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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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jlacosta
Apr 2, 2025
0 replies
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Hard limits
Snoop76   4
N 3 minutes ago by wh0nix
$a_n$ and $b_n$ satisfies the following recursion formulas: $a_{0}=1, $ $b_{0}=1$, $ a_{n+1}=a_{n}+b_{n}$$ $ and $ $$ b_{n+1}=(2n+3)b_{n}+a_{n}$. Find $ \lim_{n \to \infty} \frac{a_n}{(2n-1)!!}$ $ $ and $ $ $\lim_{n \to \infty} \frac{b_n}{(2n+1)!!}.$
4 replies
Snoop76
Mar 25, 2025
wh0nix
3 minutes ago
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Old or new
sqing   1
N 4 minutes ago by lbh_qys
Source: ZDSX 2025 Q845
Let $ a,b,c>0 $ and $ a^2+b^2+c^2+ abc=4 $ . Prove that $$1\leq \frac{1}{2a+bc }+ \frac{1}{2b+ca }+ \frac{1}{2c+ab }\leq \frac{1}{\sqrt{abc} }$$
1 reply
1 viewing
sqing
4 hours ago
lbh_qys
4 minutes ago
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<DPA+ <AQD =< QIP wanted, incircle circumcircle related
parmenides51   40
N 7 minutes ago by ihategeo_1969
Source: IMo 2019 SL G6
Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$.

(Slovakia)
40 replies
parmenides51
Sep 22, 2020
ihategeo_1969
7 minutes ago
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GCD of x^2-y, y^2-z and z^2-x
EmersonSoriano   2
N 12 minutes ago by atdaotlohbh
Source: 2018 Peru Southern Cone TST P5
Find all positive integers $d$ that can be written in the form
$$ d = \gcd(|x^2 - y| , |y^2 - z| , |z^2 - x|), $$where $x, y, z$ are pairwise coprime positive integers such that $x^2 \neq y$, $y^2 \neq z$, and $z^2 \neq x$.
2 replies
EmersonSoriano
Apr 2, 2025
atdaotlohbh
12 minutes ago
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BMO2 2016 Q.1
sbealing   8
N Mar 21, 2016 by NgoNgang
Source: BMO2 2016
Three circles of radius $r_{1},r_{2},r_{3}$ touch each other externally and share a common tangent. The tangent touches the three circles at $A,B,C$ with $B$ between $A$ and $C$. Show that:
$16(r_{1}+r_{2}+r_{3})\geq9(AB+BC+CA)$
8 replies
sbealing
Jan 31, 2016
NgoNgang
Mar 21, 2016
J
BMO2 2016 Q.1
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Reveal topic tags
geometry
inequalities
BritishMathematicalOlympiad
L
G H BBookmark kLocked kLocked NReply
Source: BMO2 2016
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sbealing
307 posts
sbealing
#1 h  Jan 31, 2016, 1:45 PM • 2 Y
Y by Adventure10, Mango247
Three circles of radius $r_{1},r_{2},r_{3}$ touch each other externally and share a common tangent. The tangent touches the three circles at $A,B,C$ with $B$ between $A$ and $C$. Show that:
$16(r_{1}+r_{2}+r_{3})\geq9(AB+BC+CA)$
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mjuk
196 posts
mjuk
#2 h  Jan 31, 2016, 3:03 PM • 4 Y
Y by sbealing, chenguk2001, Adventure10, Mango247
By Pythagoras Theorem:
$AB=\sqrt{(r_1+r_3)^2-(r_1-r_3)^2}=\sqrt{4r_1r_3}$
$BC=\sqrt{(r_2+r_3)^2-(r_2-r_3)^2}=\sqrt{4r_2r_3}$
$AC=\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=\sqrt{4r_1r_2}$
$AC=AB+BC\rightarrow r_3=\frac{r_1r_2}{(\sqrt{r_1}+\sqrt{r_2})^2}$

$16(r_1+r_2+r_3)=16r_1+16r_2+16\cdot \frac{r_1r_2}{(\sqrt{r_1}+\sqrt{r_2})^2}$
(1) $2r_1+2r_2+16\cdot \frac{r_1r_2}{(\sqrt{r_1}+\sqrt{r_2})^2}\stackrel{AG}{\geq} (\sqrt{r_1}+\sqrt{r_2})^2+16\cdot \frac{r_1r_2}{(\sqrt{r_1}+\sqrt{r_2})^2}\stackrel{AG}{\geq} 8\sqrt{r_1r_2}$
(2) $14r_1+14r_2\stackrel{AG}{\geq} 28 \sqrt{r_1r_2}$
(1)+(2) $16(r_1+r_2+r_3)=16r_1+16r_2+16\cdot \frac{r_1r_2}{(\sqrt{r_1}+\sqrt{r_2})^2}\geq 36\sqrt{r_1r_2}=18\cdot AC=9(AB+BC+AC)$
Attachments:
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Reason: s
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Lonesan
2170 posts
Lonesan
#3 h  Jan 31, 2016, 3:55 PM • 3 Y
Y by adityaguharoy, Adventure10, Mango247

An amazing result (I bet you didn't know! :-) - just joking! )
The following identity I found in a japanese book:

$\frac{1}{ \sqrt{r_3}} = \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}} \ \ ; $


Greetings! :-)
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AndrewTom
12750 posts
AndrewTom
#4 h  Jan 31, 2016, 8:07 PM • 1 Y
Y by Adventure10
How do we prove this result posted by Lonesan?
This post has been edited 1 time. Last edited by AndrewTom, Jan 31, 2016, 8:26 PM
Reason: Added a question mark?
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mjuk
196 posts
mjuk
#5 h  Jan 31, 2016, 8:30 PM • 3 Y
Y by AndrewTom, Adventure10, Mango247
AndrewTom wrote:
How do we prove this result posted by Lonesan?

It's equivalent to my post: $r_3=\frac{r_1r_2}{(\sqrt{r_1}+\sqrt{r_2})^2}$
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jlammy
1099 posts
jlammy
#6 h  Jan 31, 2016, 8:43 PM • 2 Y
Y by Adventure10, Mango247
http://www.artofproblemsolving.com/community/c6h1191264
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AndrewTom
12750 posts
AndrewTom
#7 h  Jan 31, 2016, 8:43 PM • 2 Y
Y by Adventure10, Mango247
Thanks mjuk.

Are there other results concerning this configuration of three circles?
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NgoNgang
62 posts
NgoNgang
#8 h  Mar 21, 2016, 7:09 PM • 2 Y
Y by Adventure10, Mango247
The similar problem was here (Baltic Way 1991 problem 16)

http://artofproblemsolving.com/community/c6h530461p3027186
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NgoNgang
62 posts
NgoNgang
#9 h  Mar 21, 2016, 7:09 PM • 2 Y
Y by Adventure10, Mango247
Lonesan wrote:
An amazing result (I bet you didn't know! :-) - just joking! )
The following identity I found in a japanese book:

$\frac{1}{ \sqrt{r_3}} = \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}} \ \ ; $


Greetings! :-)

Right here http://artofproblemsolving.com/community/c6h530461p3027186
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