21 bis. Some problems with complex numbers.

by Virgil Nicula, Apr 22, 2010, 4:20 PM

PP1 (OLM Bucuresti 2006 , Dinu Serbanescu). Let $\{x,y\}\subset\mathbb C$ with the property $\forall\ n\in\mathbb N^*$ we have $|x^n - y^n| = 1$ . Prove that $xy = 0\ .$

Proof. I"ll use twice the identity of the parallelogram $:\ \boxed{|u+v|^2+|u-v|^2=2\left(|u|^2+|v|^2\right)}\ (*)\ .$

$\odot\ \left\{\begin{array}{c} |x-y|=1\\\\ \left|x^2-y^2\right|=1\end{array}\right\|\ \Longrightarrow\ |x+y|=1\ \implies\ |x-y|^2$ $+|x+y|^2=2\left(|x|^2+|y|^2\right)\ \Longrightarrow\ |x|^2+|y|^2=1\ (1)\ .$

$\odot\begin{array}{ccc} \nearrow & \left|x^2-y^2\right|=1 & \searrow\\\\ \searrow & \left|x^4-y^4\right|=1 & \nearrow\end{array}\odot \Longrightarrow\ \left|x^2+y^2\right|=1\ \implies\ \left|x^2-y^2\right|^2+$ $\left|x^2+y^2\right|^2=2\left(|x|^4+|y|^4\right)\ \Longleftrightarrow\ |x|^4+|y|^4=1\ (2)\ .$

$\odot\ \ \left(|x|^2+|y|^2\right)^2=|x|^4+|y|^4+2|xy|^2\ \stackrel{(1)\wedge (2)}{\ \ \Longrightarrow\ \ }\ |xy|=0\ \Longrightarrow\ xy=0\ .$


PP2. Let $ z\in\mathbb C$ be a complex number for which exists $ n\in\mathbb N$ so that $ n\ge 2$ and $ 2^{n - 1}\cdot\left(z^n + 1\right) = (z + 1)^n$ . Prove that $ |z| = 1$ .

Proof. Observe that $ z\ne 0\ (z=0\implies n=1)$ . Suppose w.l.o.g. $ z\ne 1$ . Denote $ |z| = r$ . Thus, $ \overline z = \frac {r^2}{z}$ and $ 2^{n - 1}\cdot\left(z^n + 1\right) = (z + 1)^n$ $ \Longleftrightarrow$ $ 2^{n - 1}\cdot\left(\overline z^n + 1\right) = (\overline z + 1)^n$ $ \Longleftrightarrow$

$ 2^{n - 1}\cdot\left(r^{2n} + z^n\right) = \left(r^2 + z\right)^n$ . Suppose $ \mathrm {ad\ absurdum}$ that $ r\ne 1$ . Therefore, $ \left\{\begin{array}{c} 2^{n - 1}\cdot\left(z^n + 1\right) = (z + 1)^n \\ \\ 2^{n - 1}\cdot\left(r^{2n} + z^n\right) = \left(r^2 + z\right)^n\end{array}\ \right\|\ \boxed - \ \uparrow$ $ \implies$ $ 2^{n - 1}\cdot\left(r^{2n} - 1\right) = \left(r^2 + z\right)^n - (z + 1)^n$

$ \implies$ $ \frac {2^{n - 1}\cdot\left(r^{2n} - 1\right)}{r^2 - 1} = \left|\sum_{k = 1}^n\left(r^2 + z\right)^{n - k}(z + 1)^{k - 1}\right|\le$ $ \sum_{k = 1}^n\left(r^2 + r\right)^{n - k}(r + 1)^{k - 1} =$ $ (r + 1)^{n - 1}\cdot\sum_{k = 1}^nr^{n - k}$ $ \implies$ $ \frac {2^{n - 1}\cdot\left(r^{2n} - 1\right)}{r^2 - 1}\le (r + 1)^{n - 1}\cdot\frac {r^n - 1}{r - 1}$ $ \implies$

$ \boxed {\ 2^{n - 1}\cdot \left(r^n + 1\right)\le (r + 1)^{n}\ \ (*)\ }$ because $ \frac {r^n - 1}{r - 1} > 0$ . From the relation $ (*)$ and the well-known inequality $ 2^{n - 1}\cdot \left(r^n + 1\right)\ge (r + 1)^{n}$ obtain $ 2^{n - 1}\cdot \left(r^n + 1\right) = (r + 1)^{n}$ ,

i.e. $ r = 1$ , what is $ \mathrm {absurd}$ . In concluzion, $ 2^{n - 1}\cdot\left(z^n + 1\right) = (z + 1)^n$ $ \implies$ $ |z| = 1$ . Here is an equivalent enunciation :

An easy extension. Let $ \{u,v\}\subset \mathbb C$ be two complex numbers for which exists $ n\in\mathbb N$ so that $ n\ge 2$ and $ 2^{n - 1}\cdot\left(u^n + v^n\right) = (u+ v)^n$ . Prove that $ |u| = |v|$ .


