AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
real numbers 
, we can select a real number 
and transform the sequence into 
. This transformation can be performed multiple times, with each chosen real number potentially being different
where 
are lengths of parallel sides and 
are other two sides.
+2 w
be a cyclic quadrilateral. Let 
, 
, 
be the feet of the perpendiculars from 
to the lines 
, 
, 
, respectively. Show that 
if and only if the bisectors of 
and 
are concurrent with 
.
be reals such that 
. Prove that
. Let and 
be the feet of the altitudes from 
to and from 
to respectively.
and 
as the reflections of across lines and , respectively. Let 
be the circumcircle of 
. Denote by the second intersection of line 
with , and by the intersection of ray 
with .
is the circumcenter of , prove that , , and are collinear if and only if quadrilateral 
can be inscribed within a circle.
be a triangle. A circle passing through and intersects segments and at and , respectively. Lines and 
intersect at 
, while lines 
and 
intersect at 
. Prove that 
if and only if 
.
be real numbers such that 
Prove that
of positive integers satisfying the following two conditions:
, if 
is in 
, then so are 
and 
; and contains an integer 
such that 
is not divisible by 
.
is a cyclic quadrilateral whose diagonals intersect at . The circumcircle of 
meets segment at points and . The circumcircle of 
meets segment at points and . Let 
and 
be the incenters of 
and 
, respectively. Segments 
and meet at 
. Prove that the points 
are cyclic.
is a cyclic quadrilateral whose diagonals intersect at . The circumcircle of meets segment at points and . The circumcircle of meets segment at points and . Let and be the incenters of and , respectively. Segments and meet at . Prove that the points are cyclic.
to the bisector of angle 
intersect 
at 
,and define 
analogously.
and 
is cyclic.
and are collinear;
(which can be gotten easily)
and 
are concyclic.Now define new point 
.Again chase some angles to get 
and are concyclic.But since 
are collinear 
.
to the bisector of angle intersect at ,and define analogously. should be distinguished from the one in the Question. is sort of vague.
is perpendicular to bisector of angle 
. Then by some simple angle chasing we are done.
intersects 
at 
respectively, and be the intersection of 
and 
which implies that lies on the perpendicular bisector of 
Meanwhile we have 
which follows that 
is the angle bisector of 
is parallel to 
are collinear, this is because 
Notice that 
Now that bisects 
we get![\[\frac{SQ}{SR}=\frac{PQ}{PR}=\frac{QA}{RB}=\frac{QI}{RJ}.\]](http://latex.artofproblemsolving.com/2/8/c/28c187e921197b9dca57a61269566a97cea00ff3.png)
Hence is parallel to as desired. 
so are concyclic.![[asy]
size(7cm); pointpen=black; pathpen=black; defaultpen(fontsize(9pt));
pair A,B,C,D,P,I,J,Q,R,E,F,S;
A=dir(-160); B=dir(-20); C=dir(30); D=dir(110);
P=IP(A--C,B--D);
E=OP(circumcircle(A,P,D),A--B);
F=IP(circumcircle(B,P,C),A--B);
I=incenter(A,D,E); J=incenter(C,F,B);
Q=OP(L(E,I,5,5),circumcircle(A,P,D));
R=OP(L(F,J,5,5),circumcircle(B,P,C));
S=extension(I,E,J,F);
filldraw(anglemark(A,Q,P,3),magenta);
filldraw(anglemark(A,D,P,3),magenta);
filldraw(anglemark(A,C,B,3),magenta);
filldraw(anglemark(P,R,B,3),magenta);
D(unitcircle);
D(circumcircle(A,P,D),blue);
D(circumcircle(B,P,C),blue);
D(A--B--C--D--cycle);
D(A--C); D(B--D); D(D--E); D(C--F);
D(I--J,red+linewidth(1.2)); D(Q--S--R, dashed);
D(Q--R,red+linewidth(1.2)); D(Q--A, dashed);
D(R--B,dashed); D(E--P--F,deepgreen+linewidth(1.1)); D(S--P,dashed);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$P$",P,dir(85)*2);
dot("$E$",E,dir(-90));
dot("$F$",F,dir(-65));
dot("$I$",I,dir(60));
dot("$J$",J,dir(-90));
dot("$Q$",Q,dir(Q));
dot("$R$",R,dir(R));
dot("$S$",S,dir(S));
[/asy]](http://latex.artofproblemsolving.com/b/0/9/b09186dca6c23686ad44501f9b5e5b151c631b99.png)
be the circumcenters of 
and 
. Let 
. Note that 
is the bisector of angle 
. Moreover, 
. Hence 
. Consequently, 
proving 
. Now it is 
proving that are concyclic. 
and 
be the midpoints of shorter arcs 
and in the circles 
and 
respectively.
is the intersection point of 
and 
.
are collinear.
and 
are similar.
