AoPS Academy
+1 w
be a convex pentagon with 
and 
For each point 
in the plane, define 
The least possible value of 
can be expressed as 
where 
and 
are positive integers and 
is not divisible by the square of any prime. Find 
be the set of positive integers. A function 
satisfies the equation ![\[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]](http://latex.artofproblemsolving.com/2/e/7/2e73ff2f75b103e9d6a4004a42e4ed7f4ee64a67.png)
for all positive integers . Given this information, determine all possible values of 
.
denote the set of positive integers. Find all functions 
such that for positive integers 
and 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of , how many orderly colorings are there?
is selected inside acute 
so that 
and 
. Point 
is chosen on ray 
so that 
. Let 
be the midpoint of 
.
is tangent to the circumcircle of triangle 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
and 
are erected outside an acute triangle 
Suppose that ![\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]](http://latex.artofproblemsolving.com/3/3/7/3378f0c4409d14bc0afddf2616992c13f624be3b.png)
Prove that lines 
and 
are concurrent.
and 
meet again at 
. belongs to the circumcircle of .
hence belongs to circumcircle of 
as well 
. Now,
so belongs to 
and similarly it belongs to and 
, so we are done.
![[asy]
size(8cm); defaultpen(fontsize(9pt)); pair A = dir(120), B = dir(180+30), C = dir(-30), P = (A+B+C)/2, D = circumcenter(B,P,C), E = circumcenter(A,P,C), F = circumcenter(A,P,B), B2=2*D-C, C1=2*D-B, C2=2*E-A, A1=2*E-C, A2=2*F-B, B1=2*F-A; draw(A--B--C--cycle, red+linewidth(1)); draw(A--A2--B1--B--B2--C1--C--C2--A1--A, lightblue); draw(circumcircle(B, P, C)^^circumcircle(A, P, C)^^circumcircle(A, P, B), deepcyan); draw(A1--B2, orange+dashed); string[] names = {"$A$", "$B$", "$C$", "$A_1$", "$B_1$", "$C_1$", "$A_2$", "$B_2$", "$C_2$", "$P$"}; pair[] points = {A, B, C, A1, B1, C1, A2, B2, C2, P}; pair[] ll = {dir(110), B, dir(-25), A1, B1, C1, A2, B2, C2, dir(-90)}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i]));
[/asy]](http://latex.artofproblemsolving.com/d/7/1/d718bfab968437a9e7d36319eb70aed160ac0a6b.png)
, 
, and concur at a point .
and 
intersect again at 
. We will show that also lies on 
. Indeed, we have![\[
\angle APC = 180^\circ - \angle CA_1A \quad \text{and} \quad \angle BPA = 180^\circ - \angle AB_1B.
\]](http://latex.artofproblemsolving.com/5/4/d/54d37261cab37ac4dcfe15a4b5cdda3df970546a.png)
Therefore,
as claimed. 
are collinear; similarly, so are 
and 
.
and analogous angle chases prove the other two collinearities. is our desired point of concurrency, so we are done. 
