40. Lema lui Haruki.

by Virgil Nicula, May 29, 2010, 6:04 PM

See : here and here.

Lemma 1. Let $ABCD$ be a cyclic quadrilateral with $I\in AC\cap BD$ , $J\in AB\cap CD$ , $K\in AD\cap BC$ and $AB=a$ , $BC=b$ , $CD=c$ ,

$DA=d$ , $AC=e$ , $BD=f$ . Then $\boxed {\frac {IA}{ad}=\frac {IB}{ab}=\frac {IC}{bc}=\frac {ID}{cd}}$ , $\boxed {\frac {JA}{de}=\frac {JB}{bf}=\frac {JC}{be}=\frac {JD}{df}}$ and $\boxed {\frac {KA}{ae}=\frac {KB}{af}=\frac {KC}{ce}=\frac {KD}{cf}}$ .

Haruki's lemma. Let $ABCD$ be a convex cyclical quadrilateral and let $P$ be a variable point on the arc $AB$ which doesn't contain the points $C$ and $D$ .

Denote $E\in PD\cap AB$ and $F\in PC\cap AB$ . Prove that the ratios $\frac {EA\cdot FB}{EF}$ and $\frac {EB\cdot FA}{EF}$ don't depend on the position of the variable point $P$ on the arc $AB$ .

Original proof. Define $G$ where $AB$ cut again the circumcircle of $\triangle PEC$ . Observe that $\widehat{ EGC}\equiv \widehat{EPC}$ and so these angles remain constant as $P$

varies on the arc $AB$ . Hence, for all positions of $P$ the value of $\widehat{EGC}$ remains constant. Hence $G$ remains fixed on $AB$ . So the length of $[BG]$ is constant.

Applying the intersecting chords theorem to $PC$ and $AG$ in the two circles, we obtain $AF\cdot FB = PF\cdot FC$ and $EF\cdot FG = PF\cdot FC$ $\implies$

$(AE + EF)\cdot FB =$ $EF\cdot (FB + BG)$ $\implies$ $AE\cdot FB = EF\cdot BG$ . In conclusion we have obtained $\frac {EA\cdot FB}{EF}=BG$ which is constant.

Proof 1 - metric (own). Denote $\left\{\begin{array}{c} PA=x\\\\ PB=y\end{array}\right\|$ . Apply first lemma in the cyclical quadrilaterals $:\ \left\{\begin{array}{cccc} APBD\ : & \frac {EA}{dx}=\frac {EB}{fy}=\frac {a}{dx+fy} & \implies & \left|\begin{array}{c} EA=\frac {adx}{dx+fy}\\\\ EB=\frac {afy}{dx+fy}\end{array}\right|\\\\ BPAC\ : & \frac {FB}{by}=\frac {FA}{ex}=\frac {a}{by+ex} & \implies & \left|\begin{array}{c} FA=\frac {aex}{ex+by}\\\\ FB=\frac {aby}{by+ex}\end{array}\right|\end{array}\right\|$ $\implies$

$EF=AF-AE=\frac {aex}{ex+by}-\frac {adx}{dx+fy}=$ $ax\cdot\left(\frac {e}{ex+by}-\frac {d}{dx+fy}\right)=$ $\frac {axy(ef-bd)}{(dx+fy)(by+ex)}\ \implies$ $\boxed {EF=\frac {a^2cxy}{(dx+fy)(by+ex)}}$ .

I used the Ptolemy's relation $ef=ac+bd$ . From the upper relations obtain easily that $\boxed {\begin{array}{c} \frac {EA\cdot FB}{EF}=\frac {bd}{c}=\frac {AD\cdot BC}{DC}\\\\ \frac {EB\cdot FA}{EF}=\frac {ef}{c}=\frac {CA\cdot DB}{CD}\end{array}\ }$ . Observe that $\boxed{\frac {EA\cdot FB}{AD\cdot BC}=\frac {EB\cdot FA}{AC\cdot BD}=\frac {EF}{CD}}$ .

Proof 2 - trigonometric (own). Denote $\left\{\begin{array}{c} m(\angle EPA)=\alpha\\\\ m(\angle FPE)=\beta\\\\ m(\angle BPF)=\gamma\end{array}\right|$ and $\left\{\begin{array}{c} m(\angle PAB)=x\\\\ m(\angle PBA)=y\end{array}\right|$ . Suppose w.l.o.g. $2R=1$ . Thus, $\left\{\begin{array}{c} PA=\sin y\\\ PB=\sin x\end{array}\right|$ . Apply the Sinus' theorem in triangles :

$\blacktriangleright\ \triangle APE\ :\ \frac {AE}{\sin\alpha}=\frac {EP}{\sin x}=\frac {\sin y}{\sin (x+\alpha )}\implies$ $\left\{\begin{array}{c} AE=\frac {\sin \alpha\sin y}{\sin (x+\alpha )}\\\\ EP=\frac {\sin x\sin y}{\sin (x+\alpha )}\end{array}\right\|$ .

$\blacktriangleright\ \triangle BPF\ :\ \frac {FB}{\sin\gamma}=\frac {\sin x}{\sin (y+\gamma )}\implies$ $FB=\frac {\sin \gamma\sin x}{\sin (y+\gamma )}$ .

