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.
be a triangle with points 
lie on the perpendicular bisector of 
such that 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
to the circumcircle of 
intersects at point 
Prove that: 
are parallel.
, 
, 
, 
be four points in the plane, with and on the same side of the line 
, such that 
and 
. Find the ratio![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles and are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles and at the point are perpendicular.)
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
. Prove that 
for every 
.
, 
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
of the acute-angled triangle 
is tangent to its side 
at a point 
. Let 
be an altitude of triangle , and let 
be the midpoint of the segment . If 
is the common point of the circle and the line 
(distinct from ), then prove that the incircle and the circumcircle of triangle 
are tangent to each other at the point .
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide . (Note: it is possible that has other prime divisors not among .) Prove that for every prime number 
, there exists a lucky number such that 
.
be a non-isosceles triangle with incenter 
Let 
be the smaller circle through 
tangent to 
and 
(the addition of indices being mod 3). Let 
be the second point of intersection of 
and 
Prove that the circumcentres of the triangles 
are collinear.
such that for every 
is satisfied? Explain the answer.
be a scalene oblique triangle, and 
a point on the orthocentroidal circle of (
)., trilinear polar of the polar conjugate (
-isoconjugate) of , Droz-Farny axis of are concurrent. with respect to is as follows:
, there exists a pair of orthogonal lines 
, 
through such that the midpoints of the 3 segments cut off by , from the sidelines of are collinear. The line through these 3 midpoints is the Droz-Farny axis of wrt .
for which ![$\dfrac{n^2+1}{[\sqrt{n}]^2+2}$](http://latex.artofproblemsolving.com/7/e/d/7ed6c67114c0c7d262f67e5b7a733daab7997415.png)
is an integer. Here ![$[r]$](http://latex.artofproblemsolving.com/2/f/c/2fcabe2c2b8ebe2c428830acb9874cb8a8df790f.png)
denotes the greatest integer less than or equal to 
.
.
where 
. Then we require ![\[
\frac{(m^2+k)^2+1}{m^2+2} = m^2 + (2k-2) + \frac{(k-2)^2+1}{m^2+2}
\]](http://latex.artofproblemsolving.com/5/1/e/51e7271de40a69d05f3f617965dbb517563c68eb.png)
to be an integer. Now 
, and hence 
must be either 
, 
or 
. We will examine each case.
, then 
, which is impossible unless 
and 
, so no solution in this case.
, then 
. Modulo we get that 
which is enough to force 
, and yet this implies 
, contradiction.
, then 
. Again, taking modulo , we get 
which is not possible..
.
divides 
then is of the form 
. So all prime divisors of 
are of the form 
. However, this would imply that 
, which can't happen.
divides then is of the form . So all prime divisors of are of the form . However, this would imply that , which can't happen.
is even, then it's possible that 
, which is definitely permissible. In particular, 
is a solution here.
, and after some calculations, we have 
. Combine these two and use the inequality method, we easily found that there is no solution.
be the greatest square less than or equal to , then 
. Now let 
where 
. So ![$[\sqrt{n}]^2=m^2$](http://latex.artofproblemsolving.com/0/2/b/02b56d681735b1c9051095e44745be0e374355ef.png)
and thus![\begin{align*}\frac{n^2+1}{[\sqrt{n}]^2+2} &=\frac{(m^2+r)^2+1}{m^2+2}\\
&=\frac{m^4+2m^2r+r^2+1}{m^2+2} \\
&=m^2+\frac{2m^2r+r^2+1-2m^2}{m^2+2} \\
&=m^2+(2r-2)+\frac{r^2+1-4r+4}{m^2+2} \\
&=m^2+2r-2+\frac{r^2-4r+5}{m^2+2}
\end{align*}](http://latex.artofproblemsolving.com/3/b/f/3bfb9c193466f7c460138efc9ab9e049eb1d9a5f.png)
![\[r^2-4r+5\leq (2m+1)^2-4(2m+1)+5=4m^2-4m+2< 4(m^2+2).\]](http://latex.artofproblemsolving.com/6/9/8/698670f0858a9ed415b9c41b76caf209cd35211f.png)
is impossible since there are no two consecutive numbers that are squares. The second case is impossible by taking modulo 
since the quadratic residues modulo are 
and 
, then 
can be congruent to 
, none of which are quadratic residues. And the last case take modulo and see that 
is congruent to modulo which is not a quadratic residue. Therefore there are no such integers . 
