JBMO 2011

mircea.lascu
Mesaje: 350
Membru din: Lun Iul 12, 2010 9:02 pm

JBMO 2011

Mesaj de mircea.lascu »

Problem 1. Let $a,b$ and $c$ be positive real numbers such that $abc=1$. Prove the inequality

$(a^{5}+a^{4}+a^{3}+a^{2}+a+1)\cdot$$(b^{5}+b^{4}+b^{3}+b^{2}+b+1)\cdot$$(c^{5}+c^{4}+c^{3}+c^{2}+c+1)\geq 8(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1).$

Problem 2. Find all the prime numbers $p$ for which there exist positive integers $x$ and $y$ that satisfy the equation

$x(y^{2}-p)+ y(x^{2}-p)=5p.$

Problem 3. Let $n>3$ be a positive integer. An equilateral triangle $ABC$ is divided into $n^{2}$ identical equilateral "small" triangles using lines parallel to its sides. The figure below illustrates the case $n=4$. Let $m$ be the number of rhombi consisting of 2 "small" triangles. Let $d$ be the number of rhombi consisting of 8 "small" triangles. Find the difference $m-d$ in terms of $n$.
Figure 1
Figure 1
fig1.png (20.35 KiB) Vizualizat de 4320 ori
Problem 4. Let $ABCD$ be a convex quadrilateral. Let $E$ and $F$ be points on the sides $AB$ and $CD$, respectively, such that $AB:AE=CD:DF=n$. If $S$ is the area of the quadrilateral $AEFD$, show that

$S\leq\displaystyle\frac{AB\cdot CD+n(n-1)DA^{2}+n\cdot DA\cdot BC}{2n^{2}}.$
Ultima oară modificat Joi Iun 23, 2011 5:49 am de către mircea.lascu, modificat de 4 ori în total.
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Mr. Ady
Mesaje: 307
Membru din: Mie Dec 08, 2010 10:55 pm
Localitate: Targoviste

Re: JBMO 2011

Mesaj de Mr. Ady »

1 si 2 imi par foarte cunoscute
Catană Adrian,
Elev la CNIV, Targoviste, clasa a X-a
Ciprian.
Mesaje: 155
Membru din: Vin Mar 11, 2011 11:13 am

Re: JBMO 2011

Mesaj de Ciprian. »

La prima problema folosim doar un mic artificiu si inegalitatea lui holder:$\prod(a^5+a^4+a^3+a^2+a^1+1)=$$\prod(a^3+1)(a^2+a+1)\ge(\sqrt[3]{(abc)^3}+1)^3*\prod(a^2+a+1)=$$8\prod(a^2+a+1)$.
Ultima oară modificat Joi Iun 23, 2011 9:35 am de către Ciprian., modificat 1 dată în total.
Ciprian.
Mesaje: 155
Membru din: Vin Mar 11, 2011 11:13 am

Re: JBMO 2011

Mesaj de Ciprian. »

La a2a restrangem astfel $(x+y)(xy-p)=5p$si apoi e simplu.
aecksteinul
Mesaje: 45
Membru din: Dum Iun 26, 2011 9:01 pm

Re: JBMO 2011

Mesaj de aecksteinul »

Official results:

Gold medalists:
1. BUL3 10+10+10+10=40
1. BUL6 10+10+10+10=40
1. ROM4 10+10+10+10=40 (+special prize for original solution)
4. ROM5 10+10+9+10=39
5. SRB3 10+10+10+6=36
6. ROM3 10+10+10+5=35
6. SRB5 10+10+10+5=35
Silver medalists:
8. BUL2 10+10+10+2=32
8. MCD1 7+10+10+5=32
10. BUL1 10+10+10+1=31
10. MLD3 10+10+10+1=31
10. MLD6 10+10+10+1=31
10. ROM2 10+8+8+5=31
10. SRB2 10+10+10+1=31
15. BIH3 10+10+10+0=30
15. BIH6 10+10+10+0=30
15. BUL4 10+10+10+0=30
15. BUL5 10+10+10+0=30
15. CYP6 10+10+10+0=30
15. MCD4 10+10+10+0=30
15. GRE5 10+10+10+0=30
15. ROM1 10+7+8+5=30
15. SRB4 10+10+10+0=30
15.SRB6 10+10+10+0=30
Bronze medalists:
25. ROM6 10+10+8+0=28
26. MLD5 10+9+7+0=26
27. GRE1 10+5+10+0=25
27. MNE1 9+10+5+1=25
29. GRE3 10+3+10+1=24
30. BIH2 9+10+3+0=22
30. MCD3 10+9+3+0=22
32. MLD4 10+7+4+2=21
33. MLD2 10+7+3+0=20
34. CYP1 5+3+10+1=19
34. MCD5 1+8+10+0=19
36. GRE4 3+5+10+0=18
37. CYP3 10+4+3+0=17
the rest of the participants obtained:
38. BIH4 10+4+1+0=15, honorable mention
39. BIH5 9+5+0+0=14
39. CYP2 5+3+5+1=14
39. MCD2 7+1+5+1=14
39. MCD6 4+0+10+0=14, honorable mention
43. GRE2 10+1+2+0=13, honorable mention
43. GRE6 10+3+0+0=13, honorable mention
45. CYP4 5+3+4+0=12
46. CYP5 5+3+1+0=9
47. MLD1 1+2+3+0=6
48. MNE1 1+0+0+0=1
48. MNE3 1+0+0+0=1
48. MNE4 0+0+1+0=1
51. BIH1 0+0+0+0=0
51. SRB1 0+0+0+0=0 (absent)

Rezults for invited teams:
Gold:
TJK1 10+10+10+6=36
KAZ1 10+10+10+6=36
Silver:
KAZ3 10+10+10+1=31
AZE1 10+10+10+0=30
TJK4 9+10+10+1=30
KAZ5 10+10+10+0=30
Bronze:
KYP2 7+7+10+1=25
TJK2 10+10+0+2=22
TJK3 9+10+2+0=21
KAZ2 1+10+10+0=21
AZE2 9+10+0+1=20
TJK5 6+6+7+0=19
TJK6 9+10+0+0=19
AZE3 10+7+0+0=17
KYP4 6+7+4+0=17
KAZ6 1+6+10+0=17
other participants:
KYP3 5+3+3+0=11
KYP6 1+4+0+0=5
KYP5 1+1+1+0=3
KYP1 1+1+0+0=2
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