437. An equivalence relation over R.

by Virgil Nicula, Dec 27, 2015, 8:02 AM

Define in $\mathbb R$ the following equivalence relation : for any $\{x,y\}\subset\mathbb R$ we"ll mean $\boxed{\ x\ .s.s.\ y\ \iff\ x=y=0\ \vee\ xy>0\ }$ , i.e. $x$ and $y$ have "same sign".

Some examples. Consider $0<a,b\ne 1$ , $\{x,y\}\subset \mathbb R$ .

$\blacktriangleright\ \frac xy\ .s.s.\ xy$ , where $y\ne 0\ ;\ |x|\ .s.s.\ x^2\ ;\ |x|-|y|\ .s.s.\ x^2-y^2$ .

$\blacktriangleright\ \sqrt[n]x\ .s.s.\ x$ , where $x\in \mathbb D(n)=\left\{\begin{array}{ccc} \mathbb R & , & n\in 2\mathbb N+1\\\\ \mathbb R_+ & , & n\in 2\mathbb N^*\end{array}\right\|$ . Convention. $\sqrt[1]x=x$ for any $x\in\mathbb R$ .

$\blacktriangleright\ \left(\sqrt[n]x-\sqrt [n]y\right)\ .s.s.\ (x-y)$ , for any $\{x,y\}\subset D(n)$ .

$\blacktriangleright\ \left(a^x-a^y \right)\ .s.s.\ (a-1)(x-y)\ \ ;\ \ a^x-b^x\ .s.s.\ x(a-b)$ . Particular cases. $a^x-1\ .s.s.\ x(a-1)$ and $a^x-a\ .s.s.\ (a-1)(x-1)$ .

$\blacktriangleright\ \log_ax-\log_ay\ .s.s.\ (a-1)(x-y)\ \ ;\ \ \log_ax\ .s.s.\ (a-1)(x-1)$ , where $x>0$ and $y>0$ .

$\blacktriangleright\ \log_ax-\log_bx\ .s.s.\ (a-1)(b-1)(x-1)(b-a)$ , where $x>0$ .

$\blacktriangleright\ \arcsin x\ .s.s.\ x\ ;\ (\arcsin x-\arcsin y)\ .s.s.\ (x-y)$ , where $\{x,y\}\subset [-1,1]$ .

$\blacktriangleright\ \arccos x\ .s.s.\ (1-x)\ ;\ (\arccos x-\arccos y)\ .s.s.\ (y-x)$ , where $\{x,y\}\subset [-1,1]$ .

$\blacktriangleright\ \arctan x\ .s.s.\ x\ ;\ (\arctan x-\arctan y)\ .s.s.\ (x-y)$ , where $\{x,y\}\subset\mathbb R$ .

$\blacktriangleright$ Generally. Let a real function $f:\mathbb E\rightarrow F$ . Then for any $\{x,y\}\subset E\ ,\ f(x)-f(y)\ .s.s.\ \left\{\begin{array}{cccc} (x-y) & \iff & f\nearrow & \mathrm{i.e.\ f\ is\ strict\ increasing} \\\\ (y-x) & \iff & f\searrow & \mathrm{i.e.\ f\ is\ strict\ decreasing}\end{array}\right\|$ .
Exercises.

$1.\blacktriangleright$ Prove that for any $0<a\ne 1$ , $0<b\ne 1$ have the relation $\left(a^b-a\right)\left(b^a-b\right)> 0$ .

Proof. $E(a,b)\equiv \left(a^b-a\right)\left(b^a-b\right)\ .s.s.\ (a-1)(b-1)\cdot(b-1)(a-1)=(a-1)^2(b-1)^2\implies$ $E(a,b)>0$ .

$2.\blacktriangleright$ Solve the inequation $E(x)\equiv (2x-1)\left(|x-2|-|x|\right)\lg|x-1|\ge 0$ .

Proof. $E(x)\equiv (2x-1)\left(|x-2|-|x|\right)\lg|x-1|\ .s.s.\ (2x-1)\cdot \left(|x-2|^2-|x|^2\right)\cdot\left(|x-1|-1\right)(10-1)\ .s.s.\ (2x-1)\cdot 4(1-x)\cdot\left(|x-1|^2-1\right)\ .s.s.$

$(2x-1)(1-x)(x-2)x$ . In conclusion, $E(x)\ge 0\iff$ $(2x-1)(1-x)(x-2)x\ge 0$ , where $x\ne 0$ , i.e. $x\in\mathbb S=\left[0,\frac 12\right]\cup (1,2]$ .

$3.\blacktriangleright$ Solve the inequation $\boxed{\max\left\{a^x,a^{-x}\right\}\ge a^{\max\left\{x,\frac 1x\right\}}}\ (*)$ , where $0<a\ne 1$ .

Proof. $a^x-a^{-x}\ .s.s.\ (a-1)x$ and $x-\frac 1x\ .s.s.\ x\left(x^2-1\right)\iff$ $\max \left\{x,\frac 1x\right\}=\left\{\begin{array}{ccc} \frac 1x & \iff & x\not\in D\\\\ x & \iff & x\in D\end{array}\right\|$ , where $D=[-1,0)\cup [1,\infty )$ and $0<a\ne 1$ and $x\ne 0$ .

