AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, an arbitrary point 
is chosen on the side 
. In triangles 
and 
, the angle bisectors 
and 
are drawn, respectively. The point 
is the circumcenter of 
. Prove that the second intersection point of the circumcircles of triangles 
and 
lies on the line 
.
be a point in the plane of triangle 
such that the segments 
, 
, and 
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to . Prove that 
is acute.
+1 w
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray at 
and ray at 
. Let 
be a point on segment 
. The line through parallel to intersects line at . Let 
be the intersection of lines 
and , and let 
be the intersection of line and the line through parallel to . Prove that line 
is tangent to . be an acute triangle with point 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of over 
with 
, let 
denote the incenter and the midpoint of side 
. The line through 
perpendicular to 
intersects the tangent from to the incircle (different from line ) at a point > Show that is tangent to the circumcircle of triangle 
.
and 
. At each step, it is allowed to add to one of the polynomials and to multiply another one by the polynomial 
. Can the polynomials become equal at some point?
such that for any positive integer number 
and any arrangement of the numbers from 1 to 
on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than .
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
.
make the equation ![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be the binary representation of 
. Prove that for any positive integer , at least one of the digits 
, 
, 
, 
equals 
.
such that![\[
4k^3 + 4k + 1
\]](http://latex.artofproblemsolving.com/f/3/5/f355e456cd47ecfdb13597d52fec3d848b67c594.png)
numbers, none of which appears more than 
times. Prove that they can be divided into 
groups of numbers each in such a way that the sums of the numbers in these groups are pairwise distinct.
be a cyclic quadrilateral. and intersect at . Let and 
be the projections of onto and and and be the midpoints of and 
, respectively. Prove that 
is a kite.
. At each move, Bob erases arbitrary two numbers 
and 
, where 
, and writes the single number 
. After 
such moves the blackboard will contain a single number. What is its maximum possible value? (Reminder that 
is the maximum integer not exceeding 
.)
satisfying ![\[f\big(f(m)+n\big)+f(m)=f(n)+f(3m)+2014\]](http://latex.artofproblemsolving.com/2/8/d/28d562062da13d43f91b1b308082b4d73c0b3cb6.png)
for all integers 
and 
,
and the functional equation becomes ![\[ g(g(m)+n+c) + g(m) = g(3m) + g(n) + 2014. \qquad (1) \]](http://latex.artofproblemsolving.com/e/c/b/ecb06100733a52fc35e7d416aa8838a96ccfad9e.png)
With 
in (1) we have ![\[ g(n+c) = g(n) + 2014. \qquad (2) \]](http://latex.artofproblemsolving.com/0/f/c/0fc43d766702d32e3b45e24e0eda00035ffb39bc.png)
In particular, 
and 
for all 
.
in (1) we have ![\[ g(g(m)) + g(m) = g(3m) \qquad (3) \]](http://latex.artofproblemsolving.com/e/3/0/e300ae8309868daf670470da6e6d8140894d2f7c.png)
This implies ![\[ g(g(m)+n+c) = g(g(m)) + g(n) + 2014 \]](http://latex.artofproblemsolving.com/2/0/3/2035447d93ac6d2c2336bfa2ab3931517b119f3f.png)
or shifting 
,![\[ g(g(m)+n) = g(g(m)) + g(n) \qquad (4) \]](http://latex.artofproblemsolving.com/d/4/e/d4efcb9bc58818ac66c4b8b8225c9c1ffd1ddda8.png)
If we apply 
in (4), we then obtain ![\[ g(n+2014) = g(n) + g(2014). \qquad (5) \]](http://latex.artofproblemsolving.com/c/7/1/c71a9c2599a73681a3dc553b4abfa4822ba2b3f2.png)
In particular, this implies that 
.![\[ g(n+2014c) = g(n)+2014^2 = g(n) + c \cdot g(2014) \implies c \cdot g(2014) = 2014^2. \]](http://latex.artofproblemsolving.com/2/b/5/2b50fee1264989c668fb712f6340061b5334cef9.png)
But if we take 
in (1) we find that ![\[ g(2014+2c) = 4 \cdot 2014 \]](http://latex.artofproblemsolving.com/0/1/5/015142ef19ef6e6ea74013d280eff0dcf1623016.png)
thus 
,
.![\[ g(n+1007) = g(n) + 2014. \qquad (6) \]](http://latex.artofproblemsolving.com/d/6/7/d679fa89f4f2d93e0b496002caba26a77cce780b.png)
From (6) we see in particular 
.![\[ 2014g(m) = g(1007g(m)) = 1007g(g(m)) \implies g(g(m)) = 2g(m). \]](http://latex.artofproblemsolving.com/0/e/e/0eecaed3feafecac40d5bd24358f34b11bf1c505.png)
Inserting this into (3) informs us that ![\[ g(3m) = 3g(m) \]](http://latex.artofproblemsolving.com/b/d/6/bd60099a7790c78def3aa211401255cf2a19823f.png)
for any integer ![\[ 3^{\varphi(1007)}g(n) = g(3^{\varphi(1007)}n) = g\left( 1007 \cdot \tfrac{3^{\varphi(1007)}-1}{1007} n + n\right) = g(n) + 2014 \cdot \tfrac{3^{\varphi(1007)}-1}{1007} n \]](http://latex.artofproblemsolving.com/0/1/d/01dcd5853eef440f304f84d8a5befec56d8e063d.png)
and hence 
.![\[ f(n) = 2n + 1007. \]](http://latex.artofproblemsolving.com/c/a/d/cada6bdc8e1d2c8fc57987bb754e2a5e8b70a5c3.png)
It is trivial to check that this solution works.
