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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Wednesday at 3:18 PM
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Wednesday at 3:18 PM
0 replies
1 area = 2025 points
giangtruong13   0
14 minutes ago
In a plane give a set $H$ that has 8097 distinct points with area of a triangle that has 3 points belong to $H$ all $ \leq 1$. Prove that there exists a triangle $G$ that has the area $\leq 1 $ contains at least 2025 points that belong to $H$( each of that 2025 points can be inside the triangle or lie on the edge of triangle $G$)X
0 replies
giangtruong13
14 minutes ago
0 replies
Burak0609
Burak0609   0
20 minutes ago
So if $2 \nmid n\implies$ $2d_2+d_4+d_5=d_7$ is even it's contradiction. I mean $2 \mid n and d_2=2$.
If $3\mid n \implies d_3=3$ and $(d_6+d_7)^2=n+1,3d_6d_7=n \implies d_6^2-d_6d_7+d_7^2=1$,we can see the only solution is$d_6=d_7=1$ and it is contradiction.
If $4 \mid n d_3=4$ and $(d_6+d_7)^2-4d_6d_7=1 \implies d_7=d_6+1$. So $n=4d_6(d_6+1)$.İt means $8 \mid n$.
If $d_6=8 n=4.8.9=288$ but $3 \nmid n$.İt is contradiction.
If $d_5=8$ we have 2 option. Firstly $d_4=5 \implies 2d_2+d_4+d_5=17=d_7 d_6=16$ but $10 \mid n$ is contradiction. Secondly $d_4=7 \implies d_7=2.2+7+8=19 and d_6=18$ but $3 \nmid n$ is contradiction. I mean $d_4=8 \implies d_7=d_5+12, n=4(d_5+11)(d_5+12) and d_5 \mid n=4(d_5+11)(d_5+12)$. So $d_5 \mid 4.11.12 \implies d_5 \mid 16.11$. If $d_5=16 d_6=27$ but $3 \nmid n$ is contradiction. I mean $d_5=11,d_6=22,d_7=23$. The only solution is $n=2024$.
0 replies
Burak0609
20 minutes ago
0 replies
Good Partitions
va2010   25
N 43 minutes ago by lelouchvigeo
Source: 2015 ISL C3
For a finite set $A$ of positive integers, a partition of $A$ into two disjoint nonempty subsets $A_1$ and $A_2$ is $\textit{good}$ if the least common multiple of the elements in $A_1$ is equal to the greatest common divisor of the elements in $A_2$. Determine the minimum value of $n$ such that there exists a set of $n$ positive integers with exactly $2015$ good partitions.
25 replies
va2010
Jul 7, 2016
lelouchvigeo
43 minutes ago
An inequality on triangles sides
nAalniaOMliO   7
N 2 hours ago by navier3072
Source: Belarusian National Olympiad 2025
Numbers $a,b,c$ are lengths of sides of some triangle. Prove the inequality$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \geq \frac{a+b}{2c}+\frac{b+c}{2a}+\frac{c+a}{2b}$$
7 replies
nAalniaOMliO
Mar 28, 2025
navier3072
2 hours ago
No more topics!
Geometric Sequence Squared
scls140511   5
N Mar 30, 2025 by Anthony2025
Source: China Round 1 (Gao Lian)
2 Let there be an infinite geometric sequence $\{a_n\}$, where the common ratio $0<|q|<1$. Given that

$$\sum_{i=1}^\infty a_n = \sum_{i=1}^\infty a_n^2$$
find the largest possible range of $a_2$.
5 replies
scls140511
Sep 8, 2024
Anthony2025
Mar 30, 2025
Geometric Sequence Squared
G H J
G H BBookmark kLocked kLocked NReply
Source: China Round 1 (Gao Lian)
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scls140511
106 posts
#1 • 1 Y
Y by Fah2008
2 Let there be an infinite geometric sequence $\{a_n\}$, where the common ratio $0<|q|<1$. Given that

$$\sum_{i=1}^\infty a_n = \sum_{i=1}^\infty a_n^2$$
find the largest possible range of $a_2$.
This post has been edited 1 time. Last edited by scls140511, Sep 8, 2024, 12:52 PM
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sami1618
881 posts
#2
Y by
We claim that we must have $a_2\in (-1/4,0)\cup (0,2)$. First recall the following fact.
If a infinite geometric sequence has starting term $a$ and ratio $-1<r<1$ then the sum is given by $$S=\frac{a}{1-r}$$
Thus we must have that $$\frac{a_1}{1-q}=\frac{a_1^2}{1-q^2}\iff 1+q=a_1$$Thus we have that $$a_2=a_1q=q^2+q$$As $q\in (-1,0)\cup (0,1)$ thus $a_2=q^2+q=(q+\frac{1}{2})^2-\frac{1}{4}$ and the finish is easy.
@ below thanks for fixing my mistake (misread $|q|$ as $q$)
This post has been edited 1 time. Last edited by sami1618, Sep 8, 2024, 11:34 PM
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Bluesoul
890 posts
#3 • 1 Y
Y by sami1618
You missed [-1/4,0)
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AshAuktober
958 posts
#4
Y by
Note that $$\frac{a_1}{q - 1} = \frac{a_1^2}{q^2-1}$$$$\implies q+1= a_1.$$Now we just have to find the range of $q(q+1) = \left(q+\frac 1 2 \right)^2 - \frac 1 4$ as $q$ ranges over $(-1, 1) \setminus 0$. Over $(-1, 1)$ the expression takes values in the interval $(-1/4, 2)$, and at $q = 0$ it takes the value $0$, and nowhere else does it take this value. So the answer is $(-1/4, 2) \setminus \{0\}$. $\square$
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Wangguanhua
4 posts
#5
Y by
I failed to exclude 0 from the range during the contest.
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Anthony2025
3 posts
#6
Y by
a horrible $0$ made me lost eight points!
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