Friday, April 21, 2023

Math 2nd Day Phnom Penh Cambodia 20/04/2023

Phnom Penh is a capital city of Cambodia where many clever students, great teachers which make students from Phnom Penh are mostly good at Math If compare to countryside students. In 2023, For 1st day of out standing student examination, I just got problems paper on social media and I will share with all of you here. Some of those problems I used to do when I was in high school.

Math Out Standing Problem 2012, 2013, 2014

Math out standing student whole country in Cambodia 2012, 2013, 2014 .

Math 1st Day Phnom Penh Cambodia 19/04/2023

Phnom Penh is a capital city of Cambodia where many clever students, great teachers which make students from Phnom Penh are mostly good at Math If compare to countryside students. In 2023, For 1st day of out standing student examination, I just got problems paper on social media and I will share with all of you here. Some of those problems I used to do when I was in highschool.

Prove that $(\frac{1}{2}) . (\frac{3}{4}) ... ... (\frac{2 n - 1}{2 n}) \leq \frac{1}{\sqrt{3 n + 1}}$ $(1/2).(3/4)......((2n-1)/(2n))\leq1/sqrt(3n+1)$

This is the problem that I picked up from book : 101 Solved Problem in Algebra from The USA IMO with the 43th problem.

What is the coefficient of $x^{2}$ $x^2$ when $(1 + x) (1 + 2 x) ... (1 + 2^{n} x)$ $(1+x)(1+2x)...(1+2^nx)$ is expanded?

This is the problem that I picked up from book : 101 Solved Problem in Algebra from The USA IMO with the 32th problem.

Vietnamese Mathematical Olympiad 2017

This is the 2007 Vietnamese Mathematical Olympiad Problem number 6. This is kind of problem that you need to use Permutation Formular . Moreover, you have to know about Sum of Sigma as well.

Math For Out-Standing Student 2023

This is the sequence combined with logarithm function which lead us to be understood of method to find the general formula of sequence. It is the basic of method statement for you to know how to find the general form of sequence when you have the power of number in our sequence.

Prove that ${(a_{n} + \frac{3}{2^{n + 2}})}_{n}^{\frac{1}{n}} (m - {(\frac{2}{3})}^{n \frac{m - 1}{m}}) < \frac{m^{2} - 1}{m - n + 1}$ $(a_n+3/2^(n+2))^(1/n)(m-(2/3)^(n(m-1)/m))<(m^2-1)/(m-n+1)$

This is the problem that I picked up from Math Book Around The World which is written by Mr. Lim Phalkun And Mr. Sen Piseth. This is the problem which combined many methods, many theory such Bernualli.

Prove that $\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + ... + \frac{1}{a_{20}}$ $1/a_1+1/a_2+1/a_3+...+1/a_20$ is a natural number

Find function $f (x)$ $f(x)$ satisfied $f (x) f (y) - f (x y) - 90 = 10 \frac{x + y}{x} y$ $f(x)f(y)-f(xy)-90=10(x+y)/xy$

Math PreyVeng Cambodia 2023

Day 01

Solution Here: Download PDF

Solution

If $x_{1}, x_{2}$ $x_1, x_2$ are the root of equation $x^{2} - x - 3 = 0$ $x^2-x-3=0$ Find the value of $A = 7 x_{1}^{5} + 19 x_{2}^{4}$ $A=7x_1^5+19x_2^4$

Solution

Problem 000

Prove that $a^{2} + b^{2} + c^{2} \geq a^{\frac{4}{3}} + b^{\frac{4}{3}} + c^{\frac{4}{3}}$ $a^2+b^2+c^2\geqa^(4/3)+b^(4/3)+c^(4/3)$

Prove that $x^{2023} + y^{2023} + z^{2023} = 0$ $x^2023+y^2023+z^2023=0$ If $\frac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}}$ $(x^2+y^2+z^2)/(a^2+b^2+c^2)=x^2/a^2+y^2/b^2+z^2/c^2$

This is the problem that I picked from Vietnamese collection problems which is the basic for kind of this question.
From that origin problem is $x^{2019} + y^{2019} + z^{2019} = 0$ $x^2019+y^2019+z^2019=0$ but, I have changed it to $2023$ $2023$ .

