26. Outside of a triangle.

by Virgil Nicula, Apr 22, 2010, 11:01 PM

P1 (with Ceva). Construct outside of $\triangle ABC$ the similar isosceles $\triangle BA'C$ , $\triangle CB'A$ , $\triangle AC'B$ in $A'$ , $B'$ , $C'$ respectively. Prove that $AA'\cap BB'\cap CC'\ne\emptyset$ .

Proof. Denote $\left\|\begin{array}{c} D\in BC\cap AA'\\\ E\in CA\cap BB'\\\ F\in AB\cap CC'\\\ \phi =m(\angle BCA')\end{array}\right\|$ .Observe that $\frac {DB}{DC}=\frac {[ABA']}{[ACA']}=\frac {ac\cdot\sin (B+\phi )}{ab\cdot\sin (C+\phi )}$ $\implies$ $\frac {DB}{DC}=\frac {c\cdot\sin (B+\phi )}{b\cdot\sin (C+\phi )}$ . Obtain

analogously $\frac {EC}{EA}=\frac {a\cdot\sin (C+\phi )}{c\cdot\sin (A+\phi )}$ and $\frac {FA}{FB}=\frac {b\cdot\sin (A+\phi )}{a\cdot\sin (B+\phi )}$ $\implies$ $\frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1$ $\implies$ $AA'\cap BB'\cap CC'\ne\emptyset$ .

Extension. Construct outside of $\triangle ABC$ the triangles $\left\{\begin{array}{c} BA'C\\\\ CB'A\\\\ AC'B\end{array}\right\|$ so that $\widehat{BAC'}\equiv \widehat {CAB'}$ and $\widehat{ABC'}\equiv\widehat {ACB'}\equiv\widehat {A'BC}\equiv\widehat{A'CB}$ . Prove that $AA'\cap BB'\cap CC'\ne\emptyset$ .

Denote $m\left(\widehat{BAC'}\right)=m\left(\widehat {CAB'}\right)=y$ , $m\left(\widehat{ABC'}\right)=m\left(\widehat {ACB'}\right)=m\left(\widehat {A'BC}\right)=m\left(\widehat{A'CB}\right)=z$

and $\left\|\begin{array}{c} D\in BC\cap AA'\\\ E\in CA\cap BB'\\\ F\in AB\cap CC'\\\ \end{array}\right\|$ .Observe that $\frac {DB}{DC}=\frac {[ABA']}{[ACA']}=\frac {ac\cdot\sin (B+z)}{ab\cdot\sin (C+z)}$ $\implies$ $\frac {DB}{DC}=\frac {c\cdot\sin (B+z )}{b\cdot\sin (C+z)}$ . Obtain analogously

$\frac {EC}{EA}=\frac {a\cdot\sin y\cdot\sin (C+z )}{c\cdot\sin z\cdot\sin (A+y )}$ and $\frac {FA}{FB}=\frac {b\cdot\sin z\cdot\sin (A+y )}{a\cdot\sin y\cdot\sin (B+z )}$ $\implies$ $\frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1$ $\implies$ $AA'\cap BB'\cap CC'\ne\emptyset$ .[/color]

P2. Outside of $\triangle ABC$ construct $\left\|\begin{array}{cc} \triangle BMC\ : & m(\widehat{MBC})=m(\widehat{MCB})=x\\\\ \triangle CNA\ : & m(\widehat{NCA})=m(\widehat{NAC})=y\\\\ \triangle APB\ : & m(\widehat{PAB})=m(\widehat{PBA})=z\end{array}\ \right\|$ and $x+y+z=90^{\circ}$ , i.e. $m(\widehat {BMC})+m(\widehat {CNA})+m(\widehat {APB})=360^{\circ}$ . Prove that

$\left\|\ \begin{array}{c}m(\widehat {PMN})=90^{\circ}-x=\frac 12\cdot m(\widehat {BMC})\\\\ m(\widehat {MNP})=90^{\circ}-y=\frac 12\cdot m(\widehat {CNA})\\\\ m(\widehat {NPM})=90^{\circ}-z=\frac 12\cdot m(\widehat {APB})\end{array}\ \right\|$ $\implies$ $\frac {MN}{\cos z}=\frac {NP}{\cos x}=\frac {PM}{\cos y}$ , i.e. $\triangle MNP$ has the angles $90^{\circ}-x$ , $90^{\circ}-y$ , $90^{\circ}-z$ . See here the Amo's proof (syntetically!).