$\blacktriangleright\ \triangle EPF\ :\ \frac {EF}{\sin\beta}=\frac {EP}{\sin (y+\gamma )}=\frac {\sin x\sin y}{\sin (x+\alpha )\sin (y+\gamma )}\implies$ $EF=\frac {\sin x\sin y\sin \beta}{\sin (x+\alpha )\sin (y+\gamma )}$ .

Therefore, $\frac {EA\cdot FB}{EF}=\frac {\sin\alpha \sin y}{\sin (x+\alpha )}\cdot\frac {\sin \gamma \sin x}{\sin (y+\gamma )}\cdot$ $\frac {\sin (x+\alpha )\sin (y+\gamma )}{\sin x\sin y\sin\beta}\iff$ $\frac {EA\cdot FB}{EF}=\frac {\sin\alpha\sin\gamma}{\sin\beta}=\frac {bd}{c}=\frac {BC\cdot AD}{DC}$ .

Observe that $\frac {EB\cdot FA}{EF}=\frac {(EF+FB)(FE+EA)}{EF}=$ $\frac {EA\cdot FB}{EF}+(AE+EF+FB)=$ $\frac {bd}{c}+a=\frac {bd+ac}{c}=\frac {ef}{c}=\frac {CA\cdot DB}{DC}$ .

Lemma 2. Let $ABC$ be a triangle and a point $M$ which belongs to the sideline $BC$ . Then exists the relation $\boxed {\frac {MB}{MC}=\frac {AB}{AC}\cdot\frac {\sin \widehat {MAB}}{\sin\widehat {MAC}}}$ .

An easy extension. Let a convex cyclical $ABCD$ and let a variable $P$ on its circumcircle. Denote $\left\{\begin{array}{c} E\in PD\cap AB\\\\ F\in PC\cap AB\end{array}\right\|$ . Prove that $\frac {EB\cdot FA}{EF}=\frac {ef}{c}$ and $\frac {EA\cdot FB}{EF}=\frac {bd}{c}$ .

Proof. $\left\{\begin{array}{cccc} \triangle APF\ ,\ E\in AF\ : & \frac {EA}{EF}=\frac {PA}{PF}\cdot\frac {\sin \widehat {EPA}}{\sin\widehat {EPF}} & \implies & \frac {EA}{EF}=\frac {PA}{PF}\cdot \frac dc\\\\ \triangle BCF\sim PAF\ : & \frac {FB}{BC}=\frac {FP}{PA} & \implies & FB=\frac {PF}{PA}\cdot b\end{array}\right\|\bigodot$ $\implies\ \boxed {\frac {EA\cdot FB}{EF}=\frac {bd}{c}=\frac {AD\cdot BC}{DC}}$ .

$\left\{\begin{array}{cccc} \triangle APE\ ,\ F\in AE\ : & \frac {FA}{FE}=\frac {PA}{PE}\cdot\frac {\sin \widehat {FPA}}{\sin\widehat {FPE}} & \implies & \frac {FA}{FE}=\frac {PA}{PE}\cdot\frac ec\\\\ \triangle BDE\sim PAE\ : & \frac {EB}{BD}=\frac {EP}{PA} & \implies & EB=\frac {PE}{PA}\cdot f\end{array}\right\|\bigodot$ $\implies\ \boxed {\frac {EB\cdot FA}{EF}=\frac {ef}{c}=\frac {CA\cdot DB}{CD}}$ .

Proposed problem.

Let $ABCD$ be a cyclical quadrilateral which is inscribed in the circle $w$ . Denote $M\in DD\cap AB$ , $N\in CC\cap AB$ si $P\in CD\cap AB$ .

Prove that the relations $\frac {AM\cdot AN}{BM\cdot BN}=\left(\frac {PA}{PB}\right)^2$ and $\frac {MA\cdot MB}{NA\cdot NB}=\left(\frac {PM}{PN}\right)^2$ . I denote $XX$ - the tangent in the point $X\in w$ .

Proof 1. Apply a well-known relation in a triangle $ABC$ , i.e. $M\in BC\ \implies\ \frac {MB}{MC}=\frac {AB}{AC}\cdot\frac {\sin\widehat {MAB}}{\sin\widehat{MAC}}$ .

