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 scalene triangle with circumcircle 
, and suppose the incircle of touches 
at 
. The angle bisector of 
meets and at 
and 
. The circumcircle of 
intersects the 
-excircle at 
, 
, and at 
. Prove that line 
passes through either or .
be positive real numbers such that : 
. Prove that : 
![$P, Q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
be relatively prime nonconstant polynomials. Show that there can be at most three real numbers 
such that 
is the square of a polynomial.
be real numbers less than 2.Then prove that 
and 
intersect each other at 
and 
. The common tangent to two circles nearer to touch and at 
and 
respectively. Let 
and 
be the reflection of and respectively with respect to . The circumcircle of the triangle 
intersect circles and respectively at points 
and 
(both distinct from ). Show that the line 
is the second tangent to and .
has circumcircle and circumcenter 
. A circle 
with center intersects the segment at points and , such that 
, , , and 
are all different and lie on line in this order. Let and 
be the points of intersection of and , such that , , , , and lie on in this order. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
.
and 
are different and intersect at the point 
. Prove that lies on the line 
.
be the diameter of semicircle 
. Circle is internally tangent to and tangent to at . Let be a point on and a point on such that 
and is tangent to . Prove that 
bisects 
of lowest possible degree such that has integer coefficients, are integers,
,
.
such that for any odd integer 
, we have 
such that for any odd integer , we have 
such that for every positive real 
, there exists an integer 
satisfying![\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]](http://latex.artofproblemsolving.com/4/4/4/4443cf7166cc65e67e2e504a826c4ac5f91aab2b.png)
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
, 
.
of real numbers that satisfy the system of equations![\[ \begin{cases}x^3 = 3x-12y+50, \\ y^3 = 12y+3z-2, \\ z^3 = 27z + 27x. \end{cases}\]](http://latex.artofproblemsolving.com/d/4/d/d4d6c485b4653577e732ff235448bb779914a576.png)
then multiply all of the equations,
.
. Substituting them, we get the following equations:![\[ \begin{cases}a^3-3a=50-4b, \\ b^3-3b=\frac{9c-2}8, \\ c^3-3c=a. \end{cases}\]](http://latex.artofproblemsolving.com/a/1/9/a19ab62ed146a7761cf5f004cb55d3ed05ecde31.png)
Note that the function 
is a cubic polynomial function. 
has two solutions, namely 
.
. If 
then 
and if 
, then 
. If 
then 
.
then 
and if 
then 
.
, then 
, giving 
. If , then 
, which means 
. If then 
, contradiction.
, then 
, giving 
. If , then 
, which means 
. If then , contradiction.
is forced, giving 
and 
, which clearly work.
is the only solution.
.![\[(x, y, z) = (\alpha + 2, \beta + 4, \gamma + 6)\]](http://latex.artofproblemsolving.com/7/c/f/7cf0f39abd7d2a24b930c04cb64bee8709eb9504.png)
![\[(\alpha + 2)^3 = 3\alpha - 12\beta + 8\]](http://latex.artofproblemsolving.com/8/c/6/8c62e932e8e9496b78fb6a3f9d8b4febdb27e9b5.png)
![\[(\beta + 4)^3 = 12\beta + 3\gamma + 64\]](http://latex.artofproblemsolving.com/6/f/b/6fb540c563317ed5e6f39777f0e987f4da1413c1.png)
![\[(\gamma + 6)^3 = 27\alpha + 27\gamma + 216\]](http://latex.artofproblemsolving.com/3/c/e/3ce31b2e4c33f0b38b184d030c5bf951b843c7ff.png)
![\[\implies\]](http://latex.artofproblemsolving.com/c/3/0/c30ac6e41a136bd85e681667b55d9a3c825ee00e.png)
![\[\alpha(\alpha +3)^2 = -12\beta\]](http://latex.artofproblemsolving.com/f/f/5/ff57086005e680583660cd538c57b045386440bc.png)
![\[\beta(\beta + 6)^2 = 3\gamma\]](http://latex.artofproblemsolving.com/8/0/a/80ae7ff7bb124e175315ce8e9483e8bd88f0c25e.png)
![\[\gamma(\gamma + 9)^2 = 27\alpha\]](http://latex.artofproblemsolving.com/c/6/d/c6d7ca9a2db52675f269bf445181d4e774dc902f.png)
![\[\alpha\beta\gamma(\alpha +3)^2(\beta + 6)^2(\gamma + 9)^2 = -972\alpha\beta\gamma\]](http://latex.artofproblemsolving.com/2/4/f/24fa185f9a62545d5196015c7c035f5194c1a8a0.png)
FTSOC, assume that 
.
results in![\[(\alpha +3)^2(\beta + 6)^2(\gamma + 9)^2\ = -972\]](http://latex.artofproblemsolving.com/3/f/3/3f316a19d805b53ac059182b3ede1662bd4d0d6a.png)
which is a contradiction due to the trivial inequality.
. Plugging 
into 
, 
, or 
, we can clearly see that it makes the rest of the variables , so our answer for 
is .
. Motivated by this answer, we notice our equations can be rearranged to be
, 
, or 
to avoid a trivial inequality contradiction. However, any one of the these imply the other two, which finishes. 
.
Multiplying we have,
Clearly if all of the factors 
, 
and 
are nonzero we have no solutions. However if we have 
, 
or 
we find the claimed solution so we are done.
