[/quote]
,
we are given :
and one side length of the triangle equals 
,
be a trapezoid with parallel sides 
and 
,
.
and 
again at 
respectively. Prove that 
are concurrent.
and 
Prove that
in it. Prove that there are an infinite number of perfect cubes in this progression. (A perfect cube is an integer of the form 
,
is an integer. For example, 
,
and 
?
there are elements 
in an arithmetic progression in the order 
such that 
.
days later (but perhaps earlier, as well). Hedvig received the trophy today. Determine all integers 
for which we can be absolutely certain that she cannot receive the trophy again in 
days, given the above information.
such that 
.
are positive integers pairwise distinct and 
.![$\sqrt[n]{p},\sqrt[n]{q},\sqrt[n]{r}$](http://latex.artofproblemsolving.com/3/1/3/3133b60c921168020d149a9d28ef69b756a704d9.png)
form terms of a same arithmetic progression.
. Prove that
Let 
be reals such that 
. Prove that
real numbers 
,
and transform the sequence into 
.
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the 
be four positive integers in arithmetic progression (that is, 
)
.
ever be a number of the form 
for some 
and some 
,
?
Y by HWenslawski, Retemoeg
Given a triangle 
with circumcenter 
and orthocenter 
. Line 
meets 
at 
respectively.
Define
as the circumcenter of 
. The circumcircle of meets the circumcircle of 
again at 
, 
. Line meets circumcircle of 
again at 
, 
and circumcircle of meets circumcircle of again at 
, 
. Define 
as the intersection of 
and and 
meets circumcircle of at points 
.
Prove that circumcircle of
and circumcircle of are tangent to each other.
Proposed by 郝敏言, China
with circumcenter 
and orthocenter 
. Line 
meets 
at 
respectively.Define
as the circumcenter of 
. The circumcircle of meets the circumcircle of 
again at 
, 
. Line meets circumcircle of 
again at 
, 
and circumcircle of meets circumcircle of again at 
, 
. Define 
as the intersection of 
and and 
meets circumcircle of at points 
.Prove that circumcircle of
and circumcircle of are tangent to each other.Proposed by 郝敏言, China
be the anti-steiner point of the Euler line. It suffices to prove that 
is concyclic.
(spiral transform) 
Also note that 
.
.
(Directed angle 
)
is the circumcircle of 
be the circumcircle of 
.
are the reflections of 
.
is the antipode of 
in circle 
is the altitude of triangle 
.
are isogonal conjugate in 
.
are collinear. Then, 
.
are concyclic. Hence, 
.
.
.
are isogonal conjugate in 
are isogonal conjugate in 
are collinear. Moreover, we can easily see that 
.
.
.
are concyclic. Hence, 
and 
is a point on 
.
.
are collinear.
.
are concyclic. 
.
.
are concyclic. Then, 
.
be the antipode of 
.
,
.
,
is the bisector of 
.
in 
are collinear. Moreover, applying the Pascal's Theorem to the set of 6 points 
,
are collinear.
.
is tangent to cirlce 
,
.
.
is tangent to circle 
is the inversion with center 
.
Since, 
is tangent to 
.
.
be tangent of 
.
.
is the perrpendicular bisector of 
so
are conjugate 
so 
so 
concyclic with 
and 
intersect at 
.
Now
We have 
so by Newton we have 
And so WRT 
are concurrent at 
.
is the perpendicular bisector of 
,
we have
Hence 
.
tagents both 
and 
.
