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.
be positive integers.Define injective function 
from 
to itself such that 
.Prove that 
divides 
.
board that can be colored so that for each square in the board, at most one square adjacent to it is colored.
such that 
.
that divides 
for the positive integer
. Prajit then tries to place n points such that no four points are concyclic and no 
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
of positive integers such that 
in all of this cases :
of positive integers such that : 
3-Every prime number less than 
can be represented in the form 
by taking an Appropriate pair cannot be represented in the form where 
and are positive integers
. The number 
is the least prime number that cannot be represented as a sum or an difference of a power of and a power of . Prove that the number cannot be represented in any of the forms 
, 
, where and are positive integers
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
satisfying
of injective functions from the set of real numbers to itself satisfying
and 
. (We say a function 
is injective if 
for any distinct real numbers 
and 
.)
, find the smallest integer 
such that any graph with vertices and at least edges must contain at least two disjoint cycles (cycles with no common vertices).
, 
and 
intersect at 
, 
and 
intersect at 
. Let 
be the circumcenters of 
and 
, respectively. Prove that the 
, 
and 
are concurrent.
, and intersect at , and intersect at . Let be the circumcenters of and , respectively. Prove that the , and are concurrent.
is given by 
Similarly 
.
and 
is 
and Equation of 
is 
can be found by the determinant
which comes out as 
(I hope I haven't made any mistake in calculations 
)
and we obtain 
and are collinear.
. That leads us to 
and 
being collinear.
. Consider the inversion 
at 
fixing the circle. Angle chase shows 
(where 
), i.e. 
. So just swaps 
and 
, implying points 
are collinear. 
![[asy]
size(200);
pair A=dir(110),B=dir(50),C=dir(-20),D=dir(-60),E=dir(-140),F=dir(170),T=extension(F,C,E,B),G=extension(A,B,C,D),H=extension(A,F,E,D),N=circumcenter(F,H,E),M=circumcenter(G,B,C);
draw(unitcircle);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$G$",G,dir(G));
dot("$H$",H,dir(H));
dot("$T$",T,dir(-90));
dot("$M$",M,dir(M));
dot("$N$",N,dir(N));
draw(A--G--D--H--A,red);
draw(E--B^^F--C,blue);
draw(N--F--E--N^^M--B--C--M,brown);
draw(N--M,dotted);
[/asy]](http://latex.artofproblemsolving.com/c/5/2/c52f56adf9f9904e20bd3251748765a16d0cf962.png)
.
we will mean angle subtended by minor arc 
on circumference.
Since 
and 
, so our Claim follows. 
and only keep in mind that 
lie on perpendicular bisectors of segments 
(respectively) and our Claim 1 is satisfied.
, 
, 
and 
be circumcenter of 
. Observe 
as 
![[asy]
size(200);
pair F=dir(110),B=dir(70),C=dir(0),E=dir(-150),O=(0,0),T=extension(E,B,C,F),N=1.2*(E+F),M=extension(O,1/2*(B+C),N,T),K=extension(N,F,M,B),X=extension(O,M,F,C),Y=extension(O,N,E,B);
draw(unitcircle);
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$O$",O,dir(-90));
dot("$N$",N,dir(N));
dot("$M$",M,dir(M));
dot("$T$",T,dir(T));
dot("$K$",K,dir(K));
dot("$X$",X,dir(-90));
dot("$Y$",Y,dir(-90));
draw(E--F--C--B--E^^F--Y^^B--X,red);
draw(B--O--N--K--M--O--F,brown);
draw(circumcircle(F,O,B),blue);
draw(N--M,dotted);
[/asy]](http://latex.artofproblemsolving.com/4/9/e/49ef1258ab29b6104a9135e928f52c7bed6461fd.png)
.
proving our Claim. 
w.r.t. 
, we obtain points are collinear, as desired. 