it's actually than you would expect though 
![[asy]
defaultpen(fontsize(10pt));
dotfactor*=1.2;
size(300);
pair A,B,C,A1,A2,B1,B2,C1,C2,X;
A = dir(120);
B = dir(210);
C = dir(330);
A1 = extension(dir(150),A,dir(10),C);
C1 = extension(dir(-70),B,dir(30),C);
X = intersectionpoints(circumcircle(A,C,A1),circumcircle(B,C,C1))[0];
A2 = intersectionpoints(C1--extension(C1,X,(10,10),(-10,10)),circumcircle(A,B,X))[1];
B1 = rotate(180,(B+A2)/2)*A;
B2 = rotate(180,(B+C1)/2)*C;
C2 = rotate(180,(C+A1)/2)*A;
pen r = red;
filldraw(A--B--C--cycle,r+white+white+white+white,r);
filldraw(B--B2--C1--C--cycle,r+white+white+white+white+white+white+white,r); filldraw(A--A1--C2--C--cycle,r+white+white+white+white+white+white+white,r); filldraw(A--A2--B1--B--cycle,r+white+white+white+white+white+white+white,r);
draw(circumcircle(A,B,B1)^^circumcircle(A,C,A1)^^circumcircle(B,C,C1),orange+orange+red);
draw(A1--B2^^B1--C2^^A2--C1,red+linewidth(1.05)); draw(A--X^^B--X^^C--X,fuchsia); draw(A--A2--B1--B--B2--C1--C--C2--A1--A, r);
dot("$A$",A,dir(125));
dot("$B$",B,dir(215));
dot("$C$",C,dir(315));
dot("$A_1$",A1,dir(A1));
dot("$B_1$",B1,dir(B1));
dot("$C_1$",C1,dir(C1));
dot("$X$",X,dir(280));
dot("$A_2$",A2,dir(A2));
dot("$B_2$",B2,dir(B2));
dot("$C_2$",C2,dir(C2));
[/asy]](http://latex.artofproblemsolving.com/e/4/5/e457b391fb976943787f57fa637bf4cffff413f9.png)
concur at a point . and be . Note that 
and 
, implying that 
.
.
, implying the collinearity.
, as needed.
, 
, and 
, then the angle condition implies that 
(complex bash didn't actually pan out for me lol i went with a synthetic solution).
, lies on the circumcircles of , 
, and . We first prove that the three circumcircles are indeed concurrent, and then show that the concurrency implies the problem.![[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 = -16.852914910450526, xmax = 34.390095350576566, ymin = -8.12314938684212, ymax = 18.42519130633257; /* image dimensions */
pair A = (3.,8.), B = (2.,0.), C = (10.,0.), B_2 = (2.,-4.), C_1 = (10.,-4.), C_2 = (17.908680279445264,6.920095244514606), A_1 = (10.908680279445264,14.920095244514606), B_1 = (-0.7171913168911064,0.33964891461138835), A_2 = (0.2828086831088937,8.33964891461139), P = (4.993661449478577,2.357439927495782);
/* draw figures */
draw(A--B, linewidth(1.));
draw(B--C, linewidth(1.));
draw(C--A, linewidth(1.));
draw(B--B_2, linewidth(1.));
draw(B_2--C_1, linewidth(1.));
draw(C_1--C, linewidth(1.));
draw(A--A_1, linewidth(1.));
draw(A_1--C_2, linewidth(1.));
draw(C_2--C, linewidth(1.));
draw(B_1--B, linewidth(1.));
draw(A--A_2, linewidth(1.));
draw(A_2--B_1, linewidth(1.));
draw(circle((1.141404341554447,4.169824457305695), 4.257302257233597), linewidth(1.));
draw(circle((6.,-2.), 4.47213595499958), linewidth(1.));
draw(circle((10.454340139722632,7.460047622257303), 7.473870181432778), linewidth(1.));
draw(A_2--C_1, linewidth(1.));
draw(B_2--A_1, linewidth(1.));
draw(B_1--C_2, linewidth(1.));
draw(B--P, linewidth(1.));
draw(P--A, linewidth(1.));
draw(P--C, linewidth(1.));
/* dots and labels */
dot(A,dotstyle);
label("$A$", (2.3,8.465636277058753), NE * labelscalefactor);
dot(B,dotstyle);
label("$B$", (2,-0.8), NE * labelscalefactor);
dot(C,dotstyle);
label("$C$", (9.852516556056544,0.3911705277736686), NE * labelscalefactor);
dot(B_2,dotstyle);
label("$B_2$", (0.9,-4.6), NE * labelscalefactor);
dot(C_1,dotstyle);
label("$C_1$", (10.166698102721334,-4.415807136197607), NE * labelscalefactor);
dot(C_2,dotstyle);
label("$C_2$", (18.398254625338808,6.674801461069453), NE * labelscalefactor);
dot(A_1,linewidth(4.pt) + dotstyle);
label("$A_1$", (10.763643041384432,15.3), NE * labelscalefactor);
dot(B_1,linewidth(4.pt) + dotstyle);
label("$B_1$", (-1.395182814542903,-0.6), NE * labelscalefactor);
dot(A_2,linewidth(4.pt) + dotstyle);
label("$A_2$", (-0.20129293721670466,8.81123597839002), NE * labelscalefactor);
dot(P,linewidth(4.pt) + dotstyle);
label("$P$", (4.6,1.3), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/8/f/3/8f3de0395c93b8cd5e77f90564f749eaadfa5e29.png)
be 
, and we wish to show that does indeed lie on 
. But note that
which is what we wanted.