Prove that $x, y, z > 0$ $x, y, z >0$ Then $M = \frac{x}{2 x + y + z} + \frac{y}{x + 2 y + z} + \frac{z}{x + y + 2 z} \leq \frac{3}{4}$ $M=x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)\leq3/4$

Find all number which its square has 4 digits number and divisible 33

In order to solve type of this problem, you have to understand about divisible theory of natural number. As GCD (Greatest Common Divisor) and LCM (Least Common Divisor). When you know about GCD and LCM, you surely can simplify the problem to be more easier to solve.
As example of our problem, $33 = 3.11$ $33=3.11$ and $G C D (3; 11) = 1$ $GCD(3;11)=1$ Then, our problem can be found as divisible with $3$ $3$ and $11$ $11$

Solution

Saturday, March 18, 2023

Solve the equation ${sin}^{2012} x + {cos}^{2012} x = \frac{1}{2^{1005}}$ $sin^2012x+cos^2012x=1/2^1005$

I will solve this problem by Khmer language and in order to solve this problem, you have to use derivative of function which you call find the minimum value of function .
This is the problem 02 of Vietnamese Out-Standing Student grade 12.

Vietnamese Mathematics Provincial 2011

This is the problems of Vietnamese Mathematics Provincial examination in 2011. I just picked problem number 04 to share all of you which is related to sequence. As you know, sequences are the problems which need more strategy to solve where you have to combine all your understanding.

Cambodia National Math 2019, 22/04/2019 Day 02

$Math Book Cambodia$

Math Cambodia 2019 Day 02

This was the problem that released for Out Standing Student in Cambodia in 2019 for 2nd day of testing.
There were two days of the testing. This is the first day of exam.
You all can Click here to download

Cambodia Grade 12, 22/04/2019 Day 01

Cambodia National Math 2019, 22/04/2019

This was the problem that released for Out Standing Student in Cambodia in 2019.
There were two days of the testing. This is the first day of exam.

You can watch solution here:

Tuesday, February 28, 2023

Mathematics Out Standing Student Phnom Penh 2020, Cambodia

This was the problem that released for Out Standing Student in Phnom Penh, Cambodia in 2020.
There were two days of the testing. This is the first day of exam.

Prove that the polynomial $x^{9999} + x^{8888} + x^{7777} + ... + x^{1111} + 1$ $x^9999+x^8888+x^7777+...+x^1111+1$ is divisible by $x^{9} + x^{8} + x^{7} + ... . + x + 1$ $x^9+x^8+x^7+....+x+1$

Problem: 01

Prove that the polynomial

$x^9999+x^8888+x^7777+...+x^1111+1$ is divisible by

$x^9+x^8+x^7+....+x+1$

Solution

Thursday, December 22, 2022

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 $(x^2+1/x^2)(x+1/x)=4$ $leftrightarrowx^3+x+1/x+1/x^3=4$

$leftrightarrowx^3+1/x^3=2$

Vietnamese Olympiad 2022: Find the value of $1/(a^2023)+1/(b^2023)+1/(c^2023)$

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: $P(x-1).P(x+1)=P(x^2-1)$

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 $11$

Prove that: $forall x\inR$ : $:|acosx+bsinx|\leqsqrt(a^2+b^2)$

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 $A=X_1^2+X_1X_2+X_2^2$ .

$X_1;X_2$ 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)

Find all Polynomials $P(x)$ which is satisfied that:

$(x-2010)P(x+67)=xP(x)$

(2010 Baltic Way)

Prove that: $tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)$

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:

$2/((2k+1)(2k+3))=1/(2k+1)-1/(2k+3)$

If $g(x)=f(x)+1-x$ , find the value of $g(2020)$

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 $g(2020)$ $\forallx,y\inR$

If $(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)$ then find the coefficients of $a_1$ ; $a_2$ ; $a_3$ and $a_20$

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, $a_20=3^10$

Prove $(21n+4)/(14n+3)$ Is Irreducible For Every Natural Number $n$

Prove

$(21n+4)/(14n+3)$ Is Irreducible For Every Natural Number

$n$ .

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

Vietnam Math Out Standing Student 2012-13

Problem 01: Solve the equation of:

$x^(4n)$ +

$\sqrt{x^{2n}+2012}$

$=2012$ for all

$n\in N$ .

Problem 02: It is given

$(U_n)$ which is determined by:

$U_1=3$

$U_(n+1)=1/3(2U_n+3/U_n^2)$ ; for all

$n\in N$ .

Let's finding the limit of:

$\lim_{n\rightarrow\infty}(u_n)$ .

Problem 03: It is given three non-negative real numbers

$x,y,z$ , Prove that:

$1/x+1/y+1/z$

$>36/(9+x^2y^2+y^2^2+z^2x^2)$ .

