[/quote]
,
be a sequence of positive integers with 
,
and![\[a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1\]](http://latex.artofproblemsolving.com/1/b/f/1bf1a88979e7d2b33530de979d66a4e81032953d.png)
for all 
.
is a prime less than 
for some positive integer 
,
such that 
.
problems in a meeting. For each 
,
-
so that the chair can ensure that the longest idle time for any leader does not exceed 
be real numbers such that 
and 
Find the value of 
and 
Find the value of 
?
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，
is a regular polygon with 
diagonals. Randomly select a triangle whose three vertices are the three vertices of polygon 
lie in a square of area 
.
that intersect in a region of area not less than 
.
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
for every 
.
,![$P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$](http://latex.artofproblemsolving.com/0/5/1/0514f85850830af9a9e1a4dd5e7a4e8e68b2dff9.png)
.
polynomial with only one root.
be a finite set of real numbers, 
be a real number, and 
be 2025 non-zero real numbers. Define![\[A =
\left\{
(x_1, x_2, \cdots, x_{2025}) : x_1, x_2, \cdots, x_{2025} \in X \text{ and } \sum_{i=1}^{2025} \lambda_i x_i = d
\right\},\]](http://latex.artofproblemsolving.com/2/e/1/2e18b6d25bfaa3e3d52f7efe5e4cba7a9cb87db9.png)
![\[B =
\left\{
(x_1, x_2, \cdots, x_{2024}) : x_1, x_2, \cdots, x_{2024} \in X \text{ and } \sum_{i=1}^{2024} (-1)^i x_i = 0
\right\},\]](http://latex.artofproblemsolving.com/4/1/f/41f6b126cc3cf24757b37147906c767d49a0fdd9.png)
![\[C =
\left\{
(x_1, x_2, \cdots, x_{2026}) : x_1, x_2, \cdots, x_{2026} \in X \text{ and } \sum_{i=1}^{2026} (-1)^i x_i = 0
\right\}.\]](http://latex.artofproblemsolving.com/d/e/6/de6586b88004e3f5aa7774525b00ee975c072ff9.png)
Show that 
.
and 
.
![$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$](http://latex.artofproblemsolving.com/2/2/e/22e6ec2f98777db80e39333542664e903663bf81.png)
![$$a+b+\frac{8}{5}c\geq \frac{9\sqrt [3]{25}-4}{20}$$](http://latex.artofproblemsolving.com/f/c/a/fca343ee701eeff4c309b31beb3426ec96b49b55.png)
Let 
.
![$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$](http://latex.artofproblemsolving.com/6/9/5/695cc35986a391fdc15e4863d361ce373fbbf64c.png)
.
be a point on side 
,
be a point on the extension of 
such that 
.
,
be a point on the side extension of 
satisfying 
.
.
,
,
,
jump is equal to 
.
over all pairs of positive integers 
.
Y by narutomath96, Viswanath, lephuongnam1568, anantmudgal09, Adventure10, GeoKing, Funcshun840, Rounak_iitr, and 5 other users
Let 
be a triangle and let 
and 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle ; the circles 
and 
are defined similarly. The circles and meet at a point 
distinct from 
; the points 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.
(Russia) Fedor Ivlev
be a triangle and let 
and 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle ; the circles 
and 
are defined similarly. The circles and meet at a point 
distinct from 
; the points 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.(Russia) Fedor Ivlev
and 
, 
similarly . Also suppose that 
is tangent to 
at 
. Solution consist of two steps , first I'm going to show 
are concurrent at 
.And then show that 
which finish the proof .
cut 
, then 
and so 
which is a fix line. On the other hand 
WRT 
, according to properties of pole and polar we deduce that 
and 
.
