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.
and 
Find the maximum value of
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
Prove that
with 
, point 
lies on the circumcircle of such that 
. The line through parallel to 
intersects 
in 
and 
in 
. Prove that the center of the circumcircle of triangle 
lies on the circumcircle of triangle 
.
be a rearrangement of the integers from 1 to 2025.
, define 
to be the number of integers 
such that 
and 
.
are given such that 
and 
for all .
satisfying this condition. be a acute triangle with 
. 
are the orthocenter and circumcenter. Let 
be a point on the same side of 
as 
, such that 
is equilateral. Let 
be a point on the same side of 
as , such that 
is equilateral. Let 
be a point on the same side of 
as , such that 
is equilateral. Let 
be the midpoint of 
. Prove that 
are collinear.
, show that![\[\left(1+\frac{a}{b}\right)\left(1+\frac{a}{b}+\frac{b}{c}\right)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq 16\]](http://latex.artofproblemsolving.com/c/1/7/c17cc42f0feafca5ecbeacde7076daa5b2615c60.png)
. By Vieta we have
Let WLOG 
. Using 
we have 
. Using 
, 
and equation 
in equation 
we have ![\[ 62(r_1+r_2)-32(r_3+r_4)=-200\tag{$5$} \]](http://latex.artofproblemsolving.com/c/1/c/c1c24e8ffcdadc268e0b9cb490b09d39846f2725.png)
Rewriting the equation we have ![\[ (r_1+r_2)(r_3+r_4)+30=k\tag{$6$} \]](http://latex.artofproblemsolving.com/7/5/1/7519a0798962f665f8ed987da0174beb1b77a95a.png)
Solving the equation and 
in terms of variable 
and 
we get 
. Putting these value in the equation 
we get out answer 
.
for some 
. Equating the 
and 
coefficients yields
so 
, and 
.
by vieta's we have 
, and 
are 
Given product of any two roots of this equation is -32. By Vietas Relation we get 
Lets assume 
from product of roots we get 
From these we also get 
Therefore, 
Is the correct answer...
. By Vieta, if , then 
. Additionally, Vieta gives 
But notice that 
again by Vieta. Then, solving we get that 
and 
. Finally, by Vieta, we see have 
and we're done.
, 
, 
and 
. WLOG, let 
, and by Vieta's, we have that 
, meaning that 
. Let 
and 
. By Vieta's, we have that![\[18=a+b+c+d=(a+b)+(c+d)=x+y,\]](http://latex.artofproblemsolving.com/c/9/6/c96f4716894eeeb77ba664f249fdd9fa36438fb1.png)
and![\[-200=abc+abd+acd+bcd=cd(a+b)+ab(c+d)=62x-32y.\]](http://latex.artofproblemsolving.com/1/c/1/1c19e783b755b05bfa2548d7013a00ab624f0b04.png)
Solving the system of equations gives us that 
and 
. Again using Vieta's, we have that![\[k=ac+ad+bc+bd+ab+cd=(a+b)(c+d)+ab+cd=xy-32+62,\]](http://latex.artofproblemsolving.com/e/a/d/eada5dc32b39f0ec39f12eba269545985560097b.png)
or 
, giving us that 
is , finishing the problem.
, 2) 
, 3) 
, 4) 
. Now we can assume 
= -32 
= 62. We now plug this into 3) and we get 
and after using 1) we get 
. Let 
, we get 
a = 4 
and 
. Now we plug in everything we know in 2) so 
k = 86 the answer is 86.
. Since we know the product of one pair of roots, we can find the product of the other pair of roots as 
. We can split the original polynomial into two quadratic polynomials:![\[
x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + 62)
\]](http://latex.artofproblemsolving.com/9/8/7/9876a06d8427ffc083c9b6b2d3ed976efdde5798.png)
and :![\[
(a, b) = (-4, -14).
\]](http://latex.artofproblemsolving.com/8/7/b/87b47dbc46bf7e1bcb055349f144b96baf31fd8b.png)
, we find:![\[
k = \boxed{86}.
\]](http://latex.artofproblemsolving.com/6/1/a/61a14b71f7ec8ba8dd9aa51fdf1f082c8b028f9b.png)
with 
By Vieta's, we get the following:
The last equation implies that 
and 
Thus,
,
Prove that
Let 
Prove that
Prove that
Where 
is 
.
.
*
.
.
and as a result, 
.
.
and 
.
as 
s
.
,
,
,
.
.
,
.![\[a(c+d)+b(c+d)=(a+b)(c+d)\]](http://latex.artofproblemsolving.com/a/8/3/a83ac093e8d99e425c2585b08196d0ea5b8608a4.png)
![\[-32c-32d+62a+62b=-32(c+d)+62(a+b)\]](http://latex.artofproblemsolving.com/e/d/0/ed00fbabdf6f2e16f848b7478c37862dc0ae34e2.png)
and 
.
.
and 
,
Without loss of generality, assume that 
Letting 
and 
,
,
,
and 
.
.
Note that:
We can factor 
as 
.
Note that:
We know that 
so 
.
and 
.
(or vice versa).
.
.
,
.
,
,
.
,
,
![\[abc+abd+acd+bcd=62(a+b)-32(c+d)=-200,\]](http://latex.artofproblemsolving.com/9/f/9/9f99c9235d8e31ea14afe8608bda592f11d42599.png)
and that 
,
and 
.![\[k=ab+ac+ad+bc+bd+cd=-32+62+(a+b)(c+d)=30+4\cdot 14=\boxed{86}.\]](http://latex.artofproblemsolving.com/0/3/4/0344796137ab70bc67e65774bf966ca6338fd3a3.png)
which is equivalent to finding 
.![\[ab(c+d)+(a+b)cd=62(a+b)-32(c+d)=-200\]](http://latex.artofproblemsolving.com/a/0/0/a00a431a73c63a4fa64bbd13cf027911fc6e9731.png)
![\[\implies 31(a+b)-16(c+d)=-100\]](http://latex.artofproblemsolving.com/a/8/0/a8036f2f37e8ac127422da80876dec5ff915ac46.png)
and 
,
,
,
.
has roots 
and 
we get
and 
in equation 
we get![$\implies -32(c + d) + 62[18 - (c + d)] = -200$](http://latex.artofproblemsolving.com/b/3/5/b35c3817b3571f26eaf9328a1ab128c9ae9ef0f2.png)
![\begin{align*}
\sigma_4=&-1984 \underbrace{\implies}_{ab=-32}cd=62\\
\sigma_3=&-200=ab(c+d)+cd(a+b)=-32(c+d)+62(a+b)\\
\sigma_2=k=&(a+b)(c+d)+\underbrace{ab+cd}_{-32+62=30}\\
\sigma_1=&18 \implies a+b=18-(c+d). \\
-200=& -32\left [18-(a+b)\right ]+62(a+b) \implies a+b=4; \; c+d=14\\
\implies & k=4 \cdot 14+30=\boxed{86}.
\end{align*}](http://latex.artofproblemsolving.com/2/c/a/2cae37a4ecc53401a680e2f76c16f5154de012e5.png)
