AoPS Academy
is an acute angle triangle such that 
and 
. Let's denote by 
the center of the circumscribed circle of the triangle and 
the intersection of altitudes of this triangle. Line 
intersects 
in point 
and 
in point 
. Find the value of the ration 
.
. Let 
be a point on side 
, and 
be a point on the extension of 
such that 
. Let be the circumcenter of 
, and be a point on the side extension of 
satisfying 
. Line BP intersects AC at point Q. Prove that 
of positive integers with 
prime and ![\[ a^p=b!+p. \]](http://latex.artofproblemsolving.com/9/f/6/9f674dabcb364872cef4153198e2e362ae40a2fd.png)
be positive reals such that 
. Prove that ![\[18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)\ge 15. \]](http://latex.artofproblemsolving.com/5/d/5/5d5c47a52d2d6ea7d13ce1c2fa14a08425bbfa5d.png)
Proposed by David Stoner
such that 
. Find the value of 
consisting of a prime number and two positive integers 
and 
such that 
and 
are both powers of .
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
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. 
satsfying 
and:![\[ f(2n)f(2n+1)=9f(n)^2+3f(n)\]](http://latex.artofproblemsolving.com/b/c/1/bc18d6ad72589f012dcab7836228b3ed7ee061c2.png)
and let 
be reals such that 
satisfying
Prove
is a parallelogram. Prove that 
because of similar triangles. Use 
and 
and rewrite to 
and we are done by PoP.