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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
Newton's Sum
P162008   3
N a minute ago by lbh_qys
If $\sum_{cyc} a = 7, \sum_{cyc} a^2 = 15, \sum_{cyc} a^3 = 19$ and $\sum_{cyc} a^4 = 25$ where each summation runs over $4$ variables $a,b,c$ and $d.$ Then find the value of $\sum_{cyc} a^5.$
3 replies
1 viewing
P162008
Today at 1:51 AM
lbh_qys
a minute ago
A nice inequality
KhuongTrang   0
2 hours ago
[quote=KhuongTrang]Problem. Let $a,b,c\ge 0: ab+bc+ca>0.$ Prove that$$\color{blue}{\frac{\left(2ab+ca+cb\right)^{2}}{a^{2}+4ab+b^{2}}+\frac{\left(2bc+ab+ac\right)^{2}}{b^{2}+4bc+c^{2}}+\frac{\left(2ca+bc+ba\right)^{2}}{c^{2}+4ca+a^{2}}\ge \frac{8(ab+bc+ca)}{3}.}$$[/quote]

0 replies
KhuongTrang
2 hours ago
0 replies
Inequalities
sqing   6
N 2 hours ago by lbh_qys
Let $x,y\ge 0$ such that $ 13(x^3+y^3) \leq 125(1+xy)$. Prove that
$$  k(x+y)-xy\leq  5(2k-5)$$Where $k\geq 5.6797. $
$$  6(x+y)-xy\leq 35$$
6 replies
sqing
Apr 20, 2025
lbh_qys
2 hours ago
Inequalities
sqing   0
2 hours ago
Let $ a,b \in [0 ,1] . $ Prove that
$$\frac{a}{ 1-ab+b }+\frac{b }{ 1-ab+a } \leq 2$$$$ \frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 }+\frac{ab }{2+ab }  \leq 1$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+ab  }\leq \frac{5}{2}$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+2ab  }\leq \frac{7}{3}$$$$\frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 } +\frac{ab }{1+ab }\leq \frac{7}{6 }$$
0 replies
sqing
2 hours ago
0 replies
Inequalities from SXTX
sqing   20
N 2 hours ago by sqing
T702. Let $ a,b,c>0 $ and $ a+2b+3c=\sqrt{13}. $ Prove that $$ \sqrt{a^2+1} +2\sqrt{b^2+1} +3\sqrt{c^2+1} \geq 7$$S
T703. Let $ a,b $ be real numbers such that $ a+b\neq 0. $. Find the minimum of $ a^2+b^2+(\frac{1-ab}{a+b} )^2.$
T704. Let $ a,b,c>0 $ and $ a+b+c=3. $ Prove that $$ \frac{a^2+7}{(c+a)(a+b)} + \frac{b^2+7}{(a+b)(b+c)} +\frac{c^2+7}{(b+c)(c+a)}  \geq 6$$S
20 replies
sqing
Feb 18, 2025
sqing
2 hours ago
Theory of Equations
P162008   3
N 3 hours ago by P162008
Let $a,b,c,d$ and $e\in [-2,2]$ such that $\sum_{cyc} a = 0, \sum_{cyc} a^3 = 0, \sum_{cyc} a^5 = 10.$ Find the value of $\sum_{cyc} a^2.$
3 replies
P162008
Wednesday at 11:27 AM
P162008
3 hours ago
Fun & Simple puzzle
Kscv   7
N 3 hours ago by vanstraelen
$\angle DCA=45^{\circ},$ $\angle BDC=15^{\circ},$ $\overline{AC}=\overline{CB}$

$\angle ADC=?$
7 replies
Kscv
Apr 13, 2025
vanstraelen
3 hours ago
A problem involving modulus from JEE coaching
AshAuktober   7
N 5 hours ago by Jhonyboy
Solve over $\mathbb{R}$:
$$|x-1|+|x+2| = 3x.$$
(There are two ways to do this, one being bashing out cases. Try to find the other.)
7 replies
1 viewing
AshAuktober
Apr 21, 2025
Jhonyboy
5 hours ago
FB = BK , circumcircle and altitude related (In the World of Mathematics 516)
parmenides51   5
N Today at 4:05 AM by jasperE3
Let $BT$ be the altitude and $H$ be the intersection point of the altitudes of triangle $ABC$. Point $N$ is symmetric to $H$ with respect to $BC$. The circumcircle of triangle $ATN$ intersects $BC$ at points $F$ and $K$. Prove that $FB = BK$.

