354. Gradul II - Univ. B.B. Cluj, aug. 1998 Prof I sau II.

by Virgil Nicula, Aug 18, 2012, 2:42 PM

Profesori I.

PPI1. Fie $R$ , $r$ raza sferei circumscrise si raza sferei inscrise pentru o piramida patrulatera regulata $VABCD$ . Sa se arate ca $\frac Rr\ge 1+\sqrt 2$ ,

Demonstratie. Notam centrul $O$ al patratului $ABCD$ , mijlocul $M$ al muchiei $[AB]$ si $m\left(\widehat{VMO}\right)=2\phi$ . Presupunem fara a restrange generalitatea ca $AB=2$ . Se arata usor

ca $OA=\sqrt 2$ , $OM=1$ si din triunghiul $VMO$ obtinem $VO=\tan 2\phi$ si $\boxed{r=\tan\phi}$ . Notam centrul $S$ al sferei circumscrise. Se observa ca $SA^2=SO^2+AO^2$ , adica

$R^2=(\tan 2\phi -R)^2+2\iff$ $\boxed{R=\frac {2+\tan^22\phi}{2\tan 2\phi}}$ . Notam $\boxed{\tan\phi=t\in (0,1)}$ . In concluzie, $\frac Rr\ge 1+\sqrt 2\iff$ $\frac {2+\tan^22\phi}{2\tan 2\phi\tan\phi}\ge 1+\sqrt 2\iff$ $\frac {2+\left(\frac {2t}{1-t^2}\right)^2}{2\cdot\frac {2t}{1-t^2}\cdot t}\ge 1+\sqrt 2$

$\iff$ $\frac {t^4+1}{2t^2\left(1-t^2\right)}\ge 1+\sqrt 2\iff$ $\frac {2t^2-2t^4}{t^4+1}\le \sqrt 2-1\iff$ $\left(\sqrt 2+1\right)t^4-2t^2+\left(\sqrt 2-1\right)\ge 0\iff$ $t^4-2\left(\sqrt 2-1\right)t^2+\left(\sqrt 2-1\right)^2\ge 0\iff$

$\left[t^2-\left(\sqrt 2-1\right)\right]^2\ge 0$ , ceea ce este adevarat. Avem egalitate daca si numai daca $\tan\phi=\sqrt {\sqrt 2-1}$ .

PPI2. Sa se rezolve ecuatiile $\left\{\begin{array}{c} f(x)=x^5-5x^4+3x^3+11x^2-6x+4=0\\\\ g(x)=x^5-5x^4+6x^3+2x^2-12x+8=0\end{array}\right\|$ stiind ca au cel putin doua radacini comune.

Demonstratie 1. $f(1)=g(1)=0$ si astfel $\left\{\begin{array}{ccc} f(x)=(x-1)u(x) & ; & u(x)=x^4-4x^3-x^2+10x+4\\\\ g(x)=(x-1)v(x) & ; & v(x)=x^4-4x^3+2x^2+4x-8\end{array}\right\|$ . Se observa ca $v(x)-u(x)=3(x^2-2x-4)$ .

Incercam (de ce ?!) daca $w(x)=x^2-2x-4$ este divizor comun pentru $u$ si $v$. Intr-adevar se confirma. Asadar, $\left\{\begin{array}{c} f(x)=(x-1)(x^2-2x-4)(x^2-2x-1)\\\\ g(x)=(x-1)(x^2-2x-4)(x^2-2x+2)\end{array}\right\|$ etc etc.

Demonstratie 2. C.m.m.d.c. $d=(u,v)$ al celor doua polinoame pentru care $\mathrm{gr}(d)\ge 1$ deoarece cele doua polinoame au cel putin o radacina comuna:

$\left\{\left|\begin{array}{ccccc} x^4 & -4x^3 & -x^2 & +10x & +4\\\\ x^4 & -4x^3 & +2x^2 & +4x & -8\end{array}\right|\begin{array}{cc} \odot & (-1)\\\\ \odot & (+1)\end{array}\right\|\bigoplus\implies$ $3x^2-6x-12\implies$ $x^2-2x-4\implies$ $\left\{\left|\begin{array}{ccccc} x^2 & -2x & -4 & &\\\\ x^4 & -4x^3 & +2x^2 & +4x & -8\end{array}\right|\begin{array}{cc} \odot & \left(+x^2\right)\\\\ \odot & (-1)\end{array}\right\|\bigoplus\implies$

