178. Some nice and simple geometry problems.

by Virgil Nicula, Nov 25, 2010, 2:04 PM

PP1. Let $\triangle ABC$ with $C=30^{\circ}\ ,\ \sin B=\frac {1}{2\sqrt{\lambda}}$ and $D\in (AC)$ so that $AD=\lambda\cdot AC$ , where $\frac 14\le \lambda$ . Ascertain $m\left(\angle ABD\right)$ .

Proof. $\frac {AC}{AB}=\frac {\sin B}{\sin C}=\frac {\frac {1}{2\sqrt{\lambda}}}{\frac 12}=$ $\frac {1}{\sqrt {\lambda}}$ $\implies$ $\lambda\cdot AC^2=AB^2$ $\implies$ $AB^2=AD\cdot AC$ $\implies$ $\triangle ABD\sim \triangle ACB$ $\implies$ $\widehat{ABD}\equiv \widehat{ACB}$ $\implies$ $m\left(\angle ABD\right)=30^{\circ}$ .

Remark. For $\lambda =\frac 12$ obtain the proposed problem from here. For $\lambda =\frac 13$ obtain the following simple problem :

Given are an $A$ -right $\triangle ABC$ with $BC=2\cdot AB$ . Denote $D\in (AC)$ for which $DC=2\cdot AD$ . Ascertain the measure of the angle $\angle ABD$ .

PP2. Let $ABC$ be a triangle with the incircle $w=C(I,r)$ which touches the sidelines $BC$ , $CA$ at $D$ , $E$ .

Denote the midpoints $M$ , $N$ of the sides $[AB]$ , $[AC]$ respectively . Prove that $MN\cap BI\cap DE\ne\emptyset$ .

Proof. Denote $X\in BI\cap MN$ and $Y\in DE\cap MN$ . Since the triangle $BMX$ is $M$ -isosceles obtain that $NX=MX-MN=$

$MB-MN=\frac c2-\frac a2\implies \boxed{NX=\frac {c-a}{2}}$ . Since the triangle $ENY$ is $N$ -isosceles obtain that $NY=NE=NC-CE=$

$\frac b2-(s-c)\implies \boxed {NY=\frac {c-a}{2}}$ . In conclusion, $NX=NY=\frac {c-a}{2}$ $\implies$ $X\equiv Y$ $\implies$ $MN\cap BI\cap DE\ne\emptyset$ .

PP3. Let $ABC$ be a triangle with the circumcircle $w=C(O,R)$ . Denote $D\in AA\cap BC$ , where I am using often the notation

$AA$ - the tangent to the circle $w$ in $A\in w$ . Denote the points $E$ , $F$ for which $\left\{\begin{array}{ccc} EB\perp BC & ; & OE\perp AB\\\\ FC\perp CB & ; & OF\perp AC\end{array}\right\|$ . Prove that $D\in EF$ .

Proof. Show easily that $\left\{\begin{array}{ccc} m\left(\widehat{OEB}\right)=B & \implies & EB=\frac {c}{2\sin B}\\\ m\left(\widehat{OFC}\right)=C & \implies & FC=\frac {b}{2\sin C}\end{array}\right\|\ \implies\ \frac {EB}{FC}=\left(\frac cb\right)^2$ and $\frac {DB}{DC}=\left(\frac cb\right)^2\implies$ $\frac {DB}{DC}=\frac {EB}{FC}$ $\implies$ $\triangle DBE\sim\triangle DCF$ $\implies$ $D\in EF$ .

PP4 (USA, 1975). Let $w_1$ , $w_2$ be two secant circles in the points $A$ , $B$ . Consider two mobile points $M\in w_1$ , $N\in w_2$ so that $A\in MN$ .

Ascertain the maximum value of the product $AM\cdot AN$ and the position of the line $MN$ when the maximum of this product is touched.