. It's well know that 
are midpoint of their corresponding arc .So 
.Besides , As we know 
are concurrent we deduce 
and 
are perspective 
)
or equivalently 
are concurrent at radical axis of 
it suffice to show that 
is cyclic .
be a line and 
be circle.Suppose 
be inscribed in 
be the circumcevian triangle of 
.
we get 
are collinear.
and 
are perspective.So 
are collinear.Hence we conclude that 
are collinear.But this line is nothing but the polar of 
.
,
is the radical axis of 
and 
.
are concurrent.
is its intouch triangle
respectively.Now let tangents at 
at 
respectively.Then its easy to show that 
is the radical axis of 
.
wrt 
concur at the pole of 
.
are concurrent ,say at 
.
wrt 
are collinear, their polars i.e 
are concurrent.So 
are perspective wrt the perspective axis of 
.
.
we conclude that the pole of the radical axis of 
of the three circles . To show that this radical centre is on 
, consider the inversion w.r.t to a circle centered at 
and then the reflection w.r.t 
are themselves and the incircle becomes the larger circle tangent to them ( internally ) . From the property of this transformation ( inversion + homothety ) , 
and the centre of this circle are collinear . We now show that this centre is actually 
!
be the mid-pt of arc 
that does not intersect the incircle . 
are defined in similar mannar . It is well-known that the power at 
to the incircle is equal to 
and 
, so 
and 'circle' 
lie on the radical axis of the incircle and 'circle' 
be the mid-pts of arc 
( not containing the third vertex of 
. ) of 
respectively . Then , we deduce that 
is homothetic to 
and the centre of homothety is the intersection of the perpendicular bisectors of 
, which is 
and the centres of these two circles lie on the perpendicular bisector of 
which passes through 
. In other words , 
is pedal triangle of 
is antipedal triangle of 
cut pedal circle 
again at 
.
intersect again at 
.
.
are concurrent.
be the intersections of 
with 
are concurrent.
on 
are concurrent if and only if 
. It can easily proved by trigonometric Ceva theorem .
,
,
,
.
,
at 
,
,
,
passes through 
,
,
are concurrent.
.
,
intersect at 
.
and 
intersect at 
,
is the polar of 
are all concurrent. Call this point 
.
,
and 
the same way. Since 
is cyclic, 
.
and 
,
is homothetic to 
wrt 
to 
bisects arc 
and 
similarly; then note that from the homothety from 
is the incircle of the triangle formed by the intersections of 
be the circumcenter of 
,
is the perpendicular bisector of 
and 
bisect 
.
passes through 
and 
and 
intersect on the radical axis of (O) and (I), which means that if M is the pole of this radical axis with respect to (I), then M (which clearly lies on 
.
,
and 
,
is on 
.
,
be the points where the incircle touches 
.
.
and similar, so 
are homothetic about 
concur at 
and similar.
,
the inverted image of 
is the 
.
is a line through 
,
.
are tangent to this mixtilinear excircle at 
respectively, so by Brianchon's theorem on circumscribed hexagon 
,
intersect at some point ![\[\frac{\sin BAA'}{\sin CAA'}=\frac{\sin (F)_V(A)_VP_A}{\sin (E)_V(A)_VP_A}=\frac{\sin A(F)_VC}{\sin (E)_V(F)_VC}\frac{\sin (F)_V(E)_VB}{\sin A(E)_VB}=\frac{\sin ACF}{\sin EFC}\frac{\sin FEB}{\sin ABE}=\frac{EF}{EC}\frac{FB}{EF}=\frac{y}{z}.\]](http://latex.artofproblemsolving.com/c/a/a/caa98e4c813c9b366939e527fc042b116b3ca9cc.png)
.
be the foot of the altitude from 
to 
.
so by the homothety, 
are collinear. Now it suffices to show that 
.![\[\frac{FD'}{ED'}=\frac{DF}{DE}\frac{\cos EFD}{\cos FED}=\frac{DF}{DE}\frac{\sin \frac{C}{2}}{\sin \frac{B}{2}}=\frac{2y}{2z}=\frac{y}{z}\]](http://latex.artofproblemsolving.com/e/7/3/e738d2bd1ec222e2ed2137fdeede4cb45b36a560.png)
,
collinear (by, for example, the ratio lemma and phantom pointing), so 
pass through 
.
