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 an acute scalene triangle with circumcenter 
. Let 
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of 
. Let 
be the circle with diameter 
. Let 
be the intersection of 
and 
. Let 
be the orthocenter of . Let 
be the intersection of 
and . Let 
be the lines through 
tangent to 
respectively. Let 
be the intersection of 
and 
. Let 
be the intersection of 
and 
. Let 
be the line through parallel to 
and let 
be the reflection of across . Prove that 
is tangent to 
.
+1 w
be a positive integer, and let its positive divisors be![\[
d_1 < d_2 < \cdots < d_k.
\]](http://latex.artofproblemsolving.com/6/3/e/63ec82aba0d6c4eece89af6a1247a89498616281.png)
Define 
to be the number of ordered pairs 
with 
such that 
.
. when![\[
n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},
\]](http://latex.artofproblemsolving.com/2/7/2/272dc2cc33ebbf695311cb61567fa0297e1299b6.png)
where the 
are distinct primes and the 
are positive integers. be the circumcenter of an acute-angled triangle with 
. The line 
meets the side at . The circumcenters of the triangles 
and 
are 
and 
, respectively. Extend the sides 
and 
beyond 
, and choose on the respective extensions points 
and such that 
and 
. Prove that the quadrilateral 
is a rectangle if and only if 
.
is a prime number
is a positive integer. Call all positive divisors of which are different from 
and beautiful divisors.We call a special number when it has at least 
beautiful divisors and difference of any beautiful divisors divides as well. Find all special numbers.
such that 
.
is defined by 
and for all positive integers 
, 
. Show that for every positive integer 
, there exists a suitable positive integer 
such that 
is a multiple of .
be a positive integer. Determine all sequences 
of positive integers such that
for all positive integers 
.
be an odd prime. Define 
over the positive integers as follows:
be chosen such that there exists an ordered pair of positive integers 
where 
such that 
. Prove that there exists at least three integers 
such that 
and 
is a perfect square.
be the nonnegative integers in 
where 
. Define 
. Prove that 
. such that 
where 
, and the statement is still true when 
? such that 
where 
, and the statement is still true when ?.
satisfying :![\[n^{2p}=m^2+n^2+p+1\]](http://latex.artofproblemsolving.com/7/4/e/74ee592cb8b3f3a3996ea34f40b78fde9d3672fc.png)
and are integers and is a prime number.
such that 
divides 
.
of on 
is on circle 
with diameter 
Further, it is the midpoint of its arc 
because bisects 
Since 
is clearly external bisector of 
then becomes B-excenter of 
is equidistant from 
(valid for any triangle), i.e. triangle 
is isosceles right with legs 
is the external bisector of the angle 
, hence is the 
excenter of . If is the incenter of , it follows that 
is cyclic, for any triangle, hence 
.
.
erected at 
intersects the side 
.
.
,
.
we have![\[ \frac{\sin{(45-\beta)}}{AD} = \frac{\sin{45}}{CD} \Longrightarrow \frac{AD}{CD} = \frac{\sin{(45-\beta)}}{\sin{45}} = \cos{\beta} - \sin{\beta} \]](http://latex.artofproblemsolving.com/9/3/3/933f2b09f2cda5afc8c4d5315d3047936c7564fe.png)
Since 
we have![\[ \frac{BD}{CD} \&= \frac{\cos{\beta}-\sin{\beta}}{\sin{\beta}} \Longrightarrow \frac{BD}{CD} + 1 = \cot{\beta} \Longrightarrow \frac{CD}{BC} = \tan{\beta}\]](http://latex.artofproblemsolving.com/e/9/5/e95364beb8b64053f769bce60906ef42134fd731.png)
So ![\[\frac{CD}{BC} = \frac{AD}{AB} \Longrightarrow \frac{AB}{BC} = \frac{AD}{CD} = \frac{AE}{CE} \]](http://latex.artofproblemsolving.com/a/c/a/acab7eb663e4c52bb4bf08dbdf23f3f5e1bf3af7.png)
From Angle Bisector, 
bisects 
,![\[\angle BED = 180 - \angle DBE - \angle BDE = 180 - \left(\frac{\beta}{2}\right) - \left(90 - \beta + 45 + \frac{\beta}{2}\right) = \boxed{45^{\circ}}\]](http://latex.artofproblemsolving.com/e/b/3/eb36d3a14918c643b3c5a4c260d5d8b5b3e7d349.png)
be the point on 
such that 
.
be the point on 
such that 
.![\[\frac{CD}{BC} = \frac{DH_2}{BH_1} = \frac{AD}{AB} \Longrightarrow \frac{AD}{CD} = \frac{AB}{BC}\]](http://latex.artofproblemsolving.com/b/4/d/b4de57501beac88b421e10502329092a2283fb58.png)
By Angle Bisector on 
we have 
,
and we get