324. Some determinants.

by Virgil Nicula, Oct 26, 2011, 8:11 AM

PP1. Prove that $\Delta\equiv \left|\begin{array}{ccc} (a+x)^2 & (a+y)^2 & (a+z)^2\\\\ (b+x)^2 & (b+y)^2 & (b+z)^2\\\\ (c+x)^2 & (c+y)^2 & (c+z)^2\end{array}\right|=-2\cdot \left|\begin{array}{ccc} 1 & 1 & 1\\\\ x & y & z\\\\ x^2 & y^2 & z^2\end{array}\right|\cdot\left|\begin{array}{ccc} 1 & 1 & 1\\\\ a & b & c\\\\ a^2 & b^2 & c^2\end{array}\right|$ .

Proof. Denote $V_0=\left(\begin{array}{c} 1\\\\ 1\\\\ 1\end{array}\right)\ ,\ V_1=\left(\begin{array}{c} a\\\\ b\\\\ c\end{array}\right)\ ,\ V_2=\left(\begin{array}{c} a^2\\\\ b^2\\\\ c^2\end{array}\right)$ and $V(a,b,c)\equiv\left|\begin{array}{ccc} 1 & 1 & 1\\\\ a & b & c\\\\ a^2 & b^2 & c^2\end{array}\right|=(c-a)(c-b)(b-a)$ .

Therefore, $\Delta=\left|\begin{array}{ccc} a^2+2xa+x^2 & a^2+2ya+y^2 & a^2+2za+z^2\\\\ b^2+2xb+x^2 & b^2+2yb+y^2 & b^2+2zb+z^2\\\\ c^2+2xc+x^2 & c^2+2yc+y^2 & c^2+2zc+z^2\end{array}\right|=$

$\left|\begin{array}{ccc} x^2\cdot V_0+2x\cdot V_1+V_2 & y^2\cdot V_0+2y\cdot V_1+V_2 & z^2\cdot V_0+2z\cdot V_1+V_2\end{array}\right|=$ $\left|\begin{array}{ccc} V_2 & 2y\cdot V_1 & z^2\cdot V_0\end{array}\right|+$ $\left|\begin{array}{ccc} V_2 & y^2\cdot V_0 & 2z\cdot V_1\end{array}\right|+$

$\left|\begin{array}{ccc} 2x\cdot V_1 & V_2 & z^2\cdot V_0\end{array}\right|+$ $\left|\begin{array}{ccc} 2x\cdot V_1 & y^2\cdot V_0 & V_2\end{array}\right|+$ $\left|\begin{array}{ccc} x^2\cdot V_0 & V_2 & 2z\cdot V_1\end{array}\right|+$ $\left|\begin{array}{ccc} x^2\cdot V_0 & 2y\cdot V_1 & V_2\end{array}\right|=$

$2yz^2\cdot \left|V_2\ V_1\ V_0\right| +2zy^2\cdot\left|V_2\ V_0\ V_1\right|+$ $2xz^2\cdot\left|V_1\ V_2\ V_0\right|+2xy^2\cdot\left|V_1\ V_0\ V_2\right|+$ $2zx^2\cdot\left|V_0\ V_2\ V_1\right|+2yx^2\cdot\left|V_0\ V_1\ V_2\right|=$

$2\cdot \left[(y-z)\cdot x^2-\left(y^2-z^2\right)\cdot x+yz(y-z)\right]\cdot\left|V_0\ V_1\ V_2\right|=$ $2(y-z)(x-y)(x-z)\cdot V(a,b,c) =$ $-2\cdot V(x,y,z)\cdot V(a,b,c)$ .


PP2. Prove that $\left|\begin{array}{ccc} x & y & z\\\\ x^2 & y^2 & z^2\\\\ yz & zx & xy\end{array}\right|=\left|\begin{array}{ccc} 1 & 1 & 1\\\\ x^2 & y^2 & z^2\\\\ x^3 & y^3 & z^3\end{array}\right|$ and $\left|\begin{array}{ccc} x+a & x+b & x+c\\\\ x^2+a^2 & x^2+b^2 & x^2+c^2\\\\ x^3+a^3 & x^3+b^3 & x^3+c^3\end{array}\right|=f(x)\cdot V(a,b,c)$

where $V(a,b,c)= \left|\begin{array}{ccc} 1 & 1 & 1\\\\ a & b & c\\\\ a^2 & b^2 & c^2\end{array}\right|$ and $f(x)=(x-a)(x-b)(x-c)+2abc$ .

Proof. Denote $D=\left|\begin{array}{ccc} x & y & z\\\\ x^2 & y^2 & z^2\\\\ yz & zx & xy\end{array}\right|$ . Multiply the columns $1$ , $2$ , $3$ cu $x$ , $y$ , $z$ respectively and divide implicitly the determinant

by $xyz\ :\ D=\frac {1}{xyz}\cdot \left|\begin{array}{ccc} x^2 & y^2 & z^2\\\\ x^3 & y^3 & z^3\\\\ xyz & xyz & xyz\end{array}\right|$ . From the third line appears the factor $xyz\ :\ D= \left|\begin{array}{ccc} x^2 & y^2 & z^2\\\\ x^3 & y^3 & z^3\\\\ 1 & 1 & 1\end{array}\right|$ .

