311. Some problems from USAMO and other contests.

by Virgil Nicula, Aug 17, 2011, 1:47 AM

USAMO 1979. Find for a fixed $P\in \left(\widehat{AOB}\right)$ the points $X\in (OA$ and $Y\in (OB$ so that $P\in (XY)$ and $\frac{1}{PX}+ \frac{1}{PY}$ is maximum.

Proof 1 (trigonometric). Denote $\left\{\begin{array}{ccc} a=m(\angle POX) & ; & x=m(\angle OXP)\\\\ b=m(\angle POY) & ; & y=m(\angle OYP)\end{array}\right|$ , where $a+b+x+y=\pi$ . Apply the Sinus' theorem in the triangles:

$\left\{\begin{array}{cc} \triangle POX\ : & \frac {PO}{PX}=\frac {\sin x}{\sin a}\\\\ \triangle POY\ : & \frac {PO}{PY}=\frac {\sin y}{\sin b}\end{array}\right|\ \bigoplus\ \implies$ $PO\cdot \left(\frac {1}{PX}+\frac {1}{PY}\right)=\frac {\sin x}{\sin a}+\frac{\sin y}{\sin b}$ . Since $PO$ and $\sin a$ , $\sin b$ are constant obtain that

$\frac{1}{PX}+ \frac{1}{PY}$ is maximum $\iff$ $\sin x\sin b+\sin y\sin a$ is maximum $\iff$ $\cos (x-b)-\cos (x+b)+\cos (y-a)-\cos (y+a)$

is maximum $\iff$ $cos (x-b)+\cos (y-a)$ is maximum because $(x+b)+(y+a)=\pi$ $\iff$ $2\cos\frac {(x+y)-(a+b)}{2}\cos\frac {x+a-y-b}{2}$

is maximum $\iff$ $2\sin(a+b)\cos\frac {x+a-y-b}{2}$ is maximum $\iff x+a=y+b=\frac {\pi}{2}$ $\iff$ $XY\perp OP$ .

Proof 2 (trigonometric). Denote $\left\{\begin{array}{c} a=m(\angle POX)\\\ b=m(\angle POY)\\\ \phi =m(\angle OPX)\end{array}\right|$ . Apply the Sinus' theorem in the triangles:

$\left\{\begin{array}{cccc} \triangle POX\ : & \frac {PX}{\sin a}=\frac {PO}{\sin(\phi +a)} & \iff & \frac {1}{PX}=\frac {1}{PO}\cdot\frac {\sin (\phi +a)}{\sin a}\\\\ \triangle POY\ : & \frac {PY}{\sin b}=\frac {PO}{\sin (\phi -b)} & \iff & \frac {1}{PY}=\frac {1}{PO}\cdot\frac {\sin (\phi -b)}{\sin b}\end{array}\right|\ \bigoplus\ \implies$ $\frac {1}{PX}+$ $\frac {1}{PY}=$

$\frac {1}{PO}\cdot (\cos\phi +\cot a\cdot\sin\phi + \cot b\sin\phi -\cos\phi )=$ $\frac {1}{PO}\cdot (\cot a+\cot b)\cdot\sin\phi$ .

In conclusion, $\frac {1}{PX}+$ $\frac {1}{PY}$ is maximum $\iff\sin\phi$ is maximum $\iff\phi=90^{\circ}\iff XY\perp OP$ .

Proof 3 (synthetic). Denote $\left\{\begin{array}{ccc} C\in (OY & ; & PC\parallel OX\\\\ Z\in (OP & ; & CZ\parallel XY\end{array}\right|$ . Since $\left\{\begin{array}{ccc} \triangle OZC\sim \triangle OPY & \implies & \frac {ZC}{PY}=\frac {OZ}{OP}\\\\ \triangle PZC\sim\triangle OPX & \implies & \frac {ZC}{PX}=\frac {PZ}{OP}\end{array}\right|\ \bigoplus\implies$ $\frac {1}{PX}+\frac {1}{PY}=\frac {1}{ZC}$ .