$2x^3-6x^2-4x+8\implies$ $x^3-3x^2-2x+4\implies$ $\left\{\left|\begin{array}{cccc} x^3 & -3x^2 & -2x & +4\\\\ x^2 & -2x & -4 & \end{array}\right|\begin{array}{cc} \odot & (-1)\\\\ \odot & (+x)\end{array}\right\|\bigoplus\implies$ $x^2-2x-4\implies$ $\left\|\left|\begin{array}{ccc} x^2 & -2x & -4\\\\ x^2 & -2x & -4\end{array}\right|\begin{array}{cc} \odot & (-1)\\\\ \odot & (+1)\end{array}\right\|\bigoplus\implies 0$ .

In concluzie, c.m.m.d.c. $d=(u,v)$ al celor doua polinoame este $d(x)=x^2-2x-4$ si obtinem usor $\left\{\begin{array}{c} u(x)=(\underline{x^2-2x-4})(x^2-2x-1)\\\\ v(x)=(\underline{x^2-2x-4})({x^2-2x+2})\end{array}\right\|$ etc etc.


Profesori II.

PPII1. Sa se determine multimea numerelor (intregi) $z\in\mathbb Z$ pentru care $\frac {z^3-3z+2}{2z+1}\in\mathbb Z$ .

Demonstratie 1. Notam $2z+1=y\in\mathbb Z$ ($y$ este impar). Asadar, $\frac {z^3-3z+2}{2z+1}\in\mathbb Z\iff$ $\frac {y^3-3y^2-9y+27}{8y}\in\mathbb Z\implies$ $\frac {y^3-3y^2-9y+27}{y}\in\mathbb Z\iff$ $\frac {27}{y}\in\mathbb Z$ , unde $y$ este

impar $\iff$ $y=2z+1\in\left\{\pm 1,\pm 3,\pm 9,\pm 27\right\}\iff$ $\boxed{\ z\in B=\{0,-1,1,-2,4,-5,13,-14\}\ }$ . In concluzie, $A=\left\{\left|\frac {z^3-3z+2}{2z+1}\right| z\in B\right\}=\{2,-4,0,6,12,80,100\}$ .

Demonstratie 2. $\frac {z^3-3z+2}{2z+1}\in\mathbb Z\iff$ $\frac {2z^3-6z+4}{2z+1}\in\mathbb Z\iff$ $z^2-\frac {z^2+6z-4}{2z+1}\in\mathbb Z\iff$ $\frac {z^2+6z-4}{2z+1}\in\mathbb Z\iff$ $\frac {2z^2+12z-8}{2z+1}\in\mathbb Z\iff$ $z+\frac {11z-8}{2z+1}\in\mathbb Z\iff$

$\frac {11z-8}{2z+1}\in\mathbb Z\iff$ $\frac {22z-16}{2z+1}\in\mathbb Z\iff$ $11-\frac {27}{2z+1}\in\mathbb Z\iff$ $(2z+1)$ divide $27\iff$ $2z+1\in\left\{\pm 1,\pm 3,\pm 9,\pm 27\right\}\iff$ $\boxed{\ z\in B=\{0,-1,1,-2,4,-5,13,-14\}\ }$ .


PPII2. Fie $\triangle ABC$ cu $A>90^{\circ}$ . Sa se determine pozitiile punctelor $X\in (BC)$ pentru care $XA^2=XB\cdot XC$ . Fain !

Demonstratie. Daca $O$ este circumcentrul $\triangle ABC$ , atunci se arata usor ca pozitiile cautate sunt $\left\{X_1,X_2\right\}=(BC)\cap w$ , unde $w$ este cercul de diametru $[AO]$ .

Exprimarea metrica a acestor pozitii se obtine usor. Notam $\frac {XB}{XC}=m$ . Aplicam relatia Stewart $\ :\ AX^2=XB\cdot XC\iff$ $\left(c^2+b^2m\right)(m+1)=2a^2m\iff$

$m_{1,2}=\frac {2a^2-\left(b^2+c^2\right)\pm\sqrt {4a^4+\left(b^2-c^2\right)^2-4a^2\left(b^2+c^2\right)}}{2b^2}$ .
This post has been edited 41 times. Last edited by Virgil Nicula, Nov 17, 2015, 7:19 AM

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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 OlympusHero, Dec 30, 2020, 6:08 PM

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

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

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    -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

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    by OlympusHero, May 14, 2020, 8:43 PM

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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
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