360. Some problems for the middle school.

by Virgil Nicula, Oct 23, 2012, 9:07 AM

P1. Let an $A$ -isosceles $\triangle ABC$ with $AB=AC=k$ such that its orthocenter $H$ belongs to its incircle $w=\mathbb C(I,r)$ , i.e. $HI=r$ . Ascertain its area $S=f(k)$ .

Proof. Is evidently that $H\in (AI)$ . Denote the midpoint $D$ of $[BC]$ and $E\in w\cap AB\ ,\ F\in w\cap AC\ ,\ AF=AE=x$ . Thus, $BH\parallel IE$ and $\triangle BDH\sim\triangle AEI\iff$

$\frac {BD}{AE}=\frac {DH}{EI}=\frac {2r}r=2\implies$ $BF=BD=2x\implies$ $k=AB=AF+BF=x+2x\implies$ $\boxed{k=3x}$ . Therefore, $BC=2\cdot BD\implies$ $\boxed{BC=4x}$ and

$AD^2=AB^2-BD^2=9x^2-4x^2=5x^2\implies$ $\boxed{AD=x\sqrt 5}$ . In conclusion, $S=\frac 12\cdot BC\cdot AD=\frac 12\cdot 4x\cdot x\sqrt 5=2x^2\sqrt 5\implies$ $S=2\left(\frac k3\right)^2\sqrt 5\implies$ $S=\frac {2k^2\sqrt 5}9$ .

P2. Let a square $ABCD$ with $AB=a$ and a circle $w=\mathbb C(A,a)$ . For an interior point $P$ of the square and $P\in w$ we know that $PB=b$ and $m\left(\widehat{BPC}\right)=\phi$ . Find $PC=f(b,\phi )$ .

Proof. Let $m\left(\widehat{PBC}\right)=x$ and $S\in PB\ ,\ CS\perp BP$ . Thus, $\left\{\begin{array}{ccc} b & = & 2a\sin x\\\\ a\sin x & = & CS\\\\ CS & = & PC\sin\phi\end{array}\right\|\ \bigodot\implies$ $\boxed{PC=\frac b{2\sin\phi}}.$

P3. Let two secant circles $w_1=\mathbb C\left(O_1,8\right)$ and $w_2=\mathbb C\left(O_2,6\right)$ , where $O_1O_2=12$ and $P\in w_1\cap w_2$ . Consider $Q\in w_1$

and $R\in w_2$ so that $P\in (QR)$ and the line $O_1O_2$ doesn't separate $Q$ and $R$ . Prove that $PQ=PR\implies PQ^2=130$ .

Proof. Let the midpoints $M$ , $X$ , $Y$ of $[O_1O_2]$ , $[PQ]$ , $[PR]$ respectively and $PX=PY=x$ , $PM=m$ . Thus, $PQ=PR=2x$ and observe that

$O_2P=O_2M=6$ , i.e. the triangle $PO_2M$ is $O_2$ -isosceles. Denote the midpoint $N$ of $[PM]$ , i.e. $O_2N\perp PM$ . Denote $U\in O_1X\cap O_2N$ and $XO_1=u$ ,

$YO_2=v$ . Thus, $\boxed{u-v=m}$ , i.e. $O_1UPM$ is a parallelogram , i.e. $PU\parallel O_1O_@$ and $R\in O_1O_2\cap w_2$ . Thus, $\boxed{u^2-v^2=8^2-6^2=2\cdot 14}\ (1)$ . Apply

theorem of median $[PM]$ in $\triangle O_1PO_2$ and obtain $PM=\sqrt {14}$ i.e. $\boxed{m=\sqrt {14}}\ (*)$ . Therefore, from $(1)$ obtain that $u^2-v^2=2m^2\iff$ $u+v=2m\iff$

$2u=3m\ \wedge\ 2v=m\iff$ $PQ^2=XY^2=O_1O_2^2-UO_1^2\iff$ $PQ^2=12^2-(u-v)^2=$ $12^2-m^2=144-14\iff$ $PQ^2=130$ .

P4. Let the equilateral triangles $\triangle ABC$ and $\triangle ECD$ so that $C\in (BD)$ and $BD$ doesn't separate $A$ and $E$ . Denote $F\in BE\cap AD$ . Prove that $FA\cdot FE=FC^2$ .

