220. Some nice geometry problems from contest.

by Virgil Nicula, Feb 4, 2011, 6:43 PM

Proposed problem 1 (Poland, 2000). Let $ABC$ be an $A$ -isosceles triangle and $P$ be an interior point such

that $\widehat{PBC}\equiv\widehat{PCA}$ . Let $M$ be the midpoint of $[BC]$ . Prove that $\left(\widehat{BPM}\right) +\left(\widehat{APC}\right) = 180^{\circ}$ .

Proof 1 (synthetic). Observe that $AB=AC\implies \widehat{PBA}\equiv\widehat {PCB}$ . Thus $AB$ , $AC$ are tangent to the circumcircle $w$ of $\triangle BPC$ . Denote the second intersection

$R$ of $AP$ with $w$ . From an well-known property (or prove easily) results that $BPCR$ is harmonically. Indeed, $\frac {BP}{BR}=$ $\frac {AP}{AR}=$ $\frac {CP}{CR}\implies$ $BP\cdot CR=BR\cdot CP$ .

The ray $[PR$ is $P$ -symmedian in $\triangle BPC$ . Indeed, $\frac {NB}{NC}=\frac {BP\cdot BR}{CP\cdot CR}=\left(\frac {PB}{PC}\right)^2$ . Therefore, $\widehat{MPB}\equiv\widehat{RPC}$ $\implies$ $\left(\widehat{BPM}\right) +\left(\widehat{APC}\right) = 180^{\circ}$ .

Proposed problem 2 (own). Let $ABC$ be an $A$ -right-angled triangle. Denote $\left\|\begin{array}{ccc} D\in BC & ; & AD\perp BC\\\ M\in (AD) & ; & MA=MD\\\ E\in CM & ; & EB\perp BC\end{array}\right\|$ . Prove that $EA=EB$ .

Proof. Let $F\in AB\cap CM$ . Thus, that $\triangle CDA\sim \triangle CAB$ and $MA=MD$ $\implies$ the ray $[CF$ is $C$ -symmedian in $\triangle ABC$ . Since $EB\perp BC$ obtain that $EB$

is tangent to the circumcircle of $\triangle ABC$ . In conclusion (from well-known property of symmedian), and $EA$ is tangent to the circumcircle of $\triangle ABC$ , i.e. $EA=EB$ .

Remark. $M$ is the symmedian point (Lemoine) of $\triangle ABC$ because $AD$ is $A$ -symmedian in given triangle.

Proposed problem 3. Let $ABC$ be a triangle with $b\ne c$ . Consider the points $\{P,Q,R,S\}\subset BC$ such that $P$ is the

midpoint of $[BC]$ , the ray $[AQ$ bisects $\angle BAC$ , $AR\perp BC$ and $AS\perp AQ$ . Prove that $PR \cdot QS=AB\cdot AC$ .

Proof 1 (metric). $\left\|\begin{array}{ccc} RC^2-RB^2=b^2-c^2 & \implies & PR=\frac {|b^2-c^2|}{2a}\\\\ \frac {QC}{b}=\frac {QB}{c}=\frac {a}{b+c} & \implies & QB=\frac {ac}{b+c}\\\\ \frac {SC}{b}=\frac {SB}{c}=\frac {a}{|b-c|} & \implies & SB=\frac {ac}{|b-c|}\end{array}\right\|$ $\implies$ $QS=\frac {2abc}{|b^2-c^2|}\implies$ $PR\cdot QS=bc$ .

Proof 2 (synthetic). Denote the second intersection $T$ of $AQ$ with the circumcircle of $\triangle ABC$ . Prove easily that $\frac {AQ}{QS}=\cos\widehat {AQS}=\frac {PR}{AT}\implies$

$PR\cdot QS=AQ\cdot AT$ and $\triangle ABT\sim\triangle AQC\implies$ $\frac {AB}{AQ}=\frac {AT}{AC}$ $\implies$ $AQ\cdot AT=bc$ . In conclusion, $PR\cdot QS=AQ\cdot AT=bc$ .