and 
be the antipodals of 
and 
respectively, and let 
and 
Then and 
are the midpoints of 
and 
respectively.![[asy]
import geometry;
import graph;
size(250);
pair A,B,C,D,E,F,G,H, Ar,Dr,P,Q,M,N,R;
A = dir(130);
B=dir(73);
C = dir(11.1);
D = dir(-70);
E=dir(-115.69);
F = dir(179.97);
G = extension(A,B,C,D);
H = extension(A,F,D,E);
Ar = -A;
Dr = -D;
Q = extension(F,Ar,E,Dr);
P = extension(B,Ar,C,Dr);
M= P*0.5+G*0.5;
N=Q*0.5+H*0.5;
R=extension(B,E,C,F);
draw(unitcircle);
draw(circumcircle(H,F,E),blue);
draw(circumcircle(B,C,G),green);
draw(A--H);
draw(D--H);
draw(A--G);
draw(C--G);
draw(H--Q, dotted+cyan);
draw(G--P, dotted + cyan);
draw(Q--P, dotted + red);
draw(G--H, dotted + red);
draw(E--Dr, dotted +magenta);
draw(F--Ar, dotted +magenta);
draw(B--Ar, dotted +magenta);
draw(C--Dr,dotted +magenta);
draw(B--E);
draw(C--F);
dot("$R$", R, dir(R));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(-125));
dot("$D$", D, dir(D));
dot("$E$", E, dir(-70));
dot("$F$", F, dir(135));
dot("$M$", M, dir(135));
dot("$N$",N, dir(-80));
dot("$A^{*}$", Ar, dir(Ar));
dot("$D^{*}$", Dr, dir(Dr));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$G$", G, dir(G));
dot("$H$", H, dir(H));
[/asy]](http://latex.artofproblemsolving.com/0/e/0/0e0dbc40e4f5f0181ed5cd96f83f08c9c079979f.png)
Applying Pascal's Theorem at yelds that 
and are colinear. On the other hand, by applying Pascal's Theorem again at 
we get that 
and 
are colinear.
and are colinear, it is sufficient(and necessary) to show that 
That is just angle chasing; we have
similarly,
Hence 
and indeed 
and we shall get 
and as collinear.
. That leads us to 
and being collinear.
and are concurrent.
and let be the center of the circle, so our goal is to show 
collinear. We use moving points. Begin by keeping all points but fixed, and extend all definitions to degenerate cases through continuity. projectively on the circle, we define to be the intersection of the perpendicular bisector of 
and of 
. Note that varies projectively, so the perpendicular bisector of varies projectively in the pencil through 
, so varies projectively on the perpendicular bisector of . A similar argument shows that varies projectively on the perpendicular bisector of 
. Thus, 
and 
both vary projectively on the pencil through 
, so it suffices to check the problem for three values of .
, as 
. Therefore, it suffices to solve it when 
, as the case 
will then follow by symmetry. Thus, we have entirely reduced to the case of . reads as follows. Let 
be a cyclic quadrilateral with center , and let . Let 
vary on the circle. Let be the intersection of the perpendicular bisector of and the line through 
perpendicular to 
. Let 
and let be the intersection of the perpendicular bisectors of and 
. Show that 
collinear. projectively. By the same logic as before, varies projectively, and varies projectively as the line through perpendicular to varies projectively on the pencil through . Thus, it suffices to check three values of .
is trivial, since . When 
, we see that 
and is the circumcenter of 
(as 
), so the problem is true by the fact that 
and that the circumcenter and orthocenter are isogonal conjugates.
, we see that is the intersection of the line through perpendicular to and the perpendicular bisector of , and is the intersection of the line through perpendicular to and the perpendicular bisector of . Thus, by the same similarity idea as above, the problem reduces to the following:
be a triangle, and let be the intersection of the line through perpendicular to 
and the perpendicular bisector of , and let 
be defined similarly with 
and swapped. Show and 
are isogonal in 
.
inversion at , since the circle through and tangent to 
gets mapped to the circle through and tangent to , so their corresponding circumcenters, i.e. and , are isogonal, as desired.
denote directed angles modulo 
.![\[\measuredangle BGC+\measuredangle EHF=\measuredangle BAF+\measuredangle EDC=\measuredangle BCX+\measuredangle XBC=\measuredangle BXC.\]](http://latex.artofproblemsolving.com/2/7/5/275048cf6f1e19eee0759d81b8e410234e2abd37.png)
Since 
, we can compose a reflection over a line through and a dilation at such that the resulting transformation maps to and 
to . Suppose that this transformation maps to . It suffices to show that 
and 
are isogonal with respect to . Observe the following lemma.
and be points on the perpendicular bisector of 
. Then, and are inverses with respect to the circumcircle of 
if and only if 
and 
are isogonal with respect to 
. be the circumcenter of . For the if direction, we will prove that 
, which suffices. Since the -altitude and 
are isogonal with respect to , the directed angle between the -altitude and is equal to the directed angle between and . Since 
is parallel to the -altitude, we have , as desired. The converse is clear by uniqueness. is the circumcenter of 
, we have ![\[\measuredangle BMO=\measuredangle BGC=\measuredangle BXC+\measuredangle FHE=\measuredangle BOM+\measuredangle ON'B=\measuredangle OBN',\]](http://latex.artofproblemsolving.com/f/b/7/fb71f22697dee0066b0f2bd6aaec87b565fd8672.png)
so and are inverses with respect to the circumcircle of , as desired. 
