AoPS Academy
satisfy the equality, 
prove that the triangle must have a right angle.
such that for any real numbers 
and 
, the following inequality holds:![\[
f\left(x^2+2y f(x)\right) + (f(y))^2 \leq (f(x+y))^2
\]](http://latex.artofproblemsolving.com/8/4/3/8437449cd3f8f0da24d17a60b55dcb563e6b5655.png)
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
and 
be positive real numbers such that![\[
\log_x 64 + \log_{y^2} 16 = \frac{5}{3} \quad \text{and} \quad \log_y 64 + \log_{x^2} 16 = 1
\]](http://latex.artofproblemsolving.com/d/7/9/d79facd90d9c5b53cdc4df9db18923ad34f472df.png)
What is the value of 
?
+1 w
. Find the probability of selecting two real numbers ![$a, b \in [0, 10]$](http://latex.artofproblemsolving.com/e/b/7/eb738a5d732aaf4e2ae46ebefe6a57ae8382f86f.png)
such that the equation has at least one real root.
to be an infinite polynomial where ![\[P(x) = \lim_{n\to \infty}\prod_{k=0}^n \left(x-\left(\cos\left(\dfrac{2\pi k}{n}\right)+i\sin\left(\dfrac{2\pi k}{n}\right)\right)\right).\]](http://latex.artofproblemsolving.com/c/a/8/ca8d0747b7186eb097eb7bd71a95f75d2e38af32.png)
, 
. 
, and 
terms. 
be real numbers with 
that satisfy the equations
} can be written in the form 
, where , and are positive integers. What is 
?
more than the first term, and the fourth term is 
more than the second term. Find the first term.
satisfies the following conditions:
Find the value of 
.
satisfies the following conditions:![\begin{align*}
&f(4xy)=2y[f(x+y)+f(x-y)]\text{ for all }x,y\in\mathbb R\\
&f(5)=3
\end{align*}](http://latex.artofproblemsolving.com/6/2/5/625b8046a00d07680dd2a8d2aa3231d5daa4fcf2.png)
.
is a function satisfying 
and 
, 
.
then minimum possible number of values of satisfying 
, for ![$x\in[0,170]$](http://latex.artofproblemsolving.com/3/d/2/3d2c6ead3059f96f853bba6cfc7440e394227ab9.png)
is?
$ is symmetrical about which line?
.
equations in real variables 
:![\[v_i = 1 + \frac{6v_i^2}{v_1^2 + v_2^2 + \cdots + v_{10}^2} \qquad (i = 1, \ldots, 10).\]](http://latex.artofproblemsolving.com/a/7/5/a75a8d8b5d2f134372e154f12225d9e4ae820cbf.png)
Find all -tuples 
that are solutions of this system.
can only be one of two values (since the given equation is a quadratic whose coefficients are the same for all ). You can let the solutions be 
. Let the sum of the 
be 
, the sum of squares of the be 
. We get in fact 
, 
.
. Suppose 
of the equal 
, the other 
equal 
. So we have 
and 
. Note that we have 
variables 
, and equations of degree at most 
, and we can solve by simple elimination/substitution. Actually it turns out really nicely, you get what appears to be an equation of degree 
in , but its actually of degree 1, and you find 
, 
,
, so all the 's are equal to 
.'s are not equal to zero or 1. This is obvious since each is at least 1, and if any equals 
, then 
(look at the original equation).
is a solution.
Let me see where I made a mistake...
., satisfy the quadratic 
. From this we can deduce 
. We know is not so no problem. Thus
. Solve it:
is real, the discriminant is nonnegative, so 
OR 
. Of course the cases are symmetric (swap with 
), so we only need to consider 
and ., all the 's are equal, and we easily find 
. If , we get 
. So these are the only two solutions.
, the s are a solution of 
.
. Say there are 
s and 
s.
, 
, 
.
and 
and solve for . Discriminant gives 
or 
.. If 
, we have 
, 
, so 
instantly, giving 
.
, we have 
, so we have 
.
. Rewrite all equations as![\[(v_i - 1)Q = 6v_i^2\]](http://latex.artofproblemsolving.com/6/e/c/6ec1fc183072b95a5501f2528b172795fe835935.png)
for all ten 
. Clearly 
is non-negative hence all 
, so we can divide both side to get![\[Q = \frac{6v_i^2}{v_i - 1}\]](http://latex.artofproblemsolving.com/b/5/c/b5cae6ec7251befe455c0ddae7547306c9181973.png)
for all ten . Let 
.
, we have 
if and only if 
or 
.
and 
and expand to get 
which holds if and only if we have or 
for 
. 
that are equal. Suppose not, and any three 
are all pairwise distinct. However, all 
are equal to each other hence 
, a contradiction.
where is some real value greater than . Then, for any other 
, we must either have 
or 
. Hence, we can partition the set of 
into two 
and 
size multisets such that one of them consists only of 's and the other consists only of 
's.![\[S = ka^2 + (10-k)\left(\frac{a}{a-1}\right)^2 = \frac{6a^2}{a-1} \implies (10-k)r^2 -6r + k = 0\]](http://latex.artofproblemsolving.com/f/a/3/fa3f8c5b93cadeec13870867e75b5038c76ced76.png)
where we let 
. Since 
is real, the discriminant of this quadratic is nonnegative, hence 
. Since the 
and 
cases are the same, it only suffices to check for 
and 
., then all of them are the same, and thus we get 
., then we get that 
hence 
and 
. Thus, in this case 
or one of its permutations.
equations in real variables :Find all -tuples that are solutions of this system.