Problem 04: Find roots positive numbers of equation:

$\sqrt{x+2\sqrt3}=\sqrt y+\sqrt z$

Solution

Find all functions $f:R\rightarrow R$ such that: $f(x)f(y)=f(xy-1)+xf(y)+yf(x)$ $\forall$ $x,y$ $\inR$

Find all functions $f: R \rightarrow R$ Such that:
$f(x)f(y)=f(xy-1)+xf(y)+yf(x)$ $\forall$ $x ,y$ $\inR$

Solution

Replace :

$y=0$ we will see that:

$f(x).f(0)=f(-1)+x.f(0)$

Notice that:

$f(0)$ not equal to

$0$ then,

$f(x)=x+c$ which

$c$ is a constant. We replaced to our hypothesis unsatisfactory. Then,

$f(0)=0$ from that we can see that:

$f(-1)=0$ .

1970 IMO Problems And Solutions

Problem 01

Let $M$ be a point on the side $AB$ of $\triangle ABC$ . Let $r_1, r_2$ , and $r$ be the inscribed circles of triangles $AMC, BMC$ , and $ABC$ . Let $q_1, q_2$ , and $q$ be the radii of the exscribed circles of the same triangles that lie in the angle $ACB$ . Prove that

$\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}$ .

Solution

1979 IMO Problems And Solutions

Problem 01

If $p$ and $q$ are natural numbers so that $\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319},$ Prove that $p$ is divisible with $1979$ .

Solution

1963 IMO Problems And Solutions

Problem 01

Find all real roots of the equation

$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$ ,

where $p$ is a real parameter.

Solution

Problem 01

Real numbers $a_1, a_2, \dots , a_n$ are given. For each $i$ ( $1\le i\le n$ ) define

$d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}$

and let

$d=\max\{d_i:1\le i\le n\}$ .

(a) Prove that, for any real numbers $x_1\le x_2\le \cdots\le x_n$ ,

$\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} (*)$

(b) Show that there are real numbers $x_1\le x_2\le x_n$ such that equality holds in (*)

Pages

Friday, April 21, 2023

Thursday, April 20, 2023

Wednesday, April 19, 2023

Tuesday, April 18, 2023

Monday, April 17, 2023

Wednesday, April 12, 2023

Saturday, April 8, 2023

Monday, April 3, 2023

Saturday, April 1, 2023

Friday, March 31, 2023

Saturday, March 25, 2023

Math PreyVeng Cambodia 2023

Day 01

Solution

Friday, March 24, 2023

Solution

Problem 000

Thursday, March 23, 2023

Wednesday, March 22, 2023

Tuesday, March 21, 2023

Monday, March 20, 2023

Solution

Saturday, March 18, 2023

Solve the equation sin2012x+cos2012x=121005sin2012x+cos2012x=121005sin^2012x+cos^2012x=1/2^1005

Friday, March 17, 2023

Vietnamese Mathematics Provincial 2011

Thursday, March 16, 2023

Cambodia National Math 2019, 22/04/2019 Day 02

Sunday, March 12, 2023

Cambodia National Math 2019, 22/04/2019

Tuesday, February 28, 2023

Mathematics Out Standing Student Phnom Penh 2020, Cambodia

Friday, December 23, 2022

Problem: 01

Solution

Thursday, December 22, 2022

Sunday, September 11, 2022

Friday, September 9, 2022

Sunday, August 29, 2021

Saturday, August 28, 2021

Thursday, July 15, 2021

Wednesday, July 14, 2021

Monday, July 12, 2021

Thursday, July 8, 2021

Wednesday, July 7, 2021

Tuesday, July 6, 2021

Vietnam Math Out Standing Student 2012-13

Solution

Monday, July 5, 2021

Find all functions f:R →Rf: R \rightarrow R Such that:f(x)f(y)=f(xy-1)+xf(y)+yf(x)f(x)f(y)=f(xy-1)+xf(y)+yf(x) ∀\forall x ,yx ,y ∈R\inR

Solution

Sunday, April 4, 2021

Problem 01

Solution

Wednesday, March 31, 2021

Problem 01

Solution

Monday, March 29, 2021

Problem 01

Solution

Problem 01

Solution

Solve the equation ${sin}^{2012} x + {cos}^{2012} x = \frac{1}{2^{1005}}$ $sin^2012x+cos^2012x=1/2^1005$

Find all functions $f: R \rightarrow R$ Such that:
$f(x)f(y)=f(xy-1)+xf(y)+yf(x)$ $\forall$ $x ,y$ $\inR$