[/quote]
,
is lovely if there exist a positive integer 
and (not necessarily distinct) positive integers 
,
,
,
such that 
and 
for 
.
,
be an odd prime. Prove that:![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
,
,
,
'![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
,
,
,
.
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
be a function such that
for all 
.
.
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
with integer coefficients such that 
and 
is a square of an integer for all nonnegative integers 
.
is defined as follows: 
if 
and 
if 
.
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
and 
such that 
.
is a benefactor of 
because 
.
benefactors.
denotes the least natural such that
Find all naturals 
such that 
.
square is divided into unit squares. Is it possible to fill each unit square with a number 
in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?
be an odd prime number. How many 
of 
are there, the sum of whose elements is divisible by 
be a natural number. The numbers 
are written in a row in some order. For each pair of adjacent numbers, their greatest common divisor (GCD) is calculated and written on a sheet. What is the maximum possible number of distinct values among the 
GCDs obtained?
Y by Davi-8191, mathematicsy, Adventure10, Mango247, and 1 other user
Circles 
and 
meet at 
and 
. Segments 
and 
are chords of and respectively, such that segment 
and ray 
meet at . Ray and segment meet at 
. Point 
lies on such that 
. Point 
lies on such that 
. Prove that points 
are collinear.
and 
meet at 
and 
. Segments 
and 
are chords of and respectively, such that segment 
and ray 
meet at . Ray and segment meet at 
. Point 
lies on such that 
. Point 
lies on such that 
. Prove that points 
are collinear.
and 
so 
so 
is cyclic quadrilateral. 
and 
so 
are collinear. Analogously 
are collinear
,
is a cyclic quad.
.
,
and 
are parallel, it follows that 
are collinear. Similarly, 
are collinear.
is the unique center of the spiral similarity sending 
to 
,
to 
;
and 
are concyclic. Let 
;![\[ \measuredangle CAB = \measuredangle XAB = \measuredangle XQB = \measuredangle Z'QB = \measuredangle Z'PB \implies PZ' \parallel AC \implies Z=Z' \]](http://latex.artofproblemsolving.com/a/f/d/afdd25fefbd842354fa970f2854228017510cd8c.png)
and 
is similar.![[asy]/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */
import olympiad;
import cse5;
size(11cm);
real lsf=0.8000;
real lisf=2011.0;
defaultpen(fontsize(10pt));
/* Initialize Objects */
pair A = (-2.0, 2.0);
pair B = (4.031165686870916, 3.133944005249443);
pair C = (1.0, -2.0);
pair D = (1.0, 0.0);
pair P = IntersectionPoint(A--B,Line(C,D,0,lisf));
pair X = IntersectionPoint(A--C,Line(B,D,0,lisf));
pair O_1 = circumcenter(A,P,C);
pair O_2 = circumcenter(P,D,B);
pair Q = 2*foot(P,relpoint(O_1--O_2,0.5-10/lisf),relpoint(O_1--O_2,0.5+10/lisf))-P;
pair Y_prime = (2)*(foot(O_1,X,Q))-Q;
pair Z_prime = (2)*(foot(O_2,Q,X))-Q;
/* Draw objects */
draw(circumcircle(A,P,C), rgb(0.7,0.8,0.9) + linewidth(1.0));
draw(circumcircle(B,P,D), rgb(0.7,0.8,0.9) + linewidth(1.0));
draw(A--B, rgb(0.4,0.4,0.4));
draw(C--P, rgb(0.4,0.4,0.4));
draw(A--C, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4"));
draw(B--X, rgb(0.5,0.5,0.5) + linetype("4 4"));
draw(P--Y_prime, rgb(0.5,0.5,0.5) + linetype("4 4"));
draw(P--Z_prime, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4"));
draw(Y_prime--Z_prime, rgb(0.4,0.8,1.0));
draw(circumcircle(C,Q,D), rgb(0.7,0.8,0.9) + linewidth(1.0));
draw(circumcircle(Q,A,B), rgb(0.7,0.8,0.9) + linewidth(1.0));
/* Place dots on each point */
dot(A);
dot(B);
dot(C);
dot(D);
dot(P);
dot(X);
dot(Q);
dot(Y_prime);
dot(Z_prime);
/* Label points */
label("$A$", A, lsf * dir(150));
label("$B$", B, lsf * dir(45));
label("$C$", C, lsf * dir(-90));
label("$D$", D, lsf * dir(60) * 1.414);
label("$P$", P, lsf * dir(100));
label("$X$", X, lsf * dir(225));
label("$Q$", Q, lsf * dir(-45));
label("$Y'$", Y_prime, lsf * dir(225));
label("$Z'$", Z_prime, lsf * dir(-45));
[/asy]](http://latex.artofproblemsolving.com/a/d/f/adfb0ab42556228a00682835a794b2f5831b70ab.png)
![\[ \langle CAB = \langle XAB = \langle XQB = \langle Z'QB = \langle Z'PB \implies PZ' \parallel AC \implies Z=Z' \]](http://latex.artofproblemsolving.com/d/e/1/de115cf864aeb20c2c327fedfbcd6a14ec2f87d3.png)
...
