258. Some difficult and nice problems with complex numbers.

by Virgil Nicula, Apr 2, 2011, 9:47 PM

Proposed problem 1. Suppose that for $n\in\mathbb N^*$ and $z\in\mathbb C$ exists $\lambda\in \mathbb C$ so that

$\lambda =\frac {z^{n+1}+1}{z^n+z}$ . Prove that $\lambda\in \mathbb R$ and $|\lambda | \le 1\ \implies\ |z|=1$ (Al. Halanay).


An easy extension. Suppose that for $\{k,n\}\subset\mathbb N^*$ , $k<n$ and $z\in\mathbb C$ exists $\lambda\in \mathbb C$ so that

$ \lambda =\frac {z^n+1}{z^{n-k}+z^k}$ . Prove that $\lambda\in \mathbb R$ and $|\lambda |\le 1\ \implies\ |z|=1$ (Fl. Vulpescu Jalea).

Proof. Let $a$ be a complex number such that $|a| \leq 1$ and let $z$ be an arbitrary complex number. Prove easily that $\left\{\begin{array}{ccc} |z| \leq 1 & \implies & \left|a +z\right| \leq \left|1+\bar{a}z\right|\\\\ |z| \geq 1 & \implies & \left|a +z\right| \geq \left|1+\bar{a}z\right|\end{array}\right\|\ (*)$ .

In both cases the equality holds iff $|a|=1$ or $|z|=1$ . Suppose that $ \lambda =\frac {z^n+1}{z^{n-k}+z^k}$ , i.e. $z^{n-k}\left(z^k-\lambda\right)=\lambda z^k-1$ . If $z^k=\lambda$ , then $\lambda z^k=1\implies$

$\lambda =\pm 1$ and $|z|=1$ . Assume that $z^k\ne \lambda$ . Thus, $z^{n-k} =\frac{\lambda z^k-1}{z^k-\lambda}$ . If $\lambda =\pm 1$ , then the result is established immediately. We suppose for now that $|\lambda | \neq 1$ .

If $|z| < 1$ by $(*)$ with $a:=-\lambda$ and $z:=z^k$ we have $1 > |z|^{n-k} = \left|\frac{\lambda z^k-1}{z^k-\lambda}\right| > 1$ which is absurd. Likewise, $|z| > 1$ is not a possibility. Therefore, $|z|=1$ .

Remark. For $k:=1$ and $n:=n+1$ obtain an old and very nice Al. Halanay's problem.


Proposed problem 2. Let $a,b,c \in \mathbb{C}, a \neq b \neq c$ and $|bc|a + |ca|b + |ab|c = 0$ . Prove that $|(b+c)(a+c)(a+b)| \geq |abc|$ .

Proof (algebraic).

Proof (geometric). If $abc = 0$ then the inequality is clearly. Suppose from this point that $abc \neq 0$ . Write $x:=\frac{a}{|a|}$ , $y:=\frac{b}{|b|}$ and $z:=\frac{c}{|c|}$ . Therefore, $x$ , $y$ and $z$

are points on the unit circle with the property that $x+y+z=0$ , i.e. the centroid of the triangle generated by $x$ , $y$ and $z$ is exactly the circumcenter. Thus, $x$ , $y$ and $z$

form an equilateral triangle. We may assume without loss of generality that $x$, $y$ and $z$ are located in this order if viewed counterclockwise. Geometrically, if $A$ , $B$ , $C$

and $O$ are points representing $a$ , $b$ , $c$ and $0$ , then $\angle{BOC}=\angle{COA}=\angle{AOB}=\frac{2\pi}{3}$ . Let $P$ , $Q$ and $R$ be the midpoints of $BC$ , $CA$ and $AB$ respectively.

The inequality we have to prove is equivalent to $OP \cdot OQ \cdot OR \geq $ $\frac{OA \cdot OB \cdot OC}{8}$ . We may instead show $OP \geq \sqrt{\frac{OB\cdot OC}{4}}$ . The Stewart's theorem

and the Law of Cosine give us $OP^2 = \frac{OB^2+OC^2}{2}-\left(\frac{BC}{2}\right)^2 = \frac{OB^2+OC^2}{2}-\frac{OB^2+OC^2+OB\cdot OC}{4}$ .

That is $OP^2 = \frac{OB^2+OC^2-OB\cdot OC}{4} \geq \frac{OB \cdot OC}{4}$ , where we have used AM-GM. The proof is now complete.


Proposed problem 3. Let $P$ be a point in an acute $\triangle ABC$. Prove that $\left\{\begin{array}{c} PA\cdot PB\sin C+PB\cdot PC\sin A+PC\cdot PA\sin B\ge 2S\\\\ PA^2\sin A+PB^2\sin B+PC^2\sin C\ge 2S\\\\ \frac {PA^3}{bc}+\frac {PB^3}{ca}+\frac {PC^3}{ab}\ge 3\cdot PG\end{array}\right\|$ , where $S=[ABC]$ .

