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.
lies inside triangle 
such that 
and 
. Point 
is the midpoint of segment 
. Point 
lies on segment 
with 
. Prove that 
.
the number of positive integer 
satisfying 
is less than or equal to 
where 
is a constant independent of 
on the board. For each procces, the stutent deletes the number x and replaces it with either 
or 
or 
. Is this possible to make the number 
on the board?
be different real numbers. Prove that![\[\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}=\left\{\begin{array}{ll}
0, & \text { if } n \text { is even; } \\
1, & \text { if } n \text { is odd. }
\end{array}\right.\]](http://latex.artofproblemsolving.com/6/5/2/6525fc6b451ff5a3c83ff2a0541b063a7cf065c3.png)
satisfying
for all real numbers 
and 
.
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
. Let 
be the second intersection of the circumcircles of triangles 
and 
. Prove that the lines 
, 
, and 
-median of 
are concurrent.
that satisfy the condition 
for the natural number 
Here 
and 
are natural numbers. Is there a natural number 
such that each of the numbers 
is the last digit of one of the numbers written by the teacher on the board? Justify your opinion.
we have 
.
. has orthocenter . Let be a point distinct from the vertices on the circumcircle of . Suppose that circle 
meets 
at 
, and circle 
meets at 
. Prove that as moves on the circumcircle, the reflection of across line 
also moves on a fixed circle. there exists an -digit number divisible by 
all of whose digits are odd.
and 
.
. Hence, 
.
. Similarly, let 
.
and 
.
and 
. Hence; 
.
.
and 
, we can find that 
and 
.
and 
, we can find that 
Contradiction. Thus, 
.
:![\[
1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n + 1) \mid (4n + 3)(4n + 5) \cdot \ldots \cdot (8n + 3).
\]](http://latex.artofproblemsolving.com/3/1/a/31a0eac3c7ca298c8fa221083dc94e127e705784.png)
such that
be a parallelogram. Points 
lie on the sides 
respectively,
and 
.
.
intersect 
,
intersect 
at 
,
and 
are cyclic, so use PoP and the parralel lines 
and 
which is obvious AM-GM)
, reflection of 
meet at a point, say, 
on 
.
imply 
.
,
,
,
![\[\frac{AB}{BC}=\frac{BE}{BC}+ \frac{EA}{AD}= \frac{\sin\theta}{\sin \alpha}+ \frac{\sin\phi}{\sin \beta}\]](http://latex.artofproblemsolving.com/6/c/8/6c84f1a4eeb01fd2328cfe91d016aba5ab511e69.png)
![\[\frac{CD}{BC}=\frac{AB}{BC}=\frac{FC}{BC}+ \frac{FD}{AD}= \frac{\sin\alpha}{\sin \theta}+ \frac{\sin\beta}{\sin \phi}\]](http://latex.artofproblemsolving.com/8/2/4/824f18b23473afb169b0e83822381356fa629df0.png)
![\[\left(\frac{AB}{BC}\right)^2=\left(\frac{\sin\theta}{\sin \alpha}+\frac{\sin\phi}{\sin \beta}\right)\left(\frac{\sin\alpha}{\sin \theta}+ \frac{\sin\beta}{\sin \phi}\right)\underbrace{\geq}_{\text{By AM-GM}}4\cdot\sqrt{\frac{\sin\theta}{\sin \alpha}\frac{\sin\phi}{\sin \beta}\frac{\sin\alpha}{\sin \theta}\frac{\sin\beta}{\sin \phi}}=4\]](http://latex.artofproblemsolving.com/0/e/e/0ee96525d2b3e3c031d9b7f658928cde0c949bb5.png)
![\[\left(\frac{AB}{BC}\right)^2\geq 4\]](http://latex.artofproblemsolving.com/e/e/c/eec36cc13209aeeeac2a754a65989681f2003a18.png)
![\[\frac{AB}{BC}\geq 2\Rightarrow AB\geq 2BC,\text{As desired.}\]](http://latex.artofproblemsolving.com/3/b/f/3bf53657205a86238bc0c8591dc7466056b5176d.png)