Proof. Observe that $\triangle ACD\ \stackrel{s.a.s}{\equiv}\triangle BCE$ $\implies$ $AD=BE$ and $\left\{\begin{array}{ccc} \widehat{CAD} & \equiv & \widehat{CBE}\\\\ \widehat{CDA} & \equiv & \widehat{CEB}\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} \widehat{CAF} & \equiv & \widehat{CBF}\\\\ \widehat{CDF} & \equiv & \widehat{CEF}\end{array}\right\|\implies$ the quadrilaterals

$ABCF$ and $CDEF$ are cyclically $\implies m\left(\widehat{AFC}\right)=\left(\widehat{CFE}\right)=120^{\circ}\ \stackrel{a.a}{\implies}\ \triangle AFC\sim\triangle CFE\implies$ $\frac {FA}{FC}=\frac {FC}{FE}\implies$ $FA\cdot FE=FC^2$ .

P5. Let $\triangle ABC$ and $D\in (BC)$ so that $AD\perp BC\ .$ Suppose that exist $M\in (BD)$ and $N\in (DC)$ so that $MB=MD=2\cdot NC$ and $NM=NA\ .$ Prove that $A=90^{\circ}\ .$

Proof 1. $\left\{\begin{array}{ccccc} NC=x\ ;\ ND=a-5x\\\\ MB=MD=2x\\\\ NA=NM=a-3x\end{array}\right\|$ $\implies$ $AB^2=AN^2+DB^2-DN^2=(a-3x)^2+(4x)^2-(a-5x)^2=$

$16x^2+2x\cdot (2a-8x)=4ax\implies$ $AB^2=BD\cdot BC\implies$ $\left\{\begin{array}{c} \triangle DBA\sim\triangle ABC\\\\ (AD\perp BC)\end{array}\right\|\implies AB\perp AC\ .$

Proof 2. $\left\{\begin{array}{ccccc} NC=x\ ;\ ND=a-5x\\\\ MB=MD=2x\\\\ NA=NM=a-3x\end{array}\right\|$ $\implies$ $AD^2=NA^2-ND^2=(a-3x)^2-(a-5x)^2=4x(a-4x)=DB\cdot DC\implies$ $AD^2=DB\cdot DC\ \stackrel{AD\perp BC}{\implies}\ AB\perp AC.$

Remark. $\boxed{AB^2=4ax}\ ;\ AC^2=$ $AD^2+DC^2=AN^2-DN^2+DC^2=(a-3x)^2-(a-5x)^2+(a-4x)^2\implies$ $\boxed{AC^2=a^2-4ax}$ $\implies$ $AC^2+AB^2=BC^2$ .

Proof 3. Let the midpoints $(P,Q,R)$ of $[MD]$ , $[AM]$ , $[AD]$ respectively. Prove easily that $PQRD$ is a rectangle and $CRQN$ is a parallelogram. Thus, $CR\parallel NQ$ where $NQ$ is

the $N$ -median in the $N$ -isosceles $\triangle AMN$ , i.e. $NQ\perp AM$ . Hence and $CR\perp AM$ , i.e. $R$ is the orthocenter of $\triangle AMC$ . In conclusion, $AB\parallel MR\perp AC\implies AB\perp AC$ .

P6 (test Elvetia). Find all values of the parameter $a\ne 0$ so that $\{x,y,z\}\subset\mathbb{R}$ verify the system $\left\{\begin{array}{ccc} x+y & = & a\\\ x^3+y^3 & = & a\\\ x^5+y^5 & = & a\end{array}\right\|\ .$

Proof. $x^3+y^3=(x+y)\left(x^2+y^2\right)-xy(x+y)\implies$ $\boxed{1+xy=x^2+y^2}\ (1)$ and $x^5+y^5=$ $\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2(x+y)$ $\implies$

$1+x^2y^2=$ $x^2+y^2$ $\implies$ $\boxed{\left(x^2-1\right)\left(y^2-1\right)=0}\ (2)\ .$ From the relation $(2)$ obtain that $x\in \{\pm 1\}\ \vee\ y\in\{\pm 1\}$ . Can suppose that $x\in \{\pm 1\}$ .

