97. Problem of butterfly.

by Virgil Nicula, Sep 2, 2010, 3:50 PM

$\blacksquare\ P.B.$ Given are a circle $w(O)$, a line $l$ and the point $F\in l$. Let $\{M,N,P,R\}\subset w$ be four points so that $F\in MN\cap PR$.
Denote the points $A\in MR\cap l$, $B\in NP\cap l$. Prove that $\boxed{\ OF\perp l\Longleftrightarrow FA=FB\ }\ .$

Proof. Denote the projections $U$, $V$ of the center $O$ to the lines $MR$, $PN$ respectively. Thus, $\triangle FNP\sim\triangle FRM$ $\Longrightarrow$ $\widehat{FUM}\equiv\widehat{FVP}$ because the sides $[FM]$, $[RM]$ of the triangle $FRM$ are similarly with the sides $[FP]$, $[NP]$ of the triangle $FNP$ $\Longrightarrow$ $\widehat{FUA}\equiv\widehat{FVB}\ \ (1)$.
$\triangleright$ Suppose that $OF\perp l\ .$ The quadrilaterals $AFUO$, $BFVO$ are cycclically with the diameters $[AO]$, $[BO]$ respectively $\Longrightarrow$ $\widehat{FUA}\equiv\widehat{FOA}$ and $\widehat{FVB}\equiv\widehat{FOB}$. From the relation $(1)$ obtain $\widehat{FOA}\equiv\widehat{FOB}$. From $OF\perp l$ results that the triangle $AOB$ is isosceles, i.e. $FA=FB\ .$
$\triangleleft$ Suppose that $FA=FB\ .$ Denote the intersections $X$, $Y$ of the bisector line of the segment $[AB]$ with the lines $OU$, $OV$ respectively. The quadrilaterals $AFXU$, $BFYV$ are cyclically with the diameters $AX$, $BY$ respectively. Thus, $\widehat{FXA}\equiv\widehat{FUA}$ and $\widehat{FYB}\equiv\widehat{FVB}$. From the relation $(1)$ obtain $\widehat{FXA}\equiv\widehat{FYB}$, i.e. $X\equiv Y\equiv O$ $\Longrightarrow$ $OF\perp l\ .$
Quote:
Let $ABCD$ be a convex quadrilateral inscribed in the circle $w$ . For the point $P\in AB\cap CD$ consider a line $d$ for which $P\in d$ .
Denote $\left\|\begin{array}{c} U\in BC\cap d\ ,\ V\in AD\cap d\\\ X\in AC\cap d\ ,\ Y\in BD\cap d\end{array}\right\|$ . Prove that $\boxed {\ PU=PV\ \Longleftrightarrow\\ PX=PY\ \Longleftrightarrow\ OP\perp d\ }$ .

Particular case 1. Consider four points $A$ , $B$ , $C$ , $D$ which belong to the circle $w=C(O)$ and a line $d$ so that $O\in d$ and
$d\cap (AC)\ne\emptyset$ , $d\cap (AD)\ne\emptyset$ , $d\cap [AB]=\emptyset$ . Denote $\left\|\begin{array}{ccccc} M_1\in AB\cap d & , & M_2\in AD\cap d & , & M_3\in AC\cap d\\\ N_1\in CD\cap d & , & N_2\in BC\cap d & , & N_3\in BD\cap d\end{array}\right\|$ .
Prove that $\boxed {\ OM_1=ON_1\ \Longleftrightarrow\ OM_2=ON_2\ \Longleftrightarrow\ OM_3=ON_3\ }$ .

Particular case 2. Let $ABC$ be a triangle inscribed in the circle $w=C(O)$ . A line $d$ for which $O\in d$ cut the lines
$AB$, $AC$ , $BC$ and $AA$ in the points $M$ , $N$ , $P$ , $R$ respectively. Prove that $\boxed {\ OM=ON\ \Longleftrightarrow\ OP=OR\ }$ .

An equivalent enunciation. Let $ABC$ be a triangle inscribed in the circle $w=C(O)$ with diameter $[AD]$ . Denote
$P\in BC\cap DD$ . Consider $M\in OP\cap AB$ and $N\in OP\cap AC$ . Prove that $\boxed {\ OM=ON\ \Longleftrightarrow\ P\in MN\ }$ .

Remark. I denoted the tangent $XX$ in the point $X\in w$ to the circle $w$.


