AoPS Academy
be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer 
the numbers 
leave different remainders upon division by ..
![$\begin{cases} x^3+x^2-xy^2=\sqrt{(x-y^2)^3} \\ 56x^2+20(x^2-y^2)=\sqrt[3]{4x(8x+1)}-2\end{cases}$](http://latex.artofproblemsolving.com/2/4/7/2472f62c2e2e6c684bc262e61da80cba4d0d263c.png)
there exists positive integers 
such that
be a scalene triangle with incircle 
. Denote by 
the midpoint of arc 
in the circumcircle of , and by 
the point where the 
-excircle touches 
. Suppose the circumcircle of 
meets again at 
and intersects at two points 
, 
.
or 
is tangent to .
such that ![\[ f(x + f(y)) = f(f(x)) + y \]](http://latex.artofproblemsolving.com/9/8/e/98ec8b2933798a3b3b7b022751d7125e97b2c14b.png)
for all 
.[/quote]
s on the page, by the way. I hope it's solvable.)
. Then we have 
. Then 
. ..."
in the equation and replace it with, say, 
or 
or something else depending on , and then do something else with that equation; or you might want to swap and 
in the equation. But writing "Let 
" or anything resembling that is horrible... if doesn't that just mean 
? How do we know whether the two 's on the two sides of your equation are the same, and if not, which one is which?
", and then end up with an equation involving 
that you do more things with, but sometimes this requires extra foresight into what you're going to do with the equation and can be annoying for a different host of reasons, e.g. you accidentally reuse a variable name used elsewhere.
denote [the given functional equation]": for example, "Let denote the statement ![\[ f(x + f(y)) = f(f(x)) + y. \]](http://latex.artofproblemsolving.com/3/1/e/31ea2839d8ccca37907b63269743ccc16b030365.png)
" with 
" with no ambiguity just by writing "
".
we get 
for all . Plugging back into we get ![\[ f(x + f(y)) = f(f(0)) + x + y \]](http://latex.artofproblemsolving.com/e/1/6/e1660e012aea29f7a6219169c42c65f893594fef.png)
for all 
. Let 
denote this statement.
; then by comparing with 
we see ![\[ f(f(x)) + y' = f(x + f(y)) = f(x + f(y')) = f(f(x)) + y, \]](http://latex.artofproblemsolving.com/7/7/6/776d8e0029c5d410d233a10ba4ac278ce19489b4.png)
so 
, and so is injective. with 
, we see... (rest of solution left as exercise for reader)
is the right letter to use, but that was the letter I learned at whatever point I picked this up and it seems pretty common on the fora — some people even use the in their solution to functional equations without even defining . (I don't recommend that.)
to a ring 
is a homomorphism 
(what a great letter) between the additive groups that also satisfies 
and 
for any 
. to a ring is a homomorphism between the additive groups that also satisfies and 
for any . is a domain, then the Jordan homomorphism is either a homomorphism or an anti-homomorphism.
denote the statement .![\[P(a, 1) \Longrightarrow \eta(a^2) = \eta(a)^2 \qquad \forall a \in R\]](http://latex.artofproblemsolving.com/a/c/2/ac2e2b944ea650ee7644d36068ba2e390d3046d0.png)
![\[P(a, a) \Longrightarrow \eta(a^3) = \eta(a)^3 \qquad \forall a \in R\]](http://latex.artofproblemsolving.com/8/f/a/8fa511a4062299feab8c756babf5fd4ff8a952ee.png)
and 
from 
yields![\[ \eta(abc + cba) = \eta(a)\eta(b)\eta(c) + \eta(c)\eta(b)\eta(a) \]](http://latex.artofproblemsolving.com/3/f/d/3fd1efb21884c4000074930ff33d17f42a162615.png)
is an additive group homomorphism)
.
we have
and 
and of course the additive-group-homomorphism properties. is a domain) one of the factors in the original expression is 0, i.e. either 
or 
for any (*). is either a homomorphism or an anti-homomorphism.
with , or like an anti-homomorphism on all pairs with . This is somewhat an abuse of English but I think it gets the point across better than equation after equation. We provide two ways to finish.
, either acts as a homomorphism or as an anti-homomorphism on both pairs 
and 
. Similar remarks hold for 
and 
. acts as a homomorphism on one pair and as an anti-homomorphism on the other.
and 
. Consider 
.
then 
too.
then too. acts as a homomorphism on both or as an anti-homomorphism on both. works on some pair as only a homomorphism and on some other pair as only an anti-homomorphism, that is: 
while 
. to work as both a homomorphism and an anti-homomorphism on the pairs 
and 
. Thus:
works on 
as only a homomorphism, and on 
as only an anti-homomorphism, impossible. is fully a homomorphism or fully an anti-homomorphism (it may be partially both, on elements that commute). Q.E.D.
and 
together imply 
(additive homomorphism properties + distributivity) implies 
since obviously 
commutes with everybody multiplicatively just like 
, the set of all 
such that is a subgroup of the additive group of . and collect all , or if you look at pairs with 
, or both.
and 
of a group 
is itself then one of the subgroups is improper. Proof: If is included in or vice versa, the result is clear; otherwise, take 
and 
and observe that 
is in neither nor , a contradiction. and collect 
and 
.
