and 
Prove that
Y by NewAlbionAcademy, airplanes1, Neversmile123, bryanbr2, Yousefi, sanabhai, HMoghaddas, dantx5, Jerry37284, Wizard_32, fjm30, Functional_equation, Math-wiz, Adventure10, Mango247, and 8 other users
Problem 85 was one of IRAN TST 2011 problems.
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k a April Highlights and 2025 AoPS Online Class Information
jlacosta 0
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0 replies
Coaxal Circles
fattypiggy123 29
N
by sttsmet
Source: China TSTST Test 2 Day 1 Q3
Let 
be a quadrilateral and let 
be a line. Let intersect the lines 
at points 
respectively. Given that these six points on are in the order 
, show that the circles with diameter 
are coaxal.
be a quadrilateral and let 
be a line. Let intersect the lines 
at points 
respectively. Given that these six points on are in the order 
, show that the circles with diameter 
are coaxal.
29 replies
q(x) to be the product of all primes less than p(x)
orl 16
N
by Maximilian113
Source: IMO Shortlist 1995, S3
For an integer 
, let 
be the least prime that does not divide 
, and define 
to be the product of all primes less than . In particular, 
For having 
, define 
. Consider the sequence 
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](//latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
. Find all 
such that 
.
, let 
be the least prime that does not divide 
, and define 
to be the product of all primes less than . In particular, 
For having 
, define 
. Consider the sequence 
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
. Find all 
such that 
.
16 replies
1 viewing
inequalities
pennypc123456789 1
N
by KhuongTrang
If 
are positive real numbers, then
we can generalize this problem
are positive real numbers, then
we can generalize this problem
1 reply
No more topics!
116 problems
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Source: By Mohammad Jafari
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Y by Adventure10, Mango247
hello, nice problems, thank you.
Sonnhard.
Sonnhard.
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Y by Adventure10, Mango247
Thanks Mr.Jafari for the problems.
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Y by Adventure10, Mango247
Great collection, thanks Mr. Jafari! 
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Y by Adventure10
Mods - What's your preferred way for us to discuss these problems?
Dang it, I can't do #1 !
Merlin
Dang it, I can't do #1 !
Merlin
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Y by Yoni2b, socrates, Amir Hossein, Adventure10, Mango247, and 1 other user
Here are a few of my solutions 
from 
that satisfy:![\[f(x+y)=f(x)f(y)+xy\]](http://latex.artofproblemsolving.com/3/9/b/39b3dae6bf6d3444d5159700c13338739a546fbb.png)
be the above assertion. Then:![\[P(0, y)\implies f(y)=f(0)f(y)\]](http://latex.artofproblemsolving.com/6/1/d/61d752533bbcb4d5127d84620040d31be8078752.png)
is always 
,
)
.![\[P(x, -x)\implies f(x)f(-x)-x^2=1\]](http://latex.artofproblemsolving.com/e/3/4/e3470c72fad34e9bb1280041e49b1fd776be559a.png)
.![\[1+4a^2=f(-2a)f(2a)=(f(a)^2+a^2)(f(-a)^2+a^2)\ge a^4\]](http://latex.artofproblemsolving.com/e/f/b/efbd4f5e83e7873fc4667f90b620b4b125b88e10.png)
,
,
such that:![\[f(x+f(x)+2y)=x+f(x)+2f(y)\]](http://latex.artofproblemsolving.com/1/d/3/1d39addc737b56ec321959a07e3217b97a52ff93.png)
is bijective and that 
.![\[P(x, -\frac{f(x)}{2})\implies -\frac{x}{2}=f(-\frac{f(x)}{2})\]](http://latex.artofproblemsolving.com/3/b/7/3b75519ca6442c051dd92ebe5e42700305d7f3f0.png)
,
,
and thus 
.
