52. Harmonical division.

by Virgil Nicula, Jul 9, 2010, 8:45 PM

$\underline{\overline{|<>Lemma<>|}}$ Let $AB$ and $\{C,D\}\subset AB$, the midpoints $M$ , $N$ of $[AB]$ , $[CD]$ respectively. Then the following relations are equivalently $:$

A0. There is $m\ne 1$, such that $\overline{CA}=m\cdot \overline{CB},\ \overline{DA}=-m\cdot \overline{DB}$, i.e. $(A,B,C,D)$ is a harmonical division.

$\blacktriangleleft\blacktriangleright$ A1. $AB^2+CD^2=4\cdot MN^2$ $\blacktriangleleft\blacktriangleright$ A2. $\left\{\begin{array}{c} NC^2=\overline{NA}\cdot \overline{NB}\\\\ MA^2=\overline{MC}\cdot \overline{MD}\end{array}\right\|$ $\blacktriangleleft\blacktriangleright$ A3. $\left\{\begin{array}{c} \overline{BC}\cdot \overline{BD}=\overline{BA}\cdot \overline{BN}\\\\ \overline{CA}\cdot \overline{CB}=\overline{CM}\cdot \overline{CD}\end{array}\right\|$ $\blacktriangleleft\blacktriangleright$ A4. $\left\{\begin{array}{c} \overline{NA}\cdot \overline{NB}+\overline{NC}\cdot \overline {ND}=0\\\\ \overline{MC}\cdot \overline{MD}+\overline{MA}\cdot \overline{MB}=0\end{array}\right\|$

$\blacktriangleleft\blacktriangleright$ A5. $\left\{\begin{array}{c} \frac{2}{\overline{AB}}=\frac{1}{\overline{AC}}+\frac{1}{\overline{AD}}\\\\ \frac{2}{\overline{CD}}=\frac{1}{\overline{AD}}+\frac{1}{\overline{BD}}\end{array}\right\|$ $\blacktriangleleft\blacktriangleright$ A6. $\overline{AD}\cdot\overline {BC}+\overline {AC}\cdot\overline{BD}=0$ $\blacktriangleleft\blacktriangleright$ A7. $\left\{\begin{array}{c} \frac{2}{\overline{AB}}=\frac{1}{\overline{CB}}+\frac{1}{\overline{DB}}\\\\ \frac{2}{\overline{CD}}=\frac{1}{\overline{CA}}+\frac{1}{\overline{CB}}\end{array}\right\|$ $\blacktriangleleft\blacktriangleright$ A8. $\left\{\begin{array}{c} \frac {BA}{BD}=2\cdot\frac {CA}{CD}\\\\ \frac {CD}{CA}=2\cdot\frac {BD}{BA}\end{array}\right\|$ $\blacktriangleleft\blacktriangleright$ A9. $\left\{\begin{array}{c} \frac {CA}{CB}=\frac {AN}{NC}\\\\ \frac {BC}{BD}=\frac {BM}{MD}\end{array}\right\|$


Remark. With this lemma we will solve many proposed problems.

$P1\blacktriangleright$ Let $\triangle ABC$ , the midpoint $M$ of $[BC]$ , the orthic $\triangle DEF$ and $\left\{\begin{array}{c} L\in BC\cap EF\\\\ S\in AM\cap LH\end{array}\right\|$ . Prove that $AM\perp LH$ and $\widehat{ASB}\equiv\widehat{ASC}\Longleftrightarrow AB=AC$ .

$P2\blacktriangleright$ Let $\triangle ABC$ with incircle $w=C(I,r)$ which touches it at $X\in BC$ , $Y\in CA$ , $Z\in AB$ and $T\in BC\cap YZ$ , $V\in TI\cap AX$. Prove that $TI\perp AX$ and $\widehat{AVB}\equiv\widehat{AVC}$

$P3\blacktriangleright$ Let $\triangle ABC$ , circumcentre $O$ , the orthocentre $H$ , point $D\in BC\cap AH$ , symmetrical point $E$ of $C$ w.r.t. $D$ , i.e. $DE=DC$ and point $F\in AB\cap HE$ . Prove that $DF\perp DO$

$P4\blacktriangleright$ Let $\triangle ABC$ , $AB\ne AC$, the centroid $G$ and $D\in BC$ , $GD\perp BC$. Ascertain $X\in AB$ , $Y\in AC$ for which $G\in XY$ and $\widehat{GDX}\equiv\widehat{GDY}$ .

