AoPS Academy
-excenter of 
is 
and 
. The foot from 
to 
is 
and 
. The intersection of 
and 
is 
. The circumcenter of 
is 
.
is tangent to the -excircle if is on the incircle of .
with 
and 
. Lines and intersect at , and lines 
and 
intersect at 
. Let 
be the midpoints of sides 
, respectively. Let 
and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to 
if and only if 
divides into two triangles with equal area.
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
be a logarithmic spiral centered at the origin (ie curve satisfying for any point on it, line 
makes a fixed angle with the tangent to at ). Let 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point. on the spiral, the polar of wrt. is tangent to the spiral.
such that every positive integer appears exactly once and the product of the first 
terms is a perfect 
power ?
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be the set of black cells at the end of the game. Ali and Shayan respectively aim to minimize and maximise the perimeter of the shape by playing optimally. (The perimeter of shape is defined as the total length of the boundary segments between a black and a white cell.), assuming both players play optimally?
called ‘nice’ if and only if 
and 
are both perfect squares. How many positive integers 
such that 
are composites for all nice numbers ?
decide whether there exist pairwise distinct positive integers 
such that for every 
, 
divides 
.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
. Adding to its area the area of 
we get:![$[CRB]+[CRD]=[BCD]=R^2\ (\ 2\ )$](http://latex.artofproblemsolving.com/9/d/0/9d09cf84b632d93d68eb266f0c47d1d75612d4b9.png)
. Adding to ![$[RDS]$](http://latex.artofproblemsolving.com/2/7/e/27e6a50c0abfcae32df9fcf2b2a8d0925998515b.png)
the same ![$[CDR]$](http://latex.artofproblemsolving.com/6/4/6/6462e2fd63e60e317abb0d4831f987ba2da5dbff.png)
we get ![$[RDS]+[CDR]=[CDSR]$](http://latex.artofproblemsolving.com/3/c/6/3c68ba02871db35c8fdf828286901404cd364a77.png)
. 
having perpendicular diagonals, its area is ![$[CDSR]=\frac{CS\cdot DR}{2}\ (\ 3\ )$](http://latex.artofproblemsolving.com/6/3/9/6396ef4e79fe87988d731e007023e2e4f68aff81.png)
, so we need 
. The last relation we get from 
, done.
is the circumradius the square.
be the center of and let 
be the point at infinity of (also point at infinity of the tangent of 
at D).
![$\frac{OB}{OD} \cdot \frac{RD}{RB}=\frac{OA}{OS} \Longrightarrow OS \cdot RD=OA \cdot RB \Longrightarrow [DSR]=[ARB].$](http://latex.artofproblemsolving.com/1/1/0/1109c08461d9b93da58cc7ded47e73461e3fddfa.png)
, then 
too. Also let 
be the length of the radius of the circle.
we have 
. Consequently 
is: ![$$[ARB] = \frac{1}{2} \cdot AB \cdot BR \cdot \sin(\angle ABR)=\frac{1}{2} r\sqrt{2}\cdot r(1-tan\phi) \frac{\sqrt{2}}{2}=\frac{1}{2}r^2(1-tan\phi)$$](http://latex.artofproblemsolving.com/a/e/2/ae251524382847a6702441fef1efe00c1eb2ffe8.png)
On the other hand, in the right-angled triangle 
we have: 
, so 
is: ![$$[DSR]=\frac{1}{2}\cdot OS \cdot DR=\frac{1}{2}rtan(45^\circ-\phi)r(1+tan\phi) = \frac{1}{2}r^2tan(45^\circ-\phi)(1+tan\phi)$$](http://latex.artofproblemsolving.com/8/4/c/84cce5c5a32ebf85639c1bae46edc6dd2cc39fb6.png)
Finally, using the trigonometrical formula:
So the area of the triangle is:![$$[DSR]=\frac{1}{2}r^2tan(45^\circ-\phi)(1+tan\phi)=\frac{1}{2}r^2\frac{1-tan\phi}{1+tan\phi}(1+tan\phi)=\frac{1}{2}r^2(1-tan\phi)$$](http://latex.artofproblemsolving.com/9/d/2/9d21642f91837863d2e8444737cf6b80395a72fa.png)
Thus the areas of and are equal.
be the intersection point of lines 
and and let 
be the projection of onto line . be the circumcenter of circle . Now 
, assume that side length of square = 1.
= 
= 
= 
, 
is cyclic.
= 
= 
, implying that 
.
area of 
, 
is an isosceles right triangle. So 
= 
and 
above, ![$[DSR]=[ARB].$](http://latex.artofproblemsolving.com/a/1/c/a1cf1807f2fdd7a23d58ae91bc83e1b38dd7d710.png)
and 
,
.
.![\[\frac {DA}{DS}=\frac {\sin \angle ASD}{\sin \angle SAD}=\frac {\sin \angle OSD}{\sin \angle SAD}=\frac {\sin \angle PBR}{\sin \angle BPR}=\frac {RP}{RB}\,.\]](http://latex.artofproblemsolving.com/4/0/5/405fb8d8f8b8eb64a33716fbe5c92b863c68420b.png)
Calculating the power of 
.
and 
.
,
and 
have equal areas. But 
.
is tangent to 
.
the claim follows.
which is true by the claim, we are done.![[asy]
size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta;
pair O,A,B,C,D,P,R,S;
O=(0,0);B=dir(45);A=dir(135);D=dir(225);C=dir(315);path w=circumcircle(A,B,C);P=dir(100);R=extension(P,C,D,B);S=extension(P,D,A,C);
draw(A--B--C--D--cycle,deep);draw(w,deep);draw(A--C,deep);draw(B--D,deep);draw(P--D,deep);draw(P--C,deep);draw(D--S--R--cycle,org);draw(A--R--B--cycle,light);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$P$",P,dir(P));
dot("$R$",R,dir(R));
dot("$S$",S,dir(S));
[/asy]](http://latex.artofproblemsolving.com/d/e/6/de65c126817b1946dcb5e31f18111a80e915f020.png)
.
.![$[PAB]=[PSR]$](http://latex.artofproblemsolving.com/e/7/f/e7f7b0ce26638596ff2035e06f35a3d64441509c.png)
.![$[DSRC]=[PDC]-[PSR]=[PDC]-[PAB]=[DBC]$](http://latex.artofproblemsolving.com/8/7/4/874118f361646a6781f453351857dc5cd25c0ac5.png)
.![$[DSR]=[CRB]$](http://latex.artofproblemsolving.com/2/2/e/22e13fa7d195107bf8b935959ea5147e32c4e0b5.png)
,