447. Locuri geometrice remarcabile.

by Virgil Nicula, Jul 7, 2016, 2:38 AM

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$\boxed{L1}\blacktriangleright$ Se dau doua drepte concurente neperpendiculare $d_{1}$ , $d_{2}$ si constantele reale pozitive $k_{1}$ , $k_{2}$ , $k\ .$ Sa se determine locul geometric al punctelor $L$ din planul $\pi=(d_{1},d_{2})$ pentru care $k_{1}\cdot \delta_{d_{1}}(L)+k_{2}\cdot \delta_{d_{2}}(L)=k\ .$ Sa se stabileasca relatia intre $k_{1}$ , $k_{2}$ , $k$ si masura unghiului ascutit dintre cele doua drepte ca locul geometric sa fie nevid (relatia de compatibilitate).
Aplicatie. Sa se arate ca intr-un patrulater circumscriptibil mijloacele diagonalelor si centrul cercului inscris sunt coliniare (dreapta lui Newton).Raspuns.

$\boxed{L2}\blacktriangleright$ Se dau doua puncte fixe $A$ , $B$ si constantele reale nenule $k_{1}$ , $k_{2}$ , $k$ . Sa se determine locul geometric al punctelor $L$ pentru care $k_{1}\cdot LA^{2}+k_{2}\cdot LB^{2}=k$ . Sa se stabileasca relatia de compatibilitate. Discutie.
Aplicatie. Se considera patrulaterul convex $ABCD$ . Sa se determine locul geometric al punctelor $L$ pentru care $LA^{2}+2\cdot LB^{2}+3\cdot LC^{2}=6\cdot LD^{2}+k\cdot [ABCD]$ . Raspuns.

$\boxed{L3}\blacktriangleright$ Se da un triunghi $ABC$ , o dreapta (un cerc) $d\ (w)$ , un punct fix $F$ si un punct mobil $M\in d\ (w)$ . Sa se determine locul geometric al punctului $L$ pentru care $\triangle FML\sim\triangle ABC$ .
Aplicatie. Fiind date doua drepte $d_{1}$ , $d_{2}$ si un cerc $w$ , sa se construiasca cel putin un triunghi $XYZ$ asemenea cu un triunghi dat (in particular $XYZ$ - triunghi echilateral) avand $X\in d_{1}$ , $Y\in d_{2}$ si $Z\in w$ .Raspuns.
Notatii. $\delta_{d}(X)$ - distanta punctului $X$ la dreapta $d$ ; $[XYZT]$ - aria patrulaterului convex $XYZT$ .

$\boxed{L4}\blacktriangleright$ Fie un triunghi $ABC$ si doua puncte mobile $X\in AB,$ $Y\in AC$ astfel incat $BX=CY.$ Notam mijlocul $L$ al segmentului $[XY].$ Sa se arate ca:
$4.1\blacktriangleright$ Daca $BC$ separa punctele $X$ si $Y,$ atunci locul geometric al lui $L$ este o dreapta paralela cu $A$ -bisectoarea exterioara a $\triangle ABC.$
$4.2\blacktriangleright$ Daca $BC$ nu separa punctele $X$ si $Y,$ atunci locul geometric al lui $L$ este o dreapta paralela cu $A$ -bisectoarea interioara a $\triangle ABC.$

Generalizare (Mihai Miculita). Fie $\triangle ABC,$ un punct fix $M\in (BC),$ doua puncte mobile $X\in (AB),$ $E\in (AC)$ si $L\in (XY)$ astfel incat $\frac {LX}{LY}=\frac {BX}{CY}=\frac {MB}{MC}.$ Sa se arate ca:
$G4.1\blacktriangleright$ Daca $BC$ separa punctele $X$ si $Y,$ atunci locul geometric al lui $L$ este o dreapta paralela cu $A$ -bisectoarea exterioara a $\triangle ABC.$
$G4.2\blacktriangleright$ Daca $BC$ nu separa punctele $X$ si $Y,$ atunci locul geometric al lui $L$ este o dreapta paralela cu $A$ -bisectoarea interioara a $\triangle ABC.$