PP3 (Stan Fulger). Construct outside of $\triangle ABC$ the equilateral $\left\{\begin{array}{ccc} \triangle BCA'\ ; & \triangle CAB'\ ; & \triangle ABC'\\\\ \triangle B'C'A''\ ; & \triangle C'A'B''\ ; & \triangle A'B'C''\end{array}\right\|$

so that $BC$ separates $\{A,A'\}$ a.s.o. and $B'C'$ separates $\{A',A''\}$ a.s.o. Prove that $\left\{\begin{array}{c} B\in B'B''\ ;\ BB'=BB''\\\\ O\in O'O''\ ;\ OO'=OO''\end{array}\right\|$

Proof. Let $X(x)$ be the point $X$ from the complex plane and which has the affix $x\in\mathbb C$ and $w=\cos\frac {\pi}3+i\sin\frac {\pi}3$ for which $\left\{\begin{array}{c} w^3+1=0\ ;\ w^2-w+1=0\\\\ \overline w=\frac 1w=-w^2=1-w\end{array}\right\|$ .

Thus, $\left\{\begin{array}{ccc} a'-b=w(c-b) & \implies & a'=(1-w)b+wc\\\\ b'-c=w(a-c) & \implies & b'=(1-w)c+wa\\\\ c'-a=w(b-a) & \implies & c'=(1-w)a+wb\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccccc} a''-b'=w(c'-b') & \implies & a''=b'+w(c'-b') & \implies & a''=2a+w^2b-wc\\\\ b''-c'=w(a'-c') & \implies & b''=c'+w(a'-c') & \implies & b''=2b+w^2c-wa\\\\ c''-a'=w(b'-a') & \implies & c''=a'+w(b'-a') & \implies & c''=2c+w^2a-wb\end{array}\right\|$

and $\left\{\begin{array}{ccc} 3o=b+c+a'=b+c+[(1-w)b+wc] & \implies & 3o=(2-w)b+(1+w)c\\\\ 3o'=a'+c'+b''=[(1-w)b+wc]+[(1-w)a+wb]+[2b+w^2c-wa] & \implies & 3o'=(1-2w)a+3b+(2w-1)c\\\\ 3o''=a'+b'+c''=[(1-w)b+wc]+[(1-w)c+wa]+[2c+w^2a-wb] & \implies & 3o''=(2w-1)a+(1-2w)b+3c\end{array}\right\|$

Observe easily that $\left\{\begin{array}{ccccc} b'+b''=[(1-w)c+wa]+[2b+w^2c-wa] & \implies & b'+b''=2c & \implies & B\in B'B''\ ;\ BB'=BB''\\\\ 3(o'+o'')=2(2-w)b+2(1+w)c=3o & \implies & o'+o''=2o & \implies & O\in O'O''\ ;\ OO'=OO''\end{array}\right\|$ .

Remark. Prove easily that $w^2-w+1=0\implies$ for any $\{x,y\}\subset\mathbb C\ ,\ \boxed{[(1-w)x+wy][(1-w)u+wv]=-wxu+w^2yv+(xv+yu)}$ . For $u:=x\ ,\ v:=y$ get

$\boxed{[(1-w)x+wy]^2=-wx^2+w^2y^2+2xy}$ . With the well-known property $"\ ABC$ is equilateral $\iff \sum a^2=\sum bc\ "$ can prove easily that $A'B'C'$ is equilateral $\implies$

$ABC$ is equilateral. Suppose that $\boxed{\sum a'^2=\sum b'c'}\ (*)$ . Thus, $\sum a'^2=\sum \left[(1-w)b+wc\right]^2=$ $\sum \left[-wb^2+w^2c^2+2bc\right]=$ $\sum\left(w^2-w\right)a^2+2\sum bc\implies$