and 
we have 
.
![[asy]
/* Borrowed then adapted from v_enhance, thanks */
import olympiad;
import cse5;
size(8cm);
real lsf=0.8000;
real lisf=2011.0;
defaultpen(fontsize(11pt));
/* Initialize Objects */
pair A = (-2.0, 2.0);
pair B = (4.031165686870916, 3.133944005249443);
pair C = (1.0, -2.0);
pair D = (1.0, 0.0);
pair P = IntersectionPoint(A--B,Line(C,D,0,lisf));
pair X = IntersectionPoint(A--C,Line(B,D,0,lisf));
pair O_1 = circumcenter(A,P,C);
pair O_2 = circumcenter(P,D,B);
pair Q = 2*foot(P,relpoint(O_1--O_2,0.5-10/lisf),relpoint(O_1--O_2,0.5+10/lisf))-P;
pair Y= (2)*(foot(O_1,X,Q))-Q;
pair Z = (2)*(foot(O_2,Q,X))-Q;
/* Draw objects */
draw(circumcircle(A,P,C), green);
draw(circumcircle(B,P,D), blue);
draw(A--B);
draw(X--D--Q--C--cycle);
draw(Y--P--Z);
draw(X--A);
draw(C--P);
draw(D--B);
draw(Y--Z, dashed);
draw(circumcircle(X,D,C), red);
/* Place dots on and label each point */
dot(A); label("$A$", A, lsf * dir(150));
dot(B); label("$B$", B, lsf * dir(45));
dot(C); label("$C$", C, lsf * dir(-90));
dot(D); label("$D$", D, lsf * dir(160) * 1.414);
dot(P); label("$P$", P, lsf * dir(100));
dot(X); label("$X$", X, lsf * dir(225));
dot(Q); label("$Q$", Q, lsf * dir(-45));
dot(Y); label("$Y$", Y, lsf * dir(225));
dot(Z); label("$Z$", Z, lsf * dir(-45));[/asy]](http://latex.artofproblemsolving.com/7/e/6/7e6793366aa995aa8459ea6ee1c499a16b396b8c.png)
,
is cyclic. By Reim's theorem on 
and 
,
are collinear; by Reim's theorem on 
and 
are collinear, and the result follows.
is cyclic from Miquel, so we may proceed with Reim's theorem on 
and 
to obtain the result.
.