(V. Starodub, Kyiv)
5 replies
parmenides51
Apr 19, 2020
jasperE3
Today at 4:05 AM
Double Sum
P162008   0
Today at 3:22 AM
Let $\xi = \lim_{m \to\infty} \sum_{n=4}^{m} \sum_{k=2}^{n-2} \frac{1}{\binom{n}{k}}.$ If the value of $\xi$ can be written as $\frac{m}{n}$ where $m$ and $n$ are co-prime positive integers then compute the value of $m^3 + n^3.$
0 replies
P162008
Today at 3:22 AM
0 replies
Polynomial Limit
P162008   0
Today at 1:47 AM
Let $p = \lim_{y\to\infty} \left(\frac{2}{y^2} \left(\lim_{z\to\infty} \frac{1}{z^4} \left(\lim_{x\to\infty} \frac{((y^2 + y + 1)x^k + 1)^{z^2 + z + 1} - ((z^2 + z + 1)x^k + 1)^{y^2 + y + 1}}{x^{2k}}\right)\right)\right)^y$ where $k \in N$ and $q = \lim_{n\to\infty} \left(\frac{\binom{2n}{n}. n!}{n^n}\right)^{1/n}$ where $n \in N$. Find the value of $p.q.$
0 replies
P162008
Today at 1:47 AM
0 replies
Octagon Problem
Shiyul   4
N Today at 1:43 AM by Sid-darth-vater
The vertices of octagon $ABCDEFGH$ lie on the same circle. If $AB = BC = CD = DE = 11$ and $EF = FG = GH = HA = sqrt2$, what is the area of octagon $ABCDEFGH$?

I approached this problem by noticing that the area of the octagon is the area of the eight isoceles triangles with lengths $r$, $r$, and $sqrt2$ or 11. However, I didn't know how to find the radius. Can anyone give me a hint?
4 replies
Shiyul
Yesterday at 11:41 PM
Sid-darth-vater
Today at 1:43 AM
Polynomial Limit
P162008   2
N Today at 1:30 AM by P162008
If $P_{n}(x) = \prod_{k=1}^{n} \left(x + \frac{1}{2^k}\right) = \sum_{k=0}^{n} a_{k} x^k$ then find the value of $\lim_{n \to \infty} \frac{a_{n - 2}}{a_{n - 4}}.$
2 replies
P162008
Wednesday at 11:55 AM
P162008
Today at 1:30 AM
Sequence problem I never used
Sedro   1
N Yesterday at 9:01 PM by mathprodigy2011
Let $\{a_n\}_{n\ge 1}$ be a sequence of reals such that $a_1=1$ and $a_{n+1}a_n = 3a_n+2$ for all positive integers $n$. As $n$ grows large, the value of $a_{n+2}a_{n+1}a_n$ approaches the real number $M$. What is the greatest integer less than $M$?
1 reply
Sedro
Yesterday at 4:19 PM
mathprodigy2011
Yesterday at 9:01 PM
Phi Combo (Summer MAT 2021/7)
innumerateguy   22
N Jul 25, 2021 by samrocksnature
The function $\phi(n)$ denotes the number of positive integers less than or equal to $n$ that are relatively prime to $n.$ Determine the number of positive integers $a,$ relatively prime to $77,$ such that there exists some integer $n$ satisfying
$$\frac{\phi(n)}{n}=\frac{a}{77}.$$William Dai
22 replies
innumerateguy
Jul 24, 2021
samrocksnature
Jul 25, 2021
Phi Combo (Summer MAT 2021/7)
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innumerateguy
2178 posts
#1 • 1 Y
Y by centslordm
The function $\phi(n)$ denotes the number of positive integers less than or equal to $n$ that are relatively prime to $n.$ Determine the number of positive integers $a,$ relatively prime to $77,$ such that there exists some integer $n$ satisfying
$$\frac{\phi(n)}{n}=\frac{a}{77}.$$William Dai
This post has been edited 3 times. Last edited by innumerateguy, Jul 24, 2021, 11:30 PM
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pi271828
3368 posts
#2 • 2 Y
Y by centslordm, Aryan-23
i got 9?

probably wrong tho
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OlympusHero
17020 posts
#3 • 1 Y
Y by centslordm
What's the answer to this one?
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pi271828
3368 posts
#4 • 1 Y
Y by centslordm
OlympusHero wrote:
What's the answer to this one?

no idea :/

tho i got 9
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Arrowhead575
2281 posts
#5
Y by
I got the number of factors of 60? so 12
This post has been edited 1 time. Last edited by Arrowhead575, Jul 24, 2021, 11:44 PM
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pi271828
3368 posts
#6
Y by
Arrowhead575 wrote:
I got the number of factors of 60? so 8

number of factors of 60 is 12 tho?