Method 1. $\boxed{\\\begin{array}{ccc}\\\ & \blacktriangleleft\ \underline{\underline{ab+bc+ca\ =\ s^2+r^2+4Rr}}\ \blacktriangleright & \\\\ \nearrow & a(b+c)\ =\ s^2-(s-a)^2 & \searrow\\\\ \bigoplus\ \rightarrow & IA^2\ =\ r^2+(s-a)^2 & \rightarrow\ \bigoplus\\\\ \searrow & bc\ =\ AI\cdot AI_a\ =\ 4Rr+IA^2 & \nearrow\\\\\end{array}}$ $\implies\blacktriangleleft$ ab+bc+ca=s^2+r^2+4Rr $\blacktriangleright$

Method 2. $\boxed{\ bc=s(s-a)+(s-b)(s-c)\implies\sum bc=s\cdot \sum (s-a)+\sum (s-b)(s-c)=s^2+\sum \frac {S^2}{r_br_c}=s^2+r\cdot \sum r_a=s^2+r(4R+r)\ }$

PP1. In a trapezium with bases $a\ ,$ $b$ and lateral sides $c\ ,$ $d$ , prove that the diagonals $p\ ,$ $q$ are given by the relations $:\ \left\{\begin{array}{ccc} p & = & \sqrt{\frac {ab^2-a^2b-ac^2+bd^2}{b-a}}\\\\ q & = & \sqrt{\frac {ab^2-a^2b-ad^2+bc^2}{b-a}}\end{array}\right\|\ .$

Proof 1 (George_54). Suppose $AB\parallel CD$ and let $\left\{\begin{array}{ccccc} AB=a & ; & CD=b & ; & AC=p\\\\ AD=c & ; & BC=d & ; & BD=q\end{array}\right\|\ .$ Apply Euler's theorem $:\ \boxed{p^2+q^2=c^2+d^2+2ab}\ (1)$ and law of Cosines $:$

$\left\{ \begin{array}{ccc} p^2 & = & b^2 + c^2 - 2bc\cos D\\\\ q^2 & = & a^2+c^2- 2ac\cos A\end{array} \right\|$ $\iff\left\{ \begin{array}{ccc} ap^2 = ab^2 + ac^2 + 2abc\cos A\\\\ bq^2 = a^2b + bc^2 - 2abc\cos A\end{array} \right\|\bigoplus$ $\implies$ $\boxed{ap^2 + bq^2 = ab^2+ ac^2+ a^2b + bc^2}\ (2)\ .$ From $(1)$ and $(2)$ we get the desired result.

Proof 2 (own). Denote $\{E,F\}\subset AB$ so that $A\in (BF)\ ,$ $B\in (AE)$ and $BE=b=AF\ .$ Apply the Stewart's theorem to the rays in the mentioned triangles $:$

$\left\{\begin{array}{ccccc} CB/\triangle ACE\ : & CA^2\cdot BE+CE^2\cdot BA=CB^2\cdot AE+BA\cdot BE\cdot AE & \iff & bp^2+aq^2=(a+b)\left(d^2+ab\right) & (1)\\\\ DA/\triangle BDF\ : & DF^2\cdot AB+DB^2\cdot AF=DA^2\cdot BF+AB\cdot AF\cdot BF & \iff & ap^2+bq^2=(a+b)\left(c^2+ab\right) & (2)\end{array}\right\|\ \bigoplus$ $\implies$ $\boxed{p^2+q^2=c^2+d^2+2ab}\ (*)$

