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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
(x^2 + 1)(y^2 + 1) >= 2(xy - 1)(x + y)
parmenides51   3
N 6 minutes ago by alizoratiopour
Source: Czech-Polish-Slovak Junior Match 2017, individual p3 CPSJ
Prove that for all real numbers $x, y$ holds $(x^2 + 1)(y^2 + 1) \ge 2(xy - 1)(x + y)$.
For which integers $x, y$ does equality occur?
3 replies
parmenides51
Feb 20, 2020
alizoratiopour
6 minutes ago
A three-variable functional inequality on non-negative reals
Tintarn   6
N 21 minutes ago by jenishmalla
Source: Dutch TST 2024, 1.2
Find all functions $f:\mathbb{R}_{\ge 0} \to \mathbb{R}$ with
\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]for all $x,y,z \in \mathbb{R}_{\ge 0}$.
6 replies
Tintarn
Jun 28, 2024
jenishmalla
21 minutes ago
Holy inequality
giangtruong13   1
N 22 minutes ago by MathPerson12321
Source: Club
Let $a,b,c>0$. Prove that:$$\frac{8}{\sqrt{a^2+b^2+c^2+1}} - \frac{9}{(a+b)\sqrt{(a+2c)(b+2c)}} \leq \frac{5}{2}$$
1 reply
giangtruong13
26 minutes ago
MathPerson12321
22 minutes ago
NT function debut
AshAuktober   1
N 23 minutes ago by Burmf
Source: 2025 Nepal Practice TST 3 P2 of 3; Own
Let $f$ be a function taking in positive integers and outputting nonnegative integers, defined as follows:
$f(m)$ is the number of positive integers $n$ with $n \le m$ such that the equation $$an + bm = m^2 + n^2 + 1$$has an integer solution $(a, b)$.
Find all positive integers $x$ such that$f(x) \ne 0$ and $$f(f(x)) = f(x) - 1.$$Adit Aggarwal, India.
1 reply
AshAuktober
42 minutes ago
Burmf
23 minutes ago
No more topics!
Hanoi Open Mathematical Olympiad 2009
famousman1996   19
N Mar 11, 2013 by SherlockBond
Q1) Let $a,b,c$ be 3 distinct mumbers from {1,2,3,4,,5,6}. Show that 7 divides $abc+(7-a)(7-b)(7-c)$.

Q2) Show that there is a natural number $n$ such that the number $a=n!$ ends exacly in 2009 zeros.

Q3) Let a,b,c be positive integers with no common factor and satisfy the conditions $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$. Prove that a+b is a square.

Q4) Suppose that $a=2^b$, where $b=2^{10n+1}$. Prove that a is divisible by 23 for any positive integer n.

Q5) Prove that $m^{7}-m$ is divisible by 42 for any positive integer m.

Q6) Suppose that 4 real numbers a,b,c,d satisfy the conditions $\begin{matrix}
a^{2}+b^2=4\\ 
c^2+d^2=4\\ 
ac+bd=2
\end{matrix}$
Find the set of all possible values the number M=ab+cd can take

Q7) Let a,b,c,d be positive integers such that a+b+c+d=99.Find the smallest and the greatest values of the following product P=abcd.

Q8) Find all the pairs of the positive integers such that the product of the numbers of any pair plus the half of one of the numbers plus one third of the other number is three times less than 1004.

Q9) Give an acute-angled triangle ABC with area S, let points A',B',C' be located as follows: A' is the point where altitude from A on BC meets the outwards facing semicirle drawn on BX as diameter.Points B',C' are located similarly. Evaluate the sum
$T=(area \Delta BCA')^2+(area \Delta CAB')^2+(area \Delta ABC')^2$.

Q10) Prove that $d^2+(a-b)^2<c^2$ ,where d is diameter of the inscribed cricle of $\Delta ABC$ .

Q11) Let A={1,2, . . . ,100} and B is a subset of A having 48 elements. Show that B has two distint elements $x$ and $y$ whose sum is divisible by 11.
19 replies
famousman1996
Feb 14, 2013
SherlockBond
Mar 11, 2013
Hanoi Open Mathematical Olympiad 2009
G H J
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famousman1996
55 posts
#1 • 1 Y
Y by Adventure10
Q1) Let $a,b,c$ be 3 distinct mumbers from {1,2,3,4,,5,6}. Show that 7 divides $abc+(7-a)(7-b)(7-c)$.

Q2) Show that there is a natural number $n$ such that the number $a=n!$ ends exacly in 2009 zeros.

Q3) Let a,b,c be positive integers with no common factor and satisfy the conditions $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$. Prove that a+b is a square.

Q4) Suppose that $a=2^b$, where $b=2^{10n+1}$. Prove that a is divisible by 23 for any positive integer n.

Q5) Prove that $m^{7}-m$ is divisible by 42 for any positive integer m.

