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 the roots of the equation 
. What is the value of the expression
?
is inscribed in a circle. Given 
and 
. Let 
and 
be the orthocentre and incentre of respectively. Find the ratio of 
to 
.
is tangent to the sides 
, 
, 
of a triangle 
at points 
respectively. The lines 
and 
meet 
at points 
and 
. Prove that the circumcircle of triangle 
and points and 
lie on a line.
such that the polynomial 
has 8 linear factors of the form 
with 
for 
. Find all possible values of the constant 
.
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
.
positive integers, less than or equal to 
, such that the difference between any 
elements in the set is at least 
.
consecutive even numbers, such that the square of their product divides the sum of their squares.
triangle with 
. We take , point on AC, AB respectively such that 
, 
. 
, 
lines intersect at point 
. If 
, find 
such that
Proposed by Ngo Van Trang, Vietnam
such that 
. Find the value of 
and 
.
and 
such that 
is 
, then find the value of
.
be the set of integers. Determine all functions 
such that for all integer :
of a triangle 
are positive integers. Let
for any positive integer 
.
and 
, determine all possible perimeters of the triangle .
has two roots, where one is twice of the other, find all possible values of .
Find the value of 
such that 
is an integer.
is 
in triangle , find the ratio of the side length to the side length .
.What is the least possible positive sum of the three integers?
be a regular nonadecagon. Lines 
and 
meet at 
. Compute 
.
be a regular nonadecagon. Lines and meet at . Compute .![\[= \frac{(\omega^5 - \omega^3) - (\omega^2 - \omega)}{\omega - 1}\]](http://latex.artofproblemsolving.com/b/6/a/b6ae9d840aa9c9ed0e951a3b9a58368815192460.png)
![\[= \omega^4 + \omega^3 - \omega\]](http://latex.artofproblemsolving.com/f/0/4/f04e28b0a84252fabe6d15af67e4d7688c180810.png)
?
be the 19th-roots of unity with 
.![\[X = \frac{\omega^0 \omega^4 (\omega^2 + \omega^3) - \omega^2 \omega^3 (\omega^0 + \omega^4)}{\omega^0 \omega^4 - \omega^2 \omega^3}\]](http://latex.artofproblemsolving.com/3/b/5/3b535dd8a08df864eb062ebdb05d836ae1682cd0.png)
![\[= \frac{\omega^6 + \omega^7 - \omega^5 - \omega^9}{\omega^4 - \omega^5}\]](http://latex.artofproblemsolving.com/a/0/1/a015858628e275d220ebb89f4647fb2cdc00dfde.png)
![\[ \frac{A_7 - X}{A_5 - X} \]](http://latex.artofproblemsolving.com/d/e/7/de7218c41a3500b606805cebb420553e127d0c38.png)
![\[ = \frac{\omega^6 - \omega^4 - \omega^3 + \omega}{- \omega^3 + \omega} \]](http://latex.artofproblemsolving.com/2/a/8/2a85e59b0c3c6597f45d795f7f5deed0bdbcfe76.png)
![\[ = 1 - \omega^3 \]](http://latex.artofproblemsolving.com/8/a/4/8a49e13dd1b819e6d8e3776a5735012406636d50.png)
We plot this on the complex plane. We let 
still be 
,
be 
.
is isosceles with 
.
.![\[\angle BOA_1 = \frac{\pi - \frac{6\pi}{19}}{2} = \boxed{\frac{13\pi}{38}}\]](http://latex.artofproblemsolving.com/8/0/9/809e94631304088982e4cf5e420e73f8cddbcb18.png)
.![\[= \omega^4 + \omega^3 + \omega\]](http://latex.artofproblemsolving.com/1/4/4/144976ecce9cb7e1b2ef0b96eaf69e6d72dafc39.png)
Therefore, ![\[ = \frac{\omega^6 - \omega^4 - \omega^3 - \omega}{- \omega^3 - \omega} \]](http://latex.artofproblemsolving.com/a/0/4/a048a571a14118eac1da00fbf524685783ea5e3c.png)