Now I"ll solve the system of the relations $(1)\ \wedge\ (2)\ :\ \Delta =\left|\begin{array}{cc} b & a\\\\ a & b\end{array}\right|=b^2-a^2\ ;$ $\Delta_p=\left|\begin{array}{cc} (a+b)\left(d^2+ab\right) & a\\\\ (a+b)\left(c^2+ab\right) & b\end{array}\right|=$ $(a+b)\left[b\left(d^2+ab\right)-a\left(c^2+ab\right)\right]\ ;$

$\Delta_q=\left|\begin{array}{cc} b & (a+b)\left(d^2+ab\right)\\\\ a & (a+b)\left(c^2+ab\right)\end{array}\right|=$ $(a+b)\left[b\left(c^2+ab\right)-a\left(d^2+ab\right)\right]\ .$ In conclusion, $\left\{\begin{array}{ccccccc} p^2 & = & \frac {\Delta_p}{\Delta} & \implies & p & = & \sqrt{\frac {b\left(d^2+ab\right)-a\left(c^2+ab\right)}{b-a}}\\\\ q^2 & = & \frac {\Delta_q}{\Delta} & \implies & q & = & \sqrt{\frac {b\left(c^2+ab\right)-a\left(d^2+ab\right)}{b-a}}\end{array}\right\|\ .$

PP2 (nikolinv). Surprizing! Midline, symmedian line, orthic line (side of the orthic triangle) and antiparallel through vertex are concurent.

Proof 1 (nikolinv). Let $E$ and $D$ be points on $A'C'$ and $A'B'$ ( $A'B'C'$ is orthic triangle of $ABC$ ) so that $AEA'D$ be a parallelogram. Then $K\in BD\cap CE\ .$ Also, it is a simple proof that $K$ is between midpoint of a side and midpoint of an altitude (See: Ross Honsberger - Episodes in Nineteenth and Twentieth Century pp 65). In fact, parallelogram is actually rhombus, so $ED\parallel BC\ .$ Further, midpoints of $ED$ and $BC$ are collinear with $K\ .$

Proof: The parallelogram $AEA'D$ is a rhombus ( $AA'$ is angle bisector of angle $C'A'B'$ ) diagonals are perpendicular. It's easy to see that $AD=DG$ and $AE=EF$ ( $F\in AE\cap BC\ ,$ $G\in AD\cap BC$ ). Since $AG$ and $AF$ are antiparallels with $A'C'$ and $A'B'\ ,$ $K\in BD\cap CE\ .$ From intercept theorem, symmedian point and midpoints of $BC\ ,$ $DE$ are collinear. It's pretty obvious that points $E$ and $D$ lying on midline of triangle $ABC\ .$

PP3. Intr-un plan $\pi$ se dau doua puncte fixe $A$ si $B$ si un punct mobil $M\ .$ Pe semidreptele $(MA$ si $(MB$ se iau punctele $C$ si $D$ astfel incat $MC=p\cdot MA$ si

$MD=q\cdot MB\ ,$ unde $\{p,q\}\subset\mathbb R^*_+\ .$ sunt constante. Aratati ca dreapta care uneste punctul $M$ cu mijlocul segmentului $[CD]$ taie dreapta $AB$ intr-un punct fix $P\ .$

Demonstratie. Daca $L$ este mijlocul segmentului $[CD]\ ,$ atunci $\boxed{\frac {PA}{PB}=\frac {LC}{LD}\cdot \frac {MA}{MB}\cdot\frac{MD}{MC}}=\frac {MA}{MB}\cdot\frac {q\cdot MB}{p\cdot MA}=\frac qp$ (constant) $\Longrightarrow P$ este fix.

