AoPS Academy
![\[x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0\]](http://latex.artofproblemsolving.com/f/d/0/fd0ab11635ee44ad272a54c0f20a2bc47a3596de.png)
cannot all be real if 
.
be a triangle with 
and let 
be the other intersection point of the angle bisector of 
with the circumcircle of the triangle . Let 
and 
be points on the sides 
and 
respectively, such that 
and let 
be the point of intersection of 
and 
. Let 
be the midpoint of 
. Prove that 
and the circumcircles of the triangles 
and 
pass through a common point.
be a sequence that consists of an initial block of 
positive distinct integers that then repeat periodically. This means that 
are distinct positive integers and 
for every positive integer 
. The terms of the sequence are not known and the goal is to find the period . To do this, at each move it possible to reveal the value of a term of the sequence at your choice. is one of the first 
prime numbers, find for which values of there exist a strategy that allows to find revealing at most 
terms of the sequence.
+2 w
be a cyclic quadrilateral with 
and 
. Point 
and 
are selected on segment such that 
. Points 
and 
are selected on segment 
such that 
. Prove that 
is a cyclic quadrilateral.
and 
intersect at points 
and 
. A line through intersects and at points 
and 
, respectively. Line intersects at point 
, and line intersects at point 
. If 
is the circumcenter of 
, prove that 
.
, with 
, is inscribed in circle 
. Let be an arbitrary point inside 
such that 
. Ray intersects again at 
(other than ). Point 
(other than ) is chosen on such that 
. Line 
intersects rays and at points 
and 
, respectively. Prove that 
.
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of , how many orderly colorings are there?
and 
with real coefficients satisfying
as the solution set of 
. From the statement we know that 
.
, we have 
and 
.
for polynomials 
, we get that 
., 
.
and 
.
and 
are pairwise distinct. Now look at 
. This must have as a root with multiplicity 
, and 
as a root with multiplicity 
. This implies that 
Rearranging gives the desired lemma.
, an obvious contradiction.
work, so 
and 
for some . Then we have: 
![\[ P' \cdot P^8 \cdot \left( 10P + 9 \right) = Q' \cdot Q^{19} \cdot \left( 21Q + 20 \right) \neq 0. \]](http://latex.artofproblemsolving.com/2/b/0/2b007bc7a7313d2fee06e81185be092842719f6f.png)
Thus 
should divide 
but ![\[ \deg(10P+9) = 21n > 20n-1 = \deg(Q' \cdot (21Q+20)). \]](http://latex.artofproblemsolving.com/a/0/2/a02800e5a576930474824d3332601eafcef123c5.png)
. is disjoint with that of 
. Similarly for . Note that 
, so the number of roots of 
, counting multiplicity must not be less than the sum of multiplicities of the roots of 
decreased by 1, so the number of roots of and is at least 
. But the number of roots of 
doesn't exceed 
, which is a contradiction.
and 
, the equation rewrites as 
. It is clear that the polynomial 
is indecomposable. Then according to Corollary 2.18 in http://dept.math.lsa.umich.edu/~zieve/papers/peter.pdf (which is proven using monodromy groups) there exists a linear polynomial 
such that 
or 
, where 
is the 
th Chebyshev polynomial. The first case is impossible since contains a nontrivial coefficient of 
but not of 
, and the second case is impossible since 
but 
.
be two coprime integers. Prove that there are no non-constant polynomials and with real coefficients such that 
the problem is true.
the problem is true.
.
the problem is true..
be the set of the roots of not counting multiplicity.Is it possible that 
and 
?
and differentiating to kill the one on the LHS.
is same as IMC 2000 day 2, prob 3.
, where 
and 
as shown in rkm0959's solution. Now, we look at the Lemma given below (it is basically a generalization of the Lemma in post #5), after which we can finish in a similar fashion as in rkm0959's solution.![$F \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/d/3/d/d3d45964209a0ff752736cab92d5a43b4afbfb1d.png)
be a non-constant real valued polynomial of degree 
. Consider the 
distinct real numbers 
where 
. Then the total number of complex solutions to the equation 
for all 
is at least 
. (The previous Lemma follows on taking 
and 
)
to denote the integers from 
to . First consider a general polynomial ![$A \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/0/1/c0197711a982fd1feffd53aaa95e1a7ea416795b.png)
. Suppose that 
has degree 
, where 
is the first derivative of . Then there exist roots of (not necessarily distinct) which are roots of also, which means that the remaining roots of are distinct and correspond to the elements of 
(Just use the fact that roots of with multiplicity have multiplicity 
in ). So we have 
. Return to our Lemma. Letting denote the total number of solutions to the given equations, and using the fact that 
, we have 
Then we wish to show that 
, which is equivalent to showing that 
Consider the polynomial 
. As all these terms in product are pairwise co-prime, so we get that 
is 
is well-known (maybe not exactly in that form). Similarly, the second half of the proof of the Lemma (introducing 
) is well motivated if one catches hold of the fact that degrees get added on multiplying pairwise co-prime polynomials.
to work with.