Q6) Suppose that 4 real numbers a,b,c,d satisfy the conditions $\begin{matrix}
a^{2}+b^2=4\\ 
c^2+d^2=4\\ 
ac+bd=2
\end{matrix}$
Find the set of all possible values the number M=ab+cd can take

Q7) Let a,b,c,d be positive integers such that a+b+c+d=99.Find the smallest and the greatest values of the following product P=abcd.

Q8) Find all the pairs of the positive integers such that the product of the numbers of any pair plus the half of one of the numbers plus one third of the other number is three times less than 1004.

Q9) Give an acute-angled triangle ABC with area S, let points A',B',C' be located as follows: A' is the point where altitude from A on BC meets the outwards facing semicirle drawn on BX as diameter.Points B',C' are located similarly. Evaluate the sum
$T=(area \Delta BCA')^2+(area \Delta CAB')^2+(area \Delta ABC')^2$.

Q10) Prove that $d^2+(a-b)^2<c^2$ ,where d is diameter of the inscribed cricle of $\Delta ABC$ .

Q11) Let A={1,2, . . . ,100} and B is a subset of A having 48 elements. Show that B has two distint elements $x$ and $y$ whose sum is divisible by 11.
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Virgil Nicula
7054 posts
#2 • 3 Y
Y by famousman1996, Adventure10, Mango247
PP. Let an acute $\triangle ABC$ with the area $S$ . Let $\left\{A_1,B_1,C_1\right\}$ be located as follows: $A_1$ is the point where altitude

from $A$ on $BC$ meets the outwards facing semicirle drawn on $[BC]$ as diameter. Points $B_1$ and $C_1$ are located similarly.

Evaluate the sum $T=[BCA_1]^2+[CAB_1]^2+[ABC_1]^2$ , where denoted $[XYZ]$ - the area of $\triangle ABC$ .


Proof. $4\cdot \sum\sin A\cos B\cos C=2\cdot \sum\sin A\left[\cos (B+C)+\cos (B-C)\right]=$ $2\cdot \sum\sin A\left[-\cos A+\cos (B-C)\right]=$

$\sum \left[-\sin 2A+2\sin (B+C)\cos (B-C)\right]=$ $\sum \left(-\sin 2A+\sin 2B+\sin 2C\right)=\sum\sin 2A=4\cdot\prod\sin A=$ $\frac {2S}{R^2}$ .

Let $D\in BC\ ,\ AD\perp BC$ . So $DA_1^2=DB\cdot DC\implies$ $DA_1^2=bc\cdot \cos B\cos C\implies$

$[BCA_1]^2=\left(\frac 12\cdot BC\cdot DA_1\right)^2=$ $\frac {a^2bc}{4}\cdot\cos B\cos C\implies$ $\sum [BCA_1]^2=\frac {abc}{4}\cdot\sum a\cdot\cos B\cos C=$

$2R^2S\sum\sin A\cos B\cos C\implies$ $\boxed{\sum [BCA_1]^2=2R^2S\cdot \frac {S}{2R^2}=S^2=[ABC]^2}$ . Otherwise, $\sum a\cdot\cos B\cos C=$

$2R\sum\sin A\cos B\cos C=$ $2R\prod\cos A\cdot\sum\tan A=$ $2R\prod\cos A\cdot\prod\tan A=2R\prod\sin A=\frac {S}{R}$ .
This post has been edited 6 times. Last edited by Virgil Nicula, Feb 15, 2013, 8:20 AM
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modularmarc101
2208 posts
#3 • 2 Y
Y by famousman1996, Adventure10
Q1

Q2(error in problem?)

Q3

Problem 4 is incorrect.

Q5
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nikoma
1976 posts
#4 • 3 Y
Y by famousman1996, Adventure10, Mango247
Q7)
Click to reveal hidden text
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hemangsarkar
791 posts
#5 • 3 Y
Y by famousman1996, Adventure10, Mango247
modularmarc101 wrote:
Q3


$a= 3, b = 6, c = 2$?




let $a = c + m$

$b = c + n$

for some positive integers $m,n$

this reduces to $c^2 = mn$

here $m$ and $n$ can't have any common factors.
for example, if $d > 1$ divids both $m$ and $n$, then $d$ divides all three of $a,b,c$..


this would mean $m = k^2$ and $n = t^2$

$a + b = (k+t)^2$



10
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modularmarc101
2208 posts
#6 • 3 Y
Y by famousman1996, Adventure10, Mango247
hemangsarkar wrote:
modularmarc101 wrote:
Q3


$a= 3, b = 6, c = 2$?




let $a = c + m$

$b = c + n$

for some positive integers $m,n$

this reduces to $c^2 = mn$

here $m$ and $n$ can't have any common factors.
for example, if $d > 1$ divids both $m$ and $n$, then $d$ divides all three of $a,b,c$..


this would mean $m = k^2$ and $n = t^2$

$a + b = (k+t)^2$

Sorry I understood $\gcd(a,b)=\gcd(b,c)=\gcd(c,a)=1$ rather than $\gcd(a,b,c)=1$ from the way the problem was stated.
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sunken rock
4380 posts
#7 • 3 Y
Y by famousman1996, Adventure10, Mango247
Q9:
(I have posted this before, somewhere around)

Let $D=BC\cap AA'$ and $H$ the orthocenter of $\Delta ABC$.
We have $\frac{[A'BC]^2}{[ABC]^2}=\frac{A'D^2}{AD^2}$, but $A'D^2=BD\cdot CD$, while $BD\cdot CD=AD\cdot DH$, consequently $\frac{[A'BC]^2}{[ABC]^2}=\frac{DH\cdot AD}{AD^2}=\frac{DH}{AD}=\frac{[BCH]}{[ABC]}$. Summing up the three similar ratios: $\frac{\Sigma{[BCH]}}{[ABC]}=\frac{[ABC]}{[ABC]}=1$.

