203. An application of the Pascal's permutation theorem.

by Virgil Nicula, Jan 1, 2011, 8:01 AM

Proposed problem. Let $ABC$ be a triangle. Consider the points $\{N,P\}\subset BC$ , $\{Q,K\}\subset CA$ and $\{L,M\}\subset AB$ such that

$LK\parallel BC$ , $MN\parallel CA$ , $PQ\parallel AB$ . Prove that the points $X\in MQ\cap BC$ , $Y\in PL\cap CA$ , $Z\in KN\cap AB$ are collinearly.

Proof 1

Proof 2.

Proof 2 (proiective). Observe vthat $\left\|\begin{array}{ccc} KL\parallel BC & \iff & KL\parallel NP\\\ MN\parallel CA & \iff & MN\parallel QK\\\ PQ\parallel AB & \iff & PQ\parallel LM\end{array}\right\|$ . Hence the points $U\in KL\cap NP$ , $V\in MN\cap QK$ and $W\in LM\cap PQ$

are infinite points, so are collinearly (all lie on the line at infinity). Therefore, the points $(K, L, M, N, P, Q)$ is Pascalian. Thus, by the Pascal permutation theorem,

the permutation $(N, P, L, M, Q, K)$ of this hextuple must also be Pascalian. Hence the points $X\in NP\cap MQ$ , $Y\in PL\cap QK$ and $Z\in LM\cap KN$

are collinearly. But $NP\cap MQ=BC\cap MQ$ , $PL\cap QK=CA\cap PL$ and $LM\cap KN=AB\cap KN$ . In conclusion, the points $X\in BC\cap MQ$ ,

$Y\in CA\cap PL$ and $Z\in AB\cap KN$ are collinearly.

Darij Grinberg wrote:

Definition. Let's call the points $(A, B, C, D, E, F)$ from same plane is a Pascalian sextuple iff the points $AB\cap DE$ , $BC\cap EF$ and $CD\cap FA$ are collinearly.

Pascal's permutation theorem. Prove that if a sextuple of points is Pascalian, then each permutation of this sextuple is Pascalian.

This is, of course, trivial using the fact that a sextuple of points is Pascalian if and only if the six points lie on one conic. But how about a proof

without using conics? There is one, and I find it quite nice. I first prove the following lemma, which is, I believe, called "Carnot's theorem":

Lemma. Let $ABC$ be a triangle, let $D_1$ and $D_2$ be two points on its sideline $BC$ , let $E_1$ and $E_2$ be two points on its sideline $CA$ , and let $F_1$ and $F_2$ be two points

on its sideline $AB$ . Then, the points $E_1F_2\cap BC$ , $F_1D_2\cap CA$ and $D_1E_2\cap AB$ are collinear iff $\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}=1$ .

Proof of Lemma. Let $X=E_1F_2\cap BC$ , $Y=F_1D_2\cap CA$ and $Z=D_1E_2\cap AB$ . Then, we have to show that the points $X$ , $Y$ , $Z$ are collinearly iff

$\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}=1$ . But since the points $X$ , $Y$ , $Z$ lie on the sidelines $BC$ , $CA$ , $AB$ of triangle $ABC$ , the Menelaos' theorem

yields that they are actually collinear iff $\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$ . So all we have to do is to prove that the equations $\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$ and

$\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}=1$ are equivalent. Indeed, this will be trivial once it will be shown that we always have the identity

$\left(\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}\right)\cdot\left(\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}\right)=-1$ . Thus, in order to prove Lemma, it will be enough to establish this identity. Well,

since the points $E_1$ , $F_2$ and $X$ on the sidelines of triangle $ABC$ are collinear, the Menelaos' theorem yields $\frac{BX}{XC}\cdot\frac{CE_1}{E_1A}\cdot\frac{AF_2}{F_2B}=-1$ . Similarly, the collinearity of

the points $F_1$ , $D_2$ and $Y$ yields $\frac{CY}{YA}\cdot\frac{AF_1}{F_1B}\cdot\frac{BD_2}{D_2C}=-1$ , and the collinearity of the points $D_1$ , $E_2$ and Z yields $\frac{AZ}{ZB}\cdot\frac{BD_1}{D_1C}\cdot\frac{CE_2}{E_2A}=-1$ . Multiplying these

three equations, we get $\left(\frac{BX}{XC}\cdot\frac{CE_1}{E_1A}\cdot\frac{AF_2}{F_2B}\right)\cdot\left(\frac{CY}{YA}\cdot\frac{AF_1}{F_1B}\cdot\frac{BD_2}{D_2C}\right)\cdot\left(\frac{AZ}{ZB}\cdot\frac{BD_1}{D_1C}\cdot\frac{CE_2}{E_2A}\right)=\left(-1\right)^3$ or equivalently,

$\left(\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}\right)\cdot\left(\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}\right)=-1$ . But it is exactly required identity and thus the proof of Lemma is complete.

