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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
simple trapezoid
gggzul   1
N 24 minutes ago by Adventure1000
Let $ABCD$ be a trapezoid. By $x$ we denote the angle bisector of angle $X$ . Let $P=a\cap c$ and $Q=b\cap d$. Prove that $ABPQ$ is cyclic.
1 reply
gggzul
Monday at 4:44 PM
Adventure1000
24 minutes ago
Interesting question from Al-Khwarezmi olympiad 2024 P3, day1
Adventure1000   0
30 minutes ago
Find all $x, y, z \in \left (0, \frac{1}{2}\right )$ such that
$$
\begin{cases}
(3 x^{2}+y^{2}) \sqrt{1-4 z^{2}} \geq z; \\
(3 y^{2}+z^{2}) \sqrt{1-4 x^{2}} \geq x; \\
(3 z^{2}+x^{2}) \sqrt{1-4 y^{2}} \geq y.
\end{cases}
$$Proposed by Ngo Van Trang, Vietnam
0 replies
Adventure1000
30 minutes ago
0 replies
Inequalities
sqing   1
N 44 minutes ago by Royal_mhyasd
Let $ a,b>0 $ and $\frac{a}{a^2+3}+ \frac{b}{b^2+ 3} \geq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{a}{a^3+3}+ \frac{b}{b^3+ 3}\geq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
1 reply
sqing
4 hours ago
Royal_mhyasd
44 minutes ago
Hard Inequality
William_Mai   10
N an hour ago by sqing
Given $a, b, c \in \mathbb{R}$ such that $a^2 + b^2 + c^2 = 1$.
Find the minimum value of $P = ab + 2bc + 3ca$.

Source: Pham Le Van
10 replies
William_Mai
May 3, 2025
sqing
an hour ago
What is the limit?
Disjeje   1
N Today at 7:54 AM by Moubinool
Let’s say An=(sin(n))^n
Does An converge if n reaches infinity?
1 reply
Disjeje
Today at 5:45 AM
Moubinool
Today at 7:54 AM
Putnam 2002 B3
ahaanomegas   2
N Today at 7:36 AM by Rohit-2006
Show that for all integers $n>1$, \[ \dfrac {1}{2ne} < \dfrac {1}{e} - \left( 1 - \dfrac {1}{n} \right)^n < \dfrac {1}{ne}. \]
2 replies
ahaanomegas
Mar 12, 2012
Rohit-2006
Today at 7:36 AM
Putnam 2013 B1
Kent Merryfield   13
N Today at 5:38 AM by Rohit-2006
For positive integers $n,$ let the numbers $c(n)$ be determined by the rules $c(1)=1,c(2n)=c(n),$ and $c(2n+1)=(-1)^nc(n).$ Find the value of \[\sum_{n=1}^{2013}c(n)c(n+2).\]
13 replies
Kent Merryfield
Dec 9, 2013
Rohit-2006
Today at 5:38 AM
Can cos(√2 t) be expressed as a polynomial in cost?
tom-nowy   2
N Today at 4:54 AM by wh0nix
Source: Question arising while viewing https://artofproblemsolving.com/community/c51293h3562250
Can $\cos ( \sqrt{2}\,  t )$ be expressed as a polynomial in $\cos t$ with real coefficients?
2 replies
tom-nowy
Yesterday at 7:10 AM
wh0nix
Today at 4:54 AM
Rolles theorem
sasu1ke   7
N Yesterday at 7:27 PM by GentlePanda24

Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that
\[
f(0) = 2, \quad f'(0) = -2, \quad \text{and} \quad f(1) = 1.
\]Prove that there exists a point \( \xi \in (0, 1) \) such that
\[
f(\xi) \cdot f'(\xi) + f''(\xi) = 0.
\]

7 replies
sasu1ke
May 3, 2025
GentlePanda24
Yesterday at 7:27 PM
Convergent series with weight becomes divergent
P_Fazioli   7
N Yesterday at 3:07 PM by P_Fazioli
Initially, my problem was : is it true that if we fix $(b_n)$ positive such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$, then there exists $(a_n)$ positive such that $\displaystyle\sum_{n\geq 0}a_n$ converges and $\displaystyle\sum_{n\geq 0}a_nb_n$ diverges ?