.![\[ \measuredangle ZQB = \measuredangle ZPB = - \measuredangle BPZ = \measuredangle ZPA = \measuredangle LPA =\measuredangle PAC =\measuredangle XAP \]](http://latex.artofproblemsolving.com/9/a/a/9aa8b95fa9b97bcb3660e3ec467ad6ff0d807f6b.png)
Now, ![\[ \measuredangle YQB = \measuredangle YQP + \measuredangle PQB = \measuredangle YCP + \measuredangle PJB = \measuredangle YCP + \measuredangle DPJ = \measuredangle YCP + \measuredangle CPY = - \measuredangle PYC = - \measuredangle PAC = \measuredangle XAP \]](http://latex.artofproblemsolving.com/c/0/5/c056205b23b357dbd087522706f6494a8d250459.png)
Finally observe that 
of the complete quadrilateral formed by the vertices 
.![\[ \implies XAQB concyclic. \]](http://latex.artofproblemsolving.com/6/5/4/654ce90f0ae2f858f6be59f1c56afe51239577fc.png)
.![\[ \measuredangle XQB = \measuredangle XAB = \measuredangle XAP \]](http://latex.artofproblemsolving.com/8/2/2/8221a0e40f66bde2d4b1484702a5c7c7dbe63341.png)
.![\[ \measuredangle ZQB = \measuredangle YQB = \measuredangle XQB \]](http://latex.artofproblemsolving.com/1/c/6/1c6e4fa710ae1ed0d8af41ddbfe4ed7062667a77.png)
This leads us to the desired conclusion.
is cyclic. If 
intersects 
,
and 
are both antiparallel to 
so 
,
,
lie on 
,
and 
are cyclic.
.
.
so 
,
.
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15.cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.716666666666658, xmax = 17.6, ymin = -1.7266666666666675, ymax = 8.073333333333334; /* image dimensions */
pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen ffqqff = rgb(1.,0.,1.);
/* draw figures */
draw(shift((3.185458267893191,3.1111558596963573)) * scale(3.1559227963577987, 3.1559227963577987)*unitcircle, linewidth(1.6) + red);
draw(shift((5.7380030468626,4.431102969733766)) * scale(2.628399180397597, 2.628399180397597)*unitcircle, linewidth(1.6) + blue);
draw((3.8,6.206666666666669)--(5.453214154262895,0.9163593004666906), linewidth(1.6) + red);
draw((3.8,6.206666666666669)--(2.4181126438879694,0.04994214780186379), linewidth(1.6) + red);
draw((5.155831433496555,1.8679879773650003)--(3.2270742084802806,3.6541096492381664), linewidth(1.6) + blue);
draw((2.4181126438879694,0.04994214780186379)--(5.453214154262895,0.9163593004666906), linewidth(1.6) + red);
draw((3.8,6.206666666666669)--(6.446607586458077,6.962182158139266), linewidth(1.6) + xdxdff);
draw((6.224733389568109,3.9612472849787848)--(3.8,6.206666666666669), linewidth(1.6) + ffxfqq);
draw((5.453214154262895,0.9163593004666906)--(6.011390333566628,1.075699412915823), linewidth(1.6) + linetype("4 4") + red);
draw((6.011390333566628,1.075699412915823)--(5.155831433496555,1.8679879773650003), linewidth(1.6) + linetype("4 4") + blue);
draw((6.011390333566628,1.075699412915823)--(6.446607586458077,6.962182158139266), linewidth(1.6) + dotted + green);
draw((5.155831433496555,1.8679879773650003)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff);
draw((3.2270742084802806,3.6541096492381664)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff);
draw(shift((3.916839652865845,1.6064204600289003)) * scale(2.1607423224144324, 2.1607423224144324)*unitcircle, linewidth(1.6) + linetype("2 2") + ffqqff);
draw(shift((5.593442797870414,1.4824607634596496)) * scale(0.5832108871504142, 0.5832108871504142)*unitcircle, linewidth(1.6) + linetype("2 2") + ffqqff);
draw((5.453214154262895,0.9163593004666906)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + red);
label("X = X'",(6.133333333333323,1.0566666666666664),SE*labelscalefactor,green);
draw((2.4181126438879694,0.04994214780186379)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff);
/* dots and labels */
dot((3.8,6.206666666666669),red);
label("$P$", (3.783333333333327,6.506666666666668), NE * labelscalefactor,red);
dot((6.06666666666666,1.823333333333333),red);
label("$Q$", (6.2,1.6233333333333333), NE * labelscalefactor,red);
label("$\omega_1$", (1.1333333333333315,5.906666666666667), NE * labelscalefactor,red);
label("$\omega_2$", (5.4,7.24), NE * labelscalefactor,blue);
dot((5.453214154262895,0.9163593004666906),red);
label("$A$", (5.4,0.573333333333333), NE * labelscalefactor,red);
dot((5.155831433496555,1.8679879773650003),blue);
label("$B$", (4.833333333333325,1.69), NE * labelscalefactor,blue);
dot((2.4181126438879694,0.04994214780186379),red);
label("$C$", (2.2833333333333297,-0.2933333333333338), NE * labelscalefactor,red);
dot((3.2270742084802806,3.6541096492381664),blue);
label("$D$", (2.966666666666662,3.6233333333333335), NE * labelscalefactor,blue);
dot((6.446607586458077,6.962182158139266),xdxdff);
label("$Z$", (6.516666666666656,7.123333333333334), NE * labelscalefactor,xdxdff);
dot((6.224733389568109,3.9612472849787848),ffxfqq);
label("$Y$", (6.4,3.923333333333334), NE * labelscalefactor,ffxfqq);
dot((6.011390333566628,1.075699412915823),green);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* re-scale y/x */
currentpicture = yscale(1.1386054421768685) * currentpicture;
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/0/c/4/0c4cb3bb54c52edf39ee09850908d166390417e6.png)
are collinear 
is cyclic.