![[asy]
//20DeuXMOII3
//setup
size(8cm);
pen blu,grn,lightblu,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); lightblu=RGB(196,216,255); lightpurple=RGB(237,186,255);
//defn
pair A,B,C,O,H,E,F; A=(3.1,11); B=(0,0); C=(14,0); O=circumcenter(A,B,C); H=A+B+C-2*O; E=extension(O,H,A,B); F=extension(O,H,A,C);
pair reflect(pair P,pair A,pair B){return 2*foot(P,A,B)-P;}
pair Hb,Hc,S,J,K; Hb=reflect(H,A,C); Hc=reflect(H,A,B); S=circumcenter(A,E,F); J=reflect(A,S,O); K=extension(Hb,F,Hc,E);
pair M,L,G,D; M=extension(J,K,O,H); L=extension(J,H,K,K+H-O); G=reflect(J,O,H); D=extension(O,H,K,G);
//draw
fill(A--B--C--cycle,RGB(226,235,255)); fill(extension(K,E,B,C)--extension(K,F,B,C)--F--E--cycle,lightblu); fill(K--extension(K,E,B,C)--extension(K,F,B,C)--cycle,RGB(226,235,255)); draw(K--E--F--K,blu);
draw(B--A--C--B^^Hb--F^^Hc--E,blu); draw(circumcircle(A,B,C),blu); draw(Hb--A--Hc,blu+dotted);
draw(circumcircle(A,E,F),purple+dotted); draw(E--S--F,purple+dotted);
draw(E--D^^A--K--L^^S--J^^F--1.5*F-.5*O,purple);
draw(circumcircle(J,H,F),purple); draw(circumcircle(J,H,E),purple);
draw(D--J--K--D,magenta+linewidth(1)); draw(circumcircle(J,S,O),magenta); draw(S--O,magenta);
draw(circumcircle(J,M,H),red+dashdotted); draw(circumcircle(G,M,H),red+dotted);
draw(J--L,red);
clip((7.2,-5.2)--(-7.4,.2)--(-7.1,16.6)--(15.7,16.6)--(15.3,-1.4)--cycle);
//label
void pt(string s,pair P,pair v,pen a){filldraw(circle(P,.09),a,linewidth(.3)); label(s,P,v);}
pt("$A$",A,dir(110),blu); pt("$B$",B,dir(90),blu); pt("$C$",C,dir(90),blu);
pt("$H$",H,dir(-20),red); pt("$H_b$",Hb,dir(50),blu); pt("$H_c$",Hc,-dir(30)/2,blu);
pt("$O$",O,dir(70),blu); pt("$E$",E,dir(-40),blu); pt("$F$",F,-dir(70),blu);
pt("$S$",S,dir(50),purple); pt("$J$",J,dir(160),purple); pt("$M$",M,dir(-50),magenta);
pt("$L$",L,dir(-70),red); pt("$G$",G,-dir(80),magenta); pt("$D$",D,dir(-90),magenta);
pt("$K$",K,dir(-60),magenta);
//label ell, Omega
label("$\Omega$",(12.95,7.85),blu); label("$\ell$",1.8*F-.8*O,purple);
[/asy]](http://latex.artofproblemsolving.com/4/2/b/42b2a6f2afa38d9379e5f7262e4e47a2edc62ede.png)
,
be the respective reflections of 
,
.
(we'll see this is an equivalent definition). As 
are external angle bisectors wrt 
,
.
.
.![\[\measuredangle H_cJ'H_b=\measuredangle H_cJ'H+\measuredangle HJ'H_b=\measuredangle H_cEH+\measuredangle HFH_b =\measuredangle(\overline{H_cE},\overline{H_bF})=\measuredangle H_bKH_c =\measuredangle H_bAH_c\Rightarrow J'\in\Omega.\]](http://latex.artofproblemsolving.com/f/2/2/f22d85c22b122c2545134263c4aa198a53df4a2e.png)
The construction of 
implies that 
respectively bisect 
,![\[\angle EJ'F=\frac12\angle H_bJ'H_c=\angle BAC=\angle EAF\Rightarrow J'\in(AEF),\]](http://latex.artofproblemsolving.com/6/6/4/664952564cc855da9fffc9e53efc51b9c8517bfc.png)
finishing the claim
;
are cyclic, 
by Reim. By homothety, 
touches 
.
.
,
cyclic; as 
,
both lie on bisector of 
.
on 
.
is the perpendicular bisector of 
by symmetry, it externally bisects 
as 
.
means 
.
cyclic and 
bisects 
is the reflection of 
by symmetry. Reflecting 
" over 
![\[\measuredangle JKA = \measuredangle JOS = \measuredangle JKS\]](http://latex.artofproblemsolving.com/9/a/b/9ab93a6078f2b7742834bc24b123c2a9a9675ed6.png)
which implies that points 
.
denote the intersection of line 
as 
.
so by the Incenter-Excenter Lemma it follows that 
.![\[\measuredangle IGJ = \measuredangle IKJ = \measuredangle DKI = \measuredangle GKI = \measuredangle GJI\]](http://latex.artofproblemsolving.com/9/4/4/9446c83b1590153bddc319a67f031085408e326f.png)
from which it follows that 
.
meet the circumcircle of 
.