, where . Therefore, ![$[\sqrt{n}] = m$](http://latex.artofproblemsolving.com/4/f/0/4f0ad7c2f261b91020519bb3dc123487777bef70.png)
. We are therefore trying to see when 
is an integer. We rewrite the fraction as 
. So we want 
. But notice that since 
, we have 
. Therefore, if 
is an integer, then it is either 
. We now go through the casework:
which leads to 
. The only squares the differ by are 
and . which would give us 
, contradiction. There are no solutions for this case.
. Let 
and 
. Then, we have 
. Since 
, we see that 
but then 
, yet another contradiction. There are no solutions for this case.
which would give us , yet another contradiction.
![$[\sqrt{n}]=m$](http://latex.artofproblemsolving.com/c/4/0/c40b798ee91009ca3d321f4c55462cca48a73c5d.png)
, i.e. 
.
where 
. The rest of the solution is then standard divisibility ideas and bounding.
, i.e. . where . The rest of the solution is then standard divisibility ideas and bounding.
where 
; then 
so we want to find when 
is integral.
is integral. Since 
, we have three cases:
. In this case 
clear contradiction.
. In this case 
; taking mod 8 gives the LHS equal to 1, 5, or 6 mod 8 and the only possible one is 6 mod 8. But then 
contradiction.
. In this case 
. Taking mod 9, the LHS can be 4, 5, 8, or 2 mod 9 which are all impossible, contradiction.
. where . Then we require to be an integer. Now , and hence must be either , or . We will examine each case., then , which is impossible unless and , so no solution in this case., then . Modulo we get that which is enough to force , and yet this implies , contradiction., then . Again, taking modulo , we get which is not possible..
for which is an integer. Here denotes the greatest integer less than or equal to .
with 
.
or 
. As 
and 
we get that 
. If is odd then 
is mod but since 
has no divisor of the form 
this gives a contradiction. So let 
. Thus 
. So 
is odd so let 
for some . So we have 
. We get that 
or 
. Thus 
for 
. So for if 
then is odd, obviously 
as 
can never be equal to for positive integers 
, and 
. So we have 
. So we get that 
which is a contradiction by mod check. Thus 
if . One can easily check that no solutions exist for . Hence no solutions for 
exist. Thus no solutions for the original problem exist
divides then is of the form . So all prime divisors of are of the form . However, this would imply that , which can't happen. is even, then it's possible that , which is definitely permissible. In particular, is a solution here. is odd".
.
, where 
. Then, we need:
to be an integer. This is equivalent to saying 
is an integer, call this . We now consider the following cases on which integer is:
. Then we need 
. However, the only consecutive perfect squares are 
, which implies 
and 
. But since , we also need 
which is a contradiction.
. Then we need 
. Now take 
, which shows that the LHS is either and the RHS is either 
. This implies that 
and 
, so 
and 
for some positive integers 
. When we substitute, the equation becomes:
wherupon taking 
implies that no solutions exist.. Then we need 
. Taking , the LHS is either 
and the RHS is , so no solutions exist in this case.
. This implies 
. But for 
we have:
so equality never holds (the reason we need is because the first inequality is not true if 
and , for instance). If the LHS is still less than the RHS, since the LHS is at most 
and the RHS is at least 
. Thus there are no solutions in this case either.
for which is an integer. Here denotes the greatest integer less than or equal to .
. Assume that exists solutions:
is surjective on 
. So we can let where 
and hence ![$[\sqrt{n}]=a$](http://latex.artofproblemsolving.com/a/9/9/a99b541b9c74efd3333f797253aff43af9260c87.png)
now lets use these informations:
That means 
where 
Case 2.- 
Case 3.- 
Hence no such positive integers.
