6. Interesting problems from JBMO.

by Virgil Nicula, Apr 19, 2010, 3:01 PM

Interesting problems.

PP1 (JBMO 2012). Let $ABC$ be an acute triangle with the orthocenter $H$ . Denote the feet $E$ , $F$ of the altitudes from $B$ , $C$ respectively and the

intersection $L$ between $[AH]$ and the circle with the diameter $[BC]$ . Prove that the circles $\odot ALE$ and $\odot HLF$ are tangent in the point $L$ .

Method 1. Prove easily that :

$1\blacktriangleright\ \widehat{EAL}\equiv\widehat{EBC}\equiv\widehat{ELC}==>$ $\widehat{EAL}\equiv\widehat{ELC}==>$ $CL$ is tangent to $\odot ALE$ .

$2\blacktriangleright\ \widehat{HFL}\equiv\widehat{CBL}\equiv\widehat{HLC}==>$ $\widehat{HFL}\equiv\widehat{HLC}==>$ $CL$ is tangent to $\odot HLF$ .

Method 2. Prove easily that :

$1\blacktriangleright\ CL^2=CD.CB=CE.CA ==>$ $CL^2=CE.CA ==>$ $\widehat{EAL}\equiv\widehat{ELC} ==>$ $CL$ is tangent to $\odot ALE$ .

$1\blacktriangleright\ CL^2=CD.CB=CH.CF ==$ $CL^2=CH.CF ==>$ $\widehat{HFL}\equiv\widehat{HLC} ==>$ $CL$ is tangent to $\odot HLF$ .

Remark. Prove that $BL$ is common tangent in $L$ to $\odot AFL$ , $\odot EHL$ . Thus, $BL\perp CL\implies$ the pairs $\left\{\left|\begin{array}{ccc} \odot EAL & \wedge & \odot EHL\\\\ \odot FAL & \wedge & \odot FHL\end{array}\right|\left|\begin{array}{ccc} \odot EAL & \wedge & \odot FAL\\\\ \odot EHL & \wedge & \odot FHL\end{array}\right|\right|$ are ortogonally.

PP2 (JBMO 2013). Let $ABC$ be an acute-angled triangle with circumcircle $w=C(O,R)$ and $c<b$ . Let $D\in BC$ be the point such that

$\widehat{BAD}\equiv\widehat{CAO}$ . Denote $\{A,E\}=AD\cap w$ and the midpoints $M$ , $N$ and $P$ of $[BE]$ , $[OD]$ and $[AC]$ respectively. Show $P\in MN$ .

Proof. Are well-known that $DE=DH$ and $BH=2\cdot OP$ , what means $DM\parallel BH$ , $BH=2\cdot DM$ and $OP\parallel BH$ , $BH=2\cdot OP$ .

In conclusion, $DM=OP$ and $DM\parallel OP$ , i.e. $OPDM$ is a parallelogram $\implies$ the midpoint $N$ of $OD$ belongs to $MP$ , i.e. $P\in MN$ .

Remark. Can apply to the orthodiagonal and cyclical quadrilateral $ABEC$ the following well-known property:

RP. Let $ABCD$ be an orthodiagonal and cyclical quadrilateral with the circumcircle $w=C(O,R)$ . Denote $E\in AC\cap BD$ , the midpoints $M , N , P , Q$ of $[AB] , [BC] , [CD] , [DA]$

respectively and the projections $X , Y , Z , T$ of the point $D\in AC\cap BD$ on $[AB] , [BC] , [CD] , [DA]$ respectively. Then $D\in MX\cap NY\cap PZ\cap QT$ , $MNPQ$ is a rectangle

and the points $\{X,Y,Z,T\}$ belong to the circle with the diameter $[MP]$ and the center in the midpoint of $[OE]$ . Hint. The quadrilaterals $OMEP$ and $ONEQ$ are parallelograms.

PP3. (SBMO 2013). In $\triangle ABC$ , the $A$ -exincircle $w_a$ is tangent to $AB$ and $AC$ in $P$ and $Q$ respectively. The $B$ -exincircle $w_b$ is tangent to $BA$

and $BC$ in $M$ and $N$ respectively. Denote the projection $K$ of $C$ on $MN$ and the projection $L$ of $C$ on $PQ$ . Prove that $MKLP$ is a cyclical quadrilateral.

Proof 1. Denote $S\in KM\cap LP$ , $R\in AB\cap SC$ and $X\in AI\cap PQ$ , $Y\in BI\cap MN$ . Observe that $\frac {CK}{CL}=\frac {CN\cdot\cos \frac B2}{CQ\cdot \cos \frac A2}=$ $\frac {(s-a)\cos\frac B2}{(s-b)\cos\frac A2}=$ $\frac {\frac {s-a}{\cos\frac A2}}{\frac {s-b}{\cos\frac B2}}=$ $\frac {IA}{IB}$ .