Proof 1. Apply the generalized Pythagoras' theorem to $[PN]$ of $\triangle NAP\ :\ PN^2=AN^2+AP^2-2\cdot AN\cdot AP\cdot\cos \widehat{PAN}=$ $\frac {b^2}{4\cos^2y}+\frac {c^2}{4\cos^2z}-\frac {bc}{2\cos y\cos z}\cdot\cos \left(90^{\circ}+A-x\right)=$

$\frac 1{4\cos^2y\cos^2z}\cdot \left[b^2\cos^2z+c^2\cos^2y+2bc\cos y\cos z\sin (A-x)\right]=$ $\frac 1{4\cos^2y\cos^2z}\cdot\left[b^2\cos^2z+c^2\cos^2y+2bc\cos y\cos z\cdot (\sin A\cos x-\sin x\cos A)\right]\implies$ $4\cos^2y\cos^2z\cdot PN^2=$

$b^2\cos^2z+c^2\cos^2y+2bc\sin A\cos x\cos y\cos z-2bc\cos A\cos z\sin x\cos y=$ $b^2\cos^2z+c^2\cos^2y+4S\cos x\cos y\cos z-\left(b^2+c^2-a^2\right)\cos y\cos z\sin x=$ $4S\cos x\cos y\cos z+$

$a^2\sin x\cos y\cos z+b^2\cos z[\sin (x+y)-\cos y\sin x]+c^2\cos y[\sin (x+z)-\cos z\sin x]=$ $4S\cos x\cos y\cos z +a^2\sin x\cos y\cos z+b^2\sin y\cos x\cos z+c^2\sin z\cos x\cos y=$

$\cos x\cos y\cos z\cdot\left(4S+\sum a^2\tan x\right)\implies$ $4\cos y\cos z\cdot PN^2=\cos x\left(4S+\sum a^2\tan x\right),$ what is a symmetric in $\{a,b,c\},$ i.e. $\cos y\cos z\cdot NP=\cos z\cos x\cdot PM=\cos x\cos y\cdot MN$
$\implies\frac {NP}{\cos x}=\frac {PM}{\cos y}=\frac {MN}{\cos z}=\sqrt{\frac {4S+\sum a^2\tan x}{4\cos x\cos y\cos z}}=\sqrt{\frac {S+\sum [BMC]}{\cos x\cos y\cos z}},$ where $4\cdot [BMC]=a^2\tan x$ a.s.o. $\implies$ the angles of $\triangle MNP$ are $\left\{\begin{array}{ccccc} M & = & 90^{\circ}-x & = & y+z\\\\ N & = & 90^{\circ}-y & = & z+x\\\\ P & = & 90^{\circ}-z & = & x+y\end{array}\right\|\ .$

Proof 2 (Amo). On $[PA]$ construct outside $\triangle PAX\equiv\triangle PBM\ .$ Hence $PM=PX$ and $\triangle AXN\equiv\triangle CMN\ .$ Indeed: $\left\{\begin{array}{c} AX=CM\\\ AN=CN\end{array}\right\|$ and $m\left(\widehat{XAN}\right)=$ $2\pi-m\left(\widehat{XAP}\right)-m\left(\widehat{PAN}\right)=$
$2\pi-m\left(\widehat{PBM}\right)-m\left(\widehat{PAN}\right)=$ $m\left(\widehat{MCN}\right)\ .$ We got that $\triangle AXN\equiv \triangle CMN$ $\implies$ $XN=MN\implies$ $\triangle PXN\equiv\triangle PMN\ .$ In conclusion, $\left(\widehat{PNM}\right)=\frac{\left(\widehat{XNM}\right)}2=\frac{\left(\widehat{ANC}\right)}2\ .$