Thinking about the continuous case : if $g:\mathbb{R}_+\longrightarrow\mathbb{R}$ is continuous, positive with $g(x)\underset{{x}\longrightarrow{+\infty}}\longrightarrow +\infty$, does $f$ continuous and positive exist on $\mathbb{R}_+$ such that $\displaystyle\int_0^{+\infty}f(x)\text{d}x$ converges and $\displaystyle\int_0^{+\infty}f(x)g(x)\text{d}x$ diverges ?

To the last question, the answer seems to be yes if $g$ is in the $\mathcal{C}^1$ class, increasing : I chose $f=\dfrac{g'}{g^2}$. With this idea, I had the idea to define $a_n=\dfrac{b_{n+1}-b_n}{b_n^2}$ but it is not clear that it is ok, even if $(b_n)$ is increasing.

Now I have some questions !

1) The main problem : is it true that if we fix $(b_n)$ positive such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$, then there exists $(a_n)$ positive such that $\displaystyle\sum_{n\geq 0}a_n$ converges and $\displaystyle\sum_{n\geq 0}a_nb_n$ diverges ? And if $(b_n)$ is increasing ?
2) is it true that if we fix $(b_n)$ positive increasing such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}\underset{{n}\longrightarrow{+\infty}}\longrightarrow 1$, then $\displaystyle\sum_{n\geq 0}\left(\frac{b_{n+1}}{b_n}-1\right)$ diverges ?
3) is it true that if we fix $(b_n)$ positive increasing such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}\underset{{n}\longrightarrow{+\infty}}\longrightarrow 1$, then $\displaystyle\sum_{n\geq 0}\frac{b_{n+1}-b_n}{b_n^2}$ converges ?
4) if $(b_n)$ is positive increasing and such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}$ does not converge to $1$, can $\displaystyle\sum_{n\geq 0}\frac{b_{n+1}-b_n}{b_n^2}$ diverge ?
5) for the continuous case, is it true if we suppose $g$ only to be continuous ?

7 replies
P_Fazioli
May 5, 2025
P_Fazioli
Yesterday at 3:07 PM
Cool Integral, Cooler Solution
Existing_Human1   2
N Yesterday at 11:33 AM by ysharifi
Source: https://youtu.be/YO38MCdj-GM?si=DCn6DaQTeX8RXhl0
$$\int_{0}^{\infty} \! e^{-x^2}\cos(5x) \,dx$$
Bonus points if you can do it without Feynman
2 replies
Existing_Human1
Yesterday at 2:15 AM
ysharifi
Yesterday at 11:33 AM
Putnam 2016 A1
Kent Merryfield   15
N Yesterday at 10:51 AM by anudeep
Find the smallest positive integer $j$ such that for every polynomial $p(x)$ with integer coefficients and for every integer $k,$ the integer
\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\](the $j$-th derivative of $p(x)$ at $k$) is divisible by $2016.$
15 replies
Kent Merryfield
Dec 4, 2016
anudeep
Yesterday at 10:51 AM
Determinant is 1
Entrepreneur   2
N Yesterday at 8:27 AM by Entrepreneur
If a determinant is of $n^{\text{th}}$ order, and if the constituents of its first, second, ..., $n^{\text{th}}$ rows are the first $n$ figurate numbers of the first, second, ..., $n^{\text{th}}$ orders respectively, show that it's value is $1.$
2 replies
Entrepreneur
Monday at 7:14 PM
Entrepreneur
Yesterday at 8:27 AM
36x⁴ + 12x² - 36x + 13 > 0
fxandi   2
N Yesterday at 7:14 AM by MeKnowsNothing
Prove that for any real $x \geq 0$ holds inequality $36x^4 + 12x^2 - 36x + 13 > 0.$
2 replies
fxandi
Monday at 10:02 PM
MeKnowsNothing
Yesterday at 7:14 AM
Hard number theory
td12345   7
N Apr 10, 2025 by td12345
Let $p$ be a prime number. Find the number of elements of the set
\[
M_p = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2p^{n} x} \in \mathbb{Q} \right\}. 
\]
Edit: I edited with a variation of the same problem because I intend to give this one to an exam for my pupils.
7 replies
td12345
Apr 9, 2025
td12345
Apr 10, 2025
Hard number theory
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td12345
233 posts
#1
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Let $p$ be a prime number. Find the number of elements of the set
\[
M_p = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2p^{n} x} \in \mathbb{Q} \right\}. 
\]
Edit: I edited with a variation of the same problem because I intend to give this one to an exam for my pupils.
This post has been edited 1 time. Last edited by td12345, Apr 14, 2025, 6:17 PM
Reason: variation of the problem
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mathprodigy2011
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Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.
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td12345
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Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.
This post has been edited 1 time. Last edited by td12345, Apr 10, 2025, 1:57 AM
Reason: typo
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mathprodigy2011
325 posts
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td12345 wrote:
Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

