397. Two similar trigonometrical identities.

by Virgil Nicula, Aug 9, 2014, 1:42 PM

PP. Prove the following identities in a triangle $ABC$ (standard notations):

$1\blacktriangleright\ \left(b^2+c^2\right)\sin 2A+\left(c^2+a^2\right)\sin 2B+\left(a^2+b^2\right)\sin 2C=12S$ , where $S$ is the area of $\triangle ABC$ .

$2\blacktriangleright\ a^3\cos (B-C)+b^3\cos (C-A)+c^3\cos (A-B)=3abc$ .

Proof. I"ll use the evident identities $\sum x\left(y^2+z^2\right)=\sum x^2(y+z)$ and $\boxed{\sum x(y+z)(y+z-x)=6xyz}\ (*)$ .

$1\blacktriangleright\ \sum\left(b^2+c^2\right)\sin 2A=\sum\left(b^2+c^2\right)\cdot 2\sin A\cdot\cos A=$ $\sum\left(b^2+c^2\right)\cdot\frac aR\cdot\frac {b^2+c^2-a^2}{2bc}=$ $\frac 1{2Rabc}\cdot\sum a^2$ $\left(b^2+c^2\right)\left(b^2+c^2-a^2\right)\ \stackrel{(*)}{=}\ \frac {6a^2b^2c^2}{2Rabc}=$

$\frac {3abc}{R}=\frac {12RS}{R}\implies$ $\boxed{\sum\left(b^2+c^2\right)\sin 2A=12S}$ . Remark. $a^2\sin 2B+b^2\sin 2A=2a^2\sin B\cos B+2b^2\sin A\cos A=$ $2ab\sin A\cos B+2ba\sin B\cos A=$

$2ab(\sin A\cos B+\sin B\cos A)=$ $2ab\sin (A+B)=$ $2ab\sin C=4S\implies$ $\boxed{a^2\sin 2B+b^2\sin 2A=4S}$ . Observe that $12S=\sum\left(b^2+c^2\right)\sin 2A\ge$

$\sum 2bc\cdot 2\sin A\cos A=\sum 4bc\sin A\cdot\cos A=8S\sum\cos A\implies$ $\boxed{\cos A\le\frac 32}$ .

$2\blacktriangleright\ \sum a^3\cos (B-C)=$ $\sum a^2\cdot 2R\sin A\cdot \cos (B-C)=$ $\sum a^2\cdot 2R\sin (B+C)\cos (B-C)=$ $R\cdot\sum a^2(\sin 2B+\sin 2C)=$ $R\cdot\sum \left(b^2+c^2\right)\sin 2A\stackrel{(1)}{=}$

$R\cdot 12S=3abc\implies$ $\boxed{\sum a^3\cos (B-C)=3abc}$ , i.e. the identity $(2)$ . Otherwise. $\frac {a^2\cos (B-C)}{bc}=$ $\frac {\sin^2A\cos (B-C)}{\sin B\sin C}=$ $\frac {2\sin A\sin (B+C)\cos(B-C)}{2\sin B\sin C}=$

$\frac {\sin A(\sin 2B+\sin 2C)}{2\sin B\sin C}=$ $\frac {\sin A(\sin B\cos B+\sin C\cos C)}{\sin B\sin C}=$ $\frac {\sin A}{\sin C}\cdot \cos B+\frac {\sin A}{\sin B}\cdot \cos C=$ $\frac ac\cdot\frac {a^2+c^2-b^2}{2ac}+\frac ab\cdot\frac {a^2+b^2-c^2}{2ab}=$ $\frac {a^2+c^2-b^2}{2c^2}+\frac {a^2+b^2-c^2}{2b^2}=$

$1+\frac {a^2-b^2}{c^2}+\frac {a^2-c^2}{b^2}\implies$ $\sum \frac {a^2\cos (B-C)}{bc}=$ $\sum\left(1+\frac {a^2-b^2}{c^2}+\frac {a^2-c^2}{b^2}\right)=$ $3+\left(\frac {a^2-b^2}{c^2}+\frac {a^2-c^2}{b^2}\right)+\left(\frac {b^2-c^2}{a^2}+\frac {b^2-a^2}{c^2}\right)+$

$\left(\frac {c^2-a^2}{b^2}+\frac {c^2-b^2}{a^2}\right)=3\implies$ $\sum a^3\cos (B-C)=3abc$ .

PP1. Prove that the linear-angled inequality $\boxed{\sum a^2\cos (B-C)\le ab+bc+ca}$ in $\triangle ABC$ (standard notations).

