BEDAH SOAL OLIMPIADE MATEMATIKA SMA (OSN/IMO) TERBARU

The Christmas Day and New Year Day for All’s day

(25th December 2014)
Next 13 points aren’t mine but I believe that all maker
Problems and solver problems are supported each others
1 this point almost can do it, but I want to use ln⁡x

Find the answer in R of (x√x)^x=(x)^(x√x)
Solution from(x√x)^x=(x)^(x√x),we have
ln⁡〖(x√x)^x 〗=ln⁡〖(x)^(x√x) 〗
x(ln⁡〖x√x〗 )=x√x ln⁡x
x(ln⁡x+ln⁡√x )=x√x ln⁡x
x ln⁡x+1/2 x ln⁡x=x√x ln⁡x
x ln⁡x+1/2 ln⁡x-x√x ln⁡x=0
ln⁡x (x+1/2 x-x√x)=0
Hence, ln⁡x=0 or x+1/2-x√x=0

ln⁡x=0↔x=1 and
x+1/2-x√x=0
→2x+x-2 x√x=0
→x(3-2√x)=0
→x=0,9/4
From rechecking solution, its answer is {1,9/4 }

2 . give x,y,z>0, Prove that
x^2/y^2 +y^2/z^2 +z^2/x^2 +8(xy+yz+zx)/(x^2+y^2+z^2 )≥11
Proof As 11- 8(xy+yz+zx)/(x^2+y^2+z^2 )≥3 and
x^2/y^2 +y^2/z^2 +z^2/x^2 ≥3
But, 0< (xy+yz+zx)/(x^2+y^2+z^2 )≤1,we have
-1≤ (-(xy+yz+zx))/(x^2+y^2+z^2 )<0
-8≤ (-8(xy+yz+zx))/(x^2+y^2+z^2 )<0
3≤ 11- 8(xy+yz+zx)/(x^2+y^2+z^2 )<11
That is, x^2/y^2 +y^2/z^2 +z^2/x^2 ∈[3,∞)┤ and

11- 8(xy+yz+zx)/(x^2+y^2+z^2 )∈[3,11)┤
Hence, we write x^2/y^2 +y^2/z^2 +z^2/x^2 ≥11-8(xy+yz+zx)/(x^2+y^2+z^2 )≥3
Therefore, x^2/y^2 +y^2/z^2 +z^2/x^2 +8(xy+yz+zx)/(x^2+y^2+z^2 )≥11 OK.

3. give a,b,c>0, prove that
(a^2+2)(b^2+2)(c^2+2)≥9(ab+bc+ca)
Proof As a^2+b^2+c^2≥ab+bc+ca
Hence, (a^2+2)(b^2+2)(c^2+2)≥
(ab+2)(bc+2)(ca+2)
We will show that
(ab+2)(bc+2)(ca+2)≥9(ab+bc+ca)
Now Seeing, as a+c+1/ac≥3, we have
2(a+c+1/ac)≥6 (good seeing)
And, ab+bc+c/b+a/b+1+2/ca≥7
And, ab+ab+ab+bc+bc+bc+c/b+2/ab+
+a/b+2/bc+1+2/ca≥15
And, (ab+2)+(ab+2)+(ab+2)+(bc+2)+
(bc+2)+(bc+2)+((ca+2)/ab)+((ca+2)/bc)+((ca+2)/ca)≥27
And, (3/(ab+2))+(3/(bc+2))+((ab+bc+ca)/(ca+2))≤1+1+1
And, (3/(ab+2))(3/(bc+2))((ab+bc+ca)/(ca+2))≤1
And, (ab+2)(bc+2)(ca+2)≥9(ab+bc+ca)
That is, (a^2+2)(b^2+2)(c^2+2)≥9(ab+bc+ca)
Next, when a,b,c,x,y,z>0 and we can prove that
If a+b+c=x+y+z,where a≤b≤c,x≤y≤z
c-a<z-x,and b-a<y-z or c-b<z-y
Then, abc≥xyz
That is, (ab+2)+(bc+2)+(ca+2)=
=3+3+(ab+bc+ca)
And (bc+2)-(ab+2)=b(c-a)<
3-(ab+bc+ca),a+b+c♥
(bc+2)-(ab+2)=b(c-a)<
(ab+bc+ca)-3,a+b+c>3
Hence, (ab+2)(bc+2)(ca+2)≥9(ab+bc+ca)

4 . give a,b,c>0 and a+b+c=3, prove that
√(ab+ac)+√(bc+ba)+√(ca+cb)≥3√2abc
Proof as a+b+c=3,we have
1/a+1/b+1/c≥1+1+1
ab+bc+ca≥3abc
And, 2(ab+bc+ca)≥6abc
And, (ab+ac)+(bc+ba)+(ca+cb)≥
≥2abcv+2abc+2abc
And then √(ab+ac)+√(bc+ba)+√(ca+cb)≥
√2abc+√2abc+√2abc=3√2abc OK
 