,
be the tangency points of those circles with the incircle. Let the tangent 
through 
similarly. (Note that 
.
wrt incircle. Then, by radical axis theorem, 
.
are collinear. But we get 
are also collinear by pascals, so 
are collinear. We conclude 
.
.
.
is the angle bisector of 
.
,
(because 
). Likewise, 
is the radical axis of 
.
,
be the orthocenter of 
,
is the common radical axis of 3 circles 
,
lies on the Euler line of 
and 
respectively. Let 
,
,
.
,
is the midpoint of the arc 
not containing 
is on the radical axis of 
,
.
with center 
.
,
is cyclic. Noting 
gives 
,
.
pass through 
and 
,
. Hence 
in 
are tangent to the incircle at 
intersects 
similarly. Clearly 
is homothetic to the intouch triangle. But 
,
is tangent to 
and 
,
,
be the exsimilicenter of the two circles. Then 
and 
.![\[
\frac{SF}{SE}=\frac{SP}{SN}
\]](http://latex.artofproblemsolving.com/3/4/8/34884ab910d1690e8da3dacf6eacd7a6963942b3.png)
![\[
SY\cdot SF = SE\cdot SZ
\]](http://latex.artofproblemsolving.com/5/4/0/540bbcfbcf6cec63bcdf7dacfb5bc75a95651e44.png)
![\[
SY\cdot SP = SZ\cdot SN
\]](http://latex.artofproblemsolving.com/5/8/4/5842d7fc20511563f02883eff820327c18294aa7.png)
.
.
be the power of a point 
with respect to a circle 
( maps 
) 
is the midpoint of arc 
is the midpoint of arc 
,
in 
,
,
is the radical axis of 
pass through the midpoint of 
and 
.
is B-midline, C-midline of 
,
are homothetic .
be the homothety center of 
and 
.
are concyclic ,
is the radical center of 
.
and 
are concurrent at 
and 
are homothetic 
.
pass through the A-excenter 
, B-excenter 
, C-excenter 
of 
,
is 
( in 
Denote 
the tangents of 
and 
Since 
then 
and similarly 
and 
are perspective through the radical center 
of 
then there exists a homology sendind 
Thus in this figure, 
concur at 
their common midpoint. Hence, in the original figure 
concur at 
lying on 
(incircle) with 
be midpoints of big arcs 
.
,
be intersection of tangents to 
and 
.
,
are collinear and by Pascal at 
,
are collinear, so 
are collinear. But 
have the same power of a point to 
and so that line is the radical axis, so 
.
are collinear! So 
but 
so 
.
are homothetic! At 
too.
are the perpendicular bisectors of 
.
,
.
.
are collinear!
are concurrent, which is true!
,
.
,
be the intouch triangle of 
be the incircle. Now 
and 
touch to 
and 
respectively, and the common tangents from this points cut at 
,
we get 
and 
and 
,
,
we get 
,
(defined analogously) with respect to 
,
,
,
.
intersect 
,
and 
similarly. By homothety, we have that 
swaps line 
.
is homothetic with 
be the center of homothety of 
,
and because it's the center of homothety 
,
as the excenters of triangle 
.
at 
bisects angle 
.
and 
.
and from Fact 5, we know 
which is symmetric. It follows that 
denote the incircle, circumcircle of triangle 
,
be the intersections of 
respectively.
,
.
and 
.
.
be the intersection of 
and 
,
are collinear. Thus we only need to show that 
are concurrent, where 
are defined similarly.
.
.
be the intersection of lines 
.
similarly. Note that 
.
concur at 
to be the 
,
to be the touchpoint of 
to be the midpoint of 
.
similarly.
colinear
,
concur at a point 
,
to 
.
are collinear (they all lie on the Euler line of 
)
)
are colinear. So together, we have 
colinear, or 
.
is symmetric in 
.
for homothety reasons.