,
covers all nonzero residues modulo 
;
,
in 
.
is not a field because 
is not prime.
and then (6) gives us
.
,
.
,
is periodic and thus bounded. Assume, 
,
.
,
,
is bounded.
,
,
,
.
.
be the given assertion. Let 
and let 
![\[P(0, n) \implies f(n + c) = f(n) + 2014. \qquad (1)\]](http://latex.artofproblemsolving.com/e/4/6/e46b71d5e6cc3349e17cc1da36361f9a32c65a23.png)
Thus, by induction, we easily establish ![\[f(n + kc) = f(n) + 2014k \quad \forall k \in \mathbb{Z}. \qquad (2)\]](http://latex.artofproblemsolving.com/f/1/b/f1be6b4581ec8f91399df224010f7890c79aab42.png)
Then given any 
,
with 
Since 
,![\[-\alpha + 2014q \le f(x) \le \alpha + 2014q.\]](http://latex.artofproblemsolving.com/5/9/4/59417187b67f36d50b38aacf4f00c666178edcf9.png)
Substituting 
and using 
,![\[-\alpha + \frac{2014x}{c} - 2014 < f(x) < \alpha + \frac{2014x}{c} + 2014. \qquad (3)\]](http://latex.artofproblemsolving.com/7/e/f/7ef917bc685cd3b6d0498fa1d9fbe67621a3bb17.png)
Meanwhile, ![\[P(n, n) \implies f(f(n) + n) = f(3n) + 2014. \qquad (4)\]](http://latex.artofproblemsolving.com/8/8/f/88fa301bd183c7df0d5da8d22577f6271ce4323f.png)
but 
Then ![\[P(y, z) \implies f(f(z) + z) = f(3y) + 2014. \qquad (5)\]](http://latex.artofproblemsolving.com/e/a/6/ea66e48fb766175eea1a2794a8ceb517e38f449a.png)
Comparing (4) and (5), it follows that 
,
for all 
However, it is clear that for sufficiently large 
,
and 
,
By injectivity, we obtain 
Substituting this expression for 
.
yields 
and from that it is obvious that for any 
we have 
.
.
and see what happens. We get
Simplify more and we get...
So now we can solve for 
and 
,
if 
.
,
Hey, the 2014 disappears! Alright so we know that 
.
we can deduce that
This tells us that 
And so 
but remember that 
.
.
.
Too long not sleeping 
must be bounded. To see this, WLOG the period is ![$f : [0,1] \to \mathbb R$](http://latex.artofproblemsolving.com/0/9/4/09473f6956c387ad19f060a139e9306eba9d7208.png)
.![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
is compact, it follows that 
.
.
.
.
,
.
.
.
,
.
.
such that 
.
.
![\[ f(f(0)(f(m)-f(0)) + n) = f(n) + 2014(f(m)-f(0)) = f(n) + (f(3m)-f(m))f(0) \implies f(3m) = \frac{(2014+f(0))f(m)}{f(0)} - 2014, \qquad (3) \]](http://latex.artofproblemsolving.com/8/9/f/89f8cf861cacc56a350ea8d80d6421d4311366e6.png)
noting that 
would contradict (1).![\[ f(f(m)+n) = f(n) + \frac{2014}{f(0)}f(m) = f(n) + df(m), \qquad (4) \]](http://latex.artofproblemsolving.com/1/5/6/1565eca6ec59eed46bb1d7afdb92c46359440e3a.png)
where we let 
and 
in (4) gives 
.
