8. The first problem Shortlist ONM 2010 Romania.

by Virgil Nicula, Apr 19, 2010, 6:00 PM

Quote:
Daca $x$ , $y$ , $z\in\mathbb C$ au acelasi modul $1$, aratati ca

$|x-y|^2\ +\ |x-z|^2\ +\ 1\ \ge\ |y-z|^2\ \ (1)$ .

Met 1. Vom folosi o identitate peste multimea numerelor complexe care se demonstreaza usor si pe care va rog sa o retineti :

$\boxed {\ |a+b+c|^2+|a|^2+|b|^2+|c|^2=|a+b|^2+|b+c|^2+|c+a|^2\ }$ care in interpretare geometrica este echivalenta

cu relatia $HA^2+HB^2+HC^2=HO^2+3R^2$ . Folosind substitutia $a=y-x$ , $b=x-z$ , $c=z$ se obtine identitatea

$|y|^2+|z|^2+|x-y|^2+|x-z|^2=|x|^2+|y-z|^2+|y+z-x|^2$ din care rezulta (evident !) inegalitatea

$\boxed {\ |y|^2+|z|^2+|x-y|^2+|x-z|^2\ge |x|^2+|y-z|^2\ }$ cu egalitate iff $\boxed {\ y+z=x\ }$ . In concluzie, $|x|=|y|=|z|=1\ \Longrightarrow\ (1)$ .

Met 2. Vom folosi aceeasi identitate insa cu o substitutie mai simpla $a:=-x$ si $b:=y$ , $c:=z$ . Se obtine astfel identitatea

$|y+z-x|^2+\sum |x|^2=|y+z|^2+|x-y|^2+|x-z|^2$ . Din relatia paralelogramului $|y+z|^2+|y-z|^2=2\left(|y|^2+|z|^2\right)$ scoatem

$|y+z|^2$ si dupa introducerea in relatia precedenta se obtine identitatea $|y+z-x|^2+|x|^2= |y|^2+|z|^2-|y-z|^2+|x-y|^2+|x-z|^2$ .

Deci $|y|^2+|z|^2-|y-z|^2+|x-y|^2+|x-z|^2\ge |x|^2$ cu egalitate iff $y+z=x$ . In concluzie $|x|=|y|=|z|=1\ \implies\ (1)$ .

Observatie. Aceasta problema este exprimarea prin numere complexe a unei inegalitati cunoscute din geometria

triunghiului: $\boxed {\ b^2+c^2+R^2\ \ge\ a^2\ } (*)$ cu egalitate daca si numai daca $A=120^{\circ}$ si $B=C$ .

Met 1. Notam cercul circumscris $C(O,R)$ al $\triangle ABC$ , mijlocul $M$ al laturii $[BC]$ si simetricul $N$ al lui $O$ fata de $M$ . Exprimam mediana $[AM]$ in triunghiurile

$ABC$ si $AON\ :\ 2\left(b^2+c^2\right)-a^2=4m_a^2=$ $2\cdot\left(AN^2+AO^2\right)-ON^2=$ $2\cdot AN^2+2R^2-4\cdot OM^2=$ $2\cdot AN^2+2R^2-4R^2+a^2$ $\Longrightarrow$

$AN^2=b^2+c^2 -a^2+R^2\ge 0$ $\Longrightarrow$ $b^2+c^2+R^2\ge a^2$ .

Met 2. Ineg. $(*)$ $\Longleftrightarrow$ $2bc\cdot\cos A+R^2\ge 0\ \Longleftrightarrow\ $ $1+8\cos A\sin B\sin C\ge 0$ $\Longleftrightarrow$

$1+4\cdot\cos A\left[\cos (B-C)+\cos A\right]\ge 0$ $\Longleftrightarrow$ $\left[2\cos A+\cos (B-C)\right]^2+\sin^2(B-C)\ge 0$ .
Quote:
Extindere. $\left(\forall\right) X$ din planul $(ABC)$ exista relatia $\boxed {\ b^2+c^2-a^2\ge XA^2-XB^2-XC^2\ }\ \ (**)$

cu egalitate daca si numai daca punctul $X$ este simetricul varfului $A$ fata de mijlocul laturii $[BC]$ .

Observatie. Daca punctul $X$ este centrul cercului circumscris triunghiului $ABC$ , atunci se obtine inegalitatea $(*)$ .

Met 1 (V.N). Notam mijlocul $M$ pt. $[BC]$ si simetricul $L$ al lui $A$ in raport cu $M$ . Aplicam teorema medianei in triunghiurile

$\left\|\begin{array}{ccc} \triangle ABC & \iff & 4m_a^2=2(b^2+c^2)-a^2\\\\ \triangle BXC & \iff & 4\cdot XM^2=2\cdot\left(XB^2+XC^2\right)-a^2\\\\ \triangle AXL & \iff & 4\cdot XM^2=2\cdot\left(XA^2+XL^2\right)-4m_a^2\end{array}\right\|$ $\Longrightarrow$ $\boxed {XA^2-XB^2-XC^2+XL^2=b^2+c^2-a^2}$ $\Longrightarrow\ (**)$ .

Met 2 (V.N). Notam mijlocul $M$ pt. $[BC]$ si simetricul $Y$ al lui $X$ in raport cu $M$ . Aplicam teorema medianei in triunghiurile

$\left\|\begin{array}{ccc} \triangle ABC & \Longrightarrow & 4m_a^2=2(b^2+c^2)-a^2\\\\ \triangle BXC & \Longrightarrow & 4\cdot XM^2=2\cdot\left(XB^2+XC^2\right)-a^2\\\\ \triangle XAY & \Longrightarrow & 2m_a^2=XA^2+YA^2-2\cdot XM^2\end{array}\right\|\Longrightarrow$ $\boxed {\ \left(XA^2-XB^2-XC^2\right)+YA^2=b^2+c^2-a^2\ }\ \Longrightarrow\ (**)$ .