is collinear. But this is clear from![\[\angle A_2PB=\angle BPC_1=90^\circ\]](http://latex.artofproblemsolving.com/9/a/6/9a62bf9ef787c8833ea715a426bf3e4c8891e6d7.png)
Similarly, lies on 
and 
, so we are done. as the intersections of 2 circles, and the way you prove the cyclic is super obvious by the angle conditions. After that, it's just 2 90 degree angles, and you win.
, thinking I'd get a bit of partial. After writing it, I realized that I had a proof and, with 20 minutes left, proceeded to write the rest of the solution.
![[asy]
import olympiad;
unitsize(13);
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(0cm);
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 = -10.252427827191417, xmax = 15.150472936867494, ymin = -3.4167712881064825, ymax = 11.241299387760378; /* image dimensions */
pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
/* draw figures */
draw(circle((0.2801492142784242,1.5), 4.481884633503346));
draw(circle((3.2738086301390403,7.2491118286162095), 3.3297516312587083));
draw((-3.454001382502029,3.9785901614824524)--(4,4));
draw((4,4)--(4,-1));
draw((4,-1)--(-3.420838326674272,-1.0456127964221371));
draw((-3.420838326674272,-1.0456127964221371)--(-3.454001382502029,3.9785901614824524));
draw((-3.454001382502029,3.9785901614824524)--(0.06810107679554245,8.14949230836156));
draw((-3.454001382502029,3.9785901614824524)--(-5.709089178789237,5.8523028157505035));
draw((-5.709089178789237,5.8523028157505035)--(-2,10));
draw((0.06810107679554245,8.14949230836156)--(-2,10));
draw((0.06810107679554245,8.14949230836156)--(4,4));
draw((0.06810107679554245,8.14949230836156)--(2.3992779710964074,10.461967575808181));
draw((-3.454001382502029,3.9785901614824524)--(4,-1));
draw((4,4)--(2.3992779710964074,10.461967575808181));
draw((0.06810107679554245,8.14949230836156)--(-5.709089178789237,5.8523028157505035));
draw((4,-1)--(-2,10));
draw((2.3992779710964074,10.461967575808181)--(-3.420838326674272,-1.0456127964221371));
draw((2.3992779710964074,10.461967575808181)--(6.428589254168415,6.183933374028034));
draw((6.428589254168415,6.183933374028034)--(4,4));
draw(circle((-2.820834296659645,7.001753244155373), 3.108577255193373));
draw((-5.709089178789237,5.8523028157505035)--(6.428589254168415,6.183933374028034));
draw((-3.454001382502029,3.9785901614824524)--(0.1949568576384766,5.981074885605413));
draw((4.000284475054789,3.9995766660700363)--(0.1949568576384766,5.981074885605413));
draw((0.06810107679554245,8.14949230836156)--(0.1949568576384766,5.981074885605413));
/* dots and labels */
dot((6.428589254168415,6.183933374028034),dotstyle);
label("$C_{2}$", (6.494915365823923,6.349748653166799), NE * labelscalefactor);
dot((4,-1),dotstyle);
label("$C_{1}$", (4.074012290397943,-0.8300529335417423), NE * labelscalefactor);
dot((-3.454001382502029,3.9785901614824524),dotstyle);
label("$B$", (-3.387675270846515,4.144405440621219), NE * labelscalefactor);
dot((4,4),dotstyle);