in (4), giving 
.
for all 
,
and 
![\[ f(3^k m) = 3^k f(m) - 1007(3^k-1). \qquad (9) \]](http://latex.artofproblemsolving.com/b/0/1/b016a9223d1d06c8ca583e2b0c1e06bf8ff5d7c7.png)
,![\[ f(3^k m) = f(m) + 2(3^k - 1)m. \qquad (10) \]](http://latex.artofproblemsolving.com/4/9/5/4950467c421f1ad31ea54edd3ed9fde45e078e43.png)
![\[ f(m) = 2m+1007, \]](http://latex.artofproblemsolving.com/8/0/f/80f04d41d9329de4553af617fefccd450158d52c.png)
for all integers 
only, which clearly works.
be fixed integers. If a function 
satisfies ![\[f(n+a)=f(n)+b \text{ and } f(n+c)=f(n)+d\]](http://latex.artofproblemsolving.com/a/7/9/a79833ad0162a7fc346c1da2e3db865e82a2535e.png)
for all integers 
(this also holds for domain 
,
and 
,
as desired.
.
,
i.e 
is a constant independent of 
Substituting 
Hence, 
.![\[f(0)=(c+1)f(0) - 2014 \implies cf(0)=2014 \implies cf(f(m)) = 2cf(m) + cf(0) \implies f(f(m)) = 2f(m) + f(0)\]](http://latex.artofproblemsolving.com/a/2/2/a2234d0aca89400d47d9e9511267ab01435874cd.png)
Comparing this to equation 
.
.
.![\[f(n+f(m))-2f(m)-2n=f(n)-2n \implies g(n+f(m))=g(n)\]](http://latex.artofproblemsolving.com/8/6/8/8680e272ab379fc48fe248eca28af42dde93bf23.png)
Substituting 
.
can also be rewritten as ![\[f(3m)-6m=3f(m)-6m-2014 \implies g(3m)=3g(m)-2014\]](http://latex.artofproblemsolving.com/2/3/b/23b75b15dda0668b733da2c2bb5b4da60a91330c.png)
.
.
.
,
for all 
is not periodic? (as says Evan Chen)
![\[
f(f(0) + n) = f(n) + 2014
\]](http://latex.artofproblemsolving.com/e/3/e/e3ea0ab4d25e52fba7d3aaa3d0d1bd69a14d37d5.png)
it follows that![\[
2014^2 + 2014f(0) = 3 \cdot 2014f(0)
\]](http://latex.artofproblemsolving.com/f/5/7/f57da56461e5b93ba302ab1cf70c5c3e43594688.png)
so 
such
and![\[
g(g(m) + 2m + n) - g(n) = g(3m) - 3g(m)
\]](http://latex.artofproblemsolving.com/d/0/c/d0cbdab58be63df376aa0b9a8e775803ded13a3f.png)
Thus, either 
in which case
,![\[
g(k(g(m) + 2m) + n) - g(n) = k(g(3m) - 3g(m))
\]](http://latex.artofproblemsolving.com/b/a/a/baaa451fcd09525f13fa7bf95dad1e5e5e95059a.png)
which since the range of ![\[
g(g(m) + 2m + n) = g(n)
\]](http://latex.artofproblemsolving.com/1/7/d/17d3ec02491d74d1c68e9287970b10c172f0627f.png)
,
,
is relatively prime with 
,
.
as needed. 
forms a complete residue system modulo 
for all 
we must have that 
,
,
.
gives 
.![\[2014f(m)=f(cf(m)+n)-f(n)=c(f(3m)-f(m)+2014)\]](http://latex.artofproblemsolving.com/9/e/f/9efc613f5d60ca16f399037d093c1a78286cf0c2.png)
We will find 
gives ![\[f(f(m)+n+2014)-f(n)=f(3m)-f(m)+3\cdot 2014=f(f(m)+n)-f(n)+2\cdot 2014\]](http://latex.artofproblemsolving.com/d/5/f/d5f8ce1bd730691dec06781a7eded3001a212c7e.png)
Since 
reaches all reals, 
generally. We have 
from the previous, but 
which implies 
,
.![\[f(a3^k)=3^kf(a)-c(3^k-1)\]](http://latex.artofproblemsolving.com/e/9/8/e985867ff642e55e7e30247f8a575be22014290b.png)
We can set 
then 
.
so 
implying 
.
be the assertion
.
so 
.
whenever 
.
by 
.
for every 
,
.
and by Euler's theorem, for 
we have 
for some 
,
or 
,
Let 
Moreover, by easy induction and (1), we have that for any integer 
Choosing 
in (3) and using (2) we have
Thus,
Choosing 
Now, (2) implies that
or
Hence 
so 
Moreover, by (4) we have
or
or
Now define function 
Thus,
Moreover, (6) implies that
Thus,
Since 
This implies that
or
or 
for all integers 
.
we have 
.
for all integers 
we obtain: ![\[k^2 + 2014k = 3 \cdot 2014k\]](http://latex.artofproblemsolving.com/3/f/3/3f3c78233795e45c939f19856286748d7ab67bb3.png)
which implies that 
or 
.