$P5\blacktriangleright$ Let $C_k=C(O_k,r_k),\ k\in\overline{1,2}$ be two circles for which $O_1\in C_2$ , $\{A,B\}= C_1\cap C_2$ . Let $d$ be a line which pass

through $O_1$ , $R\in AB\cap d$ , $\{M,N\}=d\cap C_1$ and cut again $C_2$ in $P$ so that $N\in (RP)$ . Prove that $\widehat{NAP}\equiv\widehat{NAR}$ .

$P6\blacktriangleright$ Generalization. Let $\triangle ABC$ and for $P$ denote $D\in BC$ , $E\in CA$ , $F\in AB$ so that $BFPD$ , $CDPE$ are cyclically. Denote $m(\angle PDC)=$ $m(\angle PEA)=$

$m(\angle PFB)=\phi$ , $L\in EF\cap BC$ . The sideline $BC$ cut again circumcircle of $\triangle DEF$ in $K$ . Prove that $m\left(\widehat {LP,AK}\right)\in \{\phi ,\pi -\phi\}$ .

$P7\blacktriangleright$ Let $\triangle ABC$ with the circumcenter $O$ . Denote the projection $P$ of $O$ on the $A$-symmedian. Prove that $\widehat{APB}\equiv\widehat{APC}$ .


Particular cases.

$C1\blacktriangleright$ Let the "podar" $\triangle DEF$ of $\triangle ABC$ , where $D\in BC$ , $E\in CA$ , $F\in AB$ and $L\in EF\cap BC$ and $BC$ cut again circumcircle of $\triangle DEF$ in $K$ . Prove that $LP\perp AK$

$C2\blacktriangleright$ Let $ABC$ be a triangle with exincircle $C(I_a)$ which touches the sides of $\triangle ABC$ in $D\in BC$ , $E\in CA$ , $F\in AB$ . Denote $T\in BC\cap EF$ . Prove that $TI_a\perp AD$ .


$P8\blacktriangleright$ See here


PP1. For an interior point $P$ of $\triangle ABC$ denote the intersections $(D,E,F)$ of $(AP,BP,CP)$ with opposite sidelines $(BC,CA,AB)$ respectively and

$T\in EF\cap BC$ . The $D$-bisectors of the triangles $ADB\ ,\ ADC$ meet the line $\overline {EFT}$ in $U\ ,\ V$ respectively. Prove that $BU\cap CV\cap AD\ne\emptyset$ .

Proof. Denote $S\in EF\cap AD$ , $R_1\in BU\cap AD$ and $R_2\in AD\cap CV$ . The division $(B,C;D,T)$ is harmonically and $DU\perp DV\ \wedge\ \widehat{UDT}\equiv\widehat{UDS}\implies$

the division $(U,V;S,T)$ is harmonically, i.e. $\left\{\begin{array}{c} \frac {BD}{BT}=\frac {CD}{CT}\\\\ \frac {UT}{US}=\frac {VT}{VS}\end{array}\right\|\ (*)$ . Apply the Menelaus' theorem to the mentioned transversals $:$

$\left\{\begin{array}{cc} \overline{R_1UB}/\triangle DST\ : & \frac {R_1S}{R_1D}\cdot\frac {BD}{BT}\cdot\frac {UT}{US}=1\\\\ \overline{R_2VC}/\triangle DST\ : & \frac {R_2S}{R_2D}\cdot\frac {CD}{CT}\cdot\frac {VT}{VS}=1\end{array}\right\|\stackrel{(*)}{\implies}$ $\frac {R_1S}{R_1D}=\frac {R_2S}{R_2D}\implies R_1\equiv R_2\equiv R\implies$ $R\in BU\cap CV\cap AD$ .