b![\[\measuredangle GOM = \measuredangle GAJ = \measuredangle GKJ = \measuredangle GKM\]](http://latex.artofproblemsolving.com/1/e/3/1e3b996020441cb8d30725a685e8573e9db4258a.png)
so quadrilateral 
is cyclic. Since 
is also clearly cyclic, it follows by Reim's Theorem that lines 
and 
are parallel.
a
denote the image of 
,
denote the reflection of 
.![\[\measuredangle JH'H = \measuredangle JBE = \measuredangle JBA = \measuredangle JH_aH\]](http://latex.artofproblemsolving.com/0/2/a/02a7e582f4bfd1a68003329d099725be3611c498.png)
which implies that 
is cyclic. Further note,![\[\measuredangle HXH_a = 2\measuredangle HXH' = 2 \measuredangle EXB = 2\measuredangle EJB = 2\measuredangle HJH' = 2\measuredangle HH_aH' = \measuredangle HH'H_a\]](http://latex.artofproblemsolving.com/4/5/7/457a443b599412dd5564173ae45cff371643ef98.png)
which implies that 
is also cyclic. Thus, these five points are all concyclic. Further, since 
and 
in 
it follows that 
is the diameter of this circle, and thus 
.
and 
.
is the image of 
since 
and 
intersect at 
.
since both triangles are isosceles and,![\[\measuredangle JSA = \measuredangle JSK = \measuredangle JOK = \measuredangle JOK'\]](http://latex.artofproblemsolving.com/c/9/f/c9f89cd46f25774994287dd8e8ec6f3ac50213f9.png)
This allows us to finish,![\[\measuredangle JHO = \measuredangle ZHO = \frac{\pi}{2} + \measuredangle JZA = \measuredangle JAS = \measuredangle JK'O\]](http://latex.artofproblemsolving.com/3/4/f/34f0ea5669bc64f66a88991bee76918dcadb5d21.png)
which implies the claim.![\[\measuredangle HGO = \measuredangle OJH = \measuredangle OK'H = \measuredangle K'GO\]](http://latex.artofproblemsolving.com/2/4/f/24ff2c7589ec9e0d7edceb93f7b073a84605ef43.png)
so points 
are homothetic, from which it follows that 
hit 
respectively, redefine 
be reflections of 
and let 
be the diametral opposite of 
colinear, so for this note that 
therefore 
therefore in addition we have 
so by angles 
which means that 
is cyclic therefore 
which gives the colinearity, now to finish just note 
is perpendicular bisector of 
which concludes.
are symetric in 
are colinear clearly but this follows from Reim's theorem as 
which now as 
means that 
is cyclic and then from Reim's we get 
colinear which now finishes by a degenerate Reim's as 
thus we are done 
are collinear.
cut 
,
and 
.
and 
are both isosceles. So now we have![\[\angle{EAS}=90^\circ-\angle{AFE}=\angle{XHF}=\angle{HXF}=\angle{BXK"}=\angle{BAK"}\]](http://latex.artofproblemsolving.com/2/9/9/2993fab8b22e128a70fa80bbb86fbb14a0f6a7a3.png)
which implies the lemma. 
with 
.
,
.
.
.
lies on 
.![\[\angle{SPJ}=90^\circ-\angle{JAP}=90^\circ-(\angle{JAB}+90^\circ-\angle{C})=\angle{C}-\angle{JAB}=180^\circ-(180^\circ-\angle{C})-\angle{JAB}=180^\circ-\angle{BJA}-\angle{JAB}=\angle{JBA}\]](http://latex.artofproblemsolving.com/1/e/7/1e779507b58fb7166704a5810c40cca179543271.png)
but clearly 
,
.
is cyclic and 
bisects 
.
,
is cyclic or 
,
and 
are cyclic.
is cyclic. Also, we know that 
.
and 
.
is isosceles trapezoid, so it's clear that 
to prove that ![\[\angle{K'NG}=180^\circ-\angle{K'KG}=180^\circ-\angle{OKG}=\angle{OMG}\]](http://latex.artofproblemsolving.com/f/a/9/fa916d618f2fdf6592fb83e253db93e7003b88a3.png)
so 
.