First case. $a\in (0,1)$ , i.e. $a^x-a^{-x}\ .s.s.\ -x$ and the relation $(*)$ becomes: $\left\{\begin{array}{ccccccccc} 1.1 & x\le -1\ : & a^x\ge a^{\frac 1x} & \iff & x\le \frac 1x & \iff & x\left(x^2-1\right)\le 0\ ,\ x\ne 0 & \mathrm{Yes}\\\\ 1.2 & -1<x<0\ : & a^x\ge a^{x} & & & & & \mathrm{Yes}\\\\ 1.3 & 0<x\le 1\ : & a^{-x}\ge a^{\frac 1x} & \iff & -x\le \frac 1x & \iff & x\left(1+x^2\right)\ge 0\ ,\ x\ne 0 & \mathrm{Yes}\\\\ 1.4 & 1<x : & a^{-x}\ge a^{x} & \iff & -x\le x & \iff & x\ge 0\ ,\ x\ne 0 & \mathrm{Yes}\end{array}\right\|$ .

Second case. $a>1$ , i.e. $a^x-a^{-x}\ .s.s.\ x$ and the relation $(*)$ becomes: $\left\{\begin{array}{ccccccccc} 2.1 & x\le -1\ : & a^{-x}\ge a^{\frac 1x} & \iff & -x\ge \frac 1x & \iff & x\left(x^2+1\right)\le 0\ ,\ x\ne 0 & \mathrm{Yes}\\\\ 2.2 & -1<x<0\ : & a^{-x}\ge a^{x} & \iff & -x\ge x & \iff & x<0\ ,\ x\ne 0 & \mathrm{Yes}\\\\ 2.3 & 0<x\le 1\ : & a^x\ge a^{\frac 1x} & \iff & x\ge \frac 1x & \iff & x\left(x^2-1\right)\ge 0\ ,\ x\ne 0 & \mathrm{Not}\\\\ 2.4 & 1\le x : & a^{x}\ge a^{x} & & & & & \mathrm{Yes}\end{array}\right\|$ .

In conclusion, the solution of the given inequation is $x\in\mathbb S(a)\equiv\left\{\begin{array}{ccc} \mathbb R^* & \iff & a\in (0,1)\\\\ (-\infty ,0)\cup[1,\infty ) & \iff & a>1\end{array}\right\|$ .

$4.\blacktriangleright$ Show that $\log_{\frac {1}{2}} x > \log_{\frac {1}{3}} x$ only when $0 < x < 1$ .

Proof. $E(x)\equiv\log_{\frac {1}{2}} x -\log_{\frac {1}{3}} x\ .s.s.\ \left(\frac 12-1\right)$ $\left(\frac 13-1\right)(x-1)\left(\frac 13-\frac 12\right)\ .s.s.\ -(x-1)$ . Thus, $\log_{\frac {1}{2}} x > \log_{\frac {1}{3}} x\iff$ $E(x)>0\iff$ $x-1<0\iff$ $0<x<1$ .


PP1. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ . Prove that $\frac {IB}{IC}=\frac {AC}{AB}\iff AC=AB$ .

Proof 1. Observe that in $\triangle BIC$ there is the chain of relatioms $IB-IC\ .s.s.\ \frac C2-\frac B2\ .s.s.\ C-B\ .s.s.\ (c-b)\implies$ $\boxed{IB-IC\ .s.s.\ (c-b)}\ (1)$ . In conclusion,

$\frac {IB}{IC}=\frac {AC}{AB}\iff$ $\frac {IB-IC}{IC}=\frac {AC-AB}{AB}\implies$ $IB-IC\ .s.s.\ (b-c)\ \stackrel{(1)}{\iff}\ (c-b)\ .s.s.\ (b-c)$ $\iff b-c=0\iff b=c\iff AC=AB$ .

Proof 2. I"ll use the standard notations and some remarkable identities. Thus, $IA^2=\frac {bc(s-a)}s\ ,$ i.e. $\boxed{IA^2=bc-4Rr}\ (*)$ a.s.o. Observe that

$IB-IC\ .s.s.\ IB^2-IC^2\ \stackrel{(*)}{=}\ ac-ab=$ $a(c-b)\ .s.s.\ (c-b)\implies$ $\boxed{IB-IC\ .s.s.\ (c-b)}\ (1)$ . In conclusion, $\frac {IB}{IC}=\frac {AC}{AB}\iff$

$\frac {IB-IC}{IC}=\frac {AC-AB}{AB}\implies$ $IB-IC\ .s.s.\ (b-c)\ \stackrel{(1)}{\iff}\ (c-b)\ .s.s.\ (b-c)$ $\iff b-c=0\iff b=c\iff AC=AB$ .


PP2. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ where denote $\in AI\cap w$ , $E\in BI\cap w$ and $F\in CI\cap w$ . Prove that $BE=CF\iff AB=AC$ .

Proof. I"ll use the well-known identities $AD^2\equiv l_a^2=\frac {2\sqrt{bcs(s-a)}}{b+c}$ a.s.o. Thus, $l_b-l_c=\frac {2\sqrt{acs(s-b)}}{a+c}-\frac {2\sqrt{abs(s-c)}}{a+b}\ .s.s.\ \frac {\sqrt{s-b}}{(a+c)\sqrt b}-\frac {\sqrt{s-c}}{(a+b)\sqrt c}\ .s.s.$

$(a+b)\sqrt{c(s-b)}-$ $(a+c)\sqrt{b(s-c)}\ .s.s.\ c(s-b)(a+b)^2-b(s-c)(a+c)^2=$ $s\left[c(a+b)^2-b(a+c)^2\right]+bc\left[(a+c)^2-(a+b)^2\right]=$

$s\left[a^2(c-b)+bc(b-c)\right]+bc(a+2s)(c-b)=$ $(c-b)\left[abc+s\left(a^2+bc\right)\right]\ .s.s.\ (c-b)\implies$ $\boxed{l_b-\l_c\ .s.s.\ (c-b)}$ . In conclusion, $l_b=l_c\iff b=c$ .
This post has been edited 47 times. Last edited by Virgil Nicula, Dec 27, 2015, 4:15 PM

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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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