aren't these solutions also 2 times perfect squares? I expanded to get 2(2^2023-1)^2. And when I let x=2m^2; and we if i made the same difference of squares; i would get (a-m)(a+m) = 2^2025. Letting a-m=2 and a+m=2^2024 wll give us m=2^2023-1 which results in the solution you have mentioned
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td12345
233 posts
#5
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mathprodigy2011 wrote:
td12345 wrote:
Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

aren't these solutions also 2 times perfect squares? I expanded to get 2(2^2023-1)^2. And when I let x=2m^2; and we if i made the same difference of squares; i would get (a-m)(a+m) = 2^2025. Letting a-m=2 and a+m=2^2024 wll give us m=2^2023-1 which results in the solution you have mentioned

Yes, they lead to the same one when computing the square root, but the set counts the \(x\) not the set of the square root expression no? I got a different number from yours and I don't know where I went wrong, I will lay down the \(x\) I found as it goes over your claimed result and maybe one of us can figure out:

For \( M_2\):
- \(x\) can be \( (2^{2025-k}-2^{k-1})^2 \) where \(k \in \{1,2,\dots, 2025\}\) . This leads to \( [\frac{2025+1}{2}]\) values of \(x\).
- \(x\) can be \( -(2^{2 \cdot 2025-2k}+2^{2k-2}+2^{2025})\) where \(k \in \{1,2,\dots, 2025\}\) . This leads to \( [\frac{2025+1}{2}]\) values of \(x\).
- \(x\) can be $2^{2 \cdot 2025-3}+2-2^{2025},  -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. This leads to \(2\) extra values for \(x\) distinct from the above.

For \(M_{2027}\):
- \(x\) can be \( \frac{2027^k+2027^{2 \cdot 2025 - k}}{2} -2027^{2025} \) where \(k \in \{0, 1,2,\dots, 2 \cdot 2025\}\) . This leads to \( 2025\) distinct non-zero values of \(x\).
- \(x\) can be \( \frac{-2027^k-2027^{2 \cdot 2025 - k}}{2} -2027^{2025} \) where \(k \in \{0, 1,2,\dots, 2 \cdot 2025\}\) . This leads to \( 2026\) distinct non-zero values of \(x\).
I double checked with W|A to see if they lead to perfect squares under the square root and they do, so unless I double counted, the answer should be $2026+2+2025 + 2026$?


Edit: ok there is a miscount for \( M_2\) because when \(k=1013\) it leads to \(0\) for the very first family and \(x\) can't be \(0\). So for \(M_2\) we shall have $1012+1013+2=2027$. But for \(M_{2027}\) count seems solid? I got $2024+2+2025+2026$
This post has been edited 8 times. Last edited by td12345, Apr 10, 2025, 8:34 PM
Reason: tpyo
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td12345
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Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?
This post has been edited 2 times. Last edited by td12345, Apr 10, 2025, 8:40 PM
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mathprodigy2011
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td12345 wrote:
Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?

ohhhh so its not just positive integers? i just never have seen z with an asterick so thats probably where i went wrong
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td12345
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mathprodigy2011 wrote:
td12345 wrote:
Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?

ohhhh so its not just positive integers? i just never have seen z with an asterick so thats probably where i went wrong

Yes it's integers that are not 0. $\mathbb{N}$ is natural numbers (positive integers) and \( \mathbb{N}^*\) is natural numbers without \(0\). Would you like to give it another try just to see if it's really $8100$?
This post has been edited 1 time. Last edited by td12345, Apr 10, 2025, 9:30 PM
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