,
![[asy]
import graph; size(11.516142568285202cm); real xmin=5.4,xmax=9.6,ymin=-2.2,ymax=1.0; pen ds=black; real lsf=0.5; pair A=(6.490118921817331,0.5239094777632556), B=(5.813020701293997,-1.1614763102776469), C=(8.011082024176732,-1.1705687001465266), O=(6.914288488880731,-0.6252031468857658), H=(6.485644669526597,-0.5577292388893864), F=(7.621484179145215,-0.7365246095683274), J=(7.318826782833286,0.5309665701006684), D=(9.325519230243023,-1.0047610778641347), K=(7.662972931454258,-1.5946624039971133), G=(6.944073611121734,-1.8497408115754244), M=(7.5214892397836035,-0.7207841531941866), Q=(5.78107335470919,-1.0901796037468472); draw(A--B--C--cycle, orange); draw(circle((6.914288488880731,-0.6252031468857656),1.2248998513145941), red); draw((xmin,-0.1574125298203219*xmin+0.4631924960564801)--(xmax,-0.1574125298203219*xmax+0.4631924960564801), red); draw(circle((6.910976695629011,-0.23630231201757),0.8689322361953719), heavygreen); draw(circle((8.182088071375446,-0.4199423281560887),1.284308290512582), lightblue); draw(A--K, orange); draw(J--K, magenta); draw(J--(6.910976695629011,-0.23630231201757), orange); draw((6.910976695629011,-0.23630231201757)--O, orange); draw(J--O, magenta); draw(A--O, orange); draw(G--D, orange); draw(J--G, magenta); draw(J--D, orange); draw(J--H, magenta); draw(circle((6.75796493793617,-1.2083453753461084),0.9840123550837034), purple); draw(Q--(6.910976695629011,-0.23630231201757), mediumcyan); dot(A,ds); label("$A$",(6.504838448350444,0.5607082940960263),NE*lsf); dot(B,ds); label("$B$",(5.827740227827111,-1.1246774939448765),NE*lsf); dot(C,ds); label("$C$",(8.024629562894667,-1.1320372572114308),NE*lsf); dot(O,ds); label("$O$",(6.84706744024539,-0.686771579584904),SW*lsf); dot(H,ds); label("$H$",(6.501158566717166,-0.5285366693539896),dir(270)*lsf); dot((6.081154081834958,-0.4940573521923375),ds); label("$E$",(6.096371587056477,-0.46597868158827915),NE*lsf); dot(F,ds); label("$F$",(7.634562109767094,-0.7088508693845664),E*lsf); dot((6.910976695629011,-0.23630231201757),ds); label("$S$",(6.9243449545442495,-0.20838696725888356),NE*lsf); dot(J,ds); label("$J$",(7.332811815838217,0.5607082940960263),NE*lsf); dot(D,ds); label("$D$",(9.342027187608544,-0.9738023469805163),NE*lsf); dot(K,ds); label("$K$",(7.6787206893664415,-1.5662632899381261),NE*lsf); dot(G,ds); label("$G$",(6.95746388924376,-1.8201751226342449),NE*lsf); dot(M,ds); label("$M$",(7.535205305668561,-0.6904514612181811),dir(225)*lsf); dot(Q,ds); label("$Q$",(5.7946212931276,-1.0621195061791662),W*lsf);
[/asy]](http://latex.artofproblemsolving.com/2/2/0/220671e9e6fff4374d02aab35ebe5f255bb7a23a.png)
![\[ \angle JKS = \angle JOS = \frac 12 \angle JOA = \angle JKA. \qquad \square \]](http://latex.artofproblemsolving.com/b/2/1/b212ee362b9b68a9b62cb951e2e83b507452c61e.png)
Note that 
,
are reflections over 
is tangent to the circumcircle of 
.
and 
be the intersections of ![\[ \angle HJD = 180 - \angle JHM - \angle ODJ = 180 - \angle JHM - \angle OKJ = 90 \]](http://latex.artofproblemsolving.com/c/3/0/c3088103c462eb4209c6fb48388a5ab5e912e9bb.png)
which implies that 
is cyclic. Yet 
is cyclic as well, implying that 
.
and 
.
and 
would be isogonal, which implies 
;![\[ \angle OKJ = 180 - \angle OSJ = \frac 12 \angle ASJ = \angle AFJ \]](http://latex.artofproblemsolving.com/d/c/0/dc0350861015e0f9c2f3f9ba27c02d648fc6bd30.png)
It follows that ![\[ 90 - \angle OKJ = 90 - \angle AFJ = \angle JAS = \angle JAK \]](http://latex.artofproblemsolving.com/1/a/a/1aab48fcf006563b1cd1a3efadd51a89b7b5ddd1.png)
Now, by 
is cyclic. Let 
intersect 
;
,
is cyclic. So angle chasing gives 
as needed.