.
and 
we have:![\[f(x+f(x)+2y)=x+f(f(x))+2f(y)\]](http://latex.artofproblemsolving.com/f/c/9/fc9bbca777bb244e0ea603d78b1466bcbc022b60.png)
![\[P(-2a, a)\implies f(a)=a\]](http://latex.artofproblemsolving.com/6/7/5/67570295b5bfc9fb434db92fd947c3d9cf543a26.png)
such that for all real ![\[(x+y)(f(x)-f(y))=(x-y)(f(x)+f(y))\]](http://latex.artofproblemsolving.com/9/5/1/951321151c0ca86911fe89929230826d0b460413.png)
,
,
into the above. This gives 
so indeed ![\[f(x+y)(x-y)=f(x-y)(x+y)\]](http://latex.artofproblemsolving.com/4/8/5/485e4499a397e24280bf4a66fb87553ee6922fe7.png)
be arbitrary positive reals, and let ![\[P(\frac{a+b}{2}, \frac{a-b}{2})\implies \frac{f(a)}{a}=\frac{f(b)}{b}\]](http://latex.artofproblemsolving.com/d/7/2/d724e94a31ee8f6e000dfeaa14a8d2c570d0a661.png)
for a real constant 
,
into the above with 
)
be positive reals with 
.![\[\frac{(x^2+1)^2}{4x}+\frac{(y^2+1)^2}{4y}+\frac{(z^2+1)^2}{4z}\ge x+y+z\]](http://latex.artofproblemsolving.com/8/2/a/82a6ae47b24ca72af4dd27a1269acc98b14e980e.png)
,
,
as desired.
be positive reals, show that:![\[(\sum\frac{a^2}{b+c})(\sum\frac{a}{(b+c)^2})\ge \frac{9}{8}\]](http://latex.artofproblemsolving.com/7/d/9/7d97ec7b67ca95022db3bd4a87f31f06a9c4b2d4.png)
![\[LHS\ge \frac{1}{9}(\sum a)(\sum a^2)(\sum\frac{1}{b+c})(\sum\frac{1}{(b+c)^2})\ge\]](http://latex.artofproblemsolving.com/e/1/4/e14aaf53cd36b4081cb413b2cf4edb6ec5f47667.png)
![\[\frac{1}{81}(\sum a)^3(\sum\frac{1}{b+c})^3\ge \]](http://latex.artofproblemsolving.com/5/3/a/53a854220b6ae8d94d5f03b3a2dfa46c99285d5c.png)
![\[9(\sum a)^3(\frac{1}{2(a+b+c)})^3=\frac{9}{8}\]](http://latex.artofproblemsolving.com/4/2/3/4236a7df038a856377c45e709e73a7dc6864924c.png)
![\[4\sqrt{x+y+z}\ge x+y+z+3\]](http://latex.artofproblemsolving.com/c/a/4/ca48295be55ac15a64fd92dc068a17c83b140feb.png)
.![\[pq\ge 9\implies 10=p+q\ge p+\frac{9}{p}\implies (p-1)(p-9)\le 0\]](http://latex.artofproblemsolving.com/b/e/a/bea96954d9dede86e3a7a0c53846d7addfa7e4b6.png)
.![\[(\sqrt{p}-3)(\sqrt{p}-1)\le 0\]](http://latex.artofproblemsolving.com/9/d/b/9dbd0022aa8aebe44587f14df815cabafc55864f.png)
.![\[ min\{\frac{a}{1+b^2}, \frac{b}{1+c^2}, \frac{c}{1+a^2}\}\le \frac{1}{2}\]](http://latex.artofproblemsolving.com/a/2/b/a2b9a34a51358a76e0d18906583c58193e247e70.png)
![\[(\frac{a}{1+b^2})(\frac{b}{1+c^2})(\frac{c}{1+a^2}\})=(\frac{a}{1+a^2})(\frac{b}{1+b^2})(\frac{c}{1+c^2}\})\le \frac{1}{8}\]](http://latex.artofproblemsolving.com/b/f/9/bf95d92d68d6dc5dc8aaea918de4d54958b8e656.png)
by AM-GM, and the result follows.![\[ab=\frac{c^2}{1+c^2}\\bc=\frac{a^2}{1+a^2}
\\ca=\frac{b^2}{1+b^2}\]](http://latex.artofproblemsolving.com/9/f/a/9faccb9ac234b121b55ecd5874d0541d33a27643.png)
then it immediately follows 
,
.![\[(ab, bc, ca), (\frac{c^2}{1+c^2}, \frac{a^2}{1+a^2}, \frac{b^2}{1+b^2})\]](http://latex.artofproblemsolving.com/6/8/c/68c71f32fa449b64e9cec7b16d5cadcc027bb1c6.png)
.