Proof. Denote $w_1=C(O_1, r_1)$ , $w_2=C(O_2,r_2)$ and $x=m(\angle ABM)$ , $y=m(\angle ABN)$ . Observe that $x+y=k$ (constant) and $\begin{array}{c} AM=2r_1\cdot\sin x\\\ AN=2r_2\cdot \sin y\end{array}$ . Therefore, $AM\cdot AN=4r_1r_2\cdot\sin x\sin y$ is maximum iff $x=y=\frac k2$ . Denote the intersection $L$ of $O_1O_2$ with a common (exterior) tangent of $w_1$ , $w_2$ .

Define the second intersections $M_0$ , $N_0$ of $LA$ with the circles $w_1$ , $w_2$ respectively. Prove easily that $AM\cdot AN$ is maximum $\iff$ $M:=M_0$ and $N:=N_0$ .

Remark. Show easily that for $x+y=k$ (constant) the product $\sin x\sin y=\frac 12\cdot\left[\cos (x-y)-\cos k\right]$ is maximum $\iff$

$\cos (x-y)$ is maximum $\iff$ $x=y=\frac k2$ . If denote $AB=d$ , then $k=\arccos\frac {d^2-\sqrt {\left(4r_1^2-d^2\right)\left(4r_2^2-d^2\right)}}{4r_1r_2}$ .

PP5 (Lituania, 1994). Let $ABCD$ be a cyclical quadrilateral. Denote the midpoints $U$ , $V$ of the sides $[AB]$ , $[CD]$ respectively

and $I\in AC\cap BD$ . Prove that the perpendiculars from $I$ , $U$ , $V$ to the lines $AD$ , $BD$ , $AC$ respectively are concurrently.

Proof. Denote the points $\{X,M\}\subset BD$ for which $AX\perp BD$ , $UM\perp BD$ , the points $\{Y,N\}\subset AC$ for which $DY\perp AC$ , $VN\perp AC$ and the intersections

$S\in UM\cap VN$ , $H\in AX\cap DY$ . Observe that $\widehat {BXY}\equiv\widehat {CAD}\equiv\widehat {CBX}$ $\implies$ $XY\parallel BC$ . Since $UM\parallel AX$ and $VN\parallel DY$ obtain that $MN$ is

middleline in the trapezoid $BCXY$ and $MN\parallel XY$ . But and $SM\perp HX$ , $SN\perp HY$ . Therefore, since the triangles $MNS$ , $XYH$ are Desarque-

correspondently obtain that $MX\cap NY\cap SH\ne\emptyset$ , i.e. $I\in SH$ . Observe that $HI\perp AD$ . In conclusion, $S\in HI\implies SI\perp AD$ , i.e. the perpendiculars from

$I$ , $U$ , $V$ to the lines $AD$ , $BD$ , $AC$ respectively are concurrently.

PP6 (nice !). Let $ABC$ be a triangle. For three given points $D\in (BC)$ , $E\in (CA)$ and $F\in (AB)$

construct the point $L$ so that the sideline $BC$ separates $L$ , $A$ and $\begin{array}{c} \widehat{BCL}\equiv\widehat{DFE}\\\ \widehat{CBL}\equiv\widehat{DEF}\end{array}$ . Prove that $[AELF]=[ABC]$ .

Proof (oneplusone). Denote the second intersection $S$ between the circumcircle of $\triangle DEF$ and the sideline $BC$ . Observe that

$\begin{array}{ccccccc} \widehat{SBL}\equiv\widehat{DEF}\equiv \widehat{BSF} & \implies & \widehat{SBL}\equiv \widehat{BSF} & \implies & BL\parallel SF & \implies & [LFB]=[LSB]\\\\ \widehat{SCL}\equiv\widehat{DFE}\equiv \widehat{CSE} & \implies & \widehat{SCL}\equiv \widehat{CSE} & \implies & CL\parallel SE & \implies & [LEC]=[LSC]\end{array}$ .

Denote $U\in LF\cap BC$ and $V\in LE\cap BC$ . Therefore, $[AELF]=[AEVUF]+[LUV]=$ $\left([ABC]-\underline{[BFU]}-\underline {\underline{[CEV]}}\right)+$

$\left([LBC]-\underline{[LBU]}-\underline{\underline{[LCV]}}\right)=$ $[ABC]+[LBC]-[LBF]-[LCE]=[ABC]$ . In conclusion, $[AELF]=[ABC]$ .