In conclusion, $\frac {1}{PX}+\frac {1}{PY}$ is maximum $\iff$ $ZC$ is minimum ( $C$ is a fixed point and $Z$ is a mobile point ) $\iff CZ\perp OP\iff XY\perp OP$ .

Remark. The area $[XOY]$ is minimum $\iff$ the fixed point $P$ is the midpoint of the segment $[XY]$ . Indeed, denote $\left\{\begin{array}{cc} C\in (OY\ ; & CP\parallel OX\\\\ D\in (OX\ ; & DP\parallel OY\end{array}\right|$

and $[OCP]=[ODP]=\sigma\ ,\ [XDP]=u\ ,\ [YCP]=v$ . Thus, the area $[XOY]$ is minimum $\iff\ u+v$ is minimum. Observe that $\frac uv=\left(\frac {PX}{PY}\right)^2=$ $\left(\frac{u+\sigma}{v+\sigma}\right)^2\implies$ $\frac uv=\left(\frac{u+\sigma}{v+\sigma}\right)^2\iff$ $u=v\ \vee\ uv=\sigma^2$ (constant). In conclusion $u+v$ is minimum $\iff u=v$ , i.e. $P$ is the midpoint of $[XY]$ .

Can easily construct the minimum position of the line $XY$ . Indeed, construct $X_{\mathrm{min}}\in (OA$ so that $OX_{\mathrm{min}}=2\cdot CP$ , where $C\in (OY$ and $CP\parallel OX$ .

USA TST 2010/11. Let $ABC$ be a triangle. Its incircle touches the sidelines $BC$ , $CA$ , $AB$ at the points $D$ , $E$ , $F$ respectively and its exincircles

touch the sidelines $BC$ , $CA$ , $AB$ at the points $D'$ , $E'$ , $F'$ respectively. Prove that $\sum EF\ge \frac {2S}{R}$ and $\sum E'F'\ge s$ , where $2s=a+b+c$ .

I"ll use in my proof some remarkable identities which are here with their proofs.

$\blacktriangleright\ \sum a\cdot\cos A=$ $2R\cdot\sum \sin A\cos A=$ $R\cdot \sum\sin 2A=$ $4R\cdot\prod\sin A=$ $4R\cdot\frac {S}{2R^2}\implies$ $\boxed{\sum a\cdot\cos A=\frac {2S}{R}\ }\ \ (1)$ .

$\blacktriangleright\ \sum a\cdot (\cos B+\cos C)=$ $\sum (b\cdot\cos C+c\cdot\cos B)=$ $\sum a\implies$ $\boxed{\ \sum a\cdot (\cos B+\cos C)=2s\ }\ \ (2)$ .

$\blacktriangleright\ \sum\cos A=1-2\sin^2\frac A2+2\cos\frac {B+C}{2}\cos\frac {B-C}{2}=$ $1-2\sin\frac A2\left(\cos\frac {B+C}{2}-\cos\frac {B-C}{2}\right)=$ $1+4\prod\sin\frac A2=$

$1+4\sqrt{\prod \frac {(s-b)(s-c)}{bc}}=$ $1+\frac {4(s-a)(s-b)(s-c)}{abc}=$ $1+\frac {4sr^2}{4Rsr}\implies$ $\boxed{\ \sum\cos A=1+\frac rR\ }\ \ (3)$ . Otherwise,

remark that $2s\cdot \sum\cos A=$ $\sum a\cdot \cos A+\sum a\cdot (\cos B+\cos C)\stackrel{(1)\wedge (2)}{=}$ $\frac {2S}{R}+2s\implies$ $\sum\cos A=1+\frac rR$ .

