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| Math Cambodia 2019 Day 02 |
Find all function `f(x)` if `(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)`
Solution
If `x+1/x=2` Find the value of `x^5+1/x^5`
As we had: `x+1/x=2` `rightarrow(x+1/x)^2=4` `leftrightarrowx^2+1/x^2=2`
We continue with `(x^2+1/x^2)(x+1/x)=4` `leftrightarrowx^3+x+1/x+1/x^3=4`
`leftrightarrowx^3+1/x^3=2`
If `a+b+c=2022` and `1/a+1/b+1/c=1/2022` Find the value of `1/(a^2023)+1/(b^2023)+1/(c^2023)`
Solution
Find all polynomials `P(x)` such that: `P(x-1).P(x+1)=P(x^2-1)`
Solution
Suppose that `\alpha` is a root of `P(x)` then, `P(\alpha)=0`
Therefore, `P((\alpha+1)-1)=0`
Find the last two digits of : `N=(1!+2!+3!+.......+101!)^101`
Note: This is equivalent to finding `N(mod11)`.
ie: The remainder when dividing `N` by `100`.
Observation: `10!\equiv0(mod100)` Because, `10!=10...5...2`
Therefore, `N\equiv(1!+2!+3!+......+9!)^101(mod100)`
`N\equiv(1+2+6+24+20+20+40+20+80)^101(mod100)`
`N\equiv13^101(mod100)`
Prove that: `A_n=3^(n+3)-4^(4n+2)` is divided by `11`
Solution
We will prove by Mathematics Induction:
If `n=0` then `A_n=3^3-4^2=27-16=11` It is true that `A_n` is divided with `11`
Assuming that: `A_n` is divided with `11` for `n=k` `(1)`
We will prove it is true for `n=k+1` then `A_(k+1)` is divided with `11`
a. Prove that: `forall x\inR` : `:|acosx+bsinx|\leqsqrt(a^2+b^2)`
b. Find the maximum and minimum of `f(x)=20cosx+21sinx+27`
Solution
a. Prove `|acosx+bsinx|\leqsqrt(a^2+b^2)` `forall x\inR`
Assuming we have:
`acosx+bsinx=sqrt(a^2+b^2)(a/sqrt(a^2+b^2)cosx+b/sqrt(a^2+b^2)sinx)`
`X_1;X_2` are the root of : `X^2-(2cost+3sint)X-11sin^2t=0`
Find the minimum of `A=X_1^2+X_1X_2+X_2^2`.
Solution
We can see that: `A=X_1^2+X_1X_2+X_2^2`
`=(X_1+X_2)^2-X_1X_2`
Following Vieta's Formulas `X^2-SX+P=0` `(1)`
Find all Polynomials `P(x)` which is satisfied that:
`(x-2010)P(x+67)=xP(x)`
(2010 Baltic Way)
a. Prove that: `tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)`
As we knew: `tan3x=(3tanx-tan^3x)/(1-3tan^2x)`
`=(3tanx-9tan^3x+8tan^3x)/(1-3tan^2x)`
`=(8tan^3x+3tanx(1-3tan^2x))/(1-3tan^2x)`
`=(8tan^3x)/(1-3tan^2x)+3tanx`
Then, `tan3x-3tanx=(8tan^3x)/(1-3tan^2x)`
Find the limit of `S_n=2/(1.3)+2/(3.5)+2/(5.3)+.......+2/((2n+1)(2n+3))`
Solution
We can see that general term of this sequence is:
`2/((2k+1)(2k+3))=1/(2k+1)-1/(2k+3)`
It is given function `f(x)` determine on `R` , satisfied that:
`f(1)=1`
`f(x+5)\geqf(x)+5`
`f(x+1)\leqf(x)+1`
If `g(x)=f(x)+1-x` , find the value of `g(2020)` `\forallx,y\inR`
Solution
From our hypothesis we already had:
`(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)`
We will find the coefficients of `a_1` ; `a_2` ; `a_3` and `a_20`
Which meant they are the coefficients of `x` ; `x^2` ; `x^3` and `x^(20)` .
We can rewrite `(1+2x+3x^2)^10=[1+x(2+3x)]^10`
We will use the Newton's Formula
`(a+b)^n`=`\sum_{i=0}^nC(n,i)``a^(n-i).b^i`
Then, `[1+x(2+3x)]^10=C(10,0)+C(10,1)x(2+3x)+C(10,2)x^2.(2+3x)^2`
`+C(10,3)x^3.(2+3x)^3+....+C(10,10)x^10.(2+3x)^10`
We observe that:
`Coef(x)=2C(10,1)=20` then, `a_1=20`
`Coef(x^2)=2^2C(10,2)=4.45=180` then, `a_2=180`
`Coef(x^3)=12C(10,2)+8C(10,3)=45.12+120.8=1500` then `a_3=1500`
`Coef(x^20)=C(10,10).3^10=3^10` then, `a_20=3^10`
Solution
Solution 01
Denoting the greatest common divisor (GCD) of `a` and `b` `(a,b)` and we will use Euclidean Algorithm Theory.
`(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1`
It follows that `(21n+4)/(14n+3)` is irreducible. Q.E.D
Solution 02
Let 
be a point on the side 
of 
. Let 
, and 
be the inscribed circles of triangles 
, and 
. Let 
, and 
be the radii of the exscribed circles of the same triangles that lie in the angle 
. Prove that
.
and are natural numbers so that![\[\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319},\]](https://latex.artofproblemsolving.com/f/e/7/fe7a47dd9fd28d623ff2b2b7186626d70073ea29.png)
Prove that is divisible with 
.
Find all real roots of the equation
,where is a real parameter.
Real numbers 
are given. For each 
(
) define
![\[d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}\]](https://latex.artofproblemsolving.com/6/6/1/661c06518e6a3ae8ca5c84d0eb4e63ef7945723e.png)
and let
![\[d=\max\{d_i:1\le i\le n\}\]](https://latex.artofproblemsolving.com/6/7/7/677a05f003aa103531ff367920be9f9dd13797f8.png)
.
(a) Prove that, for any real numbers 
,
![\[\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} (*)\]](https://latex.artofproblemsolving.com/a/7/a/a7ac71dcf18b9336f8b3d4cbb575615c1aefca54.png)
(b) Show that there are real numbers 
such that equality holds in (*)