is cyclic.
are collinear.
are collinear.
are collinear.
are collinear. 
.
.
Now observe that 
,
,
is cyclic. We now have
Hence 
,
are collinear, as desired. 
is the Miquel Point of the quadrilateral 
,
t
,
,
an
,
are collinear.
an
quadrilateral 
we get that 
and 
are collinear. Similarly 
.
so 
are collinear.
so 
are collinear.
.
,![\[\measuredangle Y'PC=\measuredangle Y'QC=\measuredangle XQC=\measuredangle XDC,\]](http://latex.artofproblemsolving.com/8/0/6/806efddbfde6eac0f8d6acca60f8c0ede3445aa8.png)
hence 
denote the image of point 
under the given inversion. Then our problem is
are points in a plane such that 
.
and 
intersect at 
.
.
intersect 
at 
respectively. Prove that 
is cyclic
Hence 
is cyclic.
Hence 
is cyclic. Similarly we can conclude that 
is cyclic. Hence we are done.
and 
are collinear. Note that![\[\measuredangle QZP=\measuredangle QDP=\measuredangle QDC=\measuredangle QXC=\measuredangle QXA.\]](http://latex.artofproblemsolving.com/b/7/2/b72a87b0806ceb2ff979e99cd3c01724db2a1ca5.png)
Since 
and 
are parallel, 
are collinear. We have![\[\measuredangle QXB=\measuredangle QAB=\measuredangle QAP=\measuredangle QYP.\]](http://latex.artofproblemsolving.com/4/b/2/4b2518ed5b8c130e1786460e7cdc5a0b116fd5a0.png)
Since 
and 
are parallel, 
is cyclic.
.
.
.
and 
.
,
are collinear.
which obviously suffices. 
.
.
as desired. 
implying 
are collinear. Combining the two collinearities yields the desired result. 
Let 
Since 
and 
are cyclic, 
so 
Thus, 
Similarly, 
is also on 
denote the directed angle of 
This spiral similarity also takes 
Since 
and 
and 
In other words, 
and 
and 
so 
and 
so 
and 
implies 
and 
implies 
The result follows.
were collinear and my goodness, how badly stuck was I back then... 
But fortunately, spiral sim gotchme! 
and 
,
.
are cyclic. Now we have 
which gives that 
are collinear and similarly 
are also collinear and we are done.
and 
are similar
and 
are similar, and angles 
and 
,
then 
are collinear, but 
because 
,
to prove 
,
and 
.![\[ \angle QYP = \angle QAP = \angle QAB = \angle QXB. \]](http://latex.artofproblemsolving.com/3/e/e/3ee4f2be240e7c04c4d44340ca2118c9df30d7a9.png)
Since 
,![\[ \angle QZP = \angle QBP = \angle QBA = \angle QXA, \]](http://latex.artofproblemsolving.com/3/4/c/34c8207634fc63858c1be8be20186967b12e490e.png)
so points 
