264. Identities with complex numbers.

by Virgil Nicula, Apr 8, 2011, 2:51 AM

Quote:
$ 1\blacktriangleright$ Prove that for any $ \{x,y,z\}\subset\mathbb C$, $ |x-y| = |y-z| = |z-x|$ $ \Longleftrightarrow$ $ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}= 0$ .
For any $x\in\mathbb C$ I"ll denote the point $X$ which has the affix $ x$ and I"ll write $ X(x)$. It is well-known, but I use same letter, lower case and capital, for

a affix and a point respectively. For example, $ XY = |x-y|$ and the triangle $ XYZ$ is equilateral if and only if $ |x-y| = |y-z| = |z-x|$ .

Lemma I (it is obviously). For any $ \{x,y,z\}\subset\mathbb C$, the relations $ \left\|\begin{array}{cc}\sum\ (x-y)(x-z) = 0\\ \\ \boxed{x^{2}+y^{2}+z^{2}= xy+yz+zx}\\ \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}= 0\\ \\ (x-y)^{2}+(x-z)^{2}= (x-y)(x-z)\\ \\ (y-z)^{2}+(x-y)(x-z) = 0\end{array}\right\|$ are equivalently.
Quote:
Lemma II. Let $ ABC$ be a triangle for which denote $ \rho =\frac{AB}{AC}$ . Then for $\omega =\cos\phi+i\cdot\sin\phi$ ,

$ \boxed{m(\widehat{BAC}) =\phi\Longleftrightarrow (a-b)^{2}+\rho^{2}\cdot (a-c)^{2}= (a-b)(a-c)\cdot\rho\left(\omega+\overline{\omega}\right)}$ .
Proof. $ m(\widehat{BAC}) =\phi$ $ \Longleftrightarrow$ $ (a-b) = (a-c)\rho\omega\ \ \vee\ \ (a-b) = (a-c)\rho\overline{\omega}$ $ \Longleftrightarrow$

$ \left[\rho\omega (a-c)-(a-b)\right]\cdot\left[\rho\overline{\omega}(a-c)-(a-b)\right] = 0$ $ \Longleftrightarrow$ $ (a-b)^{2}+\rho^{2}\cdot (a-c)^{2}= (a-b)(a-c)\cdot\rho\left(\omega+\overline{\omega}\right)$ .

Particular cases.

$ 1\blacktriangleright\triangle ABC$ is $ A$- isosceles, i.e. $ \rho = 1$ $ \Longleftrightarrow$ $ (a-b)^{2}+(a-c)^{2}= (a-b)(a-c)\cdot (\omega+\overline{\omega})$ .

$ 2\blacktriangleright\triangle ABC$ is $ A$- right and $ A$- isosceles, i.e. $ \left\|\begin{array}{c}\rho = 1\\ \\ \phi =\frac{\pi}{2}\\ \\ \omega = i\end{array}\right\|$ $ \Longleftrightarrow$ $ (a-b)^{2}+(a-c)^{2}= 0$ $ \Longleftrightarrow$ $ 2(a-b)(a-c)+(b-c)^{2}= 0$ .

$ 3\blacktriangleright\triangle ABC$ is equilateral, i.e. $ \left\|\begin{array}{c}\rho = 1\\ \\ \phi =\frac{\pi}{3}\\ \\ \left|\begin{array}{c}\omega =\frac{1}{2}\cdot\left(1+i\cdot\sqrt 3\right)\\ \\ \omega^{3}=-1\ ;\ \omega^{2}-\omega+1 = 0\\ \\ \overline{\omega}=\frac{1}{\omega }= 1-\omega =-\omega^{2}\end{array}\right|\end{array}\right\|$ $ \Longleftrightarrow$ $ (a-b)^{2}+(a-c)^{2}= (a-b)(a-c)$ .

From the lemma I results $ |a-b| = |b-c| = |c-a|$ $ \Longleftrightarrow$ $ \triangle ABC$ is equilateral $ \Longleftrightarrow$ $ (a-b)^{2}+(b-c)^{2}+(c-a)^{2}= 0$ q.e.d.

$ 2\blacktriangleright$ Prove that for any $ \{a,b\}\subset\mathbb C$, $ |a| = |b| = |a+b|$ $ \Longleftrightarrow$ $ a^{2}+b^{2}+(a+b)^{2}= 0\ .$[/quote]

Proof. I"ll use the well-known identity $ |x+1|^{2}+|x-1|^{2}= 2\left(|x|^{2}+1\right)$ (the rule of the parallelogram)

which is equivalently with $ (x+1)(\overline x+1)+(x-1)(\overline x-1) = 2(x\overline x+1)$, what is evidently.

This identity is equivalently with the identity (seemingly more generally) $ \boxed{|a+b|^{2}+|a-b|^{2}= 2\left(|a|^{2}+|b|^{2}\right)}$.