Proof of proposed problem. Denote the projection $\mathrm{pr}_{d}([XY])$ of $[XY]$ on the line $d$ . Therefore,

$\blacksquare\ EF\ge \mathrm{pr}_{BC}([EF])=\mathrm{pr}_{BC}([EA]+[AF])=$ $(s-a)\cdot (\cos B+\cos C)$ $\implies$ $\sum EF\ge \sum (s-a)\cdot (\cos B+\cos C)=$

$2s\sum\cos A-\sum a\cdot (\cos B+\cos C)\stackrel{(3)\wedge (2)}{=}$ $2s\left(1+\frac rR\right)-2s\implies$ $\boxed{\ EF+FD+DE\ge \frac {2S}{R}\ }$ .

$\blacksquare\ E'F'\ge \mathrm{pr}_{BC}([E'F'])=a-\mathrm{pr}_{BC}(BF'+CE')=$ $a-(s-a)(\cos B+\cos C)\implies$

$\sum E'F'\ge \sum\left[a-(s-a)(\cos B+\cos C)\right]=$ $2s-2s\cdot\sum\cos A+\sum a\cdot (\cos B+\cos C)=$

$2s-2s\left(1+\frac rR\right)+2s=$ $2s\left(1-\frac rR\right)\ge 2s\left(1-\frac 12\right)$ $\implies$ $\boxed {\ E'F'+F'D'+D'E'\ge s\ }$ because $R\ge 2r$ .

USAJMO - Maryland. Prove that $a>0\ ,\ b>0\ \implies\ \frac{1}{a^2} + \frac{1}{b^2} + \frac{4}{a^2+b^2} \ge \frac{32(a^2+b^2)}{(a+b)^4}$ .

Proof. On the one hand $\frac{1}{a^2} + \frac{1}{b^2} + \frac{4}{a^2+b^2}=\frac {a^2+b^2}{a^2b^2}+\frac{4}{a^2+b^2}\ge 2\cdot\sqrt{\frac {a^2+b^2}{a^2b^2}\cdot\frac{4}{a^2+b^2}}\implies$

$\boxed{\ \frac{1}{a^2} + \frac{1}{b^2} + \frac{4}{a^2+b^2}\ge\frac {4}{ab}\ }\ (1)$ and on the other hands $\boxed {\ \frac {4}{ab}\ge\frac{32(a^2+b^2)}{(a+b)^4}\ }\ (2)\ \iff$ $(a+b)^4\ge 8ab\left(a^2+b^2\right)\iff$

$\left(a^2+2ab+b^2\right)^2\ge 8ab\left(a^2+b^2\right)\iff$ $\left(\frac ab+\frac ba+2\right)^2\ge 8\left(\frac ab+\frac ba\right)\iff$ $(y+2)^2\ge 8y$ , where $y=\frac ab+\frac ba\iff$

$(y-2)^2\ge 0$ , what is truly. In conclusion, the inequality $(2)$ is truly and so from the relations $(1)$ and $(2)$ obtain the required inequality.

Remark. $\left\{\begin{array}{c} (a+b)^2\ge 4ab\\\\ (a+b)^2\le 2\left(a^2+b^2\right)\end{array}\right|$ . But $(a+b)^4\ge 8ab\left(a^2+b^2\right)$ . Indeed, $(a+b)^4=\left[\left(a^2+b^2\right)+2ab\right]^2=$

$\left[\left(a^2+b^2\right)^2+4a^2b^2\right]+4ab\left(a^2+b^2\right)\ge$ $2\sqrt{\left(a^2+b^2\right)^2\cdot 4a^2b^2 }+4ab\left(a^2+b^2\right)=$ $4ab\left(a^2+b^2\right)+4ab\left(a^2+b^2\right)=$ $8ab\left(a^2+b^2\right)$ .