PP7 (Valentin Vornicu). Let $ABC$ be a triangle and the points $M$ and $N$ on the sides $AB$ respectively $BC$ , such that $2 \cdot \frac{CN}{BC} = \frac{AM}{AB}$ .

Let $P$ be a point on the line $AC$ . Prove that the lines $MN$ and $NP$ are perpendicular if and only if $PN$ is the interior angle bisector of $\angle MPC$ .

Lemma. Given are the triangle $ABC$ and the points $M\in (AB)$ , $N\in (BC)$ so that $2\cdot \frac{CN}{CB}=\frac{AM}{AB}$ .

Denote the intersection $R\in MN\cap AC$ . Then the point $N$ is the middlepoint of the segment $MR$ .

First proof (grade 7). Denote $S\in AC$ so that $MS\parallel BC$ . From Thales' theorem applied to the lines $MS\parallel BC$ , $NC\parallel MS$ in $\triangle ABC$ and $\triangle RMS$ respectively

we obtain the relations $\frac{MS}{BC}=\frac{AM}{AB}$ and $\frac{NC}{MS}=\frac{RN}{RM}$ . Therefore, $\frac{MS}{BC}\cdot \frac{NC}{MS}=$ $\frac{AM}{AB}\cdot \frac{RN}{RM}$ $\Longrightarrow$ $\frac{RN}{RM}=\frac{NC}{BC}\cdot \frac{AB}{AM}$ $\Longrightarrow$ $RM=2\cdot RN$ $\Longrightarrow$ $RN=MN$ .

Second proof (grade 9). I"ll apply the Menelaus' theorem to the transversal $\overline {ACR}$ for $\triangle BMN\ :\ \frac{AM}{AB}$ $\cdot\frac{CB}{CN}\cdot \frac{RN}{RM}=1$ $\Longrightarrow$ $RM=2\cdot RN$ $\Longrightarrow$ $RN=MN$ .

Remark. The first method is in point of fact the proof of the Menelaus's theorem in this particular case. The proposed problem has in addition ... a short tail !

PP8 (own). Let an $A$ -right $\triangle ABC$ with the circumcircle $w$ and the incircle $(I)$ . Denote $D\in (BC)$ for which $AD\perp BC$ . Consider the circles $w_1$ , $w_2$

which are tangent to altitude $[AD]$ , interior tangent to $w$ and to side $[BC]$ in the points $E\in (BD)$ , $F\in (DC)$ respectively. Prove that $BE\cdot CF=IA^2$ .

Proof.Let $X\in w\cap w_1$ , $N\in AD\cap w_1$ . From an well-known property $N\in XC$ and $CE^2=$ $CN\cdot CX=CD\cdot CB=$ $CA^2$ ,

i.e. $CE=b$ . Prove analogously that $BF=c$ . Thus $BE\cdot CF=(a-b)(a-c)=2(s-a)^2=IA^2$ , i.e. $BE\cdot CF=IA^2$ .

PP9. $AB$ is a chord of a circle $w$ and the tangents at $A$ and $B$ meet at $C$ . If $P$ is any point on the circle and $PL\ ,$

$PM\ ,\ PN$ are perpendiculars from $P$ to $AB\ ,\ BC\ ,\ CA$ respectively, then prove that $PL^2=PM\cdot PN$ .

Proof. Observe that $ALPN$ and $BLPM$ are cyclically and $\left\|\begin{array}{c} \widehat{CAB}\equiv\widehat {CBA}\\\ \widehat{LAP}\equiv\widehat {MBP}\\\ \widehat{LBP}\equiv\widehat {NAP}\end{array}\right\|$ . Thus $\widehat{LPN}\equiv\widehat{LPM}$ and $\widehat{LNP}\equiv\widehat{LAP}\equiv$

$\widehat{MBP}\equiv\widehat{MLP}$ $\implies$ $\widehat{LNP}\equiv\widehat{MLP}$ . In conclusion, $\triangle PLN\sim\triangle PML$ $\implies$ $\frac {PL}{PM}=\frac {PN}{PL}$ $\implies$ $PL^2=PM\cdot PN$ .

