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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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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0 replies
jlacosta
Apr 2, 2025
0 replies
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Common tangent to diameter circles
Stuttgarden   1
N a few seconds ago by jrpartty
Source: Spain MO 2025 P2
The cyclic quadrilateral $ABCD$, inscribed in the circle $\Gamma$, satisfies $AB=BC$ and $CD=DA$, and $E$ is the intersection point of the diagonals $AC$ and $BD$. The circle with center $A$ and radius $AE$ intersects $\Gamma$ in two points $F$ and $G$. Prove that the line $FG$ is tangent to the circles with diameters $BE$ and $DE$.
1 reply
Stuttgarden
Mar 31, 2025
jrpartty
a few seconds ago
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Two Functional Inequalities
Mathdreams   2
N a minute ago by kokcio
Source: 2025 Nepal Mock TST Day 2 Problem 2
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x) \le x^3$$and $$f(x + y) \le f(x) + f(y) + 3xy(x + y)$$for any real numbers $x$ and $y$.

(Miroslav Marinov, Bulgaria)
2 replies
Mathdreams
13 minutes ago
kokcio
a minute ago
J
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Beautiful problem
luutrongphuc   15
N 2 minutes ago by r7di048hd3wwd3o3w58q
Let triangle $ABC$ be circumscribed about circle $(I)$, and let $H$ be the orthocenter of $\triangle ABC$. The circle $(I)$ touches line $BC$ at $D$. The tangent to the circle $(BHC)$ at $H$ meets $BC$ at $S$. Let $J$ be the midpoint of $HI$, and let the line $DJ$ meet $(I)$ again at $X$. The tangent to $(I)$ parallel to $BC$ meets the line $AX$ at $T$. Prove that $ST$ is tangent to $(I)$.
15 replies
luutrongphuc
Apr 4, 2025
r7di048hd3wwd3o3w58q
2 minutes ago
J
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FE with a lot of terms
MrHeccMcHecc   0
3 minutes ago
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ $$f(x)f(y)+f(x+y)=xf(y)+yf(x)+f(xy)+x+y+1$$
0 replies
MrHeccMcHecc
3 minutes ago
0 replies
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No more topics!
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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
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Geometric Sequence Squared
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Reveal topic tags
geometric sequence
ratio
L
G H BBookmark kLocked kLocked NReply
Source: China Round 1 (Gao Lian)
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scls140511
106 posts
scls140511
#1 h  Sep 8, 2024, 12:52 PM • 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
883 posts
sami1618
#2 h  Sep 8, 2024, 4:20 PM
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
891 posts
Bluesoul
#3 h  Sep 8, 2024, 9:55 PM • 1 Y
Y by sami1618
You missed [-1/4,0)
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AshAuktober
967 posts
AshAuktober
#4 h  Sep 20, 2024, 9:59 AM
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
Wangguanhua
#5 h  Mar 15, 2025, 12:45 AM
Y by
I failed to exclude 0 from the range during the contest.
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Anthony2025
3 posts
Anthony2025
#6 h  Mar 30, 2025, 3:27 AM
Y by
a horrible $0$ made me lost eight points!
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