Since $\widehat{AIB}\equiv\widehat{KCL}$ obtain that $\triangle KCL\sim\triangle AIB$ . Thus, $\widehat{ABY}\equiv\widehat {ABI}\equiv\widehat{KLC}\equiv\widehat {KSC}\implies$ $BRYS$ is cyclically $\implies SC\perp AB$ (analogously $ARXS$

is cyclically). Since $RMKC$ and $RPLC$ are cyclically obtain that $\left\{\begin{array}{ccc} SC\cdot SR & = & SK\cdot SM\\\\ SC\cdot SR & = & SL\cdot SP\end{array}\right\|\implies$ $SK\cdot SM=SL\cdot SP\implies$ $MKLP$ is a cyclical quadrilateral.

Proof 2. Denote $S\in KM\cap LP$ . t\Thus, $AM=BP=s-c$ , $MP=a+b$ and $\left\{\begin{array}{c} m\left(\widehat{PMS}\right)=90^{\circ}-\frac B2\\\\ m\left(\widehat{MPS}\right)=90^{\circ}-\frac A2\\\\ m\left(\widehat{MSP}\right)=90^{\circ}-\frac C2\end{array}\right\|$ . Theorem of Sines in $\triangle MPS\ :\ \frac {a+b}{\cos\frac C2}=$

$\frac {SM}{\cos \frac A2}=\frac {SP}{\cos \frac B2}\implies$ $SM^2-SP^2=\frac {(a+b)^2}{\cos^2\frac C2}\cdot\left(\cos ^2\frac A2-\cos^2\frac B2\right)=$ $\frac {ab(a+b)^2}{2s(s-c)}\cdot(\cos A-\cos B)=$ $\frac {ab(a+b)^2}{2s(s-c)}\cdot\frac {2(b-a)s(s-c)}{abc}\implies$

$\boxed{SM^2-SP^2=\frac {(a+b)(b^2-a^2)}{c}}$ . Apply the theorem of Cosines in the triangles $CAM$ and $CBM\ :\ \left\{\begin{array}{c} CM^2=(s-c)^2+b^2+2b(s-c)\cos A\\\\ CP^2=(s-c)^2+a^2+2a(s-c)\cos B\end{array}\right\|\implies$

$CM^2-CP^2=b^2-a^2+2(s-c)(b\cos A-a\cos B)$ . Observe that $b\cos A-a\cos B=\frac 1{2c}\cdot\left[\left(b^2+c^2-a^2\right)-\left(a^2+c^2-b^2\right)\right]=$ $\frac {b^2-a^2}{c}$ si $CM^2-CP^2=$

$b^2-a^2+2(s-c)\cdot \frac {b^2-a^2}{c}=\left(b^2-a^2\right)\cdot\left(1+\frac {a+b-c}{c}\right)\implies$ $\boxed{CM^2-CP^2=\frac {(a+b)(b^2-a^2)}{c}}$ . In conclusion, $SM^2-SP^2=CM^2-CP^2\implies$ $SC\perp AB$ .

Denote $R\in SC\cap AB$ . Since $RMKC$ and $RPLC$ are cyclically obtain that $\left\{\begin{array}{ccc} SC\cdot SR & = & SK\cdot SM\\\\ SC\cdot SR & = & SL\cdot SP\end{array}\right\|\implies$ $SK\cdot SM=SL\cdot SP\implies$ $MKLP$ is cyclically.

An easy extension. Let $\triangle ABC$ , $\{M,P\}\subset AB$ so that $\left\{\begin{array}{c} A\in (BM)\\\\ B\in (AP)\end{array}\right\|$ and $N\in BC\ ,\ Q\in AC$ so that $C\in (BN)\cap (AQ)$ . Denote

$\left\{\begin{array}{c} K\in MN\ ,\ CK\perp MN\\\\ L\in PQ\ ,\ CL\perp PQ\end{array}\right\|$ . Suppose that $AP=AQ=BM=BN=x>\max\{a,b,c\}$ . Prove that $MKLP$ is cyclically $\iff 2x=a+b+c$ .

Proof. Denote $S\in PQ\cap MN$ and $R\in SC\cap AB$ . Observe that $MP=2x-c$ . Apply the theorem of Sines in $\triangle MPS\ :\ \frac {SM}{\cos\frac A2}=$ $\frac {SP}{\cos\frac B2}=$ $\frac {2x-c}{\cos\frac C2}\implies$

$\left\{\begin{array}{c} SM=\frac {(2x-c)\cos\frac A2}{\cos\frac C2}\\\\ SP=\frac {(2x-c)\cos\frac B2}{\cos\frac C2}\end{array}\right\|$ . Thus, $SM^2-SP^2=\frac {(2x-c)^2}{\cos^2\frac C2}\cdot\left(\cos^2\frac A2-\cos^2\frac B2\right)=$ $\frac {ab(2x-c)^2}{2s(s-c)}\cdot (\cos A-\cos B)=$ $\frac {ab(2x-c)^2}{2s(s-c)}\cdot\frac {2s(b-a)(s-c)}{abc}\implies$