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
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 = -11.846109032258052, xmax = 15.708729054374183, ymin = -7.27546163156413, ymax = 6.765737157794893; /* image dimensions */
pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ffvvqq = rgb(1,0.3333333333333333,0);
/* draw figures */
draw(circle((0.45269760075919907,-1.8833432923747244), 3.7801918556637735), linewidth(2));
draw(circle((0.45269760075919996,-1.8833432923747244), 3.7801918556637726), linewidth(0.4));
draw((xmin, -0.4614874235435215*xmin-5.131959835926818)--(xmax, -0.4614874235435215*xmax-5.131959835926818), linewidth(0.4)); /* line */
draw((xmin, 0.8516129032258101*xmin + 2.3321262456524723)--(xmax, 0.8516129032258101*xmax + 2.3321262456524723), linewidth(0.4)); /* line */
draw((xmin, -0.45745723368333807*xmin + 1.3680219871147685)--(xmax, -0.45745723368333807*xmax + 1.3680219871147685), linewidth(0.4)); /* line */
draw((xmin, 0.5740907879410247*xmin-6.218526721815619)--(xmax, 0.5740907879410247*xmax-6.218526721815619), linewidth(0.4)); /* line */
draw((-2.900396276482841,-0.13788864786840901)--(-2.774749537546841,-3.851447820865748), linewidth(0.4));
draw((3.9451313646220614,-0.43670689346261216)--(3.375447508457395,-4.280713402031743), linewidth(0.4));
draw(circle((-3.6958844426792545,-2.0237088502354235), 2.0467338024799657), linewidth(0.4));
draw(circle((5.03418858344353,-2.5623227488821447), 2.3883652130019946), linewidth(0.4) + fuqqzz);
draw((-3.6958844426792545,-2.0237088502354235)--(5.03418858344353,-2.5623227488821447), linewidth(0.4));
draw((-2.900396276482841,-0.13788864786840901)--(3.375447508457395,-4.280713402031743), linewidth(0.4));
draw((-2.774749537546841,-3.851447820865748)--(3.9451313646220614,-0.43670689346261216), linewidth(0.4));
draw((-3.6958844426792545,-2.0237088502354235)--(0.45269760075919996,-1.8833432923747244), linewidth(0.4));
draw((0.45269760075919996,-1.8833432923747244)--(5.03418858344353,-2.5623227488821447), linewidth(0.4));
draw((-0.7364802170295132,1.7049301898595939)--(1.0492369130972348,-5.616169475638818), linewidth(0.4));
draw((xmin, 0.03929459568935406*xmin-2.2853532031665322)--(xmax, 0.03929459568935406*xmax-2.2853532031665322), linewidth(0.4)); /* line */
draw((-5.684322766062706,-2.5087163680267555)--(-3.6958844426792545,-2.0237088502354235), linewidth(0.4));
draw((5.03418858344353,-2.5623227488821447)--(7.354527903590921,-1.9963600027088542), linewidth(0.4));
draw(circle((2.694637471341226,1.5129057014699254), 5.8334830610106), linewidth(0.4) + ffvvqq);
/* dots and labels */
dot((-0.7364802170295132,1.7049301898595939),dotstyle);
label("$A$", (-0.6456152002838303,1.9162479805451729), NE * labelscalefactor);
dot((-2.900396276482841,-0.13788864786840901),dotstyle);
label("$B$", (-2.8167254177136165,0.06978975824507017), NE * labelscalefactor);
dot((-2.774749537546841,-3.851447820865748),dotstyle);
label("$C$", (-2.6949809195399834,-3.6434174360507408), NE * labelscalefactor);
dot((1.0492369130972348,-5.616169475638818),dotstyle);
label("$D$", (0.9370632759733968,-5.286968161394788), NE * labelscalefactor);
dot((3.375447508457395,-4.280713402031743),dotstyle);
label("$E$", (3.4531162382284757,-4.069523179658457), NE * labelscalefactor);
dot((3.9451313646220614,-0.43670689346261216),dotstyle);
label("$F$", (4.021257229705429,-0.23457148718901272), NE * labelscalefactor);
dot((-5.684322766062706,-2.5087163680267555),linewidth(4pt) + dotstyle);
label("$G$", (-5.596558126011566,-2.3448094555319874), NE * labelscalefactor);
dot((7.354527903590921,-1.9963600027088542),linewidth(4pt) + dotstyle);
label("$H$", (7.430103178567149,-1.837540713141849), NE * labelscalefactor);
dot((-3.6958844426792545,-2.0237088502354235),linewidth(4pt) + dotstyle);
label("$M$", (-3.60806465584223,-1.8578314628374546), NE * labelscalefactor);
dot((5.03418858344353,-2.5623227488821447),linewidth(4pt) + dotstyle);
label("$N$", (5.116957713268125,-2.4056817046188037), NE * labelscalefactor);
dot((0.45269760075919996,-1.8833432923747244),linewidth(4pt) + dotstyle);
label("$O$", (0.5312482820612873,-1.715796214968216), NE * labelscalefactor);
dot((0.33291750702024897,-2.272271344330264),linewidth(4pt) + dotstyle);
label("$I$", (0.12543328814917776,-2.4259724543144094), NE * labelscalefactor);
dot((-1.6755196242868575,-2.3511920693724626),linewidth(4pt) + dotstyle);
label("$J$", (-1.5992804359772876,-2.1824834579671433), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/6/e/0/6e06cd9ba58138fff8af375ce10bc2f334d83f41.png)
, so 
is cyclic
with radius 
the small circles swap and hence we are done.