Proof. $\sum a^2\cos (B-C)=\sum a\cdot 2R\sin (B+C)\cos (B-C)=$ $R\cdot\sum a(\sin 2B+\sin 2C)=$ $R\cdot \sum a(2\sin B\cos B+2\sin C\cos C)=$ $\sum a(b\cos B+c\cos C)=$

$\sum a\left(b\cdot\frac {a^2+c^2-b^2}{2ac}+c\cdot\frac {a^2+b^2-c^2}{2ab}\right)=$ $\sum \left[\frac {bc}2+\frac {b\left(a^2-b^2\right)}{2c}+\frac {bc}{2}+\frac {c\left(a^2-c^2\right)}{2b}\right]=$ $\sum\left[bc-\frac {(b+c)(b-c)^2}{2a}\right]\le \sum bc\implies$ $\sum a^2\cos (B-C)\le \sum bc$ .

PP2. Prove that $x^3-x^2-2x+1=0\implies\begin{array}{ccccc} \nearrow & x_1 & = & 2\cos\frac {\pi}7 & \searrow\\\\ \rightarrow & x_2 & = & 2\cos\frac {3\pi}7 & \rightarrow\\\\ \searrow & x_3 & = & 2\cos\frac {5\pi}7 & \nearrow\end{array}\odot$ .

Proof.

$\blacktriangleright\ s_1=x_1+x_2+x_3=$ $2\left(\cos\frac {\pi}7+\cos\frac {3\pi}7+\cos\frac {5\pi}7\right)=$ $\frac {2\cos\frac {3\pi}7\sin\frac {3\pi}7}{\sin\frac {\pi}7}=\frac {\sin \frac {6\pi}7}{\sin\frac {\pi}7}=1\implies \boxed{\ s_1=1\ }\ (*)$ .

$\blacktriangleright\ s_2=x_1x_2+x_2x_3+x_3x_1=$ $2\left(2\cos\frac {\pi}7\cos\frac {3\pi}7+2\cos\frac {3\pi}7\cos\frac {5\pi}7+2\cos\frac {5\pi}7\cos\frac {\pi}7\right)=$ $2\left(\cos\frac {4\pi}7+\cos\frac {2\pi}7+\cos\frac {8\pi}7+\cos\frac {2\pi}7+\cos\frac {6\pi}7+\cos\frac {4\pi}7\right)=$

$4\left[\cos\frac {2\pi}7-\left(\cos\frac {3\pi}7+\cos\frac {\pi}7\right)\right]\ \stackrel{(*)}{=}$ $4\left[\cos\frac {2\pi}7-\left(\frac 12-\cos \frac {5\pi}7\right)\right]=4\left(\cos\frac {2\pi}7+\cos \frac {5\pi}7\right)-2=-2\implies \boxed{\ s_2=-2\ }$ .

$\blacktriangleright\ s_3=x_1x_2x_3=8\cos\frac {\pi}7\cos\frac {3\pi}7\cos\frac {5\pi}7=$ $4\cos\frac {\pi}7\left(\cos\frac {8\pi}7+\cos\frac {2\pi}7\right)=$ $4\cos\frac {\pi}7\left(\cos\frac {2\pi}7-\cos\frac {\pi}7\right)=$ $2\left[\left(\cos \frac {3\pi}7+\cos \frac {\pi}7\right)-\left(1+\cos \frac {2\pi}7\right)\right]\ \stackrel{(*)}{=}$

$2\left[\left(\frac 12-\cos \frac {5\pi}7\right)-\left(1+\cos \frac {2\pi}7\right)\right]=$ $2\left[-\frac 12-\left(\cos \frac {5\pi}7+\cos \frac {2\pi}7\right)\right]=-1\implies \boxed{\ s_3=-1\ }$ . The equation is $x^3-s_1x^2+s_2x-s_3=0$ , i.e. $x^3-x^2-2x+1=0$ .

Remark. $\cos\frac {\pi}7+\cos\frac {6\pi}7=$ $\cos\frac {2\pi}7+\cos\frac {5\pi}7=$ $\cos\frac {3\pi}7+\cos\frac {4\pi}7=0$ .

PP3. Sa se demonstreze prin inductie relatia $\boxed{\sum_{k=1}^n k\cdot\sin kx=\frac{(n+1)\sin nx-n\sin(n+1)x}{4\sin^2\frac x2}\ ,\ x\ne 2t\pi\ ,\ t\in\mathbb{Z}}\ .$

Proof 1 (inductie).

Etapa 1. Pentru $n=1$ egalitatea devine $\sin x=\frac{2\sin x-\sin 2x}{4\sin^2\frac x2}=\frac{2\sin x(1-\cos x)}{4\sin^2\frac x2}$ $\Longleftrightarrow\ \sin^2\frac x2=\frac{1-\cos x}{2}\ ,$ adevarat.