Quote:

An easy extension. Let $ABC$ be a triangle with the circumcircle $w$ and $b\ne c$ . Consider the points $\{P,Q,R,S\}\subset BC$ and $\{A,T\}=AQ\cap w$

such that $AR\perp BC$ , $PT\perp BC$ and $AS\perp AQ$ . Prove that $PR \cdot QS=bc\cdot\frac {\sin (B+y)}{\sin (B+x)}$ , where $\left\|\begin{array}{c} x=m\left(\widehat{QAB}\right)\\\ y=m\left(\widehat{QAC}\right)\end{array}\right\|$ .

Remark. For $x=y=\frac A2$ obtain the proposed problem. Nice problem !

Proposed problem 4 (INMO, 2011). Let $D$ , $E$ , $F$ be points on the sides $[BC]$ , $[CA]$ , $[AB]$ respectively of

$\triangle ABC$ such that $BD=CE=AF$ and $\widehat{BDF}\equiv \widehat{CED}\equiv\widehat{AFE}$ . Show that $\triangle ABC$ is equilateral.

Proof 1. Denote $BD=CE=AF=x$ and $m(\angle BDF)=m(\angle CED)=m(\angle AFE)=\theta$ . Observe that $\Delta ABC\sim \Delta EFD$ . Thus

exists $k\in \mathbb R^*_+$ so that $\frac {DE}{b}=\frac {EF}{c}=\frac {FD}{a}=k$ . Suppose w.l.o.g. that $a\ge b\ge c$ . Therefore $\frac{\sin A}{\sin\theta}\ge\frac{\sin B}{\sin\theta}$ and $\frac{c-x}{b-x}\ge \frac{a}{b}\ge 1\Longrightarrow$

$c\ge b$ . In conclusion, $c=b\Longrightarrow \sin A=\sin B$ and $ABC$ is equilateral.

Proposed problem 5. Let $ABC$ be a triangle. Consider the points $M\in (BC)$ , $N\in (CA)$ and $P\in (AB)$ so that $\frac {MB}{MC}=\frac {NC}{NA}=\frac {PA}{PB}$ .

Prove that for any interior point $X$ wr.t. $\triangle ABC$ there is the inequality $\boxed{\ XM^2+XN^2+XP^2\ge \frac {1}{12}\cdot\left(a^2+b^2+c^2\right)\ge r(R+r)\ }$ .

Proof. Denote $r=\frac {MB}{MC}$ . Apply the Stewart's theorem to $[XM]/\triangle BXC\ :\ XM^2=\frac {1}{r+1}\cdot XB^2+\frac {r}{r+1}\cdot XC^2-\frac {r}{(r+1)^2}\cdot a^2$ a.s.o. $\implies$

$\sum XM^2=\frac {1}{r+1}\cdot \sum XB^2+\frac {r}{r+1}\cdot\sum XC^2-\frac {r}{(r+1)^2}\cdot \sum a^2=$ $\sum XA^2-\frac {r}{(r+1)^2}\cdot \sum a^2$ . Since $\sum XA^2=3\cdot XG^2+\frac 13\cdot\sum a^2$

where $G$ is the centroid of $\triangle ABC$ obtain that $\sum XM^2=3\cdot XG^2+\left[\frac 13-\frac {r}{(r+1)^2}\right]\cdot\sum a^2$ . Thus, $\sum XM^2\ge \frac {r^2-r+1}{3(r+1)^2}\cdot\sum a^2\implies$

$XM^2+XN^2+XP^2\ge \frac {1}{12}\cdot\left(a^2+b^2+c^2\right)$ because $\frac {r^2-r+1}{(r+1)^2}\ge\frac 14$ . We have equalitity if and only if $r=1$ and $X:=G$ . On other hand

$\boxed{\sum a^2}=2s^2-2r^2-8Rr\ge$ $2\left(16Rr-5r^2\right)-2r^2-8Rr=$ $12\left(2Rr-r^2\right)\ge$ $\boxed{12r(R+r)}$ because $2R-r\ge R+r\iff R\ge 2r$ .