, we find that the degree of is 
for some 
and the degree of is 
. divides 
. Additionally, notice that 
hence we know that 
divides 
. But this is cap; the degree of the left is and the degree of the right is 
, smaller. Thus we are done.
work, so and for some . Then we have: Thus should divide but 
is an arbitrary root of either side of the equation (we will refer to this as a "root of the equation"), and let 
denote 
. We consider four cases—note that must fall in exactly one of these. is a root of , then it is a root of 
with multiplicity divisible by 
, and a root of 
with multiplicity divisible by 
. Thus is a root of the equation, it must have multiplicity divisible by 
. Thus we can say that there are distinct roots of the equation which contribute a total multiplicity of 
. is a root of 
, then it is a root of the equation with multiplicity divisible by , so we can say there are distinct roots with total multiplicity 
. is a root of 
, then we can say there are distinct roots with total multiplicity 
. is a root of 
, then we can say there are distinct roots with total multiplicity 
.
and similarly for the other pairs of variables. will have exactly 
roots with multiplicity (since each root contributes nine times) and will have exactly 
roots with multiplicity (since each root contributes once). Likewise, will have exactly 
roots and 
will have exactly 
roots. with multiplicity 
is a root of 
with multiplicity 
. Thus, has at least 
roots (with multiplicity). Thus,
and similarly 
(from looking at instead of ), so
On the other hand, this means that
which is absurd, hence no such polynomials exist. 
denote the set of roots, without multiplicity, of a polynomial . Then 
and 
for some positive integer .
by the lemma. On the other hand, 
This yields a contradiction. 
and its derivative 
. The former implies that any root of is a root of or . In particular, does not share any roots with , so 
. However this is absurd since 
, so no such polynomials exist.
and 
for some . Clearly we have that all the roots of and all the roots of 
or 
. Let be the set of distinct roots of , 
be the set of distinct roots of and and 
similarly. We have that 
by Putnam 1956 B7, thus we also get that as ,
, so we have a contradiction and thus no such polynomials exist.
and in question by mistake and Im too lazy to change, sorry.
.
and 
for some 
. Differentiate it once and see that we have the equations 
Divide the two equations and we get ![\[(Q+1)QP'(21P+20)=(P+1)PQ'(10Q+9]\]](http://latex.artofproblemsolving.com/c/5/a/c5a8d5b269e476db4902b34fc17e43899699e594.png)
But see that 
anfd so we have ![\[(Q+1)Q \mid (P+1)PQ' \implies 42d \le 20d+21d=41d\]](http://latex.artofproblemsolving.com/7/8/8/7887b454413432a90b6091bef92bd6845db2843c.png)
Which is a contradiction.
.![$X,Y,Z \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/8/a/5/8a5243f7f1c3c100a2636d38d6ad7244a0351170.png)
such that 
have no common complex roots, and for some positive integer 
.![$X_1,X_2 \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/7/0/3/7039ca2ddf2d4dd047717a2782741a4ca75c475d.png)
such that 
,
,
holds.![\[ X= \prod_{i=1}^{n} X_i^{e_i}\]](http://latex.artofproblemsolving.com/9/e/0/9e073860988da37db0d2ae49c74ebb905f63717f.png)
.
,
divides 
,
divides 
,
.
and 
,![$Q_1,Q_2 \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/e/0/e/e0e47c04af3fde7bf35109842640f73dbe2da8f1.png)
such that 
,
,
.![$P_1,P_2 \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/0/9/6/0960572f4142d9415ffcb52005b80eeb3e260c7b.png)
such that 
, 
,
.
.
so that they all have positive leading coefficients.
and 
.
,
,
for some positive real 
are all positive)
for all sufficiently large 
for each fixed constant 
.
for all sufficiently large 
for each fixed constant 
.
,
such that
,
.
such that 
,
,
a contradiction.
at the original equation, we get ![\[P_2^9(P+1)=cQ_2^{20}(Q+1)\]](http://latex.artofproblemsolving.com/1/c/5/1c50d5a0ff054b004c847ef8aea8f5774f36ecdf.png)
.
such that
.
such that 
.
,
,
, a contradiction.
as 
goes to 
.
such that 
for every real 
,