$\boxed{\sum a'^2=\sum \left(-a^2+2bc\right)}\ (1)$ . On other hand, $\sum b'c'=\sum [(1-w)c+wa][(1-w)a+wb]=$ $\sum \left[-wca+w^2ab+\left(a^2+bc\right)\right]=$ $\sum\left[\left(w^2-w+1\right)bc+a^2\right]\implies$

$\boxed{\sum b'c'=\sum a^2}\ (2)$ . In conclusion, the relations $(*)\wedge (1)\wedge (2)\implies$ $\sum a^2=\sum \left(-a^2+2bc\right)\implies$ $\sum a^2=\sum bc\implies$ the triangle $ABC$ is equilateral.


PP4. Let $ABCD$ , $EBFG$ be two squares for which denote the midpoints $M$ , $N$ , $L$ , $P$ of the segments $[BE]$ , $[CB]$ , $[DF]$ , $[AG]$ respectively. Prove that $MNLP$ is a square.

Proof. Prove easily that for $A(a)\ ,\ B(-ia)\ ,\ C(-a)\ ;\ D(ia)\ ($ can verify $A+C=B+D$ , i.e. $ABCD$ is a parallelogram and $C-B=i(A-B)$ , i.e. $BA\perp BC\ )$

obtain that $E(z)\ ,\ F[iz-(1+i)a]\ ,\ G[-a+(1+i)z]\ ($ can verify $E+F=B+G$ , i.e. $ABCD$ is a parallelogram and $F-B=i(E-B)$ , i.e. $BF\perp BE\ )$ .

Therefore, $M\left(\frac{z-ia}2\right)\ ,\ N\left[\frac{-(1+i)a}2\right]\ ,\ L\left(\frac{-a+iz}2\right)\ ,\ P\left[\frac{(1+i)z}2\right]$ . Observe that $ML=NP=(1+i)(z-a)$ , i.e. $MNLP$ is a parallelogram

and $N-M=i(P-M)=\frac {-(a+z)}2$ , i.e. $MN\perp MA$ . In conclusion, $MNLP$ is a square.


PP5. Let $ABC$ , $EBD$ be two equilateral triangles for which denote the midpoints $M$ , $N$ , $P$ of the segments $[BE]$ , $[CB]$ , $[AD]$ respectively and

the centroids $R$ , $S$ , $T$ of the triangles $EBD$ , $MNP$ , $ABC$ respectively. Prove that $MNP$ is an equilateral triangle and $S$ is the midpoint of $[RT]$ .

Proof. Prove easily that for $A(a)\ ,\ B(0)\ ,\ C(wa)\ ,\ E(z)$ obtain that $D(wz)\ ,\ M\left(\frac z2\right)\ ,\ N\left(\frac{wa}2\right)\ ,\ P\left(\frac{a+wz}2\right)$ , where $w^3+1=0=w^2-w+1$ . Prove easily that

$m^2+n^2+p^2=mn+np+pm$ , i.e. $\triangle MNP$ is equilaterally and $R\left[\frac {(1+w)z}{3}\right]$ , $S\left[\frac {(1+w)(a+z)}{6}\right]$ , $T\left[\frac {(1+w)a}{3}\right]$ , where $2S=R+T$ , i.e. $S$ is the midpoint of $[RT]$ .
This post has been edited 48 times. Last edited by Virgil Nicula, Nov 23, 2015, 6:57 AM

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Nice ones!!

by Fermat_Theorem, Jan 12, 2019, 6:51 AM

Own problems or extensions/generalizations of some problems which was posted here.

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Virgil Nicula
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132. Asupra unei inegalitati intr-un triunghi.
131. A stronger Form of the Steiner-Lehmus Theorem (own).
130. A triangle and a fixed point.
129. BX=CY and three properties.
128. Value of IG and two inequalities in neighbourhood of 0.
127. Pagina educativa.
126. Routh's theorem.
125. I like very much this Darij Grinberg's message.
124. A nice and easy problem with a right-angled triangle.
123. My extension and its proof.
122. An identity for an equilateral triangle. Extension.
121. Viete's relations with cos A , cos B , cos C.
120. Two equal neighbouring angles/sides in a quadrilateral.
119. Inspired by a Miculita's problem.
118. An usual remarkable geometrical locus.
117. Some applications of harmonic division/quadrilateral.
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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