Proof (metric). Denote the intersection $R\in AC\cap BD$. The powers of the points $R\ ,\ N_{3}\ ,\ M_{3}$ w.r.t. the circle $w$ are : $p_{w}(R)=RA\cdot RC=RB\cdot RD$ : $p_{w}(N_{3})=N_{3}D\cdot N_{3}B$ ; $p_{w}(M_{3})=M_{3}A\cdot M_{3}C$.
$\blacktriangleright$ Apply the Menelaus' theorem to the transversals $\overline{AM_{2}D}$, $\overline{BN_{2}C}$ for the triangle $M_{3}RN_{3}$ : $\frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot \frac{DN_{3}}{DR}\cdot\frac{AR}{AM_{3}}=1$ and $\frac{N_{2}N_{3}}{N_{2}M_{3}}\cdot\frac{CM_{3}}{CR}\cdot\frac{BR}{BN_{3}}=1$ $\Longrightarrow$ $\frac{N_{2}M_{3}}{N_{2}N_{3}}\cdot \frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot\frac{RA}{RD}\cdot\frac{RC}{RB}\cdot\frac{N_{3}D}{M_{3}C}\cdot\frac{N_{3}B}{M_{3}A}=1$, i.e. $\boxed{\ \frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot\frac{N_{2}M_{3}}{N_{2}N_{3}}=\frac{p_{w}(M_{3})}{p_{w}(N_{3})}\ }\ \ (1)\ .$
$\blacktriangleright$ Apply the Menelaus' theorem to the transversals $\overline{M_{1}AB}$, $\overline{N_{1}CD}$ for the triangle $M_{3}RN_{3}$ : $\frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{BN_{3}}{BR}\cdot\frac{AR}{AM_{3}}=1$ and $\frac{N_{1}M_{3}}{N_{1}N_{3}}\cdot\frac{DN_{3}}{DR}\cdot\frac{CR}{CM_{3}}=1$ $\Longrightarrow$ $\frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{N_{1}M_{3}}{N_{1}N_{3}}\cdot\frac{RA}{RB}\cdot\frac{RC}{RD}\cdot\frac{N_{3}B}{M_{3}A}\cdot\frac{N_{3}D}{M_{3}C}=1$, i.e. $\boxed{\ \frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{N_{1}M_{3}}{N_{1}N_{3}}=\frac{p_{w}(M_{3})}{p_{w}(N_{3})}\ }\ \ (2)\ .$ From the relations $(1)$, $(2)$ obtain the conclusion of the our problem.

A very interesting particular case, when $A\equiv B$. Let $ABC$ be an acute triangle inscribed in the circle $w=C(O,R)$. Denote the point $A'\in w\cap (AO$ and the intersections $P\in BC\cap A'A'$, $M\in AB\cap PO$, $N\in AC\cap PO$. Then $OM=ON\ .$

Proposed problem. Let $ABC$ be an isosceles triangle with $AB=AC$ , and let $D$ be the midpoint of $[BC]$ . Let $E$ be the foot of the perpendicular from $D$ to $AB$ and $F$ be the midpoint of $DE$ . Prove that $AF\perp CE$ .