Lemma. Let $ \{C,D\}\subset d = AB$ so that $ \left|\begin{array}{c} C\in (AB)\\\\ B\in (CD)\end{array}\right|$ . For $ P\not\in d$ let $ \left|\begin{array}{c} m(\widehat {APC}) =x\\\\ m(\widehat {CPB}) =y\\\\ m(\widehat {BPD}) =z\end{array}\right|$ . Then $ C$ , $ D$ are ".h.c." w.r.t. $ A$ , $ B$ $\iff$ $\boxed{\cos (x+z)=\cos (y+x)\cdot\cos (y+z)}$ .

Proof. $\frac {CA}{CB} = \frac {DA}{DB}\ \Longleftrightarrow\ \frac {PA}{PB}\cdot\frac {\sin \widehat {CPA}}{\sin\widehat {CPB}} =$ $ \frac {PA}{PB}\cdot\frac {\sin \widehat {DPA}}{\sin\widehat {DPB}}\ \Longleftrightarrow\ \frac {\sin x}{\sin y} =$ $\frac {\sin (x+y+z )}{\sin z}\Longleftrightarrow$ $\boxed{\sin x\cdot\sin z= \sin y\cdot\sin (x+ y+z )}$ $ \Longleftrightarrow$

$ \cos (x- z ) - \cos (x+ z) = \cos (x+ z ) - \cos (x+ 2y +z)$ $ \Longleftrightarrow$ $ 2\cdot\cos (x+z) = \cos (x-z) + \cos (x+ 2y+z )$ $ \Longleftrightarrow$ $\cos(x+z)=\cos (y+x)\cdot\cos (y+z)\ (*)$ .

Remark. Prove easily that if $ C$ , $ D$ are ".h.c." w.r.t. $ A$ , $ B$ i.e. the relation $ (*)$ is truly, then $ PA\perp PC$ i.e. $x+y=90^{\circ}\ \Longleftrightarrow$ the ray $ [PB$ is the bisector line of $ \widehat {CPD}$ i.e. $y=z$ .

PP2 (easy problem). $ \triangle\ ABC$ and $ \left\|\begin{array}{ccccccc} E\in AC & , & BE\perp AC & ; & F\in AB & , & CF\perp AB\\\\ M\in (BC) & , & BM=2\cdot MC & ; & C\in (BN) & , & CN=BC\end{array}\right\|$ $ \Longrightarrow$ $ \frac {FM}{FN}=\frac {EM}{EN}$


PP3 (E.C. Ortega). Let $\triangle ABC$ and for $M\in (AC)$ , $P\in (AB)$ denote $L\in BM\cap CP$ , $K\in AL\cap MP$ . Let $N\in (BC)$ so that $\widehat{ANP}\equiv\widehat{LNM}$ . Prove that $KN\perp BC$ .

Proof. Let $S\in AL\cap BC$ , $R\in PM\cap BC$ and $T\in BC$ so that $KT\perp BC$ . Hence: $\left\{\begin{array}{cccc} (P,M;K,R)\ \mathrm{is\ h.d.}\ : & TK\perp TR & \iff & \widehat{KTP}\equiv\widehat{KTM}\\\\ (A,L;K,S)\ \mathrm{is\ h.d.}\ : & TK\perp TS & \iff & \widehat{KTA}\equiv\widehat{KTL}\end{array}\right\|\implies$

$\widehat{ATP}\equiv\widehat{KTP}-\widehat{KTA}=\widehat{KTM}-\widehat{KTL}=\widehat{LTM}\implies$ $\widehat{ATP}\equiv\widehat{LTM}$ . In conclusion, $\left\{\begin{array}{ccc} \widehat{ATP} & \equiv & \widehat{LTM}\\\\ \widehat{ANP} & \equiv & \widehat{LNM}\end{array}\right\|\implies$ $T\equiv N\implies AN\perp BC$ .
This post has been edited 100 times. Last edited by Virgil Nicula, Dec 18, 2016, 8:36 AM

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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).
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  • Total entries: 456
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