PP6 (Japan Mathematical Olympiad, 2011). For an acute $\triangle ABC$ with the orthocenter $H$ denote $\begin{array}{ccc} M\in (BC) & ; & MB=MC\\\ P\in AM & ; & HP\perp AM\end{array}$ . Prove that $AM\cdot PM=BM^2$ .

Proof. Let $AD$ , $BE$ , $CF$ be the altitudes of $ABC$ , where $D\in BC$ , $E\in CA$ , $F\in AB$ . Prove easily that $AEPHF$ , $HPMD$ , $EFDM$

are cyclically and the radical axis $EF, PH, MD$ concur at $T$ . Since $APDT$ is cyclically obtain that $MA\cdot MP=MT\cdot MD$ .

Since the division $(B,D,C,T)$ is harmonically obtain that $MT\cdot MD=MB^2$ . In conclusion, $MA\cdot MP=MB^2$ .

Proposed problem 7. Let $\triangle ABC$ with $A=120^{\circ}$ , incenter $I$ and bisectors $AD$ , $BE$ , $CF$ , where $D\in BC$ , $E\in CA$ , $F\in AB$ . Prove that the $DE\perp DF$ .

Proof 1 (synthetic). Can show easily that $E$ is the $B$ -exincentre of $\triangle ABD$ and $F$ is the $C$ -exincentre of $\triangle ACD$ . Thus,

$DE$ and $DF$ are the $D$ -bisectors in the triangles $ABD$ , $ACD$ respectively. In conclusion, $DE\perp DF$ .

Proof 2 (metric). From the well-known relation $AD=\frac {2bc\cdot\cos\frac A2}{b+c}$ obtain for $A=120^{\circ}$ that $AD=\frac {bc}{b+c}$ , i.e. $\frac {1}{AD}=\frac 1b+\frac 1c$ . Therefore,

$\begin{array}{ccc} \frac {DA}{DC}=\frac {\frac {bc}{b+c}}{\frac {ab}{b+c}}=\frac {EA}{EC} & \implies & \widehat{EDA}\equiv\widehat{EDC}\\\\ \frac {DA}{DB}=\frac {\frac {bc}{b+c}}{\frac {ac}{b+c}}=\frac {FA}{FC} & \implies & \widehat{FDA}\equiv\widehat{FDB}\end{array}$ . In conclusion, $DE$ and $DF$ are $D$ -bisectors in triangles $ABD$ , $ACD$ respectively $\implies$ $DE\perp DF$ .

Remark. Apply generalized Pytagoras' theorem : $\left\|\begin{array}{c} DE^2=AD^2+AE^2-AD\cdot AE\\\ DF^2=AD^2+AF^2-AD\cdot AF\\\ EF^2=AE^2+AF^2+AE\cdot AF\end{array}\right\|$ . Thus, $DE\perp DF\iff$

$EF^2=DE^2+DF^2\iff$ $2\cdot AD^2=AE\cdot AF+AD\cdot (AE+AF)$ . In conclusion, $DE\perp DF\iff$

$3\cdot AD^2=(AD+AE)(AD+AF)$ . Can use the equivalence $\boxed{A=120^{\circ}\iff \frac {(a+b)(a+c)}{b+c}=a+b+c}$ .

PP8. Two circles $O$ , $O'$ meet each other at points $A$ , $B$ . A line from $A$ intersects the circle $O$ at $C$ and the circle $O'$ at $D$ ( $A$ is between $C$ and $D$ ) . Let

$M$ , $N$ be the midpoints of the arcs $BC$ , $BD$ respectively (not containing $A$ ) and let $K$ be the midpoint of the segment $[CD]$ . Show that $KM\perp KN$ .