Return to the second Oni's equivalence. Suppose w.l.o.g. $ b\ne 0$ and denote $ x =\frac{a}{b}$. Then this equivalence becomes

$ |x+1| = |x| = 1\ \left(\ \overline x =\frac{1}{x}\ \right)\Longleftrightarrow$ $ x^{2}+1+(x+1)^{2}= 0$ $ \Longleftrightarrow$ $ x^{2}+x+1 = 0$. Using the upper rule obtain $ |x+1| = |x| = 1$ $ \Longleftrightarrow$

$ |x-1|^{2}= 3$ $ \Longleftrightarrow$ $ (x-1)(\overline x-1) = 3$ $ \Longleftrightarrow$ $ (x-1)\left(\frac{1}{x}-1\right) = 3$ $ \Longleftrightarrow$ $ (x-1)^{2}+3x = 0$ $ \Longleftrightarrow$ $ x^{2}+x+1 = 0$ q.e.d.

Remark. Let $ ABC$ be a triangle with the circumcircle $ w = C(O,r)$ and the centroid $ G$. Choose the origin in the circumcenter $ O$ ,

i.e. $ o = 0$, $ |a| = |b| = |c| = r$ and $ 3g = a+b+c$. Therefore, $ \triangle ABC$ is equilateral $ \Longleftrightarrow$ $ G\equiv O$ $ \Longleftrightarrow$ $ a+b+c = 0$ .

$ 1\blacktriangleright$ I"ll prove prove that for any complex numbers $ a$, $ b$, $ c$, $ |a| = |b| = |c| = r\ne 0$, i.e. $ a\overline a = b\overline b = c\overline c = r^{2}$ exists the equivalence

$ a+b+c = 0\Longleftrightarrow ab+bc+ca = 0$. Indeed, $ a+b+c = 0\Longleftrightarrow$ $ \overline{\sum a}= 0$ $ \Longleftrightarrow$ $ \sum\overline a = 0$ $ \Longleftrightarrow$ $ r^{2}\cdot\sum\frac{1}{a}= 0$ $ \Longleftrightarrow$ $ ab+bc+ca = 0$ .

$ 2\blacktriangleright$ I"ll prove that for any complex numbers $ a$, $ b$, $ c$ exists the implication $ \left\{\begin{array}{c}a+b+c = 0\\ \ ab+bc+ca = 0\end{array}\right\|$ $ \implies$ $ |a| = |b| = |c| $.

Indeed, we can consider $ a,b,c$ as the roots of the equation $ x^{3}= k\in\mathcal C$ $ \implies$ $ |x|^{3}= |k|$ $ \implies$ $ |a| = |b| = |c| =\sqrt [3]{|k|}$ .

More precisely : if two from the relations $ \left\{\begin{array}{c}a+b+c = 0\\ \\ |a| = |b| = |c|\\ \\ ab+bc+ca = 0\end{array} \right\|$ are truly, then and the another relation is truly.

Particularly, for $ \left\{\begin{array}{c}a: = y-z\\ \ b: = z-x\\ \ c: = x-y\end{array}\right\|$ results $ |y-z| = |z-x| = |x-y|\Longleftrightarrow$ $ (y-z)(z-x)+(z-x)(x-y)+(x-y)(y-z) = 0$ $ \Longleftrightarrow$

$ \sum (x-y)(x-z) = 0$ $ \Longleftrightarrow$ $ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}= 0$ using the upper lemma I.


Proposed problem. If $w\ne 1$ is cube root of unity satisfying $\left\{\begin{array}{c} \frac{1}{a+w} + \frac{1}{b+w} +\frac{1}{c+w}=2w^2\\\\ \frac{1}{a+w^2} + \frac{1}{b+w^2} +\frac{1}{c+w^2}=2w\end{array}\right\|$ , then find the value of $E\equiv \frac{1}{a+1} + \frac{1}{b+1} +\frac{1}{c+1}$ .

Proof. $\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}\iff$ $\frac{3x^2+2(a+b+c)x+(ab+bc+ca)}{(a+x)(b+x)(c+x)}=\frac{2}{x}\iff$ $f(x)\equiv 3x^3-(ab+bc+ca)\cdot x-2abc=0$ .

Since $f(w)=f(w^2)=0$ and $1+w+w^2=0$ obtain that the number $1$ is third root of the equation $f(x)=0$ , i.e. $\frac{1}{a+1} + \frac{1}{b+1} +\frac{1}{c+1}=2$ .

PP-1. The distinct complex numbers $z_{1}$ , $z_{2}$ , $z_{3}$ are the affix of the vertices for the equilateral triangle $A_1A_2A_3$

respectively if and only if they verify the relation $z_{1} + z_{2}\epsilon +z_{3}\epsilon^2 =0$ , where $\epsilon\ne 1$ and $\epsilon^3=1$ .

