,
,
.
and 
are located on 
and 
respectively, such that 
and 
.
to 
.
be the circumcircle of a triangle 
the midpoint of the arc 
and 
intersect the perpendicular bisectors of 
and 
,
intersect at 
.
which satisfy the following relationship for all real numbers 
and 
![\[f(xf(y)-f(x))=2f(x)+xy.\]](http://latex.artofproblemsolving.com/d/0/e/d0e9e56267aa6241a0323c7b2245e4752c338785.png)
and 
.
and 
Prove that
and 
.
Equality holds when 
Equality holds when 
Equality holds when 
such that:
.
,
.
Y by Ygg, manuel153, ivanbart-15, monsterrr, amatysten, tenplusten, BogdanB, nguyendangkhoa17112003, Centralorbit, Adventure10, and 3 other users
In a triangle 
, the excircle 
opposite 
touches 
at 
and 
at 
, while the excircle 
opposite 
touches 
at 
and 
at 
. Let 
be the projection of 
onto 
and let 
be the projection of onto 
. Show that the quadrilateral 
is cyclic.
(Bulgaria)
, the excircle 
opposite 
touches 
at 
and 
at 
, while the excircle 
opposite 
touches 
at 
and 
at 
. Let 
be the projection of 
onto 
and let 
be the projection of onto 
. Show that the quadrilateral 
is cyclic.(Bulgaria)
![\[ KL\bot CI \]](http://latex.artofproblemsolving.com/2/3/6/2364d44d856a36f7291f96c821666b84078d305d.png)
is incenter of the 
.
and the rest is easily angle chasing.
as the intersection of 
with point C in its interior : you will easily get 
and 
.
be the intersection point of 
and 
and let 
.![\[AB=a+b\\ BC=b+c\\ CA=c+a\]](http://latex.artofproblemsolving.com/2/0/0/2009373b8cc4b86f360cba8cdf48ff29514253e7.png)
and ![\[\angle BAC=2\alpha\\ \angle ABC=2\beta\\ \angle BCA=2\gamma.\]](http://latex.artofproblemsolving.com/e/9/c/e9c4ed1d855a12c5b017e26a03d909c5d123e68e.png)
We have that 
and that ![\[CK=a\cos{\beta}\\ CL=b\cos{\alpha}.\]](http://latex.artofproblemsolving.com/7/e/1/7e1412e6c6836a14c124c1af4001dc20739482f1.png)
We also have that ![\[\frac{YH}{XH}=\frac{MH}{PH}=\frac{\sin{\beta}}{\sin{\alpha}}\]](http://latex.artofproblemsolving.com/3/4/1/34108dc58d5cbf4e571c9a53e3907c70c4549974.png)
because af Sines Law in 
.
is the center of the excircle 
of ![\[\tan{\alpha}=\frac{b}{TI_{C}} \text{ and } \tan{\beta}=\frac{a}{TI_{C}}\]](http://latex.artofproblemsolving.com/b/a/0/ba06cf0dc529161da5df98230855d617e3c3f7cf.png)
so we must have that ![\[a\tan{\alpha}=b\tan{\beta}\Rightarrow \frac{a\cos{\beta}}{b\cos{\alpha}}=\frac{\sin{\beta}}{\sin{\alpha}}\Rightarrow \frac{CK}{CL}=\frac{YH}{XH}\]](http://latex.artofproblemsolving.com/f/e/3/fe325765ea91d2bbdf7303a7d9bf1dd8dda38c2c.png)
but we can see that 
and 
so triangles 
and 
are under an homothetic transformation with center in 
.
is cyclic, and since 
then ![\[\boxed{MKLP \text{ is cyclic}}\]](http://latex.artofproblemsolving.com/c/c/0/cc0d62a733da2283f9f7af7d790d3fa05bb4bf8f.png)
as desired.
.
is perpendicular to 
.
then 
so pencil 
,
then 
.
we have 
sooo 
,
is perpendicular to 
is perpendicular to 
so 
.
id 
and we are done.
intersect on the bisector of 
,
along from A to B, and this is 
away from the midpoint of AB.
but this is true as both are equal to the inradius.
,
,
,
.
, the 
is tangent to 
is tangent to 
. Observe that 
, 
and 
.
