[/quote]
,
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
be a right-angled triangle where 
=
be an altitude of that triangle and points 
and 
be the midpoints of 
,
is the circumcenter of the triangle 
,
and 
,
be a point 
)
be the anticomplement of 
be the 
-
is parallel to the real asymptote of the circular pivotal isocubic with pivot 
and 
.
such that 
.
is a power of a prime number, but for arbitrary 
and, for each 
,
be the unique integer in the range 
for which 
is divisible by 
.
the obtained sequence is 
.
eventually becomes constant.
lie on sides 
,
.
meet at orthocenter 
.
such that 
and 
.
and 
intersect at point 
.
.
.
.
over 
intersects lines 
and 
and 
and 
.
lies on 
.
such that 
prove that the triangle must have a right angle.
such that for any real numbers 
and 
,![\[
f\left(x^2+2y f(x)\right) + (f(y))^2 \leq (f(x+y))^2
\]](http://latex.artofproblemsolving.com/8/4/3/8437449cd3f8f0da24d17a60b55dcb563e6b5655.png)
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
and 
.
and 
the orthocenters of triangles 
and 
,
be a triangle. A circle passing through 
and 
intersects the sides 
and 
respectively. Prove that 
,
and 
are concurrent, where 
and 
are the orthocentres of triangles 
respectively.
Y by Amir Hossein, Adventure10, Mango247, and 2 other users
Point 
lies on side 
of a convex quadrilateral 
. Let 
be the incircle of triangle 
, and let 
be its incenter. Suppose that is tangent to the incircles of triangles 
and 
at points 
and 
, respectively. Let lines 
and 
meet at 
, and let lines 
and 
meet at 
. Prove that points , , and are collinear.
Author: Waldemar Pompe, Poland
lies on side 
of a convex quadrilateral 
. Let 
be the incircle of triangle 
, and let 
be its incenter. Suppose that is tangent to the incircles of triangles 
and 
at points 
and 
, respectively. Let lines 
and 
meet at 
, and let lines 
and 
meet at 
. Prove that points , , and are collinear.Author: Waldemar Pompe, Poland
and 
satisfies 
(the segments are not oriented).
the circle tangent to the lines 
,
of the quadrilateral 
,
,
the incircles of triangles 
is the exsimilicenter of 
and 
at the insimilicenter of 
intersects 
and 
are circumscribed. Denote by 
and 
their incircles. We proceed similarly: 
is the exsimilcenter of 
is harmonic.
and 
in convex quadrilateral 
are tangent, then ![\[WX+WY-XY=2WT=WZ+WY-YZ\]](http://latex.artofproblemsolving.com/9/a/c/9acfb1f9cf82507df3bcc25e43f008269f477efc.png)
,
is externally tangent to two circles 
and 
at 
passes through the exsimilicenter of 
be the incenter of 
be the incenter of 
be the incenter of 
(this exists by Lemma 1)
be the incenter of 
(
be the intersection of 
and 
.
lie on a line.
and 
and Lemma 2, 
is a line. (as 
,
,
.
concur on the exsimilicenter of the incircles of 
and 
,
is a line, as 
,
,
.
and 
be the incircle and incenter of 
,
and 
be the incircle and incenter of 
,
,
and 
,
are tangential quadrilaterals. Let their incircles be 
and 
,
,
of 
,
are collinear. By Monge's theorem on 
of 
,
are collinear.
be the exsimilicenter of 
and 
to find that 
,
are collinear.
![\[PK=\frac{PD+PC-DC}{2}=\frac{PD+PA-DA}{2}\implies AP+CD=PC+DA\]](http://latex.artofproblemsolving.com/8/f/f/8ff257043d2a2459e2c306a0b4d114e6716e8857.png)
implies that quadrilateral 
.
be the insimilicenter of 
,
similarly. By Monge D'Alembert on 
;
.
and 
,
concur. But this follows from Monge D'Alembert on 
meet at 
.
and 
.
.![\[
\frac{PD+PC-CD}{2}=\frac{PA+PK-AD}{2}
\]](http://latex.artofproblemsolving.com/4/4/d/44d79eaaae06a8688653bdd813f91b673dcb816c.png)
and 
is circumscribed. Similarly, so is 
.
denote the exsimilicenter of two circles 
,
denote their insimilicenter.
lies on 
and similarly 
.
