Many geometry problems can be solved using a purely algebraic approach - by placing the geometric diagram on a coordinate plane, assigning each point an x/y coordinate, writing out the equations of lines and circles, and solving these equations. This method of solving geo problems (often called coordinate bashing) can be quite powerful given the right conditions, but it has some problems. Issues with coordinate bash Equations for circles are ugly Two variables are necessary for each random point Rotations are extremely painful Attempting to solve the equations may result in massive 5th degree polynomials in 8 variables... Fortunately, these problems can be fixed by replacing the Cartesian plane with the complex plane...

3 Quick Introduction to Complex Numbers A complex number (in rectangular form) is a number of the form a + bi, where a and b are real and i = 1. We define the real and imaginary parts of a complex z = a + bi as Re(z) = a and Im(z) = bi. Complex numbers can be plotted on the complex plane. The number a + bi is placed where the coordinate (a, b) is placed on the Cartesian plane. The horizontal axis is called the real axis and the vertical axis is called the imaginary axis. The conjugate of a complex number z, denoted by z, is its reflection about the real axis. For any z = a + bi we have z = a bi. ab = ā b and a + b = ā + b. Re(z) = z + z and Im(z) = z z. z is real if and only if Im(z) = 0, which occurs when z = z. Similarly a number z is pure imaginary iff z = z.

4 Quick Introduction to Complex Numbers The magnitude of z = a + bi, denoted by z, is its distance from the origin in the complex plane. If z = a + bi then z = a + b. Notice that for any complex z, z z = z. a b is the distance between a and b. A complex number z can also be expressed in polar form as r(cos θ + i sin θ) for a real r and angle θ, where r = z and θ is the angle formed by the positive real axis and the ray starting at the origin pointing towards z, measured counterclockwise. For simplicity we shall let cis θ = cos θ + i sin θ. The set of possible values of cis θ forms the unit circle on the complex plane - a circle centered at the origin with radius 1. For any angle θ we have cis θ = 1 = cis ( θ) cis θ

5 Complex Bash We can put entire geometry diagrams onto the complex plane. Each point is represented by a complex number, and each line or circle is represented by an equation in terms of some complex z and possibly its conjugate z. By standard, the complex number corresponding to a point is denoted by the lowercase character of of the point s label (for example, if A = (, 3), then a = + 3i). This is called complex bashing, and can be extremely powerful.

6 Translations The greatest advantage of using complex bash as opposed to any other type of bash is the simplicity of performing the basic transformations of translation, rotation, and dilation. To perform a translation of a units right and b units up to some complex number z, notice that complex number addition works the same way as vector addition, so one can simply add a + bi to z. To translate an object such that some point p becomes the origin, simply subtract p from all the points on the object.

7 Homothety (Dilation) To find the image of some point z after a homothety centered at the origin with scale factor k, simply multiply z by k. If the center is some point p instead, one can find the image by translating everything such that p is at the origin, then performing the homothety centered at the origin, then translating everything again such that p is back at its original place. Hence the image of z under a homothety centered at p with scale factor k is k(z p) + p.

8 Rotations A well-known theorem states that (cis θ)(cis φ) = cis (θ + φ). Thus, to rotate a complex number z = rcis φ by θ about the origin, multiply it by cis θ. To rotate z about an arbitrary point p, we can use the same manipulation used in the previous slide: translate everything by p, then perform the rotation around the origin, then translate everything +p. Hence the result of rotating z by θ about p is (z p)cis (θ) + p.

9 A Quick Example - Problem The points (0, 0), (a, 11), and (b, 37) are the vertices of an equilateral triangle. Find the value of ab. (1994 AIME #8)

10 A Quick Example - Solution Let O = 0, P = a + 11i, and Q = b + 37i on the complex plane. Since OPQ is an equilateral triangle, we know POQ = 60, so Q is the result of rotating P around O by 60. But then we know that q = p cis 60 = b + 37i = ( a 11 ) 3 + ( a ) i Notice that the imaginary part of the left side of this equation must equal the imaginary part of the right side, so ( a ) 3 37i = + 11 i = a = 1 3. Similarly we can equate the real parts, which gives us b = a 11 3 = 5 3. Thus the answer is ( 1 3 ) ( 5 3 ) = 315.

11 Lines and Collinearity The following theorems are commonly used in complex bash: Three distinct complex numbers a, b, and c are collinear if and only if c a is real. b a Since a complex number is real if and only if it is equal to its conjugate, the above means the equation for a line passing through a and b, in terms of z is z a b a = ( ) z a. b a If a point Z divides a segment AB into segments AZ and BZ in the ratio k/(1 k) then z = (b a)k + a. In particular, the midpoint of AB is a + b.

12 The Centroid - Problem Let ABC be a triangle in the complex plane. Find a formula for the centroid of ABC in terms of the complex numbers a, b, and c.

13 The Centroid - Solution Let M be the midpoint of BC. Then we know m = b + c. A well-known theorem states that the centroid G divides MA in the ratio 1 3 : 3. Hence, g = (a m) 1 ( 3 + m = a b + c ) b + c = a + b + c. 3

14 Parallelism and Perpendicularity Given lines AB and CD, one can determine whether or not they are parallel or perpendicular with the following theorems. AB CD b a ( ) b a b a is real d c d c =. d c AB CD b a d c is imaginary b a d c = ( b a d c ).

