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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
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Indonesia Regional MO 2019 Part A
parmenides51   15
N 10 minutes ago by Rohit-2006
Indonesia Regional MO
Year 2019 Part A

Time: 90 minutes Rules


p1. In the bag there are $7$ red balls and $8$ white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...


p2. Given a regular hexagon with a side length of $1$ unit. The area of the hexagon is ...


p3. It is known that $r, s$ and $1$ are the roots of the cubic equation $x^3 - 2x + c = 0$. The value of $(r-s)^2$ is ...


p4. The number of pairs of natural numbers $(m, n)$ so that $GCD(n,m) = 2$ and $LCM(m,n) = 1000$ is ...


p5. A data with four real numbers $2n-4$, $2n-6$, $n^2-8$, $3n^2-6$ has an average of $0$ and a median of $9/2$. The largest number of such data is ...


p6. Suppose $a, b, c, d$ are integers greater than $2019$ which are four consecutive quarters of an arithmetic row with $a <b <c <d$. If $a$ and $d$ are squares of two consecutive natural numbers, then the smallest value of $c-b$ is ...


p7. Given a triangle $ABC$, with $AB = 6$, $AC = 8$ and $BC = 10$. The points $D$ and $E$ lies on the line segment $BC$. with $BD = 2$ and $CE = 4$. The measure of the angle $\angle DAE$ is ...


p8. Sequqnce of real numbers $a_1,a_2,a_3,...$ meet $\frac{na_1+(n-1)a_2+...+2a_{n-1}+a_n}{n^2}=1$ for each natural number $n$. The value of $a_1a_2a_3...a_{2019}$ is ....


p9. The number of ways to select four numbers from $\{1,2,3, ..., 15\}$ provided that the difference of any two numbers at least $3$ is ...


p10. Pairs of natural numbers $(m , n)$ which satisfies $$m^2n+mn^2 +m^2+2mn = 2018m + 2019n + 2019$$are as many as ...


p11. Given a triangle $ABC$ with $\angle ABC =135^o$ and $BC> AB$. Point $D$ lies on the side $BC$ so that $AB=CD$. Suppose $F$ is a point on the side extension $AB$ so that $DF$ is perpendicular to $AB$. The point $E$ lies on the ray $DF$ such that $DE> DF$ and $\angle ACE = 45^o$. The large angle $\angle AEC$ is ...


p12. The set of $S$ consists of $n$ integers with the following properties: For every three different members of $S$ there are two of them whose sum is a member of $S$. The largest value of $n$ is ....


p13. The minimum value of $\frac{a^2+2b^2+\sqrt2}{\sqrt{ab}}$ with $a, b$ positive reals is ....


p14. The polynomial P satisfies the equation $P (x^2) = x^{2019} (x+ 1) P (x)$ with $P (1/2)= -1$ is ....


p15. Look at a chessboard measuring $19 \times 19$ square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of $k$ coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of $k$ so that the game never ends for any initial square selection is ....
15 replies
parmenides51
Nov 11, 2021
Rohit-2006
10 minutes ago
J
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A Complex Product
Saucepan_man02   10
N an hour ago by alexheinis
Calculate $\prod_{n \ge j>i \ge 1} (\epsilon^j-\epsilon^i)$ where $\epsilon = e^{\frac{2 \pi}{n}i}$.
10 replies
Saucepan_man02
Today at 2:15 AM
alexheinis
an hour ago
J
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Inequalities
lgx57   3
N 3 hours ago by lgx57
Let $0 < a,b,c < 1$. Prove that

$$a(1-b)+b(1-c)+c(1-a)<1$$
3 replies
lgx57
Mar 19, 2025
lgx57
3 hours ago
J
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Inequalities
sqing   0
5 hours ago
Let $ a,b $ be reals such that $ a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $ a,b $ be reals such that $ a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
0 replies
sqing
5 hours ago
0 replies
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No more topics!
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Help me pls
trito11   6
N Apr 13, 2025 by HuongToiVMO
Let 2017 positive integers$a_1,a_2,...a_{2017}$ and an integer $a>1$ which $a_1*a_2*...*a_{2017}|a$.Prove that $ a^{2018}+a-1$ isn't divisible by $(a+a_1-1)(a+a_2-1)(a+a_3-1)...(a+a_{2017}-1)$
6 replies
trito11
Dec 23, 2018
HuongToiVMO
Apr 13, 2025
J
Help me pls
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Reveal topic tags
High school math
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trito11
199 posts
trito11
#1 h  Dec 23, 2018, 12:13 AM • 2 Y
Y by Adventure10, Mango247
Let 2017 positive integers$a_1,a_2,...a_{2017}$ and an integer $a>1$ which $a_1*a_2*...*a_{2017}|a$.Prove that $ a^{2018}+a-1$ isn't divisible by $(a+a_1-1)(a+a_2-1)(a+a_3-1)...(a+a_{2017}-1)$
This post has been edited 9 times. Last edited by trito11, Dec 23, 2018, 12:47 AM
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trito11
199 posts
trito11
#2 h  Dec 23, 2018, 12:19 AM • 2 Y
Y by Adventure10, Mango247
Why latex is wrong with large number?
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spartacle
538 posts
spartacle
#3 h  Dec 23, 2018, 12:25 AM • 3 Y
Y by trito11, Adventure10, Mango247
In order to make a subscript or superscript with more than one letter, you have to add curly braces around the script.

So in this case, use a_{2017} instead of a_2017
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ShreyJ
1143 posts
ShreyJ
#4 h  Dec 23, 2018, 12:39 AM • 2 Y
Y by Adventure10, Mango247
False. Let all $a_i=1$ and $a=1$. Then $a^{2018}+a-1=1$ and $(a+a_1-1)(a+a_2-1)(a+a_3-1)...(a+a_{2017}-1)=(1)(1)(1)...(1)=1$.

Did you mean distinct positive integers?
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trito11
199 posts
trito11
#5 h  Dec 23, 2018, 12:48 AM • 2 Y
Y by Adventure10, Mango247
ShreyJ wrote:
False. Let all $a_i=1$ and $a=1$. Then $a^{2018}+a-1=1$ and $(a+a_1-1)(a+a_2-1)(a+a_3-1)...(a+a_{2017}-1)=(1)(1)(1)...(1)=1$.

Did you mean distinct positive integers?

Sorry i wirte wrong
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MathInfinite
187 posts
MathInfinite
#6 h  Dec 23, 2018, 9:45 AM • 1 Y
Y by Adventure10
In fact you can replace 2018 by any natural number.
Here’s an attempt at a sketch solution.
Let $a = ka_1a_2a_3...a_n$. Assume there is an integer $m$ such that $a^{n+1} + a - 1 = m(a_1+a-1)(a_2+a-1)...(a_n+a-1).$

Hint1
Hint2
Hint3
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Hint5
Hint6
Hint7
This post has been edited 1 time. Last edited by MathInfinite, Dec 30, 2018, 3:25 AM
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HuongToiVMO
1 post
HuongToiVMO
#7 h  Apr 13, 2025, 3:05 AM
Y by
Can you pls write in in details ?
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