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this guy is an absolute legend. much love wherever you are Michael
by
LeonidasTheConquerer, Aug 5, 2024, 9:37 PM
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amazing blog
by
anurag27826, Jun 17, 2023, 7:20 AM
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hi i randomly found this
by
purplepenguin2, Mar 1, 2023, 8:43 AM
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can i be a contributor please?
by
cinnamon_e, Mar 10, 2022, 6:58 PM
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orzorzorzorzorzorozo
by
samrocksnature, Jul 16, 2021, 8:25 PM
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2021 post
by
the_mathmagician, May 5, 2021, 3:28 PM
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Let
be an equilateral triangle of side length 
. Let 
be the point such that 
is the midpoint of 
, and let 
be the incenter of triangle 
. Let 
be the point on line 
such that 
and 
are perpendicular. $ \
by
ARay10, Aug 25, 2020, 5:55 PM
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Nice blog!
by
User526797, Jan 12, 2020, 4:48 PM
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oh my gosh it's been so longggggg.... contrib? what does that mean?
by
adiarasel, Dec 1, 2019, 8:31 PM
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2019 post
by
piphi, Aug 10, 2019, 6:32 AM
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hi contrib please
by
Emathmaster, Dec 27, 2018, 5:38 PM
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hihihihihi contrib plzzzzz
by
haha0201, Aug 20, 2018, 3:58 PM
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contrib please
by
Max0815, Aug 1, 2018, 12:35 AM
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contrib /charmander
by
mathmaster2000, Apr 16, 2017, 4:59 PM
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for contrib
by
SomethingNeutral, Mar 30, 2017, 7:57 PM
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all contrib requests added I usually get to posting every night after running on the treadmill and showering, posting my favorite problem(s) I did during the day.
by
shiningsunnyday, Jun 16, 2016, 4:37 AM
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Since problems is plural, he might post multiple problems through different posts in one day...
by
zephyrcrush78, Jun 15, 2016, 8:37 PM
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hi can i have contrib?
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wu2481632, Jun 15, 2016, 8:36 PM
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Hello it is called problems of the DAY
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Temp456, Jun 15, 2016, 8:17 PM
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10th shout lol
How often will this blog be updated?
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zephyrcrush78, Jun 15, 2016, 5:24 PM
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why are we counting shouts
contrib pls
by
MathAwesome123, Jun 15, 2016, 2:55 PM
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eighth shout
contrib pls
by
skipiano, Jun 15, 2016, 2:21 PM
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seventh shout
Can I be contrib?
Though probably not because it's not like I'll post anything anyways :/ :/
by
zephyrcrush78, Jun 15, 2016, 1:20 AM
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sixth shout
new goal: get the shouts that are congruent to 1 mod 5
by
DeathLlama9, Jun 15, 2016, 12:33 AM
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fifth shout >
by
budu, Jun 14, 2016, 10:04 PM
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4th yay
by
EpicSkills32, Jun 14, 2016, 9:10 PM
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third shout
by
skipiano, Jun 14, 2016, 2:35 PM
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Second shout
by
zephyrcrush78, Jun 14, 2016, 12:06 AM
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omg yesssss
by
DeathLlama9, Jun 13, 2016, 2:19 PM
by shiningsunnyday, Dec 12, 2016, 4:10 PM 
such that 
for all 
such that 
for all 
be given assertion. If we fix 
and vary 
we easily see 
is surjective. Sweet. Perhaps we can prove it's injective as well? Let 
Taking advantage of symmetry, we have ![\[P(a,b) = P(b, a) \Rightarrow a + f(b) = b + f(a) \Rightarrow a =b.\]](http://latex.artofproblemsolving.com/d/1/f/d1f81ef628cccba7e4ca32df12bfbbff55f1480a.png)
Therefore, 
In addition, there exists 
such that 
Drop in 
To finish, put 
Checking, we indeed have 
and thus the only solution is 
and 
![\[f(f(x+y))=f(x)+f(y).\]](//latex.artofproblemsolving.com/7/a/1/7a1c6497ace6ada9c6f036e13cc79db68db305aa.png)
and 
![\[f(f(x+y))=f(x)+f(y).\]](http://latex.artofproblemsolving.com/7/a/1/7a1c6497ace6ada9c6f036e13cc79db68db305aa.png)
We then have ![\[f(f(x)) = f(x) + c \Rightarrow f(f(x+y)) = f(x+y) + c.\]](http://latex.artofproblemsolving.com/f/e/a/feaacdf9b9bbbe2154772c28390d48212f96f764.png)
Plugging this back into the original equation, ![\[f(f(x+y)) = f(x+y) + c = f(x) + f(y).\]](http://latex.artofproblemsolving.com/2/c/7/2c72e743114b6db8825300ad2cc235944787d1b8.png)
We seem ALMOST there, but we would like 
as opposed to 
and so we have ![\[f(x+y) + c = f(x) + f(y) \Rightarrow g(x+y) = g(x) + g(y) \Rightarrow g(x) = kx, f(x) = kx+c.\]](http://latex.artofproblemsolving.com/f/0/4/f040f0b0fccf407961a333c292ee0c20b2d444cf.png)
To finish, we plug in to the original equation: ![\[f(f(x+y)) = f(k(x+y)+c) = k^2(x+y)+ck + c = kx+c+ky+c \Rightarrow k^2(x+y) + ck = k(x+y) + c\]](http://latex.artofproblemsolving.com/0/7/e/07e90d22878c88f986dd1b13db5bb2f6700f399a.png)
![\[(x+y)(k^2-k)+c(k-1)=0, (k-1)(k(x+y)+c)=0.\]](http://latex.artofproblemsolving.com/8/1/a/81a7eb5ffdee92cdeb2969f40b16f8852e9254c2.png)
If 
,
work. Otherwise, 
and since 
can vary, 
and we may check only 
work in this case. Therefore, the only solutions are 
May 2017
April 2017