and your overcounting some btw

or maybe not...
This post has been edited 2 times. Last edited by pi271828, Jul 24, 2021, 11:46 PM
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OlympusHero
17020 posts
#7
Y by
I guessed 16 oops now 1/9

Am crying :(
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Arrowhead575
2281 posts
#8
Y by
sry, I meant 12. (I thought I got 8, but when i checked my answer i got 12)
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pi271828
3368 posts
#9
Y by
OlympusHero wrote:
I guessed 16 oops now 1/9

Am crying :(

meh

this is probably aime problem 10-15 level and olympiad level, i wouldnt worry
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pi271828
3368 posts
#10
Y by
Arrowhead575 wrote:
sry, I meant 12. (I thought I got 8, but when i checked my answer i got 12)

the nine i got was 30,40,50,45,48,54,57,58,59
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centslordm
4740 posts
#12 • 1 Y
Y by Mogmog8
eleven :maybe:
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Bole
578 posts
#14 • 1 Y
Y by I-_-I
The answer is $8$ I believe
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pi271828
3368 posts
#15
Y by
why is everyone getting different answers lol
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innumerateguy
2178 posts
#16
Y by
The official answer is 8 from $2^3=8$.
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Pleaseletmewin
1574 posts
#17 • 1 Y
Y by mathtiger6
I didn't do MAT but this is pretty nice
Sol
#9
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mathtiger6
326 posts
#18 • 1 Y
Y by TopNotchMath
$\frac{\phi(n)}{n} = \frac{a}{77}$.
Clearly $77 \mid n$.

$\frac{\phi(n)}{n} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \frac{10}{11}$, where the first three factors of the RHS could be multiplied or not into the expression. If there is a higher prime in the denominator, there would be no way to cancel it.

We then find
$$\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} = \frac{a}{60},$$where the LHS's three factors could be multiplied in or not. Therefore, the answer is $2^3 = 8$.
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Arrowhead575
2281 posts
#19
Y by
DOH, I had 8 and then changed to 12
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samrocksnature
8791 posts
#20
Y by
I don't understand why $11,13,\ldots \not{\mid} n.$
This post has been edited 1 time. Last edited by samrocksnature, Jul 25, 2021, 12:27 AM
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TopNotchMath
1747 posts
#21 • 1 Y
Y by samrocksnature
samrocksnature wrote:
I don't understand why $11,13,\ldots \not{\mid} n.$

When you simplify the LHS, you get the following.

You have $$\left(1 - \dfrac{1}{p_1}\right)\left(1-\dfrac{1}{p_2}\right)\dots = \dfrac{a}{77}$$
Clearly, we need to have $7,11 | n$ to get the denominator of $77$. If we substitute this in, we get:

$$60\left(1-\dfrac{1}{p_3}\right)\dots = a$$
Primes greater than $5$ do not divide $60$, and we used $7$ and $11$ to clear the denominator, so that is why.
This post has been edited 2 times. Last edited by TopNotchMath, Jul 25, 2021, 12:52 AM
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asdf334
7585 posts
#22 • 1 Y
Y by centslordm
smh when you get the x1/2 x2/3 x4/5 thing then proceed to do 1 + 3 instead of 2^3
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Geometry285
902 posts
#23
Y by
If $n=77k$, we have $k=1$ makes $\phi(77)=10 \cdot 6 = 60$. Now, for $k \ge 1$, we have $\frac{60 \cdot \phi(k)}{77k} = \frac{a}{77}$. Clearly, for $a$ to be a distinct positive integer, $\frac{\phi(k)}{k}$ must be a distinct value that divides $60$, the only possible values of which $k = \{2,3,5 \}$ since $\frac{60}{k}$ is an integer, and $\phi(k)$ is positive. Now, associating any of these numbers results in a distinct output, so the total amount of sets for these $3$ numbers are $2^3=\boxed{008}$
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dchenmathcounts
2443 posts
#24
Y by
samrocksnature wrote:
I don't understand why $11,13,\ldots \not{\mid} n.$

11 actually must divide $n$. Here's why no prime greater than $11$ can divide $n$.

Consider the greatest prime that divides $n$; clearly $\phi(n)$ cannot be divisible by it, so it is in the denominator. However, the greatest prime divisor of $77$ is $11$.
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samrocksnature
8791 posts
#25
Y by
dchenmathcounts wrote:
samrocksnature wrote:
I don't understand why $11,13,\ldots \not{\mid} n.$

11 actually must divide $n$. Here's why no prime greater than $11$ can divide $n$.

Consider the greatest prime that divides $n$; clearly $\phi(n)$ cannot be divisible by it, so it is in the denominator. However, the greatest prime divisor of $77$ is $11$.

Wait oops mistype
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