The crux of the problem is the following claim.
consists of at most different numbers.
. We have:
Subtracting these equations yields to:
where 
. This means that either or 
. If we fix 
, then we get the values that are equal to and some that are not. We will prove that those, which are not equal to , are equal between themselves. Suppose there are indices such that 
and 
, then 
. This proves the claim. numbers, which attains the value and 
numbers which attains the value . From before established results we get the following equations.
First equation can be rewritten as follows:
This means that:
Plugging this back into the first equation yields to:
Treating this as a quadratic equation with respect to , gives us tht its discriminant must be nonnegative; therefore:
Thus 
, which gives us solutions 
and 
and it permutations.
of distinct indexes such that 
, we get:
and
which implies that 
. Therefore, we have at most two values for the 's, say 
.
, where is the amount of 
. Suppose 
. By subtracting 
we get:
From 
and 
we get:
and since 
is obviously a contradiction
If 
, we get 
which implies a contradiction from 
and 
.
, we get 
and 
, which is a solution.
we get 
, which is also a solution.
and 
and its permutations. 
for all . Then, 
, so 
so 
. By Jensen's, 
. This also means that 
for all . Let 
of the be 
and 
of the be 
. Then, we must have that 
or 
. Since 
, 
. If 
, we get the solution 
, corresponding to the -tuple 
. Otherwise, since 
, 
so 
. Since , 
or 
. We simplify to 
or 
. When , the discriminant is 
. When , our quadratic becomes 
so 
so 
, corresponding to -tuple 
and permutations. Therefore, the solutions are and and permutations.
and permutations of 
. First, summing over all gives us ![\[\sum_{i=1}^{10} v_i = 16.\]](http://latex.artofproblemsolving.com/5/1/9/5197467a82e8460316f042303206fbdf20bd841a.png)
Now, note that 
for all , then we have for some , ![\[\frac{v_i}{v_j}=\frac{t+6v_i^2}{t+6v_j^2}\]](http://latex.artofproblemsolving.com/9/1/3/9131f2463e36f11eb46c84ffc40a2bdfd1096539.png)
where 
, which then gives us ![\[\frac{t}{v_i} + 6v_i = \frac{t}{v_j} + 6v_j \iff t\cdot \frac{v_j-v_i}{v_iv_j}=6(v_j-v_i).\]](http://latex.artofproblemsolving.com/d/d/e/dde3f63d94570db1c95e2a117e1010acd02dd1bc.png)
Thus, either 
, or 
. In the case where for all , we have that there is only one -tuple, which must be . In the case that 
for some , it is first clear that we cannot have 
for some 
, because otherwise we would have 
, which is a contradiction. So, we can only possibly have for some s and 
for other s. WLOG the number of s is at least the number of s. We have the equations ![\[a=1+\frac{a}{b}, b = 1+\frac{b}{a} \implies ab = a+b.\]](http://latex.artofproblemsolving.com/e/6/1/e6162fdc70944652caf5a5109caa495ef153a8a2.png)
Now, 
, and since 
, we cannot have of either or , so either or 
. Substituting 
into these equations, we get 
and no solution for the second equation. Then, we can see that all permutations of the -tuple indeed works, so we are done.
comes from the fact that the RHS would be 
if the s weren't there. Intuitively, there aren't many ways the RHS which contains can become .
. Let![\[
v_i=m_i+1
\]](http://latex.artofproblemsolving.com/e/4/4/e4437d38cc0fd9aeaaba262eaa17cbe64f3abe8d.png)
Now we have 
. We have![\[
\frac{m_i}{(m_i+1)^2}=\frac{6}{(m_1+1)^2+(m_2+1)^2+\dots +(m_{10}+1)^2}
\]](http://latex.artofproblemsolving.com/c/d/f/cdf0970d6514ee13b612606e99770865218cd68c.png)
Since the rhs is equal for all 
we have![\[
\frac{m_i}{(m_i+1)^2}=\frac{m_j}{(m_j+1)^2}
\]](http://latex.artofproblemsolving.com/5/1/6/516be60e95923832bb36d2c8b07176d39df7f165.png)
And after doing some simple algebra we end up with![\[
(m_i-m_j)(m_im_j-1)=0
\]](http://latex.artofproblemsolving.com/1/3/7/137d5df702752e7c4b3f214ab7d07d6bfdb1143e.png)
Hence for any pair of 
and 
we have that they are either equal or their multiplication is . if one of the numbers is larger than and the other is smaller then . OR they are both equal to . So, since the numbers are either equal or when multiplied equal to , we know that if they are both smaller than , then they are equal. Alternatively, if they are both larger then then are also equal., let it have the value , if the number is smaller then let it have the value . numbers with value , numbers with value and 
numbers with value . Of course 
and 
.