,
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Y by Amir Hossein, Adventure10, and 1 other user
Problem 43
Problem 30
Problem 31
Problem 26
Problem 81, Stronger Version
Problem 60
Problem 85
Problem 42
Problem 6
Problem 5
Problem 2
Problem 20
Problem 4
Problem 34
Problem 33
Other problems:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=378717
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=378367
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=378164
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=378718
Problem 30
Problem 31
Problem 26
Problem 81, Stronger Version
Problem 60
Problem 85
Problem 42
Problem 6
Problem 5
Problem 2
Problem 20
Problem 4
Problem 34
Problem 33
Other problems:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=378717
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=378367
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=378164
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=378718
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Y by Amir Hossein, applepi2000, kienthuckiquai, Adventure10, Mango247
that for all 
satisfy:![\[f(x+y+f(x+y))=2f(x)+2f(y)\]](http://latex.artofproblemsolving.com/f/c/d/fcd130e403a7011632842e9f696c6aca306a045a.png)
to get ![\[f(x+f(x))=2f(x)+2f(0).\]](http://latex.artofproblemsolving.com/7/4/8/748550f1ae56393a674b2135811103a76b7aa3ca.png)
into the last equality to get ![\[f(x+y+f(x+y))=2f(x+y)+2f(0).\]](http://latex.artofproblemsolving.com/4/0/9/409cd74d58e44ef0fbd58ae46ec3bfc963e024aa.png)
![\[f(x+y)-f(0)=f(x)-f(0)+f(y)-f(0),\]](http://latex.artofproblemsolving.com/c/b/6/cb6f853b68b960239197f7cb8e9c8e0cf35d8191.png)
be a function that for all 
satisfies:![\[f(x+f(x)+y)=x+f(x)+2f(y).\]](http://latex.artofproblemsolving.com/3/b/d/3bd795029663688092f6a8d433c8c4c253cc1df0.png)
such that 
Then 
give 
So 
we get ![\[f(-f(x))=-\frac{x}{2},\]](http://latex.artofproblemsolving.com/3/2/c/32c7e1ae39b2b1b511074ca8f538e12711d24b73.png)
so 
so, since 
gives 
which is not a solution.
such that for all ![\[f(x+f(x)+2f(y))=x+f(x)+y+f(y)\]](http://latex.artofproblemsolving.com/1/d/5/1d5b336d6e3c8afbac9fee3854073edf80cb6479.png)
to get ![\[f(x+f(x)+2f(y))=x+f(x)+y+f(y)=f(y+f(y)+2f(x)),\]](http://latex.artofproblemsolving.com/8/f/6/8f6c6a68ec2e621d0abdea5443063a1f4b9bf0f0.png)
that is 
is constant: 
![\[f(2x+2f(y))=x+f(x)+y+f(y)\]](http://latex.artofproblemsolving.com/5/5/c/55c4e7cb219ce1fabe5b08d64869d97af1341d5c.png)
so ![\[f(f(x)+2f(y))=f(x)+y+f(y).\]](http://latex.artofproblemsolving.com/2/4/7/24714e47ffe537f4824239289d762103107ef3f0.png)
gives 
and so 
![\[f(x+f(x))=2x.\]](http://latex.artofproblemsolving.com/1/2/a/12a58d23a3e27754f54e763005d1486aeed429a6.png)
is a bijective function.