Proof. Let $L$ be the reflection of $M$ accross $K$ . Then $\triangle CKM\cong\triangle DKL$ . We have $DL=CM=BM$ and $DN=BN$ . Observe that

$\angle MBN=360^{\circ}-(\angle MBA+\angle ABN)=$ $\angle ACM+\angle NDA=$ $\angle KDL+\angle NDA=$ $\angle NDL$ $\implies$ $\angle MBN=\angle LDN\implies$

$\triangle MBN\cong\triangle LDN\implies$ $MN=LN$ . Since $MN=LN$ , $MK=KL\implies$ $NK\perp ML\implies$ $KM\perp KN$ .

PP9. Let $ABC$ be a triangle. For a point $E\in AD$ , where $D\in BC$ and $AD\perp BC$ define the points $F\in AC\cap BE$ and $G\in AB\cap CE$ . Prove that $\widehat{ADF}\equiv\widehat{ADG}$ .

Proof 1 (synthetic). Let $d\parallel BC$ be a line so that $A\in d$ and $K\in DF\cap d$ , $L\in DG\cap d$ . Thus, $\frac {FC}{FA}=\frac {DC}{AK}\ \ \wedge\ \ \frac {GA}{GB}=\frac {AL}{DB}\ \ (*)$ . Apply the Ceva's theorem to the point $E$ and $\triangle ABC\ :$

$\frac {DB}{DC}\cdot \frac {FC}{FA}\cdot\frac {GA}{GB}=1$ . From $(*)$ obtain that $\frac {DB}{DC}\cdot \frac {DC}{AK}\cdot\frac {AL}{DB}=1$ , i.e. $AK=AL$ , what means the triangle $KDL$ is isosceles because and $DA\perp KL$ . In conclusion, $\widehat{ADF}\equiv\widehat{ADG}$ .

Proof 2 (trigonometric). Note $\left\{\begin{array}{c} m(\angle ADF)=x\\\\ m(\angle ADG)=y\end{array}\right|$ . Apply an well-known relation to the cevians : $\left\{\begin{array}{cccc} DF/\triangle ADC\ : & \frac {FC}{FA}=\frac {DC}{DA}\cdot \frac {\sin\widehat{FDC}}{\sin\widehat{FDA}} & \implies & \frac {FC}{FA}=\frac {DC}{DA}\cdot \cot x\\\\ DG/\triangle ADB\ : & \frac {GA}{GB}=\frac {DA}{DB}\cdot \frac {\sin\widehat{GDA}}{\sin\widehat{GDB}} & \implies & \frac {GA}{GB}=\frac {DA}{DB}\cdot \tan y\end{array}\right|\ \ (1)$ .

Apply the Ceva's theorem to the point $E$ and $\triangle ABC\ :\ \frac {DB}{DC}\cdot \frac {FC}{FA}\cdot\frac {GA}{GB}=1$ . Using the relations from $(1)$ obtain easily that $\tan x=\tan y$ , i.e. $x=y\iff$ $\widehat{ADF}\equiv\widehat{ADG}$ .

Proof 3 (proiective). Let $S\in FG\cap AD$ and $P\in FG\cap BC$ . The division $(G,F;S,P)$ is harmonically, i.e. $\frac {SG}{SF}=\frac {PG}{PF}$ and from property $DS\perp DP\iff \widehat{SDF}\equiv\widehat{SDG}$ obtain the conclusion.

PP10. Let $ABC$ be a triangle with the incenter $I$ . Denote $E\in BI\cap AC$ and $F\in CI\cap AB$ . Prove that $A=60^{\circ}\iff$ $BF+CE=BC$ .

Proof. $A=60^{\circ}\iff$ $a^2=b^2+c^2-bc\iff$ $\frac {b}{a+c}+\frac {c}{a+b}=1\iff$ $\frac {ba}{a+c}+\frac {ca}{a+b}=a\iff$ $CE+BF=BC\iff$ $\frac {IE}{IB}+\frac {IF}{IC}=1$ .

This post has been edited 94 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:11 PM

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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

220. Some nice geometry problems from contest.

by Virgil Nicula, Feb 4, 2011, 6:43 PM

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