PP10. Let $ABC$ be a triangle. Prove that $\frac A4=\frac B2=\frac C1\ \implies\ \frac{1}{AB}=\frac{1}{BC}+\frac{1}{AC}$ .

Proof 1 (trigonometric). $\frac A4=\frac B2=\frac C1=\frac {\pi}{7}\implies A=4C\ \wedge\ B=2C$ and $C=\frac {\pi}{7}$ . Thus $\frac{1}{AB}=\frac{1}{BC}+\frac{1}{AC}\implies \frac{1}{\sin C}=\frac{1}{\sin 4C}+\frac{1}{\sin 2C}$ $\implies$

$\sin C(\sin 4C+\sin 2C)=\sin 4C\sin 2C$ $\implies$ $\sin C\cdot 2\sin 3C\cos C=\sin 4C\sin 2C$ $\implies$ $\sin 3C=\sin 4C\ ,\ C\ne\frac {\pi}{2}$ $\implies$ $C=\frac {\pi}{7}$ , what is truly.

Lemma. Prove that in any triangle $ABC$ there is the equivalence $\boxed {\ B=2C\ \iff\ b^2=c(c+a)\ }$ . Indeed, if denote $D\in (AC)\ ,$

$\widehat{DBA}\equiv\widehat{DBC}$ , then $DA=\frac{bc}{a+c}$ and $\triangle ABD\sim \triangle ACB\iff AB^2=AD\cdot AC\iff c^2=\frac {b^2c}{a+c}\iff b^2=c(c+a)$ .

Proof 2 of PP10. $\left\|\begin{array}{ccccc} B=2C & \iff & b^2=c(c+a) & \iff & b^2-c^2=ac\\\ A=2B & \iff & a^2=b(b+c) & \iff & b(b+c)=a^2\end{array}\right\|$ $\implies$ $\frac {b-c}{b}=\frac ca\iff$ $c(a+b)=ab\iff$ $\frac 1c=\frac 1a+\frac 1b$ .

Proof 3. Let $P_1P_2P_3P_4P_5P_6P_7$ be consecutive vertices of a regular heptagon. Note that $\triangle P_1P_2P_4$ satisfies $\angle P_1P_2P_4=$ $2\angle P_2P_1P_4=$ $4\angle P_1P_4P_2$ . By the

Ptolemy's theorem for cyclic quadrilateral $P_1P_2P_3P_5$ obtain that $P_1P_2 \cdot ( P_3P_5+P_1P_5) =P_1P_3 \cdot P_2P_5$ . Since $P_1P_4=P_1P_5=P_2P_5$ and

$P_1P_3=P_2P_4=P_3P_5$ , it follows that $P_1P_2 \cdot (P_2P_4+P_1P_4)=P_2P_4 \cdot P_1P_4$ $\Longrightarrow \ \frac{1}{P_1P_2}=\frac{1}{P_2P_4}+\frac{1}{P_1P_4}$ .

PP11. Let $\alpha$ , $\beta$ be two secant circles for which denote $\{P,Q\}=\alpha\cap\beta$ . For a common secant line $l$ denote

$\{A,B\}=l\cap\alpha$ and $\{C,D\}=\beta\cap l$ so that $C\in (AB)$ and $B\in (CD)$ . Prove that $\widehat{APC}\equiv\widehat{BQD}$ .

Proof. Denote $\{P,E\}=AP\cap\beta$ and $\{Q,F\}=DQ\cap\alpha$ . Observe that $AF\parallel DE$ and $\widehat{APC}\equiv\widehat{ADE}\equiv\widehat{FAD}\equiv\widehat{BQD}$ , i.e. $\widehat{APC}\equiv\widehat{BQD}$ .

This post has been edited 80 times. Last edited by Virgil Nicula, Dec 1, 2015, 9:20 AM

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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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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

178. Some nice and simple geometry problems.

by Virgil Nicula, Nov 25, 2010, 2:04 PM

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