Etapa 2. Presupunem egalitatea adevarata pentru $n$ numere si o vom demonstra pentru $n+1$ numere. Vom arata ca $\frac{(n+1)\sin nx-n\sin(n+1)x}{4\sin^2\frac x2}+(n+1)\sin(n+1)x=$

$\frac{(n+2)\sin(n+1)x-(n+1)\sin(n+2)x}{4\sin^2\frac x2}\ \stackrel{x\ne 2t\pi}{\iff}$ $\underline{\underline{(n+1)\sin nx}}-$ $\underline{\underline{\underline{n\sin(n+1)x}}}+(n+1)[\sin(n+1)x]\cdot 4\sin^2\frac x2=\underline{\underline{\underline{(n+2)\sin(n+1)x}}}-\underline{\underline{(n+1)\sin(n+2)x}}$ $\Longleftrightarrow$

$\underline{\underline{(n+1)}}[\sin nx+sin(nx+2x)]+\underline{\underline{(n+1)}}[\sin(n+1)x]\cdot4sin^2\frac x2=2\underline{\underline{(n+1)}}\sin(n+1)x$ $\Longleftrightarrow\ 2\underline{\underline{\sin(n+1)x}}\cdot\cos x+\underline{\underline{sin(n+1)x}}\cdot 4\sin^2\frac x2=2\underline{\underline{\sin(n+1)x}}\ .$

Daca $\sin(n+1)x=0$ atunci e evident. Altfel putem imparti prin $\sin(n+1)x\ :\ \cos+2\sin^2\frac x2=1\ \Longleftrightarrow\ sin^2\frac x2=\frac{1-\cos x}{2}\ ,\ \boxed{O.K}\ .$

Proof 2 (direct). Ne propunem sa gasim formula pentru suma $f(x)\equiv\sum_{k=1}^nk\cdot \sin kx$ din care vom obtine $\sum_{k=1}^nk\cdot \sin k=f(1)\ .$

Metoda I. Se calculeaza direct sau prin inductie suma $\boxed{\sum_{k=1}^n\sin kx=\frac {\sin\frac {(n+1)x}{2}\cos\frac {nx}{2}}{\sin \frac x2}}$ si se aplica identitatea $\sum _{k=1}k\cdot a_k=\sum _{s=1}^n\ \sum_{k=s}^na_k$ pentru $a_k=\sin kx\ ,\ k\in\overline{1,n}\ .$

Metoda II. Se calculeaza direct sau prin inductie suma $\overline{\underline{\left\|\ \sum_{k=1}^n\cos kx=\frac {\cos\frac {(n+1)x}{2}\cos\frac {nx}{2}}{\sin \frac x2}\ \right\|}}$ si prin derivare se obtine $\sum_{k=1}^nk\cdot \sin kx=-\left(\sum_{k=1}^n\cos kx\right)^{\prime}=\ \ldots\ \ldots\ .$

Metoda III (numere complexe). Notam $z=cos x+i\cdot\sin x\ .$ Atunci $z^k=\cos kx+i\cdot\sin kx\ ,$ pentru orice $k\in \overline {1,n}\ .$ Notam suma $S_n\equiv\sum_{k=1}^nk\sin kx\in \mathbb R$ si suma $C_n\equiv\sum_{k=1}^nk\cos kx\in\mathbb R\ .$ Asadar $C_n+i\cdot S_n=\sum_{k=1}^nkz^k$ care se calculeaza prin unul din procedeele mentionate mai sus dupa care se identifica partile reale/imaginare.

Comentariu. Formulata astfel, problema initiala nu este interesanta deoarece ni se da rezultatul si nu apare "ipoteza inductiei" ca un act creator, devenind doar un exercitiu de rutina, plicticos. Ceea ce ma surprinde in mod neplacut este faptul ca numeroase exercitii care apar la capitolul "Inductia matematica" sunt de aceasta forma sau eventual se cere "cat face suma ... " ca apoi sa se solicite "verificarea prin inductie", ceea ce este la fel de "nociv" pentru a intelege profund aceasta metoda fecunda a inductiei matematice. Dupa parerea mea ar fi trebuit sa se solicite calcularea sumei respective, ramanand la alegerea elevului procedeul, eventual si metoda inductiei complete. Prin acest "discurs" nu ma adresez "problemistilor", ci indeosebi profesorilor de matematica (in actul predarii).

This post has been edited 29 times. Last edited by Virgil Nicula, May 30, 2016, 4:56 PM

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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
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blog
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the visits tho
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long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
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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

397. Two similar trigonometrical identities.

by Virgil Nicula, Aug 9, 2014, 1:42 PM

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