; also, as we discussed, 
and 
are subgroups of the additive group of . Then by Lemma 2, either 
or 
(**).
and 
.
, which is a subgroup of from our initial discussion and Lemma 1. Similarly, 
is a subgroup of too. Then (**) says that for any 
, either 
or 
, i.e. 
. By Lemma 2 either 
, whence is a homomorphism, or 
, whence is an anti-homomorphism. Q.E.D.
be a point on 
such that 
is an isosceles trapezoid, we have 
and 
. be the second intersection of 
with 
.
. So, 
.
also lies on 
.
and 
. So 
.
intersects again at 
, note that 
.
, i.e. is the midpoint of .
. and 
over to get 
and 
, respectively. Also, reflect over 
to get .
and 
.
. It follows that 
is the reflection of 
over 
.
is the reflection of 
over 
.
.
. Since is the midpoint of 
, is also the midpoint of , done.
. Note that this preserve all the properties of the configuration.
. Lowercase letters will denote the corresponding uppercase letter points in the configuration, let the cyclic quadrilateral 
lie on the unit circle.
are collinear we have,![\[\frac{d-c}{q-c}=\frac{\bar{d}-\bar{c}}{\bar{q}-\bar{c}}\implies \bar{q}=\frac{c-q+d}{cd}\]](http://latex.artofproblemsolving.com/5/f/6/5f67db63556f789eeed8ff27bc40a9a71bee7907.png)
Again since 
,![\[|b-c|=|c-q|\implies (c-b)(\frac{1}{c}-\frac{1}{b})=(\frac{1}{c}-\frac{c-q+d}{cd})(c-q)\]](http://latex.artofproblemsolving.com/8/0/d/80dfe807f60e0b8766813c45bbcf1135c2fc0345.png)
![\[\implies q=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+c\]](http://latex.artofproblemsolving.com/2/6/6/26694a785e0a97a46d61d5f5d8aa5adfbb4910f6.png)
Similarly,![\[p=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+a\]](http://latex.artofproblemsolving.com/a/f/6/af651b78c203dde9d541284ef4a818deaec8b70a.png)
We can rotate the quadrilateral, so WLOG we can assume 
.
and the other has 
sign. WLOG assume 
has sign.![\[m=\frac{p+q}{2}=\frac{\sqrt{d}(a-c)+c+a}{2}\]](http://latex.artofproblemsolving.com/9/8/6/9865f8e2dde2a80af2ec2ae72b3e6a90156797a9.png)
Let 
, we wish to show 
. Now,![\[\delta=\frac{2a+\sqrt{d}(c-a)-c-a}{2c+\sqrt{d}(c-a)-c-a}=\frac{(c-a)(\sqrt{d}-1)}{(c-a)(\sqrt{d}+1)}=\frac{\sqrt{d}-1}{\sqrt{d}+1}\]](http://latex.artofproblemsolving.com/2/d/1/2d1f5feeda0bf1fec98970da2de594d2923039b8.png)
Hence,![\[\bar{\delta}=\frac{\frac{1}{\sqrt{d}}-1}{\frac{1}{\sqrt{d}}+1}=\frac{1-\sqrt{d}}{1+\sqrt{d}}\]](http://latex.artofproblemsolving.com/0/3/8/038f3e9a08b91b2fcc5a1bfed126a7ca40300187.png)
Thus, . And we are done. 
. Note that this preserve all the properties of the configuration.. Lowercase letters will denote the corresponding uppercase letter points in the configuration, let the cyclic quadrilateral lie on the unit circle. are collinear we have,Again since ,Similarly,We can rotate the quadrilateral, so WLOG we can assume . and the other has sign. WLOG assume has sign.Let , we wish to show . Now,Hence,Thus, . And we are done. 
.
" in the very first line of my proof, so I think it should be fine.
." in the very first line of my proof, so I think it should be fine.
be the reflection of point through point 
is a parallelogram.
, it follows that that 
. 
.
and 
be the spiral center. Then, since 
, 
, which means that 
passes through the midpoint 
of . Seeing this, we observe that 
or
or 
as desired.
and 
to 
respectively such that 
and 
.
.
and 
such that 
and 
.
then 
wich gives 
so 
and similarly 
.
but 
and replacing in 
gives 
so 
and so 
but we proved 
and 
thus 
bisects 
and 
bisects 
so 
as desired
be the midpoints of 
,
with ratio 
,
.
about perpendicular bisector of 
be the midpoints of 
respectively. So 
hence 
are cyclic. The fact that 
and 
finishes.