![$n=m^2+k,k\in[0,2m]\cap\mathbb Z,m\in\mathbb Z$](http://latex.artofproblemsolving.com/8/3/e/83eb3e3f5c2c0eb664856a0709d06a3ddfde6ce0.png)
then we have:
By the aforementioned bounds, we obtain 
.
, and it's well-known that the only squares which are one apart are and , but there are no solutions here after testing 
, .
, now
gives that 
and 
. Now 
while if 
, 
, so no solutions.
, by
we have 
, which enforces and . Then 
while, if 
, 
, contradiction. satisfy this condition. 
, where 
are both nonnegative integers and ![$m \in [0,2k]$](http://latex.artofproblemsolving.com/b/1/2/b12c63dba33198485915249fad9b4a30b9a02b9a.png)
.
needs to be an integer, so 
needs to be an integer. Since , we have that can only be 
or . I claim that it can't be any., we have that 
, and this clearly holds false., we have that 
. Observing this in mod , we see that 
and 
. Let 
and 
. Substituting these values in we get that 
, which clearly isn't possible again by observing in modulo ., we have that 
. The right hand side is equal to 
, hence this is absurd. can never be an integer, so there exist no solutions for . 
. Let 
and 
for some 
. Clearly 
. Due to clear size reasons the quotient of the last divisibility cannot exceed . When the quotient exceeds working mod yields contradiction and when the quotient is there is only one possibility; which fails as well.
.
, where 
.
is an integer.
.
is a positive integer.![\[(c-2)^2+1\le (2m-2)^2+1=4m^2-8m+5<4m^2+8=4(m^2+2)\]](http://latex.artofproblemsolving.com/0/c/7/0c7704d9a836a8582151da9e9b46b2528164fa91.png)
and 
are positive, we have ![\[\frac{(c-2)^2+1}{m^2+2}\in \{1,2,3\}\]](http://latex.artofproblemsolving.com/5/6/e/56e6f82ee47b114f0420c8d2608ecb976c340e99.png)
. is a perfect square, which is not possible as 
.
. is a perfect square.
, so 
, both are not QR's.
. is a perfect square. Clearly not possible as it is 
.
where 
. Our question turns into
Long division yields
Since 
, the maximum value our expression can get is
Hence we will divide into cases
Hence we have proven our claim, therefore we are done.
works.
, then 
. Let 
, then![\[\frac{n^2+1}{\left\lfloor\sqrt n\right\rfloor^2+2}=\frac{(x^2+r)^2+1}{x^2+2}\rightarrow x^2+2\mid (r-2)^2+1.\]](http://latex.artofproblemsolving.com/a/f/9/af94f74bc6c14824326af910f2ba2be7f757ed98.png)
, so![\[\frac{(r-2)^2+1}{x^2+2}\in\{1,2,3\}.\]](http://latex.artofproblemsolving.com/4/0/4/40488a7b8852be59edf809942110de8696ff0d6b.png)
. Then 
, so 
. So 
, so 
, absurd as is positive.. Then 
, but taking mod yields a contradiction.. Then 
, but taking mod yields a contradiction.
has no integer solutions. 
is either 0 or 1 mod 3, and 
is also either 0 or 1 mod 3. For the total to be a multiple of 3, both have to be 0 mod 3, so and 
are both multiplies of 3, so the LHS is a multiple of 9, contradiction.
has no integer solutions. is either 0, 1, or 4 mod 5, and 
is either 0, 2, or 3 mod 5. The only way for the total to be 0 mod 5 is if both and are multiples of 5, so the LHS is a multiple of 25, contradiction.
. From now on, suppose 
. Suppose that 
for a positive integer 
and nonnegative integer 
such that 
(essentially make as large as possible). Then, 
Thus, 
Then, note that 
so we must have one of 
The first one clearly cannot happen as a positive square plus 1 is never a square, while the other two cannot happen by Claims 1 and 2, so we are done.
where . Note that this implies that 
. Note that we require 
Note that we can rewrite this as 
Equivalently, 
so we are left with 
. This is trivial to prove. It suffices to check three cases.
from which we get 
and 
. This yields that so there are no solutions.
.