means that is on 
. Define 
, now![\[\measuredangle FYG=\measuredangle FYH=\measuredangle FEH=\measuredangle FCD=\measuredangle FCG\]](http://latex.artofproblemsolving.com/b/6/5/b654a2dafed548b431423190b49c218446dcefa7.png)
hence 
concyclic. with radius![\[\sqrt{XF\cdot XC}=\sqrt{XB\cdot XE}=\sqrt{XY\cdot XG}\]](http://latex.artofproblemsolving.com/9/0/3/903029c375dbc857f4bf015d4daa53ecc32df332.png)
hence circles 
and 
are swapped, so the circumcenters are collinear with . Done!
and . Then, let the former intersect and again at 
and 
respectively. By Reims, 
and are parallel.
, and similarly, 
, and so the two circles are homothetic, with homotethy centre 
, and we are done.
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
be real positive numbers such that: 
.
![\[
\text{Find all non-negative integers } (a, b, c, n) \text{ such that}
\]](http://latex.artofproblemsolving.com/9/5/9/9591f7c842d1b6afcba11b15b3ea3f9e01bc937a.png)
![\[
2^a + 3^b + 5^c = n!
\]](http://latex.artofproblemsolving.com/b/c/6/bc6b0754dd874f646fcab124c91583b620db273c.png)
be the point of intersection of 
are collinear.
,
are collinear.![\[\angle AHN=90^\circ-\angle HEF=90^\circ-\angle DAF\]](http://latex.artofproblemsolving.com/4/6/8/468ccd5528a3f3cc63b01ae0f496a75c0b5dd7e3.png)
so that 
and similarly 
so that 
.
since then we get that 
and 
are similar from where the claim immediately follows.
jointly any more):![\[\frac{HN}{HS}=\frac{FN}{HS}=\frac{FN/FH}{HS/FH}=\frac{\sin \angle AHN/\sin 2\angle DAF}{\sin \angle AFC/\sin \angle HSF}\]](http://latex.artofproblemsolving.com/7/0/8/708725c851cf9fed997fa371c1ba2c3ffa4864c3.png)
![\[=\sin \angle HSF \cdot \frac{\cos \angle DAF }{\sin \angle AFC\sin 2\angle DAF}=\sin \angle HSF \cdot \frac{1}{2\sin \angle DAF \sin \angle AFC}\]](http://latex.artofproblemsolving.com/c/6/0/c608bd728105e031b01d20246e53664e21ebd71d.png)
and similarly![\[\frac{GM}{GS}=\sin \angle GSC \cdot \frac{1}{2\sin \angle ADC \sin \angle DCF}\]](http://latex.artofproblemsolving.com/6/e/6/6e6c57b88d0dd79e28d481bbff44dd18748dfaf2.png)
from where the claimed equality immediately follows since 
and 
and 
.
,
,
,
.
.
and similar expressions, we get 
.
in the earlier equation, we get the desired result.
by Pascal on 
Notice 
Then by LoS on 
and extended LoS on 
as 
and 
Hence, 
,
,
and 
,
is the homothetic center of 
and 
.
.
be the circumcentre of 
.
are collinear. A similar inversion at 
are collinear. In particular, 
.
concur, therefore it suffices to show that 
concur or that triangles 
and 
are perspective. From Desargue's, this is equivalent to having 
and 
collinear. Now observe that by parallel-antiparallel stuff, 
both are perpendicular to 
.
.
are antiparallel in 
.