.
and 
. Observe that 
and 
. In conclusion,
. Since 
and 
and 
is similar, where 
.
,
,
,
is the semi-perimetr of 
with point 
:
:
is an increasing function of 
between 
and 
.
,
,
in terms of the sides of triangle ABC.) This last part is just a long computation, not really interesting !
.
.
and 
be the centers of the incircle and 
,
and 
the intersections of 
with 
is cyclic since 
,
.
lie on the circle with diameter 
.
,
is cyclic.
,
and we are done.
and 
.
,
,![\[\frac{CL}{CK}=\frac{\sin\alpha/2}{\sin\beta/2}\]](http://latex.artofproblemsolving.com/a/9/c/a9ce36f25b6fd1f4d463c2b14ab4d35269b62f61.png)
and thus 
is cyclic.
is cyclic and how to arrive at 
in school5's solution.
, as I skimmed through solutions I didn't notice anyone mentioning this properties of the figure and I think it would be such a pity if nobody said it..
the points of tangency of the 
.
.
are collinear.
mapping 
to 
,
and 
are the midpoints of 
and 
.
.
is an antiparallel to 
in triangle 
,
,
,
.
,
.
,
.
,
.
etc... If 
and assuming the problem is true, since the lines adjoining the projections of 
.
.
.
etc... Well, 
which is the external bisector at 
so 
.
,
.
we get 
.
so sine rule in 
gives 
.
.
can be found trigonometrically.Now all we have to verify is that 
which is easy. 
be the excircles opposite vertices 
.
at points 
and 
respectively. Let 
denote the exterior angle bisector of 
and 
be the centres of these excircles. Let 
and 
and 
and 
.
are collinear with 
.
which passes through all six points 
with 
"
and prove that the corresponding Ray's subtend the same cross ratio on 
.
which is symmetric in 
and so these two cross ratios are equal. Therefore, by the projective definition of a conic, we get the validity of our Lemma.
which lies on 
(by showing it lies on the other two altitudes) we get that 
which lies on 
are collinear. Notice that it is clear that 
by simple angle chasing. Now, we have 
are symmetric in 
are also symmetric in 
meaning that 
![[asy]
size(12cm);
pair C=(9,14),B=(0,0),A=(25,0);
draw(A--B--C--cycle);
dot("$A$",A,S);
dot("$B$",B,S);
dot("$C$",C,2.5*dir(95));
pair IA=extension(B,bisectorpoint(-A,B,C),A,bisectorpoint(B,A,C));
pair P=foot(IA,A,B), Q=foot(IA,A,C);
draw(arc(IA,IA+dir(-30)*(P-IA),IA+dir(30)*(Q-IA)));
draw(P--A--Q);
dot("$P$",P,S);
dot("$Q$",Q,NNE);
pair IB=extension(A,bisectorpoint(C,A,2*A-B),B,bisectorpoint(A,B,C));
pair M=foot(IB,A,B), N=foot(IB,B,C);
draw(arc(IB,IB+dir(-30)*(N-IB),IB+dir(30)*(M-IB)));
draw(M--B--N);
dot("$M$",M,S);
dot("$N$",N,WNW);
draw(P--Q^^N--M);
pair K=foot(C,M,N), L=foot(C,P,Q);
dot("$K$", K, NE);
dot("$L$",L,NW);
draw(L--C--K);
pair x=circumcenter(M,L,P);
draw(arc(x,x+dir(-30)*(M-x),x+dir(30)*(P-x)),dotted+gray);
markscalefactor *= 3;
draw(rightanglemark(C,L,Q));
draw(rightanglemark(C,K,N));
pair pq=(P+Q)/2,pl=(P+L)/2;
draw((1.5*A-.5*pq)--(1.5*pq-0.5*A),red,Arrows);