. We will prove that cannot be a square. Note that if 
then 
which is bad. So let 
such that 
. However, this means that 
which is clearly false. Therefore, cannot be a square and there are no solutions for this case.
.
. Note that since 
we know that cannot be a square and there are no solutions in this case.
s.t. m is maximal but k stays a positive integer; we want ![\[\frac{(m^2+k)^2+1}{m^2+2} = m^2 + (2k-2) + \frac{(k-2)^2+1}{m^2+2}\in\mathbb{Z}.\]](http://latex.artofproblemsolving.com/7/c/e/7ce785e512e87806d33a16a9f9652be7021d018d.png)
Since , we only need to compute if the fraction equals 1,2,3.
, which is impossible unless 
, contradiction.
, contradiction.
, contradiction. We've exhausted all cases, so there is no solution for such n.
.
, and let 
. Then the condition becomes 
is an integer and 
. Thus 
. Since 
, so 
.
, then 
, a contradiction.
, then 
. So taking mod 3 forces 
. Then taking mod 9 gives 
, a contradiction.
, then 
, so 
and 
. This contradicts .
exist. Let 
and set 
, where 
. The divisibility condition 
after some reduction. In particular, this implies that 
, then 
must be a square, which is impossible.
, then 
must be a square, which is impossible by mod .
, then 
must be a square, which is impossible by mod .
and thus no valid .
.
. Then using the floor function inequalities, we get,![\[ \dfrac{k^4+1}{k^2 + 2}\le \dfrac{n^2 + 1}{k^2 + 2} \le \dfrac{(k+1)^4 + 1}{k^2 + 2}. \]](http://latex.artofproblemsolving.com/a/c/d/acd5191f37517735d9abe976da1aa31f5f63cd99.png)
. This gives us that 
. Also, we have that 
. This gives us 
. Combining these two bounds, we get,![\[ k^2 - 1 \le \dfrac{n^2 + 1}{k^2 + 2} \le k^2 + 4k + 3. \]](http://latex.artofproblemsolving.com/d/9/0/d90577bd715bb152b7573a072e856b8901ff8bb7.png)
, that is 
. Now we first need to find all the set of possible perfect squares in the interval ![$[k^4 + k^2 -3, k^4 + 4k^3 + 5k^2 + 8k + 5]$](http://latex.artofproblemsolving.com/9/d/5/9d500ee88443bcc54364cbc627e04cf365f7353b.png)
. Now we can note that all the perfect squares in this range are 
. This gives us that 
. as 
where 
. Since 
is an integer, we get that 
, that is,![\[ k^2 + 2 \mid (k^2 + i)^2 + 1 \equiv (-2 + i)^2 + 1. \]](http://latex.artofproblemsolving.com/8/0/b/80b6bf43184b6882bef8d4192ec8d0cd42d3d9eb.png)
which gives that the only possible values for 
are 
. Now among these possibilities, 
fails due to 
. Also, 
fails due to 
.
. This is only possible if 
. So now we are working with 
which forces 
, contradiction as was a positive integer.
. Suppose otherwise, then 
where 
and .
, so by plugging in 
(i.e. long division), we have:
.
, since the numerator is positive and 
. These cases each fail by 
, modulo , modulo respectively, so there are no solutions.
, so we have 
where 
we have:
has no solutions , and so no such works. 
where 
and 
. Then we want 
to be an integer. Subtracting 
means we want 
to be an integer. Since , we must have that the fraction is , , or .
gives 
, which is not possible.
gives 
. Since squares are never 
, let 
and 
. Then, 
so 
. This is clearly not possible because 
.
gives 
. Since perfect squares are 
, let 
and 
. Then 
so 
. This is clearly not possible because 
..
such that 
(so that 
). We have that:
However, since 
we have that 
Hence, we have that 
and by difference of squares, 
Hence, 
However, neither of these satisfy 
we have that 
Hence, 
By taking modulo 8, this is impossible.
we have that 
Hence, 
By taking modulo 3, this is impossible.
,
,
such that 
coprime, 
and 
.