Therefore $\left\{\begin{array}{ccccccc} x\ \stackrel{(1)}{=}\ 1 & \implies & y^2=y & \implies & y\in\{0,1\} & \implies & a\in \{1,2\}\\\\ x\ \stackrel{(1)}{=}\ -1 & \implies & y^2=-y & \implies & y\in\{0,-1\}& \implies & a\in \{-1,-2\}\end{array}\right\|$ . In conclusion, $\boxed{a\ \in\ \{\ \pm 2\ ;\ \pm 1\ \}}$ .

P7. Let $\{a,b,c\}\subset\mathbb R^*$ so that $\left\{\begin{array}{ccc} a-\frac{1}{b} & = & 3\\\\ b-\frac{1}{c} & = & 4\\\\ c-\frac{1}{a} & = & 5\end{array}\right\|\ .$ Calculate $E\equiv abc-\frac{1}{abc}\ .$

Proof. With the substitution $\left\{\begin{array}{ccc} x & = & bc\\\ y & = & ca\\\ z & = & ab\end{array}\right\|$ in the well known identity $(x-1)(y-1)(z-1)=xyz-(xy+yz+zx)+(x+y+z)-1$ obtain the chain of the identities:

$(bc-1)(ca-1)(ab-1)=(abc)^2-abc(a+b+c)+(bc+ca+ab)-1\iff$ $\frac {bc-1}c\cdot \frac {ca-1}a\cdot\frac {ab-1}b=abc-(a+b+c)+\left(\frac 1a+\frac 1b+\frac 1c\right)-\frac 1{abc}\iff$

$\left(b-\frac 1c\right)\left(c-\frac 1a\right)\left(a-\frac 1b\right)=\left(abc-\frac 1{abc}\right)-\left(a-\frac 1b\right)-\left(b-\frac 1c\right)-\left(c-\frac 1a\right)\ .$ In conclusion, $4\cdot 5\cdot 3=\left(abc-\frac 1{abc}\right)-(3+4+5)\iff$ $\boxed{abc-\frac 1{abc}=72}\ .$

P8. Let $a,\,b,\,c\in\mathbb{R}$ such that $a-7b+8c=4$ and $8a+4b-c=7\ .$ Find the value of the expression $a^2-b^2+c^2\ .$

Proof. $\left|\begin{array}{c} a-7b+8c=4\ \odot\ 1\\\\ 8a+4b-c=7\ \odot\ 8\end{array}\right|\bigoplus\stackrel{c\uparrow}{\implies} 65a+25b=60\implies \boxed{13a+5b=12}\ (1)\ .$ Similarly $\left|\begin{array}{c} a-7b+8c=4\ \odot\ 4\\\\ 8a+4b-c=7\ \odot\ 7\end{array}\right|\bigoplus$ $\stackrel{b\uparrow}{\implies} 60a+25c=65\implies$

$\boxed{12a+5c=13}\ (2).$ Addition and difference $(1),$ $(2)\ :$ $\left\{\begin{array}{ccc} 25a+5(b+c)=25 & \implies & 5(a-1)=-(b+c)\\\\ a+5(b-c)=-1 & \implies & a+1=5(c-b)\end{array}\right\|\bigodot\implies$ $a^2-1=b^2-c^2\implies a^2-b^2+c^2=1.$

P9 (MO, Rusia). Let $\{a,b\}\subset\in\mathbb{R}^*\ ,\ a\ne b$ such that $\frac{a}{b}+a=\frac{b}{a}+b\ .$ Find the value of the sum $\frac{1}{a}+\frac{1}{b}\ .$

Proof 1. Denote $\frac ab+a=\frac ba+b=k\ .$ Therefore, $\left\{\begin{array}{c} \frac 1b+1=\frac ka\\\\ \frac 1a+1=\frac kb\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} \frac 1b-\frac 1a=k\left(\frac 1a-\frac 1b\right) & \implies & \boxed{k=-1}\\\\ \frac 1b+\frac 1a+2=k\left(\frac 1a+\frac 1b\right) & \implies & \frac 1a+\frac 1b=-1\end{array}\right\|\ .$