,
lies on 
,
is collinear with 
.
and 
and by the converse of Pithot's Theorem, the claim holds. A similar argument works for 
with center 
the circle tangent to the lines 
with center 
and let circles 
and 
the inscribed circles in the quadrilaterals 
be the exsimilicenter and insimilicenter, respectively, of the circles 
are collinear and 
.
are collinear and 
are collinear. Similar statements shall hold for 
and 
proving the result.
is the insimilicenter for circles 
passes through the exsimilicenter of circles 
are collinear and the conclusion holds.
and 
meet at 
.
be the circle centered at ![\[AP - AD = KP - KD = CP - CD,\]](http://latex.artofproblemsolving.com/3/2/7/32789152f03a98335f1f44d731c9b8dc87108151.png)
so 
.
.
,
,
(see SL 2017 G7 for another example of this lemma). 
)
,
,
the circumcircle.![[asy] size(12cm);
pair D = (11.14,-6.04); pair C = (2.,-6.); pair P = (4.444652508480686,1.8445570362182018); pair I = (5.51778691510771,-3.4234688887952602); pair K = (7.493465232949166,-1.745777572891788); pair L = (3.0432615410670767,-2.652315765041075); pair T = (5.5064439102990566,-6.015345487572424); pair A = (15.364272954643416,5.744248447444408); pair B = (-3.880008641661257,-1.1284043179454462); pair F = (5.883808240537381,-3.2775609912250143); pair J = (6.629240970901842,-2.980407478817377); pair G = (9.058135630380526,-11.847675599776444);
filldraw(B--A--G--cycle, invisible, deepcyan); filldraw(incircle(B, A, G), invisible, deepcyan); filldraw(P--C--D--cycle, invisible, red); filldraw(incircle(P, C, D), invisible, red);
draw(B--F--A, dotted); draw(A--C, dotted); draw(B--D, dotted); pair E = extension(B, D, A, C);
draw(incircle(B, P, C), deepgreen); draw(incircle(A, P, D), deepgreen); pair I_1 = extension(D, I, B, incenter(B, P, C)); filldraw(CP(I_1, foot(I_1, C, D)), invisible, orange+dashed); pair I_2 = extension(C, I, A, incenter(D, P, A)); filldraw(CP(I_2, foot(I_2, C, D)), invisible, orange+dashed);
dot("$D$", D, dir(D)); dot("$C$", C, dir(225)); dot("$P$", P, dir(120)); dot("$I$", I, dir(I)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$T$", T, dir(T)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$F$", F, dir(F)); dot("$J$", J, dir(J)); dot("$G$", G, dir(G)); dot("$E$", E, dir(E));
/* TSQ Source:
!size(12cm);
D = (11.14,-6.04) .C = (2.,-6.) R225 .P = (4.444652508480686,1.8445570362182018) R120 I = (5.51778691510771,-3.4234688887952602) K = (7.493465232949166,-1.745777572891788) L = (3.0432615410670767,-2.652315765041075) T = (5.5064439102990566,-6.015345487572424) A = (15.364272954643416,5.744248447444408) B = (-3.880008641661257,-1.1284043179454462) F = (5.883808240537381,-3.2775609912250143) J = (6.629240970901842,-2.980407478817377) G = (9.058135630380526,-11.847675599776444)
B--A--G--cycle 0.1 lightcyan / deepcyan incircle B A G 0.1 lightcyan / deepcyan P--C--D--cycle 0.1 lightred / red incircle P C D 0.1 lightred / red
B--F--A dotted A--C dotted B--D dotted E = extension B D A C
incircle B P C deepgreen incircle A P D deepgreen I_1 := extension D I B incenter B P C CP I_1 foot I_1 C D 0.1 yellow / orange dashed I_2 := extension C I A incenter D P A CP I_2 foot I_2 C D 0.1 yellow / orange dashed
*/ [/asy]](http://latex.artofproblemsolving.com/7/a/f/7af15aecbb23e3e794461bfee20e8433fd0303c9.png)
and 
,
passes through said insimilicenter. By symmetry, so does 
,
passes through said exsimilicenter. By symmetry, so does 
,
,the incircles of 
and 
. Denote by 
,the incircles of Quadrilaterals 
. Denote by 
-
.
lie on the same line .
and 
the exsimilicenter and insimilicenter of circles 
and 
respectively .
implies that 
.
.
implies 
.