.![\[P=\dfrac{a}{\sqrt{2bc+1}}+\dfrac{b}{\sqrt{2ca+1}}+\dfrac{c}{\sqrt{2ab+1}}.\]](http://latex.artofproblemsolving.com/1/f/3/1f39a54e4cf378bfd4e66d97949350b9b37dc1ec.png)
u
such that ![$f\in [0, 2m]$](http://latex.artofproblemsolving.com/b/4/f/b4fcc264e19eb861b8f41f929100f56f1c76f5c9.png)
.
.![\[m^2 + 2 | (m^2 + f)^2 + 1 \iff m^2 + 2 | (f - 2)^2 + 1\]](http://latex.artofproblemsolving.com/e/c/1/ec1b364313537e99889ea91f1dfe150e213ad631.png)
Note that![\[(f-2)^2 + 1\leq (2m - 2)^2 + 1 = 4m^2 - 8m + 5 < 4m^2 + 5 < 4(m^2 + 2)\]](http://latex.artofproblemsolving.com/6/2/3/623074c602927299c4500907fed75ff4b8c1f7c2.png)
Thus, 
.![$k = f -2, k \in [-2, 2(m - 1)]$](http://latex.artofproblemsolving.com/d/c/e/dceea117acc432dfcaec70bfca08ff12e7802842.png)
.
.
.![$k\in [-2, -2]$](http://latex.artofproblemsolving.com/3/b/8/3b87cc1fe72b9575926e220f70373616cc3ca1fc.png)
,
,
.
(since 
)
which is absurd.
.
which is impossible.
.
.
,
.
,
.
.
,
or 
.
,
,
and 
.
,
,
,
or 
,
where 
From the above bound we know that 
where 
which is impossible.
and by FLT we know that if 
then 
which is a contradiction and therefore 
but then 
which is impossible.
but 
no solution.
with 
;
.![\[a^2+2\mid (r-2)^2+1\]](http://latex.artofproblemsolving.com/4/8/6/486c669dd046ad3498b2a9e6f87ca5cca075da7b.png)
and by bounding we obtain either![\[(r-2)^2=a^2+1\]](http://latex.artofproblemsolving.com/d/5/2/d52ddf1a795a9cbebf1a0e97e5bc108920dfcd54.png)
![\[(r-2)^2=2a^2+3\]](http://latex.artofproblemsolving.com/b/b/8/bb82a107d970b88f334c88d68907174c5228fff1.png)
![\[(r-2)^2=3a^2+5\]](http://latex.artofproblemsolving.com/e/9/7/e97027756a252bac6866680d282a9882c7c5ea44.png)
and the first equation fails by DOS, the second by noting that we must have 
,
.
![$[\sqrt{n}]=k$](http://latex.artofproblemsolving.com/4/e/3/4e32b7ca7323a82482648f1606573594fe7e217b.png)
.
Note that 
and 
.
,
which yields no solutions.
,
.
,
.
,
which is impossible as 
.
for 
.![\[ \frac{(k^2 + m)^2 + 1}{k^2 + 2} = \frac{k^4 + 2k^2m + m^2 + 1}{k^2 + 2} = k^2 - 2 + 2m + \frac{m^2 - 4m + 5}{k^2 + 2} \]](http://latex.artofproblemsolving.com/8/6/e/86e11a43e9504d55822cae7352ef4344ab932cab.png)
This implies that 
for 
.
,
,
,
gives no solutions. If 
,
must divide![\[a^4+2a^2b+b^2+1 \implies 2a^2b-2a^2+b^2+1 \implies b^2-4b+5.\]](http://latex.artofproblemsolving.com/a/f/c/afc3f7099c11f5f96b8beeb1408616884e8637d7.png)
,![\[b = \sqrt{2a^2+3}+2, \quad \sqrt{3a^2+5}+2.\]](http://latex.artofproblemsolving.com/c/5/d/c5d4e981ce2f280775255a5b7ed1e09954911ff6.png)
.
If 
then 
,
.
,
,
which implies 
,
which implies 
,
which is a contradiction since the lhs is always even and the rhs is always odd. If 
then 
which is impossible. Thus no solutions exist and we are finished. 
and 
and 
.
Note that 
.
or 
then 
,
then 
which is not possible. Thus no solutions exist.