Proof 1 (synthetic).Denote the intersection $I\in CE\cap AF$ and the projection $G$ of the point $C$ to the line $AB$. Thus, $DE\parallel CG$ and $EG=EB$, i.e. the ray $[CE$ is the $C$- median in the triangle $CGB$. Prove easily that $ADE\sim CBG$. The $A$-median of the triangle $ADE$ is similarly with the $C$- median of the triangle $CGB$ and the side $[AD]$ of the triangle $ADE$ is similarly with the side $[CB]$ of the triangle $CBG$. Therefore, $\widehat{DAF}\equiv\widehat{BCE}$, i.e. $\widehat{DAI}\equiv\widehat{DCI}$, what means that the quadrilateral $ACDI$ is cyclically, i.e. $\widehat{ADC}\equiv\widehat{AIC}$. But $DA\perp DC$. In conclusion, and $IA\perp IC$, i.e. $AF\perp CE\ .$
Proof 2 (metric). $AC^{2}-AE^{2}=AC^{2}-(AD^{2}-DE^{2})=$ $(AC^{2}-AD^{2})+DE^{2}=DC^{2}+DE^{2}$ $\Longrightarrow$ $\boxed{\ AC^{2}-AE^{2}=DC^{2}+DE^{2}\ }\ \ (1)\ .$
$4\cdot (FC^{2}-FE^{2})=$ $2\cdot (CE^{2}+CD^{2})-DE^{2}-DE^{2}=$ $2\cdot (CE^{2}+DB^{2}-DE^{2})=$ $2\cdot (CE^{2}+BE^{2})=$ $4\cdot DE^{2}+CB^{2}=$ $4\cdot (DC^{2}+DE^{2})$ $\Longrightarrow$ $\boxed{\ FC^{2}-FE^{2}=DC^{2}+DE^{2}\ }\ \ (2)\ .$ From the relations $(1)$ and $(2)$ obtain $AC^{2}-AE^{2}=FC^{2}-FE^{2}$, i.e. $\boxed{\ AF\perp CE\ }\ .$
Proof 3 (trigonometric). Denote $\phi =m(\widehat{DAF})$ and $\psi =m(\widehat{BCE})$. Therefore, $1=\frac{FD}{FE}=\frac{AD}{AE}\cdot\frac{\sin\phi}{\sin\left(\frac{A}{2}-\phi\right)}=\frac{\sin\phi}{\cos\frac{A}{2}\sin\left(\frac{A}{2}-\phi\right)}$ $\Longrightarrow$
$2\sin\phi =\sin (A-\phi )-\sin \phi$ $\Longrightarrow$ $3\sin \phi =\sin (A-\phi )$ $\Longrightarrow$ $3\tan\phi =\sin A-\cos A\tan\phi$ $\Longrightarrow$ $\boxed{\ \tan\phi =\frac{\sin A}{3+\cos A}\ }\ \ (1)\ .$
$\frac{EB}{EA}=\left(\frac{DB}{DA}\right)^{2}=\tan^{2}\frac{A}{2}\ ,\ \frac{EB}{EA}=\frac{CB}{CA}\cdot\frac{\sin\psi }{\sin \left(90-\frac{A}{2}-\psi \right)}=\frac{2\sin\frac{A}{2}\sin\psi }{\cos\left(\frac{A}{2}+\psi\right)}\ .$ Therefore, $\sin\frac{A}{2}\cos\left(\frac{A}{2}+\psi \right)=2\cos^{2}\frac{A}{2}\sin\psi$ $\Longrightarrow$ $\sin (A+\psi )-\sin \psi =2\sin\psi (1+\cos A)$ $\Longrightarrow$ $\sin A\cos\psi=3\sin\psi+\sin\psi \cos A$ $\Longrightarrow$ $\sin A=\cos A\tan \psi+3\tan\psi$ $\Longrightarrow$ $\boxed{\ \tan\psi =\frac{\sin A}{3+\cos A}\ }\ \ (2)\ .$ From the relations $(1)$ and $(2)$ obtain $\phi =\psi$ $\Longrightarrow$ the quadrilateral $ACDI$ is cyclically $\Longrightarrow$ $AF\perp CE\ .$
Proof 4 (analytic). $D(0,0)\ ,\ B(0,b)\ ,\ C(-b,0)\ ,\ A(0,a)$, where $a>0$, $b>0$. The slope of the line $AB$ is $s(AB)=-\frac{a}{b}$, the slope of the line $DE$ is $s(DE)=-\frac{1}{s(AB)}=\frac{b}{a}$ and the point $E$ is the intersection between the line $AB$ with the equation $\frac{x}{b}+\frac{y}{a}=1$ and the line $DE$ with the equation $y=\frac{b}{a}x$. Thus, the point $E$ has the coordinates $x_{E}=\frac{a^{2}b}{a^{2}+b^{2}}$ and $y_{E}=\frac{b^{2}a}{a^{2}+b^{2}}$. The middlepoint $F$ of the segment $DE$ has the coordinates $x_{F}=\frac{1}{2}\cdot \frac{a^{2}b}{a^{2}+b^{2}}$ and $y_{F}=\frac{1}{2}\cdot \frac{b^{2}a}{a^{2}+b^{2}}$. Therefore, the slope of the line $CE$ is $s(CE)=\frac{\frac{b^{2}a}{a^{2}+b^{2}}}{\frac{a^{2}b}{a^{2}+b^{2}}+b}=\frac{ab}{2a^{2}+b^{2}}$ and the slope of the line $AF$ is $s(AF)=\frac{\frac{b^{2}a}{2(a^{2}+b^{2})}-a}{\frac{a^{2}b}{2(a^{2}+b^{2})}}=-\frac{2a^{2}+b^{2}}{ab}\ .$ In conclusion, $s(CE)\cdot s(AF)=-1$, i.e. $CE\perp AF\ .$
This post has been edited 11 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:35 AM

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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).
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  • Total entries: 456
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