.
are collinear as desired
,
denote the incircles of 
,
.
and ![\[\frac{PC+PD-CD}2=PK=\frac{PA+PD-CD}2\implies PC+AD=PA+CD,\]](http://latex.artofproblemsolving.com/5/1/2/512ad2efd3c37ef4380400b4b20e1479f5a66b90.png)
so by Pitot's theorem, quadrilateral 
has an incircle 
.
has an incircle 
.![[asy] size(9cm); defaultpen(fontsize(10pt)); pen pri=lightblue; pen sec=red; pen tri=purple; pen qua=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible;
pair P,C,D,I,K,L,IA,IB,A,B,EE,F,Q,O,I1,I2; P=dir(120); C=dir(205); D=dir(335); I=incenter(P,C,D); K=foot(I,P,D); L=foot(I,P,C); IA=K+0.8*(K-I); IB=extension(L,I,P,P+(C-P)*dir(90)*(IA-P)/(I-P)); A=extension(P,reflect(P,IA)*K,D,reflect(D,IA)*K); B=extension(P,reflect(P,IB)*L,C,reflect(C,IB)*L); EE=extension(A,C,B,D); F=extension(A,K,B,L); Q=extension(A,D,B,C); O=incenter(Q,A,B); I1=extension(A,IA,C,I); I2=extension(B,IB,D,I);
draw(B--F--A,tri); draw(B--D,sec); draw(A--C,sec); draw(C--foot(O,B,C),pri); filldraw(incircle(Q,A,B),qfil,qua+Dotted); filldraw(circle(I1,abs(I1-foot(I1,C,D))),sfil,sec+dashed); filldraw(circle(I2,abs(I2-foot(I2,C,D))),sfil,sec+dashed); filldraw(incircle(P,A,D),tfil,tri); filldraw(incircle(P,C,B),tfil,tri); filldraw(incircle(P,C,D),qfil,qua); filldraw(A--B--C--D--cycle,fil,pri); draw(C--P--D,pri); draw( (EE+4*(EE-O))--(O+3*(O-EE)),blue+dashed);
dot("\(P\)",P,N); dot("\(C\)",C,SW); dot("\(D\)",D,SE); dot("\(A\)",A,NE); dot("\(B\)",B,W); dot("\(K\)",K,dir(15)); dot("\(L\)",L,dir(120)); dot("\(E\)",EE,S); dot("\(F\)",F,dir(280)); dot("\(I\)",I,dir(260)); dot("\(O\)",O,SE); label("\(\omega\)",I+inradius(P,C,D)*SE,E,qua); label("\(\Gamma\)",O+inradius(Q,A,B)*SE,SE,qua); label("\(\omega_A\)",IA+inradius(A,D,P)*dir(300),dir(330),tri); label("\(\omega_B\)",IB+inradius(B,C,P)*dir(150),dir(150),tri); label("\(\Gamma_A\)",I1+abs(I1-foot(I1,C,D))*dir(-30),dir(120),sec); label("\(\Gamma_B\)",I2+abs(I2-foot(I2,C,D))*dir(150),W,sec); [/asy]](http://latex.artofproblemsolving.com/9/1/a/91a76d73de2b9fbf7e04756f16f084899a80e741.png)
,
denote the centers of 
.
centered at 
tangent to 
,
,
.
,
;
,
.
;
,
.
,
m
as the argument will be similar for 
Let 
touch 
at 
respectively and 
touch 
It is evident that 
Now 
and we are don
be 
respectivel
be the exsimilicenter of 
be the exsimilicenter of 
and 
be the exsimilicenter of 
By definitio
(since 
is the line through the centers of 
)
to 
and 
We easily see that 
and 
Thus due to the mentioned theore
we see that 
is the insimilicenter of 
and 
therefore by 
it follows that 
and 
yields that 
and 
and 
We have that 
Therefore 
tells us that 
are collinea
.
are tangent at point 
,
be the excimilienters of 
and 
,
,
are collinear. Similarly 
are collinear.
,
are collinear, and moreover, 
.
.
are concurrent. However, since 
lies on 
,
)
are in perspective, so by Desargues' Theorem, we have that 
are collinear, so we are done!
by 
Then by the three homothety centers theorem center of external homothety 
lies on 
and analogously for center of external homothety 
Applying three homothety centers theorem again, we can conclude that points 
are respectively centers of external and internal homothety 
finally implying 
.
