AoPS Academy
be a positive integer, and consider an 
grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence. grid cannot be partitioned into less than right-down or right-up paths. For example, the following partition of the 
grid uses 
paths.
![\[f(xy)+f(xf(x^2y))=f(x^2)+f(y^2)+f(f(xy^2))+x \]](http://latex.artofproblemsolving.com/d/7/b/d7bf030b047d49477eba887e8bb78c6e5b126fe7.png)
holds for all 
.
as 
. Then we may show that 
through angle chasing, redefine 
as 
. Then let 
, note that 
is cyclic and tangent to 
, likewise we may angle chase that 
is cyclic so 
and thus 
is tangent to 
, thus 
, and thus 
. Finally, the result is another painful angle chase.
lies on 
, where 
is the radius of by trig bash
and then angle/arc bash the required angle equality.
. Let the circumcircle of triangle 
intersects segments 
and 
again at 
and 
, respectively. Since 
it follows that 
is tangent to 
. lies on 
has equal length to the diameter of . Notice that 
Since 
, right triangles 
and 
are similar. Then 
Which is equal to the diameter of . Notice that the power of with respect to is equal to 
which is equal to 
. Denote the radius of as 
. The power can also be written as 
so 
Then clearly 
, finishing the claim.![[asy]
import geometry;size(12cm);pair A=(0,0);pair B=(4,0);pair D=(0,-3);pair C=D+rotate(220)*(A-D);pair E=C+.17(B-C);pair G=intersectionpoint(line(D,E),line(A,B));pair tac=circumcenter(G,E,B);line cat=line(B,B+rotate(90)*(tac-B));pair F=intersectionpoint(cat, line(E,D));pair O=circumcenter(C,D,E);pair P=intersectionpoint(circle(A,D,G),circle(C,E,D));circle w=circle(B,D,E);point[] X=intersectionpoints(w,line(A,D));point[] Y=intersectionpoints(w,line(B,F));pair X=X[1];pair Y=Y[0];pair K=circumcenter(A,B,C);pair Z=intersectionpoint(line(B,D),line(X,Y));pair Q=intersectionpoint(line(A,C),line(B,P));filldraw(A--B--D--cycle,white+palegreen);filldraw(B--X--Y--cycle,white+palegreen);filldraw(Z--X--B--cycle, palegreen);draw(A--B--C--D--cycle);draw(F--G);draw(circle(C,D,E), dotted);draw(P--F);draw(circle(A,D,G),dashed+purple);draw(w,blue);draw(B--F);draw(circle(G,E,B),red);draw(G--A);draw(A--C,grey ); draw(B--P,grey );draw(circle(A,B,C),dashed+orange);perpendicularmark(line(A,D),line(A,B));draw(A--D,StickIntervalMarker(1,1));draw(D--C,StickIntervalMarker(1,1));markangle(B,Q,A,radius=.4cm);markangle(B,Q,A,radius=.5cm);markangle(F,P,D,radius=.4cm);markangle(F,P,D,radius=.5cm);draw(G--P--D,grey);dot("A", A,dir(60));dot("B", B,dir(20));dot("C", C,dir(-30));dot("D", D,2*dir(180));dot("E",E,1.5*dir(100));dot("F",F,dir(-20));dot("G",G,dir(160));dot("P",P,dir(180));dot("O",O,dir(250));dot("X",X,dir(160));dot("Y",Y);dot("Q",Q,dir(-25));
[/asy]](http://latex.artofproblemsolving.com/8/0/4/804e42ee6ebbbf626198747fd3b000247b300d3c.png)
and 
are cyclic. 
Then to finish the problem 
such that 
.
Where 
be real numbers.Prove that
.
be a point on side 
,
such that 
.
be the circumcenter of 
,
satisfying 
.
and 
.
Let 
so that 
tangent 
and 
tangent to 
,
,