Met 3 (Moldovan). Notam mijloacele $M$ , $N$ pt. $[BC]$ si $[AX]$ respectiv . Deci $b^2+c^2-a^2 \ge XA^2-XB^2-XC^2 \Longleftrightarrow$

$\frac{2(b^2+c^2)-a^2}{4}-\frac{a^2}{4} \ge \frac{XA^2}{2}-\frac{2(XB^2+XC^2)}{4} \Longleftrightarrow$ $\frac{2(b^2+c^2)-a^2}{4}+\frac{2(XB^2+XC^2)-a^2}{4} \ge \frac{XA^2}{2}\ \Longleftrightarrow$

$AM^2+MX^2 \ge \frac{XA^2}{2} \ \Longleftrightarrow$ $\frac{2(AM^2+MX^2)-XA^2}{4}\ \ge 0\ \Longleftrightarrow\ MN^2\ \ge\ 0$ care este adevarata.
Quote:
Generalizare. $\left(\forall\right) X$ din planul $(ABC)$ si $\left(\forall\right)$ numerele reale $m\ne -1$ , $p\ne -1$ exista inegalitatea

$\boxed {p\left(c^2+mb^2\right)-\frac {m(1+p)}{1+m}\cdot a^2\ge\frac {p(1+m)}{1+p}\cdot XA^2-XB^2-m\cdot XC^2}\ \left(***\right)$ .

Demonstratie. Consideram punctele $M\in BC$ si $Y\in MX$ pentru care $\overline {BM}=m\cdot\overline {MC}$ si $\overline {XM}=p\cdot\overline {MY}$ . Aplicam relatia lui Stewart

pentru cevienele si triunghiurile mentionate : $\begin{array}{ccc} \overline{AM}/BAC & \implies & (1+m)^2AM^2=(1+m)\left(c^2+mb^2\right)-ma^2\ .\\\\ \overline{AM}/XAY & \implies & (1+p)^2AM^2=(1+p)\left(AX^2+p\cdot AY^2\right)-p\cdot XY^2\ .\\\\ \overline{XM}/BXC & \implies & (1+m)^2\cdot XM^2=(1+m)\cdot\left(XB^2+m\cdot XC^2\right)-ma^2\ .\end{array}$ .

Insa $\overline {XY}=\frac {1+p}{p}\cdot \overline {XM}$ . Eliminand $AM^2$ si $XM^2$ intre cele trei relatii de sus obtinem $-p(1+p)(1+m)\left(c^2+mb^2\right)+mp(1+p)a^2=$

$-p(1+m)^2\left(AX^2+pAY^2\right)+(1+p)(1+m)\left(XB^2+m\cdot XC^2\right)-m(1+p)a^2$ , adica $p(1+p)(1+m)\left(c^2+mb^2\right)-m(1+p)^2a^2=$

$p(1+m)^2\cdot AX^2-(1+p)(1+m)\cdot XB^2-m(1+m)(1+p)\cdot XC^2+p^2(1+m)^2\cdot AY^2$ . Din $AY^2\ge 0$ se obtine

inegalitatea $(***)$ . Avem egalitate daca si numai daca $Y\equiv A$ , adica $M\in AX$ si $\overline{XM}=p\cdot \overline {MA}$ .
Quote:
Problema propusa 1. Sa se arate ca intr-un triunghi $ABC$ exista inegalitatea $\boxed {\ \frac {b+c}{a}\ +\ \frac {a^2-bc}{b^2+c^2-bc}\ \ge\ 2\ }$ .

Indicatie. Se aplica inegalitatea $(**)$ pentru $X:=I_a$ - centrul cercului $A$ - exinscris triunghiului $ABC$ .

Observatie. Aceasta inegalitate este echivalenta cu $\boxed {\ a^3+b^3+c^3+abc\ \ge\ 2a(b^2+c^2)\ }\ \ge\ 4abc$ .
Quote:
Problema propusa 2. Sa se arate urmatoarea identitate (peste $C$ ), care in interpretare geometrica este echivalenta cu relatia $HO^2+\sum OM^2=12R^2$ ,

unde $NP\parallel BC$ , $PM\parallel CA$ si $MN\parallel AB$ : $|x+y+z|^2+|y+z-x|^2+|z+x-y|^2+|x+y-z|^2=4\cdot\left(|x|^2+|y|^2+|z|^2\right)$ .
Quote:
Problema propusa 3. Sa se arate ca daca este dat un triunghi $\Delta\equiv ABC$ , atunci exista un punct $X$ astfel incat

$XB^2+XC^2=XA^2\iff$ $\Delta$ este ascutit sau dreptunghic si $X$ este unic daca si numai daca $\Delta$ este dreptunghic.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5722

http://artofproblemsolving.com/Forum/viewtopic.php?f=47&t=345744

http://artofproblemsolving.com/Forum/viewtopic.php?f=47&t=345742

http://www.artofproblemsolving.com/viewtopic.php?t=55455&search_id=655031139

http://www.artofproblemsolving.com/viewtopic.php?p=296657&search_id=655031139#296657

http://www.artofproblemsolving.com/viewtopic.php?p=277553&search_id=1382933615#277553

See and here
This post has been edited 37 times. Last edited by Virgil Nicula, Oct 1, 2017, 7:55 AM

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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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