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.661033188860394, xmax = 9.796622553109685, ymin = -3.057994928096622, ymax = 6.113152768694817; /* image dimensions */
pen qqzzff = rgb(0,0.6,1); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1); pen qqqqcc = rgb(0,0,0.8); pen qqzzcc = rgb(0,0.6,0.8); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255);
draw((-1.5830860824502149,5.739481315434459)--(-2.692128000908574,0.507679826139392)--(-0.03245869884812526,-1.8102802942447922)--(2.9860163659991943,4.193898946519175)--cycle, linewidth(0.4) + xdxdff);
/* draw figures *//* special point *//* special point *//* special point */
draw(circle((-0.011215221393844288,1.707172869407228), 1.560930244173962), linewidth(0.4) + qqzzff); /* special point */
draw(circle((-0.7515164232739878,4.2798836439913295), 1.116173900077119), linewidth(0.4)); /* special point *//* special point */
draw(circle((-1.9888002577400228,0.7427281455339743), 0.6392961535831577), linewidth(0.4)); /* special point *//* special point */
draw((-1.5830860824502149,5.739481315434459)--(-2.692128000908574,0.507679826139392), linewidth(0.4) + xdxdff);
draw((-2.692128000908574,0.507679826139392)--(-0.03245869884812526,-1.8102802942447922), linewidth(0.4) + xdxdff);
draw((-0.03245869884812526,-1.8102802942447922)--(2.9860163659991943,4.193898946519175), linewidth(0.4) + xdxdff);
draw((2.9860163659991943,4.193898946519175)--(-1.5830860824502149,5.739481315434459), linewidth(0.4) + xdxdff);
draw((-2.692128000908574,0.507679826139392)--(2.9860163659991943,4.193898946519175), linewidth(0.4));
draw((-1.5830860824502149,5.739481315434459)--(-0.03245869884812547,-1.8102802942447926), linewidth(0.4)); /* special point *//* special point */
draw(circle((-0.6238467778454819,1.1988881998148306), 1.879983042437203), linewidth(0.4) + qqqqcc);
draw(circle((-0.0036057933673060358,2.9671269773510742), 2.120056515899163), linewidth(0.4) + qqzzcc); /* special point */
draw(circle((0.7434688451636409,1.6558376695206374), 3.122817022774604), linewidth(0.4) + yqqqyq);
draw((-0.7654371025350631,1.758476630468249)--(0.7434688451636409,1.6558376695206374), linewidth(0.4) + dotted);
draw((-1.5830860824502149,5.739481315434459)--(-0.44286019056493237,3.207234817383469), linewidth(0.4));
draw((-2.692128000908574,0.507679826139392)--(-1.4141946133352048,1.0229564906462136), linewidth(0.4));
draw((2.9860163659991943,4.193898946519175)--(-2.228813147027906,2.6933243019720856), linewidth(0.4));
draw((-2.228813147027906,2.6933243019720856)--(-0.03245869884812547,-1.8102802942447926), linewidth(0.4));
/* dots and labels */
label("$D$", (2.9471018676129757,4.137606802562069), NE * labelscalefactor);
label("$P$", (-2.301793984009765,2.6235022299785773), SW * labelscalefactor);
label("$I$", (-0.03784714690874773,1.6429392686864108), NE * labelscalefactor);
label("$\omega$", (0.30823389825319125,3.070523579979418), NE * labelscalefactor,qqzzff);
label("$K$", (-0.5137085840064138,3.142623797721489), NE * labelscalefactor);
label("$\omega_1$", (-1.2635508485239482,5.2623701993383785), NE * labelscalefactor);
label("$A$", (-1.6528920243311296,5.6805514622423905), NE * labelscalefactor);
label("$L$", (-1.451011414653332,0.979617265459357), NE * labelscalefactor);
label("$\omega_2$", (-2.2441138098161084,1.3401183541697124), NE * labelscalefactor);
label("$C$", (-0.12436740819923246,-1.8899714006750712), SE * labelscalefactor);
label("$B$", (-2.7632353775590173,0.431655610619617), SW * labelscalefactor);
label("$E$", (-0.8309495420715245,1.6717793557832392), NE * labelscalefactor);
label("$F$", (0.19287354986587826,1.6285192251379965), NE * labelscalefactor);
label("$\Omega_2$", (-1.8980327646541697,2.3927815332039497), NE * labelscalefactor,qqqqcc);
label("$\Omega_1$", (-0.758849324329454,4.757668675143881), NE * labelscalefactor,qqzzcc);
label("$O$", (0.6543149434151302,1.5996791380411681), NE * labelscalefactor);
label("$\gamma$", (0.19287354986587826,4.555788065466082), NW * labelscalefactor,yqqqyq);
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/* end of picture */[/asy]](http://latex.artofproblemsolving.com/e/9/2/e92cfff64dbbe813f0bcf017257b59098a4aa5b3.png)
and 
and 
respectively. Also, let 
be the circle tangent to rays 
and 
Notice 
Similarly, let 
and 
respectively.
and 
we see that 
lies on 
By Monge on 
we see that 
Hence, 
we see that 
lies on 
By Monge on 
Hence, 
where 
and 
are circumscriptible. Call their incircles 
and 
be 
and 
the exsimilicenters of 
and get that 
.
and get that 
.
and get that 
and 
are concurent.
and 
,
and similarly for 
and incircle of 
.