Efectuate a permutation with the lines $(\ 1\rightarrow 2\ ;\ 2\rightarrow 3\ ;\ 3\rightarrow 1\ )$ and obtain the proposed identity.

It is well-known that the Vandermonde determinant $V(a,b,c)$ is equal to $(c-a)(c-b)(b-a)$ . Define the column-matrices $X=\left(\begin{array}{c} 1\\\\ x\\\\ x^2\end{array}\right)$ , $ A=\left(\begin{array}{c} 1\\\\ a\\\\ a^2\end{array}\right)$ , $ B=\left(\begin{array}{c} 1\\\\ b\\\\ b^2\end{array}\right)$ , $C=\left(\begin{array}{c} 1\\\\ c\\\\ c^2\end{array}\right)$ . Observe that $D=\left|xX+aA\ \ xX+bB\ \ xX+cC\right|=$ $|xX\ bB\ cC|+$ $|aA\ xX\ cC|+$ $|aA\ bB\ xX|+$ $|aA\ bB\ cC|=$

$xbc\cdot V(x,b,c)+$ $axc\cdot V(a,x,c)+$ $abx\cdot V(a,b,x)+$ $abc\cdot V(a,b,c)=$ $bcx\cdot V(b,c,x)+$ $cax\cdot V(c,a,x)+$ $abx\cdot V(a,b,x)+$ $abc\cdot V(a,b,c)=$

$x\cdot\sum cb(c-b)(x-b)(x-c)+$ $abcV(a,b,c)=$ $h(x)+abc\cdot V(a,b,c)$ , where $h(x)=x\cdot\sum cb(c-b)(x-b)(x-c)$ . Observe that $h(0)=0$

and $h(a)=h(b)=h(c)=$ $abc\cdot V(a,b,c)$ . In conclusion, $h(x)=\left[(x-a)(x-b)(x-c)+abc\right]\cdot V(a,b,c)$ , i.e. $D=f(x)\cdot V(a,b,c)$ .

PP3. Let $\{a,b,c\}$ be the roots of the equation $x^3-s_1x^2+s_2x-s_3=0$ . Ascertain $\Delta_1\left(s_1,s_2,s_3\right)=\left|\begin{array}{ccc} a & b & c\\\\ b & c & a\\\\ c & a & b\end{array}\right|^2$

and $\Delta_2\left(s_1,s_2,s_3\right)=\left|\begin{array}{ccc} 1 & 1 & 1\\\\ a & b & c\\\\ a^2 & b^2 & c^2\end{array}\right|^2$ . Prove that $\{a,b,c\}\subset\mathbb R$ and $s_1=0\implies$ $4s_2^3+27s_3^2\le 0$ .

With other words, the equation $x^3+mx+n=0$ has real roots if and only if $4m^3+27n^2\le 0$ .

Proof. From the Viete's relations obtain that $\left\{\begin{array}{ccccc} s_1 & = & a+b+c\\\\ s_2 & = & ab+bc+ca\\\\ s_3 & = & abc \end{array}\right\|$ and $\left\{\begin{array}{ccccc} S_2 & = & a^2+b^2+c^2 & = & s_1^2-2s_2\\\\ S_3 & = & a^3+b^3+c^3 & = & s_1^3-3s_1s_2+3s_3\\\\ S_4 & = & a^4+b^4+c^4 & = & s_1^4-4s_1^2s_2+4s_1s_3+2s_2^2\end{array}\right\|$ .

Therefore, $\Delta_1=\left|\begin{array}{ccc} a & b & c\\\\ b & c & a\\\\ c & a & b\end{array}\right|\cdot\left|\begin{array}{ccc} a & b & c\\\\ b & c & a\\\\ c & a & b\end{array}\right|=$ $\left(a^3+b^3+c^3-3abc\right)^2=$ $\left|\begin{array}{ccc} S_2 & s_2 & s_2\\\\ s_2 & S_2 & s_2\\\\ s_2 & s_2 & S_2\end{array}\right|=\left(S_2+2s_2\right)\left(S_2-s_2\right)^2=$ $\left[s_1(s_1^2-3s_2)\right]^2$ .

$\Delta_2=\left|\begin{array}{ccc} 1 & 1 & 1\\\\ a & b & c\\\\ a^2 & b^2 & c^2\end{array}\right|^2=$ $\left|\begin{array}{ccc} 1 & 1 & 1\\\\ a & b & c\\\\ a^2 & b^2 & c^2\end{array}\right|\cdot \left|\begin{array}{ccc} 1 & a & a^2\\\\ 1 & b & b^2\\\\ 1 & c & c^2\end{array}\right|=$ $\left|\begin{array}{ccc} 3 & s_1 & S_2\\\\ s_1 & S_2 & S_3\\\\ S_2 & S_3 & S_4\end{array}\right|=$ $\left|\begin{array}{ccc} 3 & 0 & -2s_2\\\\ 0 & -2s_2 & 3s_3\\\\ -2s_2 & 3s_3 & 2s_2^2\end{array}\right|=-\left(4s_2^3+27s_3^2\right)\ge 0$ .
This post has been edited 62 times. Last edited by Virgil Nicula, Nov 19, 2015, 8:19 PM

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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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    by MrMustache, Nov 11, 2020, 4:52 PM

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    by piphi, Jun 10, 2020, 11:44 PM

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  • Joined: Jun 22, 2005
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  • Total entries: 456
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