AoPS Academy
be a triangle with circumcenter 
. Point 
is the intersection of the parallel line from to 
with the perpendicular line to 
from 
. Let 
be the point where the external bisector of 
intersects with . Let 
be the projection of onto 
. Prove that the lines 
have a common point.
, where 
and 
. Prove that 
could be expressed as product of two non-constant polynomials with integer coefficients.
, we have 
When does the equality hold?
let and 
be the circumcenter and the orthocenter. The point 
is the reflection of 
with respect to 
. Assume that is not on the same side of 
as . Points 
lie on 
respectively such that 
. Let be the intersection point of 
. Prove that 
![\[
a_7 \cdot 3^7 + a_6 \cdot 3^6 + a_5 \cdot 3^5 + a_4 \cdot 3^4 + a_3 \cdot 3^3 + a_2 \cdot 3^2 + a_1 \cdot 3^1 + a_0 \cdot 3^0
\]](http://latex.artofproblemsolving.com/0/c/e/0ce2c259010b678238f11d35fc1328ea76f945f6.png)
where 
for 
?
sides and 
sides, respectively. The total number of sides is 33, and the total number of diagonals is 243. What are the values of and ?
be a polynomial with integer coefficients such that 
and 
for some integer 
. Prove that there exists no integer 
such that 
.
for some real number 
and ![\[ a_n=na_{n-1}+(n-1)(n!(n-1)!-1) \]](http://latex.artofproblemsolving.com/4/1/f/41f0cc7b041a8bba056882b5c794503195c374fd.png)
for integers 
. Given that 
, and given that 
, where 
and 
are positive integers whose greatest common divisor is 
, compute 
be the intersection of 
and 
![[asy]
pair O=(0,0), C=(1,0), B=(cos(2pi/9), sin(2pi/9)), A=(cos(2pi/3), sin(2pi/3)), E=(cos(13pi/9), sin(13pi/9)), D=(cos(-4pi/9), sin(-4pi/9));
draw(A--B--C--D--E--cycle);
draw(circle(O, 1));
pair F=-A;
draw(A--F);
draw(C--E);
draw(D--B);
label("$C$", C, dir(0));
label("$B$", B, dir(40));
label("$A$", A, dir(120));
label("$E$", E, dir(260));
label("$D$", D, dir(-80));
label("$O$", O, W);
draw(D--O--C--O--B--O--E);
real a=sin(2pi/9), b=cos(2pi/9), c=cos(4pi/9), d=sin(4pi/9);
real x=(b*d+a*c)/(a+d+b*sqrt(3)-c*sqrt(3));
pair G=(x, -x*sqrt(3));
label("$G$", G+(0,0.1), dir(100));
[/asy]](http://latex.artofproblemsolving.com/2/4/8/248e2e816ab6d7616e9a881eff05296214182cf6.png)
and 
Then, 
and 
Thus, 
so 
Now, 
so 
and 
and 
and 
and 
so 
and since 
we have 
Also, by the inscribed angle theorem, 
and 
As such, 
and 
We prove this below.
by sum-to-product and double angle.
so 
as desired. 
satisfying:
and 
Then 
in terms of 
be a cyclic pentagon inscribed in a circle of centre 
.
and 
.
.
and 
and let 
be the point on the circumcircle such that 
is a diameter. We'll perform some angle chasing before we continue. Let ![\[a=\overarc{AB},b=\overarc{BC},c=\overarc{CD},d=\overarc{DE},e=\overarc{EA},\]](http://latex.artofproblemsolving.com/e/9/4/e9470b7a6c9283a2313703364d7de9904c1ed0c3.png)
. By pentagon properties 
.
Solving we get 
.
.
and 
so 
.
Now we'll consider 
.
.
and 
.
.
so 
.
so:
In order for 
so 
.
.
Which is true. So