This can clearly be shown to work just by plugging in.
and 
After plugging in and reducing the equation, we get the following systems of equations:
If any of 
are equal to 
then all of them must be simply by plugging in, giving us the solution claimed at the beginning of our solution. Henceforth assume that are nonnegative.![\[
-\frac{p}{12q} = \left(\frac{1}{p+3}\right)^2,
\]](http://latex.artofproblemsolving.com/a/3/4/a3473d77c997fce634847ec787644ec992ef817d.png)
so![\[
-\frac{81p}{q} = 972 \cdot \left(\frac{1}{p+3}\right)^2.
\]](http://latex.artofproblemsolving.com/2/6/c/26c9412bb67c0ec5a06851bfa7ec5f838d50956b.png)
Meanwhile, the second and third equations multiply to![\[
qr((q+6)(r+9))^2 = 81pr,
\]](http://latex.artofproblemsolving.com/d/1/5/d1591e876da0291923bc1dc6e5b289b315b5613e.png)
which, since 
rearranges to![\[
\frac{81p}{q} = ((q+6)(r+9))^2.
\]](http://latex.artofproblemsolving.com/2/1/7/2172e9ebb4934b62e9520402f78defebd57099d4.png)
Adding our two equations together, we get![\[
972 \cdot \left(\frac{1}{p+3}\right)^2 + ((q+6)(r+9))^2 = 0.
\]](http://latex.artofproblemsolving.com/9/c/b/9cbb335217865cfd3ae938c5df63c25a67e718ff.png)
Since the square of a real number is nonnegative, all of the squares in the above expression are 
In particular, 
which is impossible.
and so the only solution to our system of equations is 
as claimed.
is the only one. Now note that the equations are equivalent to:
It is easy to see that , , or are independently sufficient enough to yield as a solution, so assume 
. Then multiplying the three equations and dividing by 
gives
which is clearly impossible. So is the only solution and we are finished.
, which clearly works.
.
Now assume none of 
are zero, since if one is then clearly all are.
In particular, if 
denotes the sign of 
, 
a clear contradiction. Therefore the only working solution is 
. 
, 
, 
![\[ \begin{cases}a^3+6a^2+9a=-12b ...(1) , \\ b^3+12b^2+36b=3c ...(2) , \\ c^3+18c^2+81c=27a ...(3) \end{cases}\]](http://latex.artofproblemsolving.com/d/e/9/de93ce1bf37ee5ffea2803f03dae3597fec5936b.png)
we will prove it by a contradiction.
and 
and 
we get:
.
and 
and 
we get:
.
from 
we find that 
and finally from we find that 
. 
Then our equations become
However, observe that for each one we can complete the square:
From the second and third equations, and have the same sign, while 
and have the same sign. However, from the first equation and have opposite signs, therefore they are both Hence 
too. This yields as our only solution.
, to which we motivate the substituitions 
. Then the equations become:
or 
. If all are nonzero, then we multiply all of these equations, cancel 
, and arrive at a contradiction. If any one is zero, the rest all are so we are done.
,
,
so from 2 we have 
so x=2,y=4,z=6
,
so from 2 we have 
and take 
from the last equation into the first one, so that it turns into
and 
.
,
and 
.
and 
.
,
,
,
.
(Or 
,
.
,
.
or 
,
.
to make more of the coefficients match. Then we get 
,
.
pretty much jumps out. We note that
,
,
,
.
equals 2, then they all do. If not, multiplication yields that
,
,
and 
(this is motivated by observing 
)
Note 
and 
.
.
,
which is why I attempted the bounding in the first place.
,
,
.
which gave 
the function 
is increasing so we have 
by the first equation. Also the third equation yields that, from 
.
.
.
.
works so 
had to be all integers. I tried substituting floors and what not to prove that this was the case, but it turned out too messy. I proceeded to try to just guess integer solutions, starting from 
,
but tried dumb things first like bounding on just 
.
as otherwise we get ![\[ \begin{cases} -12(y-4)=(x-2)(x-1)^2, \\ 3(z-6)=(y-4)(y+2)^2, \\ 27(x-2)=(z-6)(z+3)^2. \end{cases}\]](http://latex.artofproblemsolving.com/e/8/e/e8ec743edbabcb686307be6aec1da3d6cbe5b19f.png)
Taking the product and cancelling the common (non-zero) factor 
we see that both sides have an opposite sign unless 
.
a contradiction! Similar arguments apply for the other two factors.
,
and similar statements with all the inequality directions reversed.
so 
It is easy to check that none of 
or 
or 
produce valid solutions. Therefore, 
.
.
,
,
.
are of opposite sign, 
are of the same sign, and 
are of the same sign. The only way this can happen is when 
.
Multiplying yields ![\[(x + 1)^2 (y + 2)^2 (z + 3)^2 (x - 2)(y - 4)(z - 6) = -972(x - 2)(y - 4)(z - 6).\]](http://latex.artofproblemsolving.com/9/7/d/97dbb06c306f58ba081db70a77d870e9635b7158.png)
Now if ![\[(x + 1)^2 (y + 2)^2 (z + 3)^2 = -972,\]](http://latex.artofproblemsolving.com/2/d/1/2d15d85201dc01a6b7fc50a61536aa826c23442b.png)
a contradiction.
We can check this works.
The equations become 
Now it is easy to see that if one of 
is zero then all are. But multiplying all together gives 
impossible if 
Thus 
so 
the equations becomes:
Now we have 
Which implies 
.
in terms of 
,
in terms of 
by looking at signs. The singular solution