P3 Outside of $\triangle ABC$ construct $\left\{\begin{array}{c} \triangle ABF\\\ \triangle ACE\\\ \triangle BCD\end{array}\right\|$ so that $DB = DC$ , $\left\|\begin{array}{c} m(\widehat {BAF}) = m(\widehat {CAE}) = x\\\ m(\widehat {ABF}) = m(\widehat {ACE}) = y\\\ m(\widehat {BCD}) = m(\widehat {CBD})=z\end{array}\right\|$ and $x + y + z = 90^{\circ}$ . Prove that $DE = DF$ si $m(\widehat {EDF}) = 2y\ .$

Particular cases. If $z = 0\ (x + y = 90^{\circ}),$ then $\triangle DEF$ is $D$ -isosceles with $m\left(\widehat{EDF}\right)=2y.$ If $y=30^{\circ}\ \left(x+z=60^{\circ}\right),$ then $\triangle DEF$ is equilateral.

Proof. Prove easily that $DB=DC=\frac {a}{2\cos z}$ and $\left\{\begin{array}{cc} \triangle ACE\ : & \frac {CE}{\sin x}=\frac {AE}{\sin y}=\frac {b}{\cos z}\\\\ \triangle ABF\ : & \frac {BF}{\sin x}=\frac {AF}{\sin y}=\frac {c}{\cos z}\end{array}\right\|$ . Apply the generalized Pythagoras' theorem in the triangles :

$EF^2=AE^2+AF^2-2\cdot AE\cdot AF\cdot \cos (A+2x)=$ $\frac {b^2\sin^2y}{\cos^2z}+\frac {c^2\sin^2y}{\cos^2z}-\frac {2bc\sin^2y}{\cos^2z}\cdot (\cos A\cos 2x-\sin A\sin 2x)$ $\implies$

$\frac {\cos^2z}{\sin^2y}\cdot EF^2=$ $\left(b^2+c^2\right)-\left(b^2+c^2-a^2\right)\cos 2x+4S\sin 2x\implies$ $\boxed{\frac {EF}{\sin y}=\frac {1}{\cos z}\cdot \sqrt{\left(b^2+c^2\right)-2bc\cdot\cos A\cos 2x+4S\sin 2x}}$ .

$DE^2=CD^2+CE^2-2\cdot CD\cdot CE\cdot \cos (C+y+z)=$ $\frac {a^2}{4\cos^2z}+\frac {b^2\sin^2x}{\cos^2z}-\frac {ab\sin x}{\cos^2z}\cdot \sin (x-C)$ $\implies$

$4\cos^2z\cdot DE^2=a^2+4b^2\sin^2x-4ab\sin x\sin (x-C)=$ $a^2+2b^2\left(1-\cos 2x\right)-4ab\sin^2x\cos C+4ab\sin x\cos x\sin C=$

$a^2+2b^2(1-\cos 2x)-(a^2+b^2-c^2)(1-\cos 2x)\implies$ $\boxed{2\cdot DE=\frac {\sqrt{\left(b^2+c^2\right)-2bc\cdot\cos A\cos 2x+4S\sin 2x}}{\cos z}}$ , what is symmetrically in $(b,c)$ .

In conclusion, $\boxed{DE=DF}$ and $2\cdot DE\cdot\sin\frac {m\left(\widehat{EDF}\right)}{2}=EF=2\cdot DE\cdot\sin y\implies$ $\boxed{m\left(\widehat{EDF}\right)=2y}$ .

P4. Let $ABC$ be a triangle. $\triangle PAB$ and $\triangle QAC$ are constructed outside of triangle $\triangle ABC$ such that

$\left\{\begin{array}{c} AP = AB\\\ AQ = AC\\\ \widehat{BAP}\equiv\widehat {CAQ}\end{array}\right\|$ . Denote $R\in BQ\cap CP$ and the circumcenter $O$ of $\triangle BCR$ . Prove that $AO \perp PQ$ .