So 
is clearly injective.
arbitrary and let 
so by injectivity of 
qed
be functions such that 
is bijective, satisfying 
for all 
so by injectivity of ![\[f(2x+2f(y))=x+f(x)+2y.\]](http://latex.artofproblemsolving.com/9/0/1/901672db98f60b83d4f4483f4e9dc8028922a26f.png)
so 
or 
from the surjectivity of 
which is indeed a solution.
such that for all 
:![\[f\left(\frac{x+f(x)}{2}+y\right)=f(x)+y.\]](http://latex.artofproblemsolving.com/b/0/f/b0f8f7e80861db53eb1fbdf64251cc32e8d5863a.png)
![\[y:=\frac{y+f(y)}{2}\]](http://latex.artofproblemsolving.com/d/d/e/dde74d206ee68ad3c18ade379101ca508b10e6ca.png)
to get ![\[f\left(\frac{x+f(x)}{2}+\frac{y+f(y)}{2}\right)=f(x)+\frac{y+f(y)}{2}.\]](http://latex.artofproblemsolving.com/2/8/e/28e428120a4dd921fcb1fdc07bb40846666780f3.png)
![\[f(y)+\frac{x+f(x)}{2}=f\left(\frac{x+f(x)}{2}+\frac{y+f(y)}{2}\right)=f(x)+\frac{y+f(y)}{2}\]](http://latex.artofproblemsolving.com/1/c/b/1cb2083eccc93db8cecf3c7099fe0fa9d7ce5cde.png)
that is 
such that for all ![\[f\left(\frac{x+f(x)}{2}+f(y)\right)=f(x)+y.\]](http://latex.artofproblemsolving.com/b/9/b/b9b509f8ba0438f6bab9116c33c6e5ac798d5125.png)
so 
respectively.
which is indeed a solution.
be a function such that
for all 
The number of the elements of the set 
is finite.
such that 
and 
so 
for some 
for each 
contradicting the second condtion.
for all 
there exists some 
such that 
that is 
we have 
and 
there exists unique 
such that 
for each 
and substituing 
:![\[f(f(f(x))+f(y)+z)=x+f(y)+f(f(z)) \]](http://latex.artofproblemsolving.com/a/4/f/a4fe456287bef1269644dc4e58ba23d769d7964f.png)
we see that 
we get 
so, by injectivity, 
and finally
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.![\[\sum\frac{x}{x^4+y+z+1}\le 1\]](http://latex.artofproblemsolving.com/a/4/9/a4903ed28c3af5c1eb1a81bf979f666d813978db.png)
,![\[x^4+3-x\ge 3x\Leftrightarrow x^4+1+1+1\ge 4x\]](http://latex.artofproblemsolving.com/2/f/2/2f22a039b782d7d564b3b50c4db7e91aaf818dc9.png)
and 
,![\[\sum\sqrt[n]{\frac{a}{(a+b)(a+c)}}\le \frac{3}{\sqrt[n]{abc}}\]](http://latex.artofproblemsolving.com/6/b/4/6b445ce7209c1df53022d02cf73b719bc9523997.png)
,![\[\sum\frac{a}{(a+b)(a+c)}\le \frac{3}{abc}\Leftrightarrow\sum a^2bc(b+c)\le 3(a+b)(a+c)(b+c)\]](http://latex.artofproblemsolving.com/9/5/e/95ea8f4f4513008a5ef023048c4818e34917d039.png)
![\[\sum_{sym}a^4b+3\sum_{sym}a^3b^2+6\sum a^3bc\ge 14\sum a^2b^2c\]](http://latex.artofproblemsolving.com/5/e/8/5e85a81a863b5cd20914bed1a72f66ab433c8eec.png)
![\[\sum\frac{a}{a^4+a^2+1}<\frac{4}{3}\]](http://latex.artofproblemsolving.com/8/9/b/89b5b6c077833f90438361b6b745d93cf158545c.png)
![\[4a^4+4a^2-9a+4>0\]](http://latex.artofproblemsolving.com/8/1/f/81ffc53e7a08a19c21b0efd82998e6a12373e00e.png)
is positive, subtract it from both sides and it suffices to show:![\[16a^3+7a>20a^2\]](http://latex.artofproblemsolving.com/5/0/7/5077fda7bf2cbc82b49731ff9a51666d5964d90d.png)