, which is evidently symmetric in 
and 
is tangent to 
similarly. Now, define 
,
,
.
and its cyclic counterparts are concurrent.
and 
as 
and the intersection of 
and 
as 
.
and 
we have that 
.
and 
.
.
.
.
of ![\[PX\cdot PX' = PX\cdot PD\cdot\frac{k+1}{2} = \frac{k+1}{2}\mathrm{Pow}_{(DEF)}(P),\]](http://latex.artofproblemsolving.com/a/2/2/a22bab6cff916315bdb60f0c3d392fb769422774.png)
which is symmetric in 
are collinear (here 
)
collinear, it suffices to show that 
collinear, or that 
,
to the intersection of the polar of 
with respect to ![\[-1=(AU;M\infty) = (SD;T'\infty),\]](http://latex.artofproblemsolving.com/8/7/1/8715a013e2e092d69567dcb94f0f1d2809d2dd9e.png)
where 
,
is the intersection of the polar of 
,
.![\[TD^2=TX^2=TB\cdot TC,\]](http://latex.artofproblemsolving.com/a/b/8/ab8ada06f7cd054cb579a50a71bbb64604370397.png)
so 
,
.
,
,
.
are the internal and external angle bisectors of 
.
is the perpendicular bisector of 
,
,
.
similarly. Furthermore, let 
and 
the midpoint of 
are defined similarly.
are collinear, where 
satisfies 
,
,
,
.
(all of them have same power of point WRT 
)
,
are concurrent. Let 
.
,
.
,
are collinear. By Pascal's Theorem on 
and 
,
lies on 
.
,
lies on 
.
similarly. It is enough to show that 
concur. Clearly 
lies on 
concur. Now the problem ought to be true no matter what configuration 
form: take a homography sending 
to the center of 
and 
.
respectively, and let 
and 
and 
respectively. Also, let 
and 
noticing 
by Radical Axis. By ISL 2002/G7, 
and 
and by Vietnam TST 2003/2, 
and 
Also, 
by Pascal on 
and 
respectively. Similarly, 
and 
![[asy]
//setup
size(13cm);
pen blu,grn,lightblu; blu=RGB(102,153,255); grn=RGB(0,204,0); lightblu=RGB(196,216,255);
//defns
pair A,B,C,I,O,D,E,F,H; A=(-1,10); B=(0,0); C=(14,0); I=incenter(A,B,C); O=circumcenter(A,B,C); D=foot(I,B,C); E=foot(I,C,A); F=foot(I,A,B); H=D+E+F-2*I;
path w=incircle(A,B,C);
pair Ka,Kb,Kc; Ka=extension(E,F,B,C); Kb=extension(D,F,A,C); Kc=extension(D,E,A,B);
path ya,yb,yc; ya=circle((Ka+D)/2,distance(D,Ka)/2); yb=circle((Kb+E)/2,distance(E,Kb)/2); yc=circle((Kc+F)/2,distance(F,Kc)/2);
pair Ja,Jb,Jc; Ja=foot(D,Ka,2I-D); Jb=foot(E,Kb,2I-E); Jc=foot(F,Kc,2I-F);
path wa,wb,wc; wa=circumcircle(B,C,Ja); wb=circumcircle(C,A,Jb); wc=circumcircle(A,B,Jc);
pair Ta=foot(D,E,F); pair U=2*circumcenter(I,Jb,Jc)-I;
pair X=extension(D,Ja,E,Jb);
//draw
filldraw(D--E--F--cycle,lightblu,blu);
draw(A--B--C--A--D--E--F--D--Ka--F,blu); draw(w,blu+dashdotted);
draw(D--Ta,purple+dotted);
draw(ya,purple); draw(yb,purple); draw(yc,purple);
draw(4*I-3*H--3*H-2*I,linewidth(1));
draw(D--Ja^^E--Jb^^F--Jc,grn+linewidth(1));
draw(wb,red); draw(wc,red);
draw(Jb--U--Jc,red);
draw(A--U,magenta+linewidth(1));
