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 inscribed in the circle 
. A line through point 
intersects 
and 
at points 
and 
, respectively. Let 
be the reflection of across the midpoint of , and 
be the reflection of across the midpoint of . Prove that:
of triangle across line 
lies on the circle .
and 
coincide.
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?
and 
.
?
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 
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 
.
.
be a point and let and be on and , respectively, such that 
is a parallelogram. The isogonal conjugate of in is . We see that 
, so 
. Now, if is on 
, then 
, so has an isogonal conjugate in 
. Hence is the isogonal conjugate of in . The problem follows from 
.
is given by![\[f = \frac{(a+b)(b+c)}{(b-c)}+a\]](http://latex.artofproblemsolving.com/6/3/0/630fdac0c5a7557bcb75d5018abbf85d906ceeb8.png)
Proof. Denote the point corresponding to 
as defined above by 
. We show lies on and 
is parallel to , so we must have 
. Firstly![\[f = \frac{(a+b)(b+c)}{(a-b)(b-c)}(a-b) + a\]](http://latex.artofproblemsolving.com/3/9/c/39cb21fd6f81112e07544fcacfa3813e55532a76.png)
and 
so 
for some 
. So lies on . Now we show 
. This holds iff![\[\frac{h-f}{a-c}=\overline{\frac{h-f}{a-c}}\]](http://latex.artofproblemsolving.com/5/c/d/5cdeff8e41fefda55abe29010bf0f01b8ff531dd.png)
We compute these separately:
and
As desired. So 
, as desired, so the expression for as above is correct. 
and w.r.t. 
and 
we have![\[e=\frac{(a+c)(b+c)}{c-b}+a\]](http://latex.artofproblemsolving.com/1/a/5/1a5dd270004536b13c315a353d7913ff74f3a9e6.png)
Since is the origin of our complex plane, we need to show 
, which is true since:
as desired, so we are done. 
. Then, 
, so 
. Therefore, 
, which means 
.
be the feet from 
to 
, and let 
. Since![\[
(B, C; D, D') = -1 = A(F, E; N, \infty_{EF})
\]](http://latex.artofproblemsolving.com/f/7/d/f7d3272ceed174bc121c8d034371b4ddbdf668e9.png)
it follows that 
. Since we are done if 
, if 
is the -antipode, scaling by two at means we want to show that 
. However, these two lines concur at the -queue point, which lies on 
so we are done.
the midpoint of 
. Since 
is a parallelogram, is also the midpoint of . Define 
as the -antipode in . It is well known that 
is a parallelogram, and since 
parallelogram is similar to .
, and since 
, there is a rotation by 
centered at that sends 
and 
. Thus, 
and rotating back gives 
. A homothety cantered at with factor 
sends 
and 
. Since 
. Thus is on the perpendicular bisector of which implies 
, as required. 
satisfying
Prove
:![\[
1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n + 1) \mid (4n + 3)(4n + 5) \cdot \ldots \cdot (8n + 3).
\]](http://latex.artofproblemsolving.com/3/1/a/31a0eac3c7ca298c8fa221083dc94e127e705784.png)
is a parallelogram. Prove that 
because of similar triangles. Use 
and 
and rewrite to 
and we are done by PoP.