Best regards,
sunken rock
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ssilwa
5451 posts
#8 • 3 Y
Y by famousman1996, Adventure10, Mango247
modularmarc101 wrote:
Q5

Isnt that assuming that m is relatively prime to both 2 and 7?
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nikoma
1976 posts
#9 • 4 Y
Y by ssilwa, famousman1996, Adventure10, Mango247
I conjecture so too ssilwa, here's another solution for Q5
Click to reveal hidden text
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ssilwa
5451 posts
#10 • 3 Y
Y by famousman1996, Adventure10, Mango247
nikoma wrote:
I conjecture so too ssilwa, here's another solution for Q5
Click to reveal hidden text

ah.. very nice!
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goran
233 posts
#11 • 3 Y
Y by famousman1996, SherlockBond, Adventure10
Question 11
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famousman1996
55 posts
#12 • 2 Y
Y by Adventure10, Mango247
hemangsarkar wrote:
modularmarc101 wrote:
Q3


$a= 3, b = 6, c = 2$?




let $a = c + m$

$b = c + n$

for some positive integers $m,n$

this reduces to $c^2 = mn$

here $m$ and $n$ can't have any common factors.
for example, if $d > 1$ divids both $m$ and $n$, then $d$ divides all three of $a,b,c$..


this would mean $m = k^2$ and $n = t^2$

$a + b = (k+t)^2$



10

I do not understand
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Virgil Nicula
7054 posts
#13 • 2 Y
Y by Adventure10, Mango247
sunken rock wrote:
Q9: Let $D=BC\cap AA'$ and $H$ the orthocenter of $\Delta ABC$. We have $\frac{[A'BC]^2}{[ABC]^2}=\frac{A'D^2}{AD^2}$, but $A'D^2=BD\cdot CD$, while $BD\cdot CD=AD\cdot DH$, consequently $\frac{[A'BC]^2}{[ABC]^2}=\frac{DH\cdot AD}{AD^2}=\frac{DH}{AD}=\frac{[BCH]}{[ABC]}$. Summing up the three similar ratios: $\frac{\Sigma{[BCH]}}{[ABC]}=\frac{[ABC]}{[ABC]}=1$.
Nice, very nice ! Thank you.
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SherlockBond
238 posts
#14 • 2 Y
Y by Adventure10, Mango247
sunken rock wrote:
Q9:
(I have posted this before, somewhere around)

Let $D=BC\cap AA'$ and $H$ the orthocenter of $\Delta ABC$.
We have $\frac{[A'BC]^2}{[ABC]^2}=\frac{A'D^2}{AD^2}$, but $A'D^2=BD\cdot CD$, while $BD\cdot CD=AD\cdot DH$, consequently $\frac{[A'BC]^2}{[ABC]^2}=\frac{DH\cdot AD}{AD^2}=\frac{DH}{AD}=\frac{[BCH]}{[ABC]}$. Summing up the three similar ratios: $\frac{\Sigma{[BCH]}}{[ABC]}=\frac{[ABC]}{[ABC]}=1$.

Best regards,
sunken rock
If we don't know the result is $area [ABC]^{2}$,so we cannot use this way,can you have the other way just in case we don't know the result?
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SherlockBond
238 posts
#15 • 2 Y
Y by Adventure10, Mango247
I think Q4 is wrong. Because 23 is an odd prime number
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SherlockBond
238 posts
#16 • 2 Y
Y by Adventure10, Mango247
I don't understand what does Q6 mean,can you explain?
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shinichiman
3212 posts
#17 • 2 Y
Y by Adventure10, Mango247
SherlockBond wrote:
I don't understand what does Q6 mean,can you explain?
Explain about what ??
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SherlockBond
238 posts
#18 • 2 Y
Y by Adventure10, Mango247
What do we have to do?
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nikoma
1976 posts
#19 • 3 Y
Y by SherlockBond, Adventure10, Mango247
SherlockBond wrote:
What do we have to do?
Find all $x$ such that $x = ab + cd$ and $a,b,c,d$ satisfy given system of equations.
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SherlockBond
238 posts
#20 • 2 Y
Y by Adventure10, Mango247
nikoma wrote:
SherlockBond wrote:
What do we have to do?
Find all $x$ such that $x = ab + cd$ and $a,b,c,d$ satisfy given system of equations.
So that we must find all of its value or the set contain it?
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