Proof. Using the relations $a=2R\sin A$ a.s.o., where $R$ is the circumradius of $\triangle ABC$ , the inequalities is equivalently with $\left\{\begin{array}{c} \sum a\cdot PB\cdot PC\ge abc\\\\ \sum a\cdot PA^2\ge abc\\\\ \sum a\cdot PA^3\ge3abc\cdot PG \end{array}\right\|$

respectively because $4RS=abc$ , i.e. $\left\{\begin{array}{c} \sum\frac{PB \cdot PC}{bc}\ge 1\\\\ \sum\frac{PA^2}{bc}\ge 1\\\\ \sum\frac{PA^3}{bc}\ge 3\cdot PG\end{array}\right\|$ . Let $x$ , $y$ , $z$ be the complex affixes of $A$ , $B$ , $C$ respectively and set $P$

as the origin of the complex plane. Then we have $\left\{\begin{array}{ccccc} \sum \frac{PB \cdot PC}{bc} & = & \sum\frac{|yz|}{|x-y||x-z|} & \ge & \left|\sum\frac{yz}{(x-y)(x-z)}\right| = 1\\\\ \sum \frac{PA^2}{bc} & = & \sum\frac{|x|^2}{|x-y||x-z|} & \ge & \left|\sum\frac{x^2}{(x-y)(x-z)}\right| = 1\\\\ \sum \frac{PA^3}{bc} & = & \sum\frac{|x|^3}{|x-y||x-z|} & \ge & \left|\sum\frac{x^3}{(x-y)(x-z)}\right| =|x+y+z|\end{array}\right\|$ as desired.


Proposed problem 4 (China Olympiad 1998). Let $P$ be an arbitrary point in the plane of triangle $[ABC]$ . Prove that

$PB\cdot PC\cdot BC+PC\cdot PA\cdot CA+PA\cdot PB\cdot AB\ge AB\cdot BC\cdot CA$ with equality iff $P$ is the orthocenter of $\triangle ABC$ .

Proof. Denote $X(x)$ - the point $X$ with affix $x\in\mathbb C$ . I"ll use the well-known identity $\sum_{\mathrm{cyc}} (b-c)(p-b)(p-c)=(a-b)(b-c)(c-a)$ .

Therefore, $\sum_{\mathrm{cyc}} BC\cdot PB\cdot PC=$ $\sum_{\mathrm{cyc}} |(b-c)(p-b)(p-c)|\ge$ $|\sum_{\mathrm{cyc}} (b-c)(p-b)(p-c)|=$ $|(a-b)(b-c)(c-a)| = AB\cdot BC\cdot CA $ .


Proposed problem 5 (Short List ONM 2002 Romania). For given $n\in \mathbb N^*$ ascertain $z\not\in\mathbb R$ for which $z^n\in\mathbb R$ and $(1+z)^n\in\mathbb R$ .

Proof. $\left\{z^n, (1+z)^n\right\}\subset\mathbb R\iff$ $\left\|\begin{array}{c} z^n=\overline z^n\\\\ (1+z)^n=\left(1+\overline z\right)^n\end{array}\right\|\iff$ $\left\|\begin{array}{c} \left(\frac {\overline z}{z}\right)^n=1\\\\ \left(\frac {1+\overline z}{1+z}\right)^n=1\end{array}\right\|\iff$ $(\exists )\ \{k,s\}\subset\overline {1,n-1}$ so that

$\left\|\begin{array}{c} \overline z=z\cdot w^k\\\\ 1+\overline z=(1+z)\cdot w^s\end{array}\right\|$ , where $w=\cos\frac {2\pi}{n}+i\cdot\sin\frac {2\pi}{n}$ and $w^n=1$ $\implies$ $z_{k,s}=\frac {w^s-1}{w^k-w^s}$ , where $k\ne s$ and $\{k,s\}\subset\overline {1,n-1}$ .


PP 6. Prove that $\{z_1 ,z_2\}\subset\mathbb C\ \ \wedge\ \ |z_1|=1\ \ \vee\ \ |z_2|=1\ \implies\ \left| {z_1 + 1} \right| + \left| {z_2 + 1} \right| + \left| {z_1 z_2 + 1} \right| \ge 2$ .

Proof. Suppose $|z_1|=1$ . Thus, $z_1\cdot\overline{z_1}=1\implies$ $|z_1 + 1| + | z_2 + 1| + | z_1 z_2 + 1| =$ $\left|\frac {1}{\overline{z_1}} + 1\right| + | z_2 + 1| + \left|\frac {1}{\overline {z_1}}\cdot z_2 + 1\right| =$

$ |\overline{z_1} + 1| + | z_2 + 1| + | \overline{z_1} + z_2| \ge$ $ |(\overline{z_1} + 1) + ( z_2 + 1) - ( \overline{z_1} + z_2)| = 2$.
This post has been edited 46 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:22 AM

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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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    by MrMustache, Nov 11, 2020, 4:52 PM

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    by piphi, Jun 10, 2020, 11:44 PM

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  • Joined: Jun 22, 2005
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  • Total entries: 456
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