Proof 2. $\frac{a}{b}+a=\frac{b}{a}+b\iff$ $\frac ab-\frac ba=b-a\iff$ $a^2-b^2=-ab(a-b)\ \stackrel{(a\ne b)}{\iff}\ a+b=-ab\iff$ $\boxed{\frac 1a+\frac 1b=-1}\ .$

Proof 3. $\frac{a}{b}+a=\frac{b}{a}+b\iff$ $\frac {a+ab}b=\frac {b+ab}a=\frac {(a+ab)-(b+ab)}{b-a}=\frac {a-b}{-(a-b)}=-1\implies$ $a+ab=-b\implies a+b=-ab$ $\implies \frac 1a+\frac 1b=-1\ .$

P10 (British M.O.,2008). Let $ABC$ be an $A$ -isosceles triangle with $A<90^{\circ},$ the orthocenter $H$ and the circumcircle $O.$ Prove that the circumcenter of $\triangle BOH$ belongs to the side $[AB].$

Proof 1. Denote $X\in (AB)$ so that $HX\perp HB.$ Thus, $m\left(\widehat{OHX}\right)=$ $m\left(\widehat{OHB}\right)-m\left(\widehat{XHB}\right)=$ $\left(180^{\circ}-C\right)-90^{\circ}=$ $90^{\circ}-C=$ $m\left(\widehat{OBX}\right)$ $\implies$

$\widehat{OHX}\equiv\widehat{OBX}$ $\implies$ $OHBX$ is cyclic $\implies$ $m\left(\widehat{XOB}\right)=$ $m\left(\widehat{XHB}\right)=$ $90^{\circ} \implies$ $OX\perp OB$ $\implies$ $O$ and $H$ belong to the circle with the diameter $[BX].$

Proof 2. Suppose w.l.o.g. $A<60^{\circ},$ i.e. $A<B=C\iff O\in (AH).$ Denote $\{X,Y\}\subset (AB)$ so that $HX\perp HB$ and $OY\perp OB.$ Observe that $:$

$\odot\begin{array}{ccccccc} \nearrow & m\left(\widehat{HBA}\right)=90^{\circ}-A & \implies & BX\cos\widehat{HBA}=BH & \implies & BX\sin A=2R\cos B & \searrow\\\\ \searrow & m\left(\widehat{OBA}\right)=90^{\circ}-C & \implies & BY\cos\widehat{OBA}=BO & \implies & BY\sin C=R & \nearrow\end{array}\odot$ Therefore, the circumcenter of $\triangle BOH$

belongs to the side $[AB]\iff$ $X\equiv Y\iff$ $BX=BY\iff$ $\frac {2\cos B}{\sin A}=\frac 1{\sin C}\iff$ $\sin A=2\cos B\sin B\iff$ $\sin A=\sin 2B\iff$

$A+2B=180^{\circ},$ what is true. In conclusion, $O$ and $H$ belong to the circle with the diameter $[BX].$ In the case $A>60^{\circ},$ i.e. $H\in (AO)$ the our proof is similarly.

P11 (Thanasis Gakopoulos). Let $ABC$ be a triangle with the incentre $I\ .$ Prove that $A=2B\iff AI+b=a\ .$ Nice remark!

Proof. Denote the circumcircle $w=\mathbb C(O,R)$ and $\{A,S\}=AI\cap w\ .$ Observe that $A=2B\iff$ $b=AC=CS=SB=SI\ ,$

i.e. $AB\parallel CS$ and $a=BC=AS=AI+SI=AI+b\ .$ In conclusion, $\boxed{\ A=2B\iff AI+b=a\ }$ (standard notations).

Remark. Apply the Ptolemy's theorem to the isosceles trapezoid $ABSC\ :\ AC\cdot BS+CS\cdot AB=BC\cdot AS\iff \boxed{\ b^2+bc=a^2\ }\iff A=2B\ .$

This post has been edited 240 times. Last edited by Virgil Nicula, Apr 11, 2018, 3:42 PM

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197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

November 2010

179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

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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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

360. Some problems for the middle school.

by Virgil Nicula, Oct 23, 2012, 9:07 AM

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