and 
be the insimilicenters and exsimilicenters of 
collinear
collinear
collinear
and 
.
collinear.![\[\triangle AUC, \overline{EIF}\]](http://latex.artofproblemsolving.com/6/e/e/6eef031520c9dbc606f281a377f82ccee8eee790.png)
![\[\triangle RAU, \overline{BVF}\]](http://latex.artofproblemsolving.com/0/2/5/025290fa9783ded168bbd149f79e8040ed7b2d90.png)
![\[\triangle RAC, \overline{EBD}\]](http://latex.artofproblemsolving.com/b/4/b/b4bbd0cf637b4fb2fee8376e8e0f6ed7539c64bb.png)
![\[\triangle RUC, \overline{DIV}\]](http://latex.artofproblemsolving.com/9/b/2/9b2132838186505fdf203a64f53d474265d337b3.png)
where the first one is what we want to prove, and the rest are known collinearities. These give equations![\[\frac{EA}{EC}\cdot \frac{IC}{IU}\cdot \frac{FU}{FA}=1\]](http://latex.artofproblemsolving.com/2/2/a/22a371b0878f76d826daca63bc51e66afbc7ebf6.png)
![\[\frac{BR}{BA}\cdot \frac{VU}{VR}\cdot \frac{FA}{FU}=1\]](http://latex.artofproblemsolving.com/f/f/7/ff77d2b9a7fb49dbb0a69495ec129712d5e60aee.png)
![\[\frac{EC}{EA}\cdot \frac{BA}{BR}\cdot \frac{DR}{DC}=1\]](http://latex.artofproblemsolving.com/c/9/5/c95edd955fc36292cb7aa7b269c6690a77f11959.png)
![\[\frac{DC}{DR}\cdot \frac{IU}{IC}\cdot \frac{VR}{VU}=1\]](http://latex.artofproblemsolving.com/f/9/2/f92e25475484e42075cd367c6ee713f88dd827ed.png)
Since these multiply to 
and three of them hold, so the fourth must also hold and 
is collinear.
and 
perspective, oops.
be the incircles of 
and 
,
be the centers of 
are concurrent, so applying Desargues's theorem on triangles 
and 
,
are collinear. Let 
.
are collinear. Thus it suffices to prove that the points 
are collinear. Let 
.
is the exsimilicenter of 
,
.
yields exsimilicenter of 
lies on line 
,
are collinear. Thus we're done. 
are inscribed quadrilateral.
we have it on 
we get insimiliscenter lies on 
is the exsimiliscenter
respectively. Then note that if we introduce 
,
and 
denote the incentres of 
and 
.
and 
are perspective, (due to perspector ![\[AI_1 \cap BI_2 \overset{\text{def}}{=} R, CI_1 \cap DI_2 \equiv I, AC \cap BD \equiv E\]](http://latex.artofproblemsolving.com/a/b/7/ab70835a84b283e5837c04477010fa4199b0cf21.png)
denote the pencil of lines through 
which swaps the following pairs ![\[(IA, II_2) \text{ and } (IB, II_1)\]](http://latex.artofproblemsolving.com/4/d/7/4d75327a446903858423c2e90ef62853135e0678.png)
From DDIT, this also swaps 
,
.
where we take two insimilicentres and one exsimilicenter, we obtain that 
and 
and 
,
.
,
collinear. Since 
collinear from before, we obtain that 
are collinear; which is precisely what we wanted to show.
,
.
,
,
,
be 
,
respectively.
to obtain that 
is the insimilicenter.
to get that the exsimilicenter of 
to get that 
as desired.
![[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(11.53303145358391cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
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[/asy]](http://latex.artofproblemsolving.com/9/f/d/9fdefb276e4d37dfd566e05d1fbf935953ef9a0f.png)
and let 
and 
.
and 
.
are collinear.
be 
respectively, now it's known that by a corolary of Pitot's theorem we get that there exists incircles 
of ![\[AD+DP-AP=2DK=CD+DP-PC \implies AD+PC=CD+AP\]](http://latex.artofproblemsolving.com/b/8/d/b8dbe8d368fbeb03ab9f638058ebce222bcd01ba.png)
And similarily for the other quadrilateral, so now by Monge d'Alembert 2 times+ notising the angle bisectors we get that 
are the insimillicenters of 
and 
respectively.
are concurrent at point 
we get that 
are colinear as desired thus we are done 
.
be the incenters of 
.
be eximilicenters of 
where 
.
are collinear.
give the result.
and 
.
be the common tangets of 
and 
are involutions thus, 
are collinear. Similarily 
are collinear.
are perspective which implies 
are collinear as desired