are positive integers, moreover, we can say that 
, so we can kill this with case work! Also, note that when we combine and along with 
, we get a quadratic whoose discriminant must be 
. In algebra, this is:![\[
(6-z)^2-4xy\geq 0
\]](http://latex.artofproblemsolving.com/d/1/9/d193cb33eaa0c2e30622fe6923fc3a49345acffc.png)
\begin{itemize}. Now the only choice for is 
which does not work. We proceed by similar reasoning, but this time we divide the case into smaller cases when 
and 
. None of them work.. No case works.
. Again, nothing.
. Here, if we let and 
we actually get something that works! (it is the only thing that works in this case).
and 
, so in turn we get 
for a single and 
for all 
However, upon putting it into the initial equation, we find out that the solution found above indeed is a correct solution.
. We now have that. Now, since 
(as can be seen in ). We have to try the cases![\[
(y,z)=(5,5),(6,4),(7,3),(8,2),(9,1),(10,0)
\]](http://latex.artofproblemsolving.com/a/1/1/a114a8b30e526a34f91a8a96276c3b35b0c204e4.png)
And after trying everything out we see that 
is the only one to produce a solution. Here we get 
which then gives us that 
for all . and 
. Then each must satisfy![\[v_i = 1 + \frac{6v_i^2}{c} \implies 6v_i^2 - cv_i + c = 0,\]](http://latex.artofproblemsolving.com/b/7/c/b7cf1b7f56e6616ca93288f31ff0f9bca3a6d788.png)
values of and values of . Notce Vieta's gives us![\[6ab = c = ka^2 + (10-k)b^2 \implies ka^2 - (6b)a + (10-k)b^2 = 0.\]](http://latex.artofproblemsolving.com/3/d/e/3dec50aad1eb4fcc8d8b1e3aecdbd3bfbc6618f5.png)
, assuming WLOG 
. The easiest way to finish from here is to note that summing our given equation across all gives 
, so 
is immediate, and substituting into the above quadratic in gives 
. Hence our solutions are the permutations of![\[\boxed{\left(\frac 85, \ldots, \frac 85\right), \left(\frac 43, \ldots, \frac 43, 4\right)}.\]](http://latex.artofproblemsolving.com/7/2/1/721a8a287adb0b0fd4ec191630f422b019568955.png)
Then 
for some values 
But by Vieta's 
Also suppose that there are 
s and 
s, WLOG 
Then 
Let 
then 
If 
we have 
So 
is a solution. Meanwhile is 
we have all the are equal so 
so 
works as well. These are the only solutions.
Hence, 
Expanding, 
Hence, either or 
The second case is equivalent to 
Thus, there are only two possible values of 
Let them be 
Let them appear 
times. Hence, we have that 
Define 
for brevity. We get that 
From the discriminant, we have that 
. However, recall that 
Clearly, only 
Clearly, we only need to check 
cases.
using 
we get 
let be the one that appears 9 times, and 
appear one time. Hence, 
Hence, 
Thus, 
and its permutations.
such that 
![\[S = v_1^2 + v_2^2 + \cdots + v_{10}^2 \implies v_i = 1+\frac{6v_i^2}{S}\]](http://latex.artofproblemsolving.com/9/d/9/9d920a33632b5704f84efe778696fe3ed91a0902.png)
Solving the quadratic gives![\[v_i = \frac{S\pm\sqrt{S^2-24S}}{12} \implies v_i^2 = \frac{S^2\pm 2S\sqrt{S^2-24S}+S^2-24s}{144}\]](http://latex.artofproblemsolving.com/1/5/9/1592aed6615b9afbbe17624e9733f1e47830cea5.png)
Now if there are 
number of 
and 
,
is![\[S = \sum_{i = 1}^{10} v_i^2 = \frac{20S^2-240S+2kS\sqrt{S^2-24S}}{144}\]](http://latex.artofproblemsolving.com/0/7/c/07c43aa54f017d2629aba1f52368e6b33b147e20.png)
![\[\implies k^2(S^2-24S) = (192-10S)^2\]](http://latex.artofproblemsolving.com/5/4/2/542b5b19a3d46128e23d90e1758f6950e40610a7.png)
where 
.![\[k^2(k^2-64) \ge 0 \implies k^2 \ge 64 \text{ or } k^2 \le 0 \implies k = 0,8,10\]](http://latex.artofproblemsolving.com/c/8/e/c8ef856ba1f605ec06f98003e839210922f744be.png)
Solving these cases for ![\[\boxed{v_i = 4 \text{ for all $i$ except one of them is } v_j = \frac{4}{3} \text{ or } v_i = \frac{8}{5} \text{ for all $i$}}\]](http://latex.artofproblemsolving.com/d/9/b/d9b8ecc31e92a707b7d7aae1c3f3b0cb1b89cb50.png)