$\left\{\begin{array}{ccc} \triangle ACP : & \left\{\begin{array}{cc} N\in AP\implies & \frac {NA}{NP}=\frac {CA}{CP}\cdot\frac {\sin\widehat {NCA}}{\sin\widehat{NCP}}=\frac {e}{CP}\cdot\frac {\sin\widehat{ABC}}{\sin\widehat {CAD}}=\frac {e}{CP}\cdot\frac ec\\\\ B\in AP\implies & \frac {BP}{BA}=\frac {CP}{CA}\cdot\frac {\sin\widehat {BCP}}{\sin\widehat{ACB}}=\frac {CP}{e}\cdot\frac {\sin\widehat{BCD}}{\sin\widehat {ACB}}=\frac {CP}{e}\cdot\frac fa\end{array}\right\|\implies & \frac {NA}{NP}\cdot BP=\frac {ef}{c}\\\\ \triangle BDP : & \left\{\begin{array}{cc} A\in BP\implies & \frac {AP}{AB}=\frac {DP}{DB}\cdot\frac {\sin\widehat {ADP}}{\sin\widehat{ADB}}=\frac {DP}{f}\cdot\frac {\sin\widehat{CAD}}{\sin\widehat {ADB}}=\frac {DP}{f}\cdot\frac ea\\\\ M\in BP\implies & \frac {MB}{MP}=\frac {DB}{DP}\cdot\frac {\sin\widehat {MDB}}{\sin\widehat{MDP}}=\frac {f}{DP}\cdot\frac {\sin\widehat{BCD}}{\sin\widehat {CAD}}=\frac {f}{DP}\cdot\frac fc\end{array}\right\|\implies & \frac {MB}{MP}\cdot AP=\frac {ef}{c}\end{array}\right\|$ $\implies\boxed {\frac {PA}{PB}\cdot\frac {PN}{PM}=\frac {AN}{BM}}$ .

$\left\{\begin{array}{ccc} \triangle BCP : & \left\{\begin{array}{cc} N\in BP\implies & \frac {NB}{NP}=\frac {CB}{CP}\cdot\frac {\sin\widehat {NCB}}{\sin\widehat{NCP}}=\frac {b}{CP}\cdot\frac {\sin\widehat{BAC}}{\sin\widehat {CAD}}=\frac {b}{CP}\cdot\frac bc\\\\ A\in BP\implies & \frac {AP}{AB}=\frac {CP}{CB}\cdot\frac {\sin\widehat {ACP}}{\sin\widehat{ACB}}=\frac {CP}{b}\cdot\frac {\sin\widehat{ACD}}{\sin\widehat {ACB}}=\frac {CP}{b}\cdot\frac da\end{array}\right\|\implies & \frac {NB}{NP}\cdot AP=\frac {bd}{c}\\\\ \triangle ADP : & \left\{\begin{array}{cc} B\in AP\implies & \frac {BP}{BA}=\frac {DP}{DA}\cdot\frac {\sin\widehat {BDP}}{\sin\widehat{ADB}}=\frac {DP}{d}\cdot\frac {\sin\widehat{BAC}}{\sin\widehat {ADB}}=\frac {DP}{d}\cdot\frac ba\\\\ M\in AP\implies & \frac {MA}{MP}=\frac {DA}{DP}\cdot\frac {\sin\widehat {MDA}}{\sin\widehat{MDP}}=\frac {d}{DP}\cdot\frac {\sin\widehat{ABD}}{\sin\widehat {CAD}}=\frac {d}{DP}\cdot\frac dc\end{array}\right\|\implies & \frac {MA}{MP}\cdot BP=\frac {bd}{c}\end{array}\right\|$ $\implies\boxed {\frac {PA}{PB}\cdot\frac {PM}{PN}=\frac {AM}{BN}}$ .

From the last two relations obtain (divide and multiply !) the required conclusion.

Proof 2. Apply a remarkable Haruki's lemma for the points $:$

$\blacktriangleright\ X:=C\ \implies\left\{\begin{array}{ccc} \frac {PA\cdot NB}{PN} & = & \frac {bd}{c}\\\\ \frac {PB\cdot NA}{PN} & = & \frac {ef}{c}\end{array}\right\|$ .

$\blacktriangleright\ X:=D\ \implies\left\{\begin{array}{ccc} \frac {MA\cdot PB}{MP} & = & \frac {bd}{c}\\\\ \frac {MB\cdot PA}{MP} & = & \frac {ef}{c}\end{array}\right\|$ .

Therefore, $\left\{\begin{array}{ccc} \frac {PA\cdot NB}{PN} & = & \frac {MA\cdot PB}{MP}\\\\ \frac {PB\cdot NA}{PN} & = & \frac {MB\cdot PA}{MP}\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c} \frac {AM\cdot AN}{BM\cdot BN}=\left(\frac {PA}{PB}\right)^2\\\\ \frac {MA\cdot MB}{NA\cdot NB}=\left(\frac {PM}{PN}\right)^2\end{array}\right\|$ .

Proposed problem. The convex quadrilateral $ABCD$ is inscribed in the circle $w$ . Let $I\in AC\cap BD$ and a line $\Delta$ for which $I\in\Delta$ cut

$AB$ , $CD$ at $M$ , $N$ respectively and cut $w$ at $P$ , $Q$ respectively, where $M\in IQ$ and $N\in IP$ . Prove that $\frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}\ .$

This post has been edited 64 times. Last edited by Virgil Nicula, Nov 27, 2015, 8:17 AM

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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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by mathMagicOPS, Jan 9, 2025, 3:40 AM
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by ihatemath123, Aug 14, 2024, 1:53 AM
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
by tigerzhang, Aug 1, 2021, 12:02 AM
Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
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Bro this blog is ripped
by samrocksnature, Apr 14, 2021, 5:16 AM
Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
godly blog. opopop
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

40. Lema lui Haruki.

by Virgil Nicula, May 29, 2010, 6:04 PM

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