draw(shift(pl-pq)*((1.5*A-.5*pq)--(1.5*pq-0.5*A)),red,Arrows);
pair mn=(M+N)/2,mk=(M+K)/2;
draw((1.5*B-.5*mn)--(1.5*mn-0.5*B),red,Arrows);
draw(shift(mk-mn)*((1.5*B-.5*mn)--(1.5*mn-0.5*B)),red,Arrows);
label("$\ell_A$",pq,2*NNW+3*dir(150),red);
label("$\ell_A'$",pl,3*WSW+4*dir(150),red);
label("$\ell_B$",mn,4*NNE,red);
label("$\ell_B'$",mk,4*E,red);
pair X=intersectionpoints(shift(pl-pq)*((1.5*A-.5*pq)--(1.5*pq-0.5*A)),(1.5*B-.5*mn)--(1.5*mn-0.5*B))[0];
pair Y=intersectionpoints(shift(pl-pq)*((1.5*A-.5*pq)--(1.5*pq-0.5*A)),shift(mk-mn)*((1.5*B-.5*mn)--(1.5*mn-0.5*B)))[0];
pair Z=intersectionpoints(((1.5*A-.5*pq)--(1.5*pq-0.5*A)),shift(mk-mn)*((1.5*B-.5*mn)--(1.5*mn-0.5*B)))[0];
dot("$X$",X,W);
dot("$I$",incenter(A,B,C),dir(90));
dot("$Y$",Y,dir(180));
dot("$Z$",Z,E);
pair Xi=foot(X,A,B),Ii=foot(incenter(A,B,C),A,B),Yi=foot(Y,A,B),Zi=foot(Z,A,B);
dot("$X'$",Xi,SW);
dot("$I'$",Ii,SSW);
dot("$Y'$",Yi,SSE);
dot("$Z'$",Zi,SE);
draw(X--Xi^^Y--Yi^^Z--Zi^^incenter(A,B,C)--Ii,dotted);
[/asy]](http://latex.artofproblemsolving.com/4/c/4/4c4097693cb390a0bd5003e4a2c7caf42b97bd3c.png)
,
,
,
.
.
and 
be the perpendicular bisectors of 
,
.
is isoceles with vertex 
.
and 
be the perpendicular bisectors of 
,
.
is isoceles with vertex 
.
,
is the incenter of 
,
,
.
,
,
,
be the projections of 
,
.
,
.
,
.
since 
,![\[I'X' = \frac{s-b}{2} \sin \frac A2 \cdot \frac{\cos \tfrac B2}{\cos \tfrac C2}.\]](http://latex.artofproblemsolving.com/3/9/6/39666a6d27136cfba6b5382865279b947c2efa3d.png)
Now we make some propositions, which can be verified easily using the half-angle formulas and the Law of Cosines: ![\[\sin \frac A2 = \sqrt{\frac{(s-b)(s-c)}{bc}}, \quad \cos \frac B2 = \sqrt{\frac{s(s-b)}{ac}}, \quad \cos \frac C2 = \sqrt{\frac{s(s-c)}{ab}}.\]](http://latex.artofproblemsolving.com/5/2/5/5251b4859e1c4c05d3693a5bc80aa217f49c91b6.png)
These formulas give ![\[I'X' = \frac{s-b}{2} \cdot \sqrt{\frac{(s-b)(s-c)}{bc}} \cdot \sqrt{\frac{s(s-b)}{ac}} \div \sqrt{\frac{s(s-c)}{ab}} = \frac{(s-b)^2}{2c}.\]](http://latex.artofproblemsolving.com/3/9/4/394024f78ede478ee75a7569c3b20ea113232226.png)
By symmetry, we can write 
,
since 
,![\[AY' = AI' + I'X' - X'Y' = (s-a) + \frac{(s-b)^2}{2c} - \frac{(s-a)^2}{2c}\]](http://latex.artofproblemsolving.com/b/6/a/b6abb159500ce5f49475a29e1e93e3b830ab8d8f.png)
and substituting 
,
.
,
.
is the perpendicular bisector of 
.
is isosceles with vertex 
.
.
.
and 
denote the centers of 
and 
respectively. Applying sine law in 
and 
and using 
and 
,
bisects 
externally.
.
and 
are the bisectors of 
,
(as 
bisects 
with radius 
.
be centers of 
so 
so 
.
.
are similar.
is cyclic and 
so 
and 
.
so 
be the centers of 
Notice 
and 
be the foot from 
noting 
and 
We see 
and 
so 
The conclusion is evident as 
where 
with 
.
and 
are parallel to 
and 
,
and 
.
,
and 
we get that 
=
.
.
,
,
.