5. give a,b,c,d>0 and a+b+c+d=1, prove
(ab(c+d))/(1-4cd)+(bc(d+a))/(1-4da)+(cd(a+b))/(1-4ab)+(da(b+c))/(1-4bc)≥128abcd/3
Proof As a+b+c+d=1, we have
4(ab+bc+cd+da)≤(a+b+c+d)^2
≤4(a^2+b^2+c^2+c^2)
So, ab+bc+cd+da≤ 1/4=1/16+1/16+1/16+1/16
And, 1/ab+1/bc+1/cd+1/cd≥64
And, ab+bc+cd+da≥64abcd
Next seeing, (c+d)+(d+a)+(a+b)+(b+c)
=2=1/2+1/2+1/2+1/2
And, ab(c+d)+bc(d+a)+cd(a+b)+da(b+c)
≥16abcd/2+16abcd/2+16abcd/2+16abcd/2
And, -ab-bc-cd-da≥-1/4=-1/16-1/16-1/16-1/16
And, -4ab-4bc-4cd-4da≥-1=-1/4-1/4-1/4-1/4
And, (1-4ab)+(1-4bc)+(1-4cd)+(1-4da)
≥3/4+3/4+3/4+3/4
And, 1/(1-4ab)+1/(1-4bc)+1/(1-4cd)+1/(1-4da)≥4/3+4/3+4/3+4/3
Therefore, (ab(c+d))/(1-4cd)+(bc(d+a))/(1-4da)+(cd(a+b))/(1-4ab)+(da(b+c))/(1-4bc)
≥4/3 (32abcd)=128abcd/3 Ok

6. Show that (444444)^2+(333333)^2=(555555)^2
And find g(-2),when fog(x)=8x^2+12x-1
And f(x)=4x-5
Solution As (555555)^2-(333333)^2
=(555555-333333)(555555+333333)
=(222222)(888888)
=(222222)(2)(444444)
=(444444)(444444)=(444444)^2 OK
And fog(x)=f(g(x) )=4g(x))-5
=8x^2+12x-1
So, g(x)=2x^2+3x+1
Therefore, g(-2)=3 OK

7. give a,b,c>0,and a+b+c=3
Show that (a+b)(b+c)(cv+a)≤8
Solution from a+b+c=3, we have
a/2+b/2+b/2+c/2+c/2+a/2=3
And, ((a+b)/2)((b+c)/2)((c+a)/2)≤1
Therefore, (a+b)(b+c)(cv+a)≤8 OK 

8 . give a,b,c>0, prove that
a/b+b/c+c/a≥(a+b)/(b+c)+(b+c)/(c+a)+(c+a)/(a+b)
Proof As 2b+2c+2d+2a=
(b+c)+(a+b)+(c+d)+(d+a)
So, 1/2b+1/2c+1/2d+1/2a≥1/(b+c)+1/(a+b)+1/(c+d)+1/(d+a)
And, 1/b+1/c+1/d+1/a≥2/(b+c)+2/(a+b)+2/(c+d)+2/(d+a)
=(1/(b+c)+1/(a+b))+(1/(b+c)+1/(c+d))+(1/(c+d)+1/(d+a))+(1/(d+a)+1/(a+b))
So, a/b+b/c+c/d+d/a≥
≥(a/(b+c)+a/(a+b))+(b/(b+c)+b/(c+d))+(c/(c+d)+c/(d+a))+(d/(d+a)+d/(a+b))
=(a/(b+c)+b/(b+c))+(b/(c+d)+c/(c+d))+(c/(d+a)+d/(d+a))+(d/(a+)+a/(a+b))
=((a+b)/(b+c))+((b+c)/(c+d))+((c+d)/(d+a))+((d+a)/(a+)): OK
Remark: a/b+b/c+c/d+d/a:A
(a+b)/(b+c)+(b+c)/(c+a)+(c+a)/(a+b):B
A≥B↔1/b+1/c+1/d+1/a≥ ( important)
(1/(b+c)+1/(a+b))+(1/(b+c)+1/(c+d))+(1/(c+d)+1/(d+a))+(1/(d+a)+1/(a+b))

9. give a,b,c>0,prove that
a^4/(a^2+ab+b^2 )+b^4/(b^2+bc+c^2 )+c^4/(c^2+ca+a^2 )≥(a^3+b^3+c^3)/(a+b+c)
Proof as 2a^2+〖2b〗^2+〖2c〗^2+ab+bc+cd≥
a^2+b^2+c^2+2ab+2bc+2cd
And a^2+ab+b^2≥a^2+ab+ac
Hence, (a^2+ab+b^2)/a^4 ≥(a+b+c)/a^3 , similar
(b^2+bc+c^2)/b^4 ≥(a+b+c)/b^3 , and
(c^2+ca+a^2)/c^4 ≥(a+b+c)/c^3
So, (a^2+ab+b^2)/a^4 +(b^2+bc+c^2)/b^4 (c^2+ca+a^2)/c^4 ≥
(a+b+c)/a^3 +(a+b+c)/b^3 +(a+b+c)/c^3
And make, a^4/(a^2+ab+b^2 )+b^4/(b^2+bc+c^2 )+c^4/(c^2+ca+a^2 )≥
a^3/(a+b+c)+b^3/(a+b+c)+c^3/(a+b+c)=(a^3+b^3+c^3)/(a+b+c):OK