Now, in order to solve the initial problem, I consider a Pascalian sextuple of points $(A, B, C, D, E, F)$ in the same plane. The problem requires me to prove that every

permutation of this sextuple is Pascalian, but, noting - like you did - that any permutation can be written as a composition of transpositions of adjacent elements, I see that it is

enough to show that every sextuple resulting from the sextuple $(A, B, C, D, E, F)$ upon a transposition of adjacent elements is Pascalian. Since it doesn't make any

difference whether we transpose $A$ with $B$ , or $B$ with $C$ , or $C$ with $D$ , or $D$ with $E$ , or $E$ with $F$ (in fact, the definition of a Pascalian sextuple is cyclically invariant),

we will restrict ourselves to considering only the transposition of $B$ with $C$ . In other words, we have to show that the sextuple of points $(A, C, B, D, E, F)$ is Pascalian.

But this is just a matter of applying Lemma : Let $U\in DE\cap FA$ , $V\in FA\cap BC$ and $W=\in \cap DE$ . Then, the points $B$ and $C$ lie on the sideline $VW$ of triangle

$UVW$ , the points $D$ and $E$ lie on its sideline $WU$ and the points $F$ and $A$ lie on its sideline $UV$ . Since the sextuple of points $(A, B, C, D, E, F)$ is Pascalian, the

points $AB\cap DE$ , $BC\cap EF$ and $CD\cap FA$ are collinearly; in other words, the points $AB\cap WU$ , $EF\cap VW$ and $CD\cap UV$ are collinearly. Thus, after

Lemma, we have $\frac{VC}{CW}\cdot\frac{VB}{BW}\cdot\frac{BE}{EU}\cdot\frac{BD}{DU}\cdot\frac{UA}{AV}\cdot\frac{UF}{FV}=1$ . But this can be rewritten in the form $\frac{VB}{BW}\cdot\frac{VC}{CW}\cdot\frac{BE}{EU}\cdot\frac{BD}{DU}\cdot\frac{UA}{AV}\cdot\frac{UF}{FV}=1$ . And thus, using

Lemma again, we see that the points $AC\cap WU$ , $EF\cap VW$ and $BD\cap UV$ are collinearly. In other words, the points $AC\cap DE$ , $CB\cap EF$ and $BD\cap FA$

are collinearly. This yields that the sextuple of points $(A, C, B, D, E, F)$ is Pascalian. And this completes the solution of the problem.

PP1. Let $ABC$ be a triangle with the circumcircle $w$ . Let $P$ be an interior point of the $\triangle ABC$ and denote $\left\| \begin{array}{ccc} X\in AP\cap w \\ Y\in BP\cap w \\ Z\in CP\cap w\end{array}\right\|$ .

Also denote 6 points such as $\left\|\begin{array}{ccc} A_1\in BC\cap ZX,\ A_2\in BC\cap XY \\ B_1\in CA\cap XY,\ B_2\in CA\cap YZ \\ C_1\in AB\cap YZ,\ C_2\in AB\cap ZX\end{array}\right\|$ . Prove that $A_1B_2\cap B_1C_2\cap C_1A_2=\{P\}$ .

Proof. Apply the Pascal's theorem to the following cyclical quadrilaterals :

$\left\{\begin{array}{cccc} ABCZYX\ : & \left|\begin{array}{c} C_1\in AB\cap ZY\\\ A_2\in BC\cap YX\\\ P\in CZ\cap XA\end{array}\right| & \implies & P\in C_1A_2\\\\ BCAXZY\ : & \left|\begin{array}{c} A_1\in BC\cap XZ\\\ B_2\in CA\cap ZY\\\ P\in AX\cap YB\end{array}\right| & \implies & P\in A_1B_2\\\\ CABYXZ\ : & \left|\begin{array}{c} B_1\in CA\cap YX\\\ C_2\in AB\cap XZ\\\ P\in BY\cap ZC\end{array}\right| & \implies & P\in B_1C_2\end{array}\right\|$ $\implies P\in A_1B_2\cap B_1C_2\cap C_1A_2$ .

PP2. Let $ABC$ be a triangle. Consider two points $B'\in (CA)$ and $C'\in (AB)$ . For a point $D$ from the

plane of $\triangle ABC$ denote $M\in DC'\cap BB'$ and $N\in DB'\cap CC'$ . Prove that $AD\cap BN\cap CM\ne\emptyset$ .

Proof. Apply the Pappus' theorem to the lines $\overline {AC'B}$ and $\overline {NB'D}\ :\ \left\{\begin{array}{c} X\in BN\cap AD\\\\ C\in AB'\cap C'N\\\\ M\in BB'\cap C'D\end{array}\right|\ \implies\ X\in MC$ .

This post has been edited 23 times. Last edited by Virgil Nicula, Nov 22, 2015, 4:41 PM

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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).

Submit

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by mathMagicOPS, Jan 9, 2025, 3:40 AM
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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
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

203. An application of the Pascal's permutation theorem.

by Virgil Nicula, Jan 1, 2011, 8:01 AM

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