.![\[h(h(m)+n)+h(m) = h(n)+h(3m)\]](http://latex.artofproblemsolving.com/4/1/5/415f55e2b5ac6d1791effe8128770277bb1fa5e4.png)
Observe that we have 
because of the displacement and thus 
.
and replacing the value of 
we get ![\[h(h(m)+n) = h^2(m)+h(n)\]](http://latex.artofproblemsolving.com/a/6/4/a64f80041414c4509e230441bc7c0023d5ebb833.png)
Repeating enough times, we obtain 
.
as we had proven earlier. Thus 
which implies that 
.![\[3^{936}h(m) = h(3^{936}m) = 2(3^{936}-1)m+h(m)\]](http://latex.artofproblemsolving.com/5/7/2/572fdb74ed6234bc5538769ae0a1525d4d8ed558.png)
which relies on 
to prove 
for all 
. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We first show the following.![\[f(n+2014)=f(n)+2\cdot 2014\]](http://latex.artofproblemsolving.com/8/f/b/8fbc562c398895ebcc8e954756cd472705a6f9f5.png)
Proof : First, let ![\[f(n+c)=f(n)+2014\]](http://latex.artofproblemsolving.com/7/2/7/727b44df5231deaca2786c9e7c59a5b03538c16c.png)
for all 
.
and 
.
yields,
as claimed.
.![\[f(n+2c)=f(n)+2\cdot 2014 = f(n+2014)\]](http://latex.artofproblemsolving.com/c/2/0/c20704a30e0c7e187bc085a7a902b7fe83d5b3bc.png)
using our previous claims. Thus, it follows that 
.![\[f(n)+T \cdot 2\cdot 2014 = f(n+2014\cdot T)=f(n)\]](http://latex.artofproblemsolving.com/6/5/e/65eb1c7fe3c18baa9346e6808254395096d87f96.png)
the first using our previous claim, and the second equality using the fact that 
is the period of this function. But this is clearly impossible since 
which is a contradiction. Thus, it follows that ![\[ f(n+1007)=f(n)+2014 \qquad (1) \]](http://latex.artofproblemsolving.com/e/d/e/edeaa387797cf419abe2b6f5979e6f42d13aeb56.png)
for all integers 
for some 
.
and 
yields,![\[f(n)+f(3\alpha)+2014 = f(f(\alpha)+n)+f(\alpha)=f(f(\beta)+n)+f(\beta)=f(n)+f(3\beta)+2014\]](http://latex.artofproblemsolving.com/0/0/3/003290ad40e3e722bc8299d7f4766fe17ddc4c64.png)
from which it follows that 
.
for all positive integers 
we have that![\[f(3^{\phi(1007)}\alpha)=f(3^{\phi(1007)}\beta)\]](http://latex.artofproblemsolving.com/3/f/7/3f7084b089bc72bfacd46ae14619e7b2f38f43c7.png)
By Euler's generalization of Fermat's Little Theorem, we have that 
which implies that 
for some 
.
for some 
.![\[f(\beta)+\frac{3^{\phi(1007)}\beta - \beta}{1007} \cdot 2014 = f(3^{\phi(1007)}\beta) = f(3^{\phi(1007)}\alpha) = f(\alpha)+\frac{3^{\phi(1007)}\alpha-\alpha}{1007} \cdot 2014\]](http://latex.artofproblemsolving.com/c/b/0/cb06344b4804b52c8bbc005d4534adc39df0ab8d.png)
which with some manipulation leads to![\[\left(3^{\phi(1007)}-1\right)(\alpha - \beta)=0\]](http://latex.artofproblemsolving.com/3/2/7/3275bfc909fef23f85e402a8642ba055eeb3bdb7.png)
which is clearly impossible since 
and 
.
and indeed 
.
where the last equality was obtain using ![\[m=3m-f(m)+1007 \implies f(m)=2m+1007\]](http://latex.artofproblemsolving.com/c/6/9/c696b03d6cb72c0a07c0ebdf6246a70d49d8ac37.png)
for all 
which was the desired answer.