10. Prove that (a^3+b^3+c^3)/(a+b+c)+(a^3+b^3+d^3)/(a+b+d)+(a^3+c^3+d^3)/(a+c+d)+
+ (b^3+c^3+d^3)/(b+c+d)≥a^2+b^2+c^2+d^2
Proof As 2(a^3+b^3+c^3+d^3)≥
a^2 b+b^2 d+d^2 c+c^2 a+a^2 c+c^2 d+d^2 b+b^2 a
And, b^3+c^3+a^3+d^3+a^3+d^3+b^3+c^3≥
a^2 b+b^2 d+d^2 c+c^2 a+a^2 c+c^2 d+d^2 b+b^2 a
And (a^3 〖+b〗^3+c^3 )+(a^3 〖+b〗^3+d^3 )+
+(a^3 〖+c〗^3+d^3 )+(b^3 〖+c〗^3+d^3 )≥a^3+a^2 b+a^2 c+
+b^2 a+b^3+b^2 d+c^2 a+c^3+c^2 d+d^2 b+d^2 c+d^3
=a^2 (a+b+c)+b^2 (a+b+d)+c^2 (a+c+d)
+d^2 (b+c+d)
Next, we can write a^3 〖+b〗^3+c^3≥a^2 (a+b+c), or
a^3 〖+b〗^3+d^3≥b^2 (a+b+d), or
a^3 〖+c〗^3+d^3≥c^2 (a+c+d),or
b^3 〖+c〗^3+d^3≥d^2 (b+c+d)
Hence, (a^3 〖+b〗^3+c^3)/(a+b+c)≥a^2,or (a^3 〖+b〗^3+d^3)/(a+b+d)≥b^2
(a^3 〖+c〗^3+d^3)/(a+c+d)≥c^2,or (b^3 〖+c〗^3+d^3)/(b+c+d)≥d^2
Therefore, (a^3+b^3+c^3)/(a+b+c)+(a^3+b^3+d^3)/(a+b+d)+
(a^3+c^3+d^3)/(a+c+d)+ + (b^3+c^3+d^3)/(b+c+d)≥a^2+b^2+c^2+d^2

11. give a,b,c>0, prove that√(a^2+8bc)
+√(b^2+8ca)+√(c^2+8ab)≤3(a+b+c)
Proof As a^2+8bc+b^2+8ca+c^2+8ab≤
3(a+b+c)^2
So, we write, a^2+8bc≤(a+b+c)^2,or
b^2+8ca≤(a+b+c)^2, or,
c^2+8ab≤(a+b+c)^2
And, √(a^2+8bc)≤a+b+c, or
√(b^2+8ca)≤a+b+c, or
√(c^2+8ab)≤a+b+c
Therefore, √(a^2+8bc)+
+√(b^2+8ca)+√(c^2+8ab)≤3(a+b+c)

12. find the solution of
x+2√(7-x)=2√(x-1)+√(-x^2+8x-7)+1
Solution give m=√(7-x),n=√(x-1), we get
x-1+2√(7-x)=2√(x-1)+√((x-1)(7-x) )
n^2+2m=2n+mn
n^2-2n=mn-2m
n(n-2)=m(n-2)
Hence, n=m,where n≠2
√(7-x)=√(x-1),√(x-1)≠2
7-x= x-1,x-1≠4
x=4,x≠5
Recheck its solution
4+2√(7-4)=2√(4-1)+√(-4^2+8x4-7)+1
4+2√3=2√3+4
13. give f(x)+f(f(x) )=2x+6,∀x∈N
Find f(2015)
Solution As f is linear equation
So, f(x)=ax+b, we get
f(f(x))=a(ax+b)+b
=a^2 x+ab+b
Seeing, f(x)+f(f(x) )=2x+6
ax+b+a^2 x+ab+b=2x+6
From solving, we get a=1,and b=2
Hence, f(x)=x+2
Recheck: f(x)+f(f(x)
x+2+x+4=2x+6
And f(2015)=2015+2=2017
Acknowledgement
This writing, if there is a mistake, and then it is mine
But if there is some profit that can make wisdom, then I
Assign this success with Pro.Dr. Narong Phannim
Who be my great teacher.
Remark: if we think that varied problems are magnificent
Food, then we will be capable to solve them happily
And important we will get new wisdom by ourselves.

Presented by Agrend Wisnu Kusuma - Bogor City

West Java - Indonesia