PP4. Let an acute $\triangle ABC,$ its circumcircle $\mathbb C(O,R)$ and the intersections $\left\{\begin{array}{c} D\in BC\cap AO\\\ E\in CA\cap BO\\\ F\in AB\cap CO\end{array}\right\|.$ Prove that $\frac {1}{AD}+\frac {1}{BE}+\frac{1}{CF}=\frac 2R$

Proof 1. Observe that $h_a=b\sin C=2R\sin B\sin C$ and $m\left(\widehat{PAD}\right)=|B-C|.$ Thus, $h_a=AD\cdot \cos (B-C)\implies$ $2R\sin B\sin C=AD\cdot \cos (B-C)\implies$

$\frac {2R}{AD}=\frac {\cos (B-C)}{\sin B\sin C}=$ $1+\cot B\cot C\implies$ $2R\cdot\sum \frac 1{AD}=\sum (1+\cot B\cot C)=4$ $\implies$ $\boxed{\frac 1{AD}+\frac 1{BE}+\frac 1{CF}=\frac 2R}\ (*)\ .$

Proof 2. Apply the theorem of Sines in $\triangle ADC\ :\ \frac {AD}{\sin C}=\frac {AC}{\sin \widehat{ADC}}\iff$ $\frac {AD}{\sin C}=$ $\frac b{\cos(B-C)}\iff$ $\frac {2R}{AD}=\frac {\cos(B-C)}{\sin B\sin C}$ a.s.o. In conclusion, $2R\cdot \sum \frac 1{AD}=$

$\sum \frac {\cos(B-C)}{\sin B\sin C}=$ $\sum\frac {\cos B\cos C+\sin B\sin C}{\sin B\sin C}=$ $\sum\left(\cot B\cot C+1\right)=4\implies$ $2R\cdot \sum \frac 1{AD}=4\implies$ $\sum\frac 1{AD}=\frac 2R\ ,$ i.e. $\boxed{\frac 1{AD}+\frac 1{BE}+\frac 1{CF}=\frac 2R}\ (*)\ .$

Proof 3. Let circumcircle $w=\mathbb C(O,R)$ and $\left\{\begin{array}{ccc} \{A,X\}=\{A,O\}\cap w\\\\ \{B,Y\}=\{B,O\}\cap w\\\\ \{C,Z\}=\{C,O\}\cap w\end{array}\right\|\ .$ Thus, $\frac {2R}{AD}=\frac {AX}{AD}=$ $\frac{AD+DX}{AD}=$ $1+\frac {DX}{DA}=1+\frac {XB\cdot XC}{AB\cdot AC}=$

$=1+\cot C\cot B\implies$ $\sum\frac {2R}{AD}=\sum\left(1+\cot C\cot B\right)=3+\sum \cot C\cot B=4\implies$ $\boxed{\frac 1{AD}+\frac 1{BE}+\frac 1{CF}=\frac 2R}\ (*)\ .$

Remark. I used two well-known identities $:\ \boxed{\sum\cot B\cot C=1}\ (1)$ and the remarkable property PP1 from here.

Proof 4. The area $S=[ABC]$ is given by $2S=BC\cdot AD\cdot\sin \widehat{ADB}=$ $a\cdot AD\cdot \cos (B-C)=$ $AD\cdot 2R\sin A\cos (B-C)=$ $AD\cdot 2R\sin (B+C)\cos (B-C)=$

$R\cdot AD\cdot (\sin 2B+\sin 2C)\implies$ $2S=R\cdot AD\cdot (\sin 2B+\sin 2C)\implies$ $\frac 1{AD}=\frac {R(\sin 2B+\sin 2C)}{2S}\implies$ $\sum\frac 1{AD}=\frac {\cancel 2R\cdot\sum\sin 2A}{\cancel 2S}=$ $\frac {4\cancel R\prod\sin A}{2R\cancel{^2}\prod\sin A}=\frac 2R\implies$

$\boxed{\frac 1{AD}+\frac 1{BE}+\frac 1{CF}=\frac 2R}\ (*)\ .$

This post has been edited 78 times. Last edited by Virgil Nicula, Nov 25, 2016, 4:13 PM

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168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

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
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
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by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
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Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
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by OlympusHero, Dec 30, 2020, 6:08 PM
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

447. Locuri geometrice remarcabile.

by Virgil Nicula, Jul 7, 2016, 2:38 AM

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