Proof. Denote $w=\cos\frac {\pi}{3}+i\sin\frac {\pi}{3}$ . Observe that $w^3=-1$ , $w^2-w+1=0$ , $\overline w=-w^2$ and $\epsilon^3=1\ ,\ \epsilon\ne 1\implies$

$\underline{\overline{\left\|\ \epsilon=w^2\ \ \wedge\ \ \epsilon^2=-w\ \right\|}}$ . Thus, $\triangle A_1A_2A_3$ is equilateral $\Longleftrightarrow$ $z_1-z_2=w(z_3-z_2)\ \ \vee\ \ z_1-z_3=w(z_2-z_3)\ \Longleftrightarrow$

$z_1+(w-1)z_2-wz_3=$ $0\ \ \vee\ \ z_1-wz_2+$ $(w-1)z_3=0\ \Longleftrightarrow$ $z_1+w^2z_2-wz_3=0\ \ \vee\ \ z_1-wz_2+w^2z_3=0\ \Longleftrightarrow$

$z_1+\epsilon z_2+\epsilon^2z_3=$ $0\ \ \vee\ \ z_1+\epsilon^2z_2+$ $\epsilon z_3=0$ , where $z^3=1\implies$ $z\in\{1,\epsilon ,\epsilon^2\}$ and $\epsilon^3=1$ , $\epsilon^2+\epsilon +1=0$ .

Remark. $\triangle A_1A_2A_3$ is equilateral $\Longleftrightarrow$ $z_1+w^2z_2-wz_3=$ $0\ \ \vee\ \ z_1-wz_2+w^2z_3=$ $0\ \Longleftrightarrow$

$\left(z_1+w^2z_2-wz_3\right)\left(z_1-wz_2+w^2z_3\right)=$ $0\ \Longleftrightarrow\ z_1^2+z_2^2+$ $z_3^2+\left(w^2-w\right)\left(z_1z_2+z_2z_3+z_3z_1\right)=0$ . Insa $w^2-w+1=0$ .

In conclusion, $\boxed{\ \triangle\ A_1A_2A_3\ \mathrm{este\ echilateral}\ \Longleftrightarrow\ z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1\ }$ .
This post has been edited 15 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:45 AM

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116. The length of AE (nine methods !).
115. An difficult equation with one radical.
114. A property of tangential triangle for ABC.
113. Range for a function.
112. The equivalence relation ".s.s."
111. Simple, but interesting ...
110. German TST 2004 - Exam V , Problem 1
109. Two equal areas ==> find A.
108. Old, short and nice proof of Euler's relation.
107. XY parallel BC and X,Y are isogonal/izotomic points.
106. OLM - Bucuresti (geometrie).
105. A problem with three circles.
104. Concursul Revistei de Matematica din Galati (RMG).
103. Metrical usual relations in triangle. Applications.
102. Characterization of a metrical relation in a triangle.
101. O problema cu numere complexe.
100. Bobiller's theorem.
99. Some simple and nice geometry problems.
98. Symmedian & median (metrical relations).
97. Problem of butterfly.
+ August 2010
96. Integrals.
95. Three circles in an angle.
94. Equations of angle bisectors (interior and exterior).
93. The Appolonius' construct problems.
92. CA , CB tangent to parabola - OIMU 2008 Problem 4.
91. JBMO 2010, Problem 3.
90. Nice and difficult inequality in ABC (own).
89. Similarity (parallel/antiparallel).
88. A nice geometric locus conditioned metrically.
87. An interesting vectorial identity.
86. A very nice maxmin.
85. Problem "slicing" : 60-24-12.
84. IRAN national math olympiad - 2010
83. A metrical characterization of concurrence.
82. AA' , BN , CM concur in Gergonne point.
81. Sawayama and Thebault’s theorem.
80. Three concurrent perpendiculars (the Carnot's lemma).
79. A quadrilateral with many perpendicularities.
78. Jakobi's theorem.
77. Two equivalent perpendicularities.
76. A parallelogram and a circle (nice !).
75. Extended Steinbart Theorem.
74. Inegalitatea Erdos-Mordell.
73. A nice problems with a square and a perpendicularity.
+ July 2010
72. Colinearity in a triangle - H , P , M .
70. Colinearity with pole/polar - H\in PQ .
71. An angle questions (synthetic/trigonometric proofs).
69*1. Slicing problem 3.
69. Position of x1, x2 w.r.t. a given real number k.
68. Some nice applications of "pole/polar".
67. Remarkable algebraic/geometrical inequalities (1).
66. Remarkable perpendicularities in a triangle.
65. Parallel to BC through I .
64. Problems of the type "instant" (at most one minute).
63. A nice problem with radical center (livetolove212).
62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
59. IMO 2010, problem 4.
58. A cevian and two incircles.
57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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  • Total entries: 456
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