$\boxed{SM^2-SP^2=\frac {(b-a)(2x-c)^2}{c}}$ . Theorem of Cosines in $\triangle CAM$ and $CBM\ :\ \left\{\begin{array}{c} CM^2=(x-c)^2+b^2+2b(x-c)\cos A\\\\ CP^2=(x-c)^2+a^2+2a(x-c)\cos B\end{array}\right\|\implies$

$CM^2-CP^2=b^2-a^2+2(x-c)(b\cos A-a\cos B)$ . Observe that $b\cos A-a\cos B=\frac 1{2c}\cdot\left[\left(b^2+c^2-a^2\right)-\left(a^2+c^2-b^2\right)\right]=$ $\frac {b^2-a^2}{c}$ si $CM^2-CP^2=$

$b^2-a^2+2(x-c)\cdot \frac {b^2-a^2}{c}=\left(b^2-a^2\right)\cdot\left(1+\frac {2x-2c}{c}\right)\implies$ $\boxed{CM^2-CP^2=\frac {(2x-c)(b^2-a^2)}{c}}$ . In conclusion, the quadrilateral $MKLP$ is cyclic $\iff$

$SC\perp AB\iff$ $SC\perp MP\iff$ $SM^2-SP^2=CM^2-CP^2\iff$ $\frac {(b-a)(2x-c)^2}{c}=\frac {(2x-c)(b^2-a^2)}{c}\stackrel{2x\ne c}{\iff} 2x-c=a+b\iff 2x=a+b+c$ .

PP4. Let f be a function $f(x)=a_0x^3+a_1x^2+a_2x+a_3,\ a_0\ne 0,\ x\in R$ and a line $d$ for which $d\cap {G_f} =(P_k |k\in \overline {1,3}).$

$(\forall ) k\in\overline{1,3}$ the tangent at $G_f$ in the point $P_k\in {G_f}$ meets once more again the graph $G_f$ in the point $Q_k.$ Prove that the point $Q_k,\ k\in \overline{1,3}$ are collinear.

Lemma. The points $X_k(x_k,f(x_k)),\ k\in \overline{1,3}$ are collinear $\Longleftrightarrow x_1+x_2+x_3=-\frac{a_1}{a_0}.$

Proof of the problem. The points $P_k,\ k\in \overline {1,3}$ are collinear $\Longleftrightarrow p_1+p_2+p_3=-\frac{a_1}{a_0}.$ But the points $P_k,P_k,Q_k$ are collinear, i.e. $2p_k+q_k=-\frac{a_1}{a_0},\ k\in \overline {1,3}.$

Thus, $2(p_1+p_2+p_3)+(q_1+q_2+q_3)=-\frac{3a_1}{a_0}$ , i.e $- \frac{2a_1}{a_0}+(q_1+q_2+q_3)=-\frac{3a_1}{a_0}$ , i.e. $q_1+q_2+q_3=-\frac{a_1}{a_0}$ , i.e. the points $Q_1,Q_2,Q_3$ are collinear.

PP5. Let an $A$ -right $\triangle ABC$ with $\frac b2<c<b$ and $M\in (AC)$ , $N\in (AB)$ so that $MC=AB$ and $NB=AM$ . Find $m\left(\widehat {BPN}\right)$ , where $P\in BM\cap CN$ .

Proof. $\left\{\begin{array}{c} MC=AB=c\\\\ AM=BN=b-c\\\\ AN=AB-BN=2c-b\end{array}\right\|$ and $\left\{\begin{array}{ccc} m\left(\widehat {ABM}\right)=x & \implies & \tan x=\frac {AM}{AB}=\frac {b-c}{c}\\\\ m\left(\widehat {ACN}\right)=y & \implies & \tan y=\frac {AN}{AC}=\frac {2c-b}{b}\end{array}\right\|$ . Thus, $m\left(\widehat {BPN}\right)=m\left(\widehat {ANC}\right)-m\left(\widehat {ABM}\right)\implies$

$m\left(\widehat {BPN}\right)=\left(90^{\circ}-y\right)-x\implies$ $\tan \left(\widehat {BPN}\right)=\tan \left[90^{\circ}-(x+y)\right]=$ $\cot (x+y)=$ $\frac {1-\tan x\tan y}{\tan x+\tan y}=$ $\frac {1-\frac {b-c}{c}\cdot \frac {2c-b}{b}}{\frac {b-c}{c}+\frac {2c-b}{b}}=\frac {b^2-2bc+2c^2}{b^2-2bc+2c^2}=1\implies$ $m\left(\widehat {BPN}\right)=45^{\circ}$

This post has been edited 192 times. Last edited by Virgil Nicula, Nov 27, 2015, 10:54 AM

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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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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

6. Interesting problems from JBMO.

by Virgil Nicula, Apr 19, 2010, 3:01 PM

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