15 Reflections - Problem Given points A, B, and C, let Z be the reflection of C about AB. Find z in terms of a, b, and c.

16 Reflections - Solution Let M be the midpoint of ZC. We know m = z + c. If Z is the reflection of C about AB, then M should lie on AB. Hence we have ( ) m b m b a b = = z + c b z + c b = a b (a b) (ā b). Also ZC should be perpendicular to AB, so ( ) z c z c a b = c z = a b ā b.

17 Reflections - Solution We now have z + c b (a b) = z + c b (ā b) and z c a b = c z ā b. This is a system of equations in z and z. Solving for z and z, we see that z = a c + bā a b b c ā b.

18 Using the Unit Circle If z is on the unit circle, z = 1, so by the theorem z z = z we get z = 1 z. This substitution is useful. Scaling a diagram does not change length ratios or angles Scale diagram so that an important circle or some important points are on the unit circle. If problem involves a central triangle, let the three vertices of the triangle be on the unit circle.

19 The Orthocenter - Problem Let ABC be a triangle inscribed in the complex unit circle, and let h = a + b + c. Prove that H is the orthocenter of ABC.

20 The Orthocenter - Solution Let us define k = h a b c Notice that k = = (a + b + c) a b c ( ) b + c = b + c b c b c = b + c b c. = 1/b + 1/c 1/b 1/c = c + b c b = b + c b c = k. Hence k is imaginary, so HA BC. Similarly, HB CA and HC AB, so H is the orthocenter of ABC, as desired.

21 Euler Line A famous theorem by Euler states that in any triangle, the circumcenter, the centroid, and the orthocenter are collinear. The line passing through these three points is called the Euler Line of the triangle. To prove this, take any ABC and inscribe it in the complex unit circle. Let H be the orthocenter, O be the circumcenter, and G be the centroid. o = 0. g = a + b + c. 3 h = a + b + c. Then g o h o = (a + b + c)/3 a + b + c = 1 R, so O, G, H are collinear. 3

22 Outer Napoleon Triangle - Problem Given any ABC, construct point E A such that BE A C is an equilateral triangle, with E A being on the opposite side of BC as A. Let N A be the centroid of BE A C. Similarly define N B and N C. Prove that N A N B N C is an equilateral triangle.

23 Outer Napoleon Triangle - Solution We assume that ABC is oriented counter-clockwise (if it is oriented clockwise, the solution is almost identical). Notice that E C is the result of rotating A by 60 about B, so e c = (a b)cis (60 ) + b = a + b Then since N C is the centroid of AE C B, Similarly, n a = b + c n c = e c + a + b 3 = a + b + a b 3i. + a b 3i. 6 + b c 3i and nb = c + a 6 + c a 3i. 6

24 Outer Napoleon Triangle - Solution To prove N A N B N C is equilateral, it suffices to show that N B is the result of rotating N A by 60 about N C. But doing so is equivalent to showing that (n a n c )cis 60 + n c = n b, which can be done with the calculations below. (n a n c )cis 60 + n c ( b + c = + b c a + b 3i a b ) ( ) 1 3 3i i ( c a = + b a c ) ( ) 1 3 3i 6 + i + a + b = c a 4 + b a c 1 (c a) (b a c) + (a + b) = 4 = c + a + c a 6 c a b a c 3i + 3i 3i = nb. 4 + a + b + a b 3i 6 + a b 3i 6 + a + b + a b 3i 4 6 (b a c) + 3(c a) + (a b) + 3i 1 And hence N A N B N C is equilateral, as desired.

25 A Cyclic Kite - Problem Quadrilateral APBQ is inscribed in circle ω with P = Q = 90 and AP = AQ < BP. Let X be a variable point on segment PQ. 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 PQ, show that M moves along a circle. (015 USAMO #)

26 A Cyclic Kite - Solution Let ω be the complex unit circle, and let O be the origin. Since P = Q = 90, AB is a diameter of ω, so we can let a = 1 and b = 1. Now notice that APBQ is a kite, so PQ AB. But AB coincides with the real axis, hence PQ is perpendicular to the real axis and thus all points on PQ have the same real part. Therefore Re(x) is constant.

27 A Cyclic Kite - Solution Since A, X, and S are collinear, we have x 1 s 1 = ( ) x 1 = x 1 s 1 1/s 1. Since XT is perpendicular to AS, we have ( ) x t x t s 1 = s 1 Solving these equations for x and x, we get x = 1 ( 1 + s + t s ) t and x = 1 x 1/t = 1/s 1. ( s + 1 t t s ).

28 A Cyclic Kite - Solution We now see that Re(x) = x + x = 1 4 ( + s + t + 1 s + 1 t s t t ). s Let z = 1. Notice that m z = s + t 1 ( ) ( ) s + t 1 s + t 1 = = 1 ( 1 4 (s + t 1) s + 1 ) t 1 = 1 4 ( 3 s t 1 s 1 t + s t + t ) = 5 s 4 Re(x). Hence all possible M are located on a circle centered at Z with 5 radius Re(x). 4

29 Good luck! A Nontrivial Mess - Challenge Let ABC be a scalene triangle. Let K a, L a and M a be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of AK a L a intersects AM a a second time at point X a different from A. Define X b and X c analogously. Prove that the circumcenter of X a X b X c lies on the Euler line of ABC. (015 USA TSTST #)