Proof. I"ll show that $OP^{2}-OQ^{2}=c^{2}-b^{2}$ , i.e. $AO\perp PQ$ . Denote $x=m(\widehat{BAP})$ , the circumcircle $C(O,\rho )$ of $\triangle BRC$ and $S\in AR\cap BC$ . Thus, $\left\{\begin{array}{c}AC=AQ\\\ AP=AB\\\ \widehat{CAP}\equiv\widehat{QAB}\end{array}\right\|$ $\implies$

$\triangle ACP\sim\triangle AQB$ $\implies$ $\left\{\begin{array}{c}CP=QB\\\ \widehat{APC}\equiv\widehat{ABQ}\\\ \widehat{ACP}\equiv\widehat{AQB}\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}\widehat{APR}\equiv\widehat{ABR}\\\ \widehat{ACR}\equiv\widehat{AQR}\end{array}\right\|$ $\implies$ the quadrilaterals $APBR$ , $AQCR$ are cyclically with the radical axis $AR$ . Observe that

$\left\{\begin{array}{c}m(\widehat{BRS})=m(\widehat{BPA})=90^{\circ}-x\\\ m(\widehat{CRS})=m(\widehat{CQA})=90^{\circ}-x\\\ m(\widehat{BRP})=m(\widehat{BAP})=2x\end{array}\right\|$ $\implies$ the ray $[RS$ is the bisector of the angle $\widehat{BRC}$ . Prove easily that $\left\{\begin{array}{c}m(\widehat{BOC})=4x\\\ a=2\rho\sin 2x\end{array}\right\|$ . Thus, $\left\{\begin{array}{c}PA=c\ ,\ QA=b\\\ PB=2c\cdot\sin x\\\ QC=2b\cdot \sin x\end{array}\right\|$

and $\left\{\begin{array}{c}m(\widehat{CBO})=m(\widehat{BCO})=90^{\circ}-2x\\\ m(\widehat{PBO})=180^{\circ}+3x-B\\\ m(\widehat{QCO})=180^{\circ}+3x-C\end{array}\right\|$ . Apply the generalized Pythagoras' theorem in $\triangle POB$ , $\triangle QOC$ to the sides $OP$ , $OQ$ respectively $:$

$\left\{\begin{array}{c} OP^{2}=BP^{2}+\rho^{2}-2\rho\cdot BP\cos \widehat{PBO}\\\\ OQ^{2}=CQ^{2}+\rho^{2}-2\rho\cdot CQ\cos \widehat{QCO}\end{array}\right\|$ $\implies$ $OP^{2}-OQ^{2}=$ $4(c^{2}-b^{2})\sin^{2}x+4c\rho\sin x\cos (B-3x)-4b\rho \sin x\cos (C-3x)=$

$4(c^{2}-b^{2})\sin^{2}x+4c\rho \sin x(\cos B\cos 3x+\sin B\sin 3x)$ $-4b\rho \sin x(\cos C\cos 3x+\sin C\sin 3x)=$ $4(c^{2}-b^{2})\sin^{2}x+4\rho\sin x\cos 3x$ $(c\cos B-b\cos C)+$

$4\rho \sin x\sin 3x(c\sin B-b\sin C)=$ $4(c^{2}-b^{2})\sin^{2}x+4\rho\sin x\cos 3x\left(\frac{a^{2}+c^{2}-b^{2}}{2a}-\frac{a^{2}+b^{2}-c^{2}}{2a}\right)=$ $4(c^{2}-b^{2})\sin^{2}x+\frac{4\rho (c^{2}-b^{2})}{a}\sin x\cos 3x=$

$\left(c^{2}-b^{2}\right)\left(4\sin^{2}x+\frac{2\sin x\cos 3x}{\sin 2x}\right)=c^{2}-b^{2}$ . I used the simple relations $c\sin B=b\sin C$ and $4\sin^{2}x+\frac{2\sin x\cos 3x}{\sin 2x}=1$ .

Tema de cercetare. Gasiti concluzia fiecarei probleme in cazul cand (cel putin) unul din cele trei triunghiuri este construit spre interiorul triunghiului dat.

This post has been edited 120 times. Last edited by Virgil Nicula, Apr 20, 2017, 9:35 AM

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177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

October 2010

170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

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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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

26. Outside of a triangle.

by Virgil Nicula, Apr 22, 2010, 11:01 PM

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