![\[\sum\frac{c}{3c+a+b}\le\frac{3}{5}\]](http://latex.artofproblemsolving.com/3/f/7/3f7f5c6f0312f9ddab7969bb90d06cd77cdbda7b.png)
and taking ![\[\sum\frac{a+b}{3c+a+b}\ge \frac{6}{5}\]](http://latex.artofproblemsolving.com/7/e/c/7ecd924ba5d741f5edf67d0fcdafad600e72b220.png)
![\[LHS=\sum\frac{(a+b)^2}{3ac+3bc+2ab+a^2+b^2}\ge \frac{4(a+b+c)^2}{8\sum ab+2\sum a^2}\ge \frac{6}{5}\]](http://latex.artofproblemsolving.com/f/3/4/f34a66d5ef2b96737b375487d2af605d8afe3856.png)
,![\[\sum\frac{1}{\sqrt{a+b}}\le\frac{1}{\sqrt{2abc}}\]](http://latex.artofproblemsolving.com/0/a/7/0a7d2bb57a932c6a6efd99d58c2242b228bf7265.png)
![\[\sum\frac{1}{\sqrt{a+b}}\le \sum\frac{1}{\sqrt{2\sqrt{ab}}}\]](http://latex.artofproblemsolving.com/7/d/3/7d3fe86fd119a23dcee1cb2379950606e2dcbec3.png)
![\[\sum \sqrt{a\sqrt{bc}}\le a+b+c\]](http://latex.artofproblemsolving.com/c/9/1/c91217d46fff913a7aaa856a0cc7d1af7e973050.png)
.![\[\sum\frac{a^4+a^2+1}{a^6+a^3+1}\le 4\]](http://latex.artofproblemsolving.com/d/4/5/d45031623b3f06cda557bef54767059519572452.png)
![\[4(x^6+x^3+1)\ge 3(x^4+x^2+1)\Leftrightarrow 4x^6-3x^4+4x^3-3x^2+1\ge 0\]](http://latex.artofproblemsolving.com/6/c/0/6c089a05319dca1654d69abbbaaaf7eccea3dbaa.png)
![\[(2x^2-1)^2x^2+(2x-1)^2x+x^4-x+1\ge 0\]](http://latex.artofproblemsolving.com/f/a/b/fab67815142b0fa0fce833c65acefba16d62d54c.png)
![$x^4+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\ge 4x\sqrt[4]{\frac{1}{27}}x>x$](http://latex.artofproblemsolving.com/7/2/0/720e14d6220e49ddecc9fc94c0950903f55eda80.png)
from AM-GM, we're done.
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Y by applepi2000, Amir Hossein, a123, Adventure10, Mango247
for all 
for all 
is finite.
for each 
Then 
contradiction.
for some 
into the first condtion to get 
and we get 
and by induction 
then third condition is false.
then 
so for each 
we have 
for some 
reduses to 
)
then 
and so 
impossible.
then 
,
then 
impossible.
Finally, 
for all 
has a finite number of elements.
and 
Then 
for some 
then 
contradiction.
![\[f(x-f(y))=f^2(y)-2xf(y)+f(x). \]](http://latex.artofproblemsolving.com/3/4/9/349abb0eeca67d2ec5275c2d1abf8835ef21875b.png)
for some 
so each 
can be written in the form 
for some 
arbitrary. Then 
for some 
and using 
we get 
which verifies.![\[f(x-y)(x+y)=(x-y)(f(x)+f(y)) .\]](http://latex.artofproblemsolving.com/3/3/8/3385c3c7f14f0d5289eb5f20d7d980a881dbd3ac.png)
we have 
that is 
to get 
and
so 
gives the general solution: 
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Y by Bumblebee60, amar_04, Wizard0001, Mango247
In fact, I didn't do it to congratulate him. I just wanted other people (newcomers to Aops) to see it
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N Quick Reply
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