clip((-7,-15)--(-7,12)--(13,12)--(15,10)--(15,-15)--cycle);
//label
label("$A$",A,dir(120)); label("$B$",B,-dir(80)); label("$C$",C,dir(-50));
label("$D$",D,dir(120)); label("$E$",E,dir(90)); label("$F$",F,dir(120));
label("$K_a$",Ka,-dir(20));label("$J_a$",Ja,dir(20));
label("$J_b$",Jb,-dir(80)); label("$J_c$",Jc,dir(70));
label("$U$",U,-dir(90)); label("$X$",X,dir(-90));
label("$\textcolor{red}{\omega_b}$",(14,1.2)); label("$\textcolor{red}{\omega_c}$",(-4,8));
label("$\textcolor[rgb]{.6,0,1}{\gamma_a}$",(-5,2.8));
label("$\textcolor[rgb]{.6,0,1}{\gamma_b}$",(6.5,10.5));
label("$\textcolor[rgb]{.6,0,1}{\gamma_c}$",(6,-4));
[/asy]](http://latex.artofproblemsolving.com/6/5/3/653514f9e4393016e2c01db39e6bbe8a987261a9.png)
,
,
denote the orthocenter and Euler line of 
is just 
,
(which exists since the former 3 circles are coaxial). We'll show this is the desired concurrency point. Also, define 
and variants. Clearly, 
is the raxis of 
,
.
denote the foot of 
,
( = variants) wrt the 
wrt all circles by orthogonality, hence also on raxis, done. 
meet @ 
is the raxis of 
.
,
,
and 
by a well-known lemma. Finally, let 
and 
similarly.
,
are all parallel, and cyclic variants hold. Thus 
;
are homothetic. Since 
is the perpendicular bisector of 
is on 
,
,
that 
,
,
.
,
be the reflection of 
is cyclic, 
,
is cyclic. Since 
,
,
,
is homothetic to 
,
is the ratio of homothety from 
as 
is a harmonic quadrilateral. Which implies that 
and take 
it becomes clear that 
it further implies that 
is tangent to 
is angle bisector of 
and midpoint of 
and 
as 
,
.
which comes from the fact that triangles 
is cyclic and
being parallel implies the result. Now the homothety form 
is tangent to incircle 
at 
,
,
concur on 
.
and 
respectively, are collinear. Note however 
,
are the same. Hence 
and 
’
and 
to 
and 
)
e
e
and 
are collinear
is the homothetic center of the intouch triangle and the excentral triangl
passes through the midpoint of arc 
,
,
(
and we know 
is on the radical axis of 
are colinear 
and 
the excentral triangle. Let 
, 
and 
respectively.
(and similarly) are collinear and point 
and 
(and similarly) and also from 
and 
intersect at the 
E
which lies on 
is also homothetic to these two triangles. Thus, we only need to show the following claim.![\[X_{57}P \cdot X_{57}L=\frac{X_{57}P \cdot X_{57}I_a \cdot X_{57}L}{X_{57}I_a} = \frac{X_{57}B\cdot X_{57}E_{Ob}\cdot X_{57}L}{X_{57}I_a} = \frac{X_{57}B \cdot X_{57}E_{Ob}\cdot X_{57}N}{X_{57}I_c} = \frac{X_{57}R \cdot X_{57}I_c \cdot X_{57}N}{X_{57}I_c}= X_{57}R \cdot X_{57}N\]](http://latex.artofproblemsolving.com/5/f/d/5fdc97e79b03da98ac09be4b2ea7f8e758d0e293.png)
which implies that 
is the Radical Axis of circles 
and 
are the other two pairwise Radical Axes. Thus, these lines 