,![\[f(n+c) = f(n)+2014.\]](http://latex.artofproblemsolving.com/4/7/a/47ab731f57bfdb02d8a92c193293e3db14829a1d.png)
![\[f(f(c)+n)+f(c) = f(n)+f(3c) + 2014.\]](http://latex.artofproblemsolving.com/b/6/e/b6e451286e36f03c55d4f45c4e7baefbbcd9983c.png)
,
.![\[f(n+2c) = f(n)+4028 = f(n+2014).\]](http://latex.artofproblemsolving.com/b/c/f/bcf0e63188faa152dbe37737c28dda6ecd6fb28f.png)
.![\[f(n)+4028p = f(n+2014p) = f(n),\]](http://latex.artofproblemsolving.com/d/c/6/dc6fdfdfab855de3c707cf3f9edc193c4e4626b2.png)
.
is false, so ![\[f(n+1007) = f(n)+2014.\]](http://latex.artofproblemsolving.com/7/6/b/76b6b48f9643f360c6eecb4a406ccc9e0c35b78f.png)
such that 
.
and 
gives![\[f(n)+f(3a)+2014 = f(n)+f(3b)+2014 \implies f(3a) = f(3b),\]](http://latex.artofproblemsolving.com/8/c/7/8c7eecf81b2ee628c45bc9c9da145ebc668cda0e.png)
.
.
,![\[f(a) + 2014 \left(\frac{(3^k-1)a}{1007} \right) = f(3^ka) = f(3^kb) = f(b)+2014 \left(\frac{(3^k-1)b}{1007} \right).\]](http://latex.artofproblemsolving.com/c/3/7/c370d0046bf7d2d887f606fb3fa2e8ea41720e5a.png)
,
![\[f(m) = f(3m-f(m))+2014 = f(3m-f(m)+1007).\]](http://latex.artofproblemsolving.com/5/2/b/52b072b00a20cd21fe369a251e5b4e58ad619a6c.png)
,
then 
So, by induction it follows that 
.
into the equation, we get 
Now putting 
.
then we have \begin{itemize} \item g(0)=0 \item g(n+1007)=g(n)+2014 \item g(g(x)+y)+g(x)=g(y)+g(3x) \qquad (1). \end{itemize} By induction we get 
.
Putting 
in 
.
into 
for some 
,
too, from 
for all 
integer. Take a 
.
then 
by induction. But this implies that 
,
.
we get 
for all 
.
for some 
.
Thus 
for all 
.
.
Then
by induction as well. However, by 
,
.
becomes 
.
.
,
,
for positive integers 
,
. Then
Similarly,
proving 
.
,
)
,
first as in the official solutions. Notice then that 
.
for any integer 
in the following ways:
Now comparing the two and cancelling gives us exactly what we need.
,
.
Plug 
into Claim 0 to get 
,
,
or 
as desired.
,
,
as desired, so plug it back into ![$[0,c-1]$](http://latex.artofproblemsolving.com/5/5/3/5537ca8ccb4cdde68f3701bbe8b008d095fdad02.png)
for all positive integers 
.
.
so 
as desired. Thus after 6 claims we have instacooked the question.
.
term on the LHS; since this is 
,
symmetry-switching method, from which we immediately get Claim 0. From here, using the fact that 
as well. As a further benefit 
then can all be computed. Now we plug this into Claim 0 and hope for the absolute best that we don’t end up having an identity in terms of 
being present in 
,
,
,![\[f(m+cn)=f(m)+2014n.\]](http://latex.artofproblemsolving.com/a/4/2/a42bd75fd38de1cf5483c1ab072b59014fb74d7d.png)
Now 
gives 
,
gives
Therefore,![\[f(0)+2014\cdot 2cn=f(2014cn)=f(0)+2014^2 n\Longrightarrow c=1007.\]](http://latex.artofproblemsolving.com/d/6/e/d6e1df932d3af8323ef92c99931ee45df113dc61.png)
Now note that by varying 
is only dependent on 
.
.![\[ch(m)=f(cf(m)+n)-f(n)=2cf(m)\Longrightarrow h(m)=2f(m).\]](http://latex.artofproblemsolving.com/2/e/a/2eaa26243bf135872d5430d04c33cd5ad64b7da4.png)
Therefore, 
,
.
.
,
and 
.![\[3^a\equiv 1\pmod c,\]](http://latex.artofproblemsolving.com/6/c/c/6cce985751d31de648129b50179ffbea3a58fc56.png)
then 
,
,