![\[ \angle BJC=90^{\circ}+ \frac{\angle BFC}{2} = 90^{\circ}+ \frac{\angle BPC}{2} =90^{\circ}+ \frac{\angle DPA}{2}=90^{\circ}+ \frac{\angle DEA}{2}= \angle DIA. \]](http://latex.artofproblemsolving.com/c/d/8/cd8de6b2cf8b36ce3357b7c363bd347df42a6924.png)
wrt and 
wrt we get ![\[ \frac{\sin \angle AJB}{\sin \angle CJA} = \frac{\sin \angle AIB}{\sin \angle BID}. \]](http://latex.artofproblemsolving.com/7/5/7/757cf5e106d85c74ef46cb8c8f453317c62c941b.png)
and thus 
is cyclic. Finally note that ![\[ \angle JBA+ \angle AIJ=180^{\circ} \implies \angle EIK = \angle EAK. \]](http://latex.artofproblemsolving.com/3/5/d/35dc1974d57dfd1af84059714dd95de95e7b7bee.png)
We conclude 
is cyclic.
and 
be the intersections of 
and 
with 
and 
respectively. Let 
. Now, we can get started., and are collinear.![\[2\measuredangle MPD = 2\measuredangle MED = \measuredangle AED = \measuredangle APD = \measuredangle CPB = \measuredangle CFD = 2\measuredangle NFB = 2\measuredangle NPB\]](http://latex.artofproblemsolving.com/6/2/5/625eaa27e2cb276b689fa9335b9e0996d56c9899.png)
This means that 
from which it is quite clear that 
as claimed.
and are parallel.
. This is beacause,![\[\measuredangle APM \sim \measuredangle CPN = \measuredangle CFN = \measuredangle NFB = \measuredangle NPB\]](http://latex.artofproblemsolving.com/1/8/5/185c25fbee05d5771f87db9125c7be3d704cf156.png)
and![\[\measuredangle PMA = \measuredangle PDA = \measuredangle BDA = \measuredangle BCA = \measuredangle BCP = \measuredangle BNP \]](http://latex.artofproblemsolving.com/1/d/a/1da256adabde3233098e85a64cd9af91eabc48f9.png)
and 
. Thus,![\[\frac{MI}{NJ} = \frac{MA}{NB} = \frac{AP}{PB} = \frac{\sin \angle ABP }{\sin \angle PAB}\]](http://latex.artofproblemsolving.com/d/c/5/dc59b40c174546a433835294831296201252c121.png)
Further,![\[\frac{TM}{TN}=\frac{\sin \angle TNM}{\sin \angle TMN}=\frac{\sin \angle FNP}{\sin \angle EMP}=\frac{\sin \angle FBP}{\sin \angle EAP}=\frac{\angle ABP}{\sin \angle PAB}\]](http://latex.artofproblemsolving.com/9/7/c/97c4a1fd6a14ed69cf93fc9c91e1a92c89f363c2.png)
![\[\frac{MI}{MJ}=\frac{TM}{TN}\]](http://latex.artofproblemsolving.com/e/a/3/ea3778d2587633737b58e8d3f9230df1ede9638e.png)
which proves the claim.![\[\measuredangle JIE = \measuredangle NME = \measuredangle PME = \measuredangle PAE = \measuredangle KAE\]](http://latex.artofproblemsolving.com/8/5/f/85fc0a834aeee81940bf10a1b42aced7825e864d.png)
which implies that points 
and are indeed concyclic which was the desired conclusion.
to 
since I'm bad and let 
. Let and denote 
and 
as arc midpoints. We angle chase 
collinear. Then, ![\[\frac{MK}{NK}=\frac{\sin MNK}{\sin NMK}=\frac{\sin ABP}{\sin BAP}=\frac{AP}{BP}\]](http://latex.artofproblemsolving.com/c/f/2/cf25934a48ba5650c366d91c999f8c4447619043.png)
by LoS. But as 
we get that 
, we get 
so 
. To finish, ![\[\measuredangle EIL=\measuredangle EIJ=\measuredangle EMN=\measuredangle EMP=\measuredangle EAP=\measuredangle EAL\]](http://latex.artofproblemsolving.com/d/2/0/d20a20e27c73065421ee75524280abd4d037fe0e.png)
?
and 
Prove that 
and 
.
Prove that
such that 
.