IMO 2003. Let an isosceles $\triangle ABC$ with $AC=BC$ , incentre $I$ and the circumcircle $w$ of the triangle $AIB$ . Let $P\in w$ lying inside $\triangle ABC$ . The lines through $P$ parallel to $CA$ and

$CB$ meet $AB$ at $D$ and $E$ respectively. The line through $P$ parallel to $AB$ meets $CA,$ $CB$ at $F,$ $G$ respectively. Prove that $DF$ and $EG$ intersect on the circumcircle of $\triangle ABC.$

Commentary. The enunciation of this problem conceals vainly in its debut that the circumcircle of $\triangle AIB$ is in fact the circle with the diameter $II_{c}\ ,$ where $I_{c}$ is the $C$ - exincenter of $\triangle ABC.$

Proof. Denote : the circumcircle $w$ of $\triangle ABC$ ; the circle $\delta$ with the diameter $[II_{c}]$ ; the circumcircles $w_{a}$ , $w_{b}$ of the isosceles trapezoids $AFPL$ , $BGPD$ respectively ;

the second intersection $R$ between $\overline{FPG}$ and the circle $\delta$ ; the intersection $L\in FD\cap GE\ .$ Prove easily that $CA$ , $CB$ are the tangents from $C$ to the circle $\delta\ .$

From the relation $FA^{2}=FR\cdot FP$ obtain $\frac{FA}{AD}=\frac{GP}{PE}\ ,$ i.e. the quadrilaterals $FADP$ , $GPEB$ are similarly as $FADP\sim GPEB\ .$ Thus,

$\begin{array}{cc}1\blacktriangleright & \begin{array}{c}\widehat{AEL}\equiv\widehat{GEB}\equiv\widehat{FDP}\equiv\widehat{AFD}\equiv\widehat{AFL}\Longrightarrow\widehat{AEL}\equiv\widehat{AFL}\Longrightarrow L\in w_{a}\\\\ \widehat{BDL}\equiv\widehat{FDA}\equiv\widehat{GEP}\equiv\widehat{BGE}\equiv\widehat{BGL}\Longrightarrow\widehat{BDL}\equiv\widehat{BGL}\Longrightarrow L\in w_{b}\end{array}\Longrightarrow\boxed{\ \widehat{DLE}\equiv\widehat{ABC}\ }\\\\ 2\blacktriangleright & \begin{array}{c}\widehat{ALD}\equiv\widehat{ALF}\equiv\widehat{APF}\\\\ \widehat{BLE}\equiv\widehat{BLG}\equiv\widehat{BPG}\end{array}\Longrightarrow \widehat{ALD}+\widehat{BLE}\equiv\widehat{APF}+\widehat{BPG}\equiv\widehat{AI_{a}B}\equiv\widehat{ABC}\Longrightarrow \boxed{\ \widehat{ALD}+\widehat{BLE}\equiv\widehat{ABC}\ }\end{array}$ $\Longrightarrow$

$\widehat{ALB}\equiv\widehat{ALD}+\widehat{BLE}+\widehat{DLE}=$ $2\cdot\widehat{ABC}=$ $180^{\circ}-\widehat{ACB}$ $\Longrightarrow$ $\boxed{\ \widehat{ALB}+\widehat{ACB}=180^{\circ}\ }$ $\Longrightarrow$ $L\in w\ .$

Variation on the same theme. Let a fixed isosceles $\triangle ABC,$ where $(AB=AC)$ and let mobile $M\in [AB,$ $N\in [AC$ so that $MN\parallel BC$ and exists $P\in (MN)$

for which $PM\cdot PN=BM^{2}.$ Construct $\{S,T\}\subset (BC)$ for which $PS\parallel BM\ ,\ PT\parallel NC\ .$ Find the locus of the intersection $L\in MS\cap NT.$ Answer.

This post has been edited 54 times. Last edited by Virgil Nicula, Mar 9, 2017, 8:46 AM

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152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

133. Some problems with projections/perpendiculars (1).

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).

Submit

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391345 views moment
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We need virgil nicula to return to aops, this blog is top 10 all time.
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blog
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the visits tho
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long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
by CSPAL, May 27, 2020, 4:17 PM
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

311. Some problems from USAMO and other contests.

by Virgil Nicula, Aug 17, 2011, 1:47 AM

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