label("$C$", (4.04084923457019,4.277057663932232), NE * labelscalefactor);
dot((0.06810107679554245,8.14949230836156),dotstyle);
label("$A$", (0.12760864689531848,8.322950474918107), NE * labelscalefactor);
dot((2.3992779710964074,10.461967575808181),dotstyle);
label("$A_{1}$", (2.4987671385796686,10.71069049451633), NE * labelscalefactor);
dot((-3.420838326674272,-1.0456127964221371),dotstyle);
label("$B_{2}$", (-3.354512215018762,-0.8797975172833719), NE * labelscalefactor);
dot((-5.709089178789237,5.8523028157505035),dotstyle);
label("$B_{1}$", (-5.908067513755754,6.3994932369084285), NE * labelscalefactor);
dot((-2,10),dotstyle);
label("$A_{2}$", (-1.7792670632004874,10.196663129186156), NE * labelscalefactor);
dot((4.000284475054789,3.9995766660700363),linewidth(4pt) + dotstyle);
dot((0.1949568576384766,5.981074885605413),linewidth(4pt) + dotstyle);
label("$X$", (0.24367934229245447,6.117607262372528), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/8/1/8/8181a7fc8527a83b9a3f23553837030ea15f333f.png)
and mutually intersect at some point 
be the intersection of two circumcircles such that 
By some simple cyclic quad angle properties, we see ![\[ \angle AXC = 180^{\circ} - \angle AA_1C, \angle BXA = 180^{\circ} - \angle AB_1B.\]](http://latex.artofproblemsolving.com/4/2/b/42be3e5be4631b4685d21ea8b884f18d9865e009.png)
It follows that 
and so our claim is proven.
and concur at ![\[\angle AA_1C_2= \angle AXC_2 = 90^{\circ}, \angle B_1A_2A = \angle AXB_1 = 90^{\circ},\]](http://latex.artofproblemsolving.com/3/8/9/38907d690d2261cf1b74300b7e780ee3af968626.png)
meaning 
are collinear. Then, we may similarly obtain ![\[\angle BB_2C_1 = \angle BXC_1 = 90^{\circ}, \angle A_2B_1B = \angle A_2XB = 90^{\circ},\]](http://latex.artofproblemsolving.com/7/9/7/797f17a71944bf29960308c038157049d25d65f1.png)
meaning 
are collinear. Finally, we have ![\[\angle A_1C_2C = A_1XC=90^{\circ}, \angle CC_1B_2 = \angle B_2XC = 90^{\circ},\]](http://latex.artofproblemsolving.com/9/c/3/9c31e3e8b89ba306fbd3ba58928bb630a60bce7a.png)
meaning 
are collinear. Putting these three results together, it is clear that and are concurrent at 
, with . Observe that 
, by the problem's condition. Hence, 
, so lies on 
.
have diameter 
, respectively (since 
and are rectangles), then 
, so lies on . Similarly, lies on 
and , so we are done.
![[asy]
unitsize(2cm);
pair X = dir(110);
pair Y = dir(210);
pair Z = dir(330);
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
draw(X--Y--Z--cycle);
pair P = (2X + 3Y + 2Z)/7;
dot("$P$", P, dir(90));
pair A = 2*foot(P, Y, Z) - P;
pair B = 2*foot(P, Z, X) - P;
pair C = 2*foot(P, X, Y) - P;
dot("$A$", A, dir(Y + Z));
dot("$B$", B, dir(X + Z));
dot("$C$", C, dir(X + Y));
pair C2 = 2Y - A;
pair B1 = 2Z - A;
pair A2 = 2Z - B;
pair C1 = 2X - B;
pair B2 = 2X - C;
pair A1 = 2Y - C;
dot("$A_1$", A1, dir(Y + Z));
dot("$A_2$", A2, dir(Y + Z));
dot("$B_1$", B1, dir(X + Z));
dot("$B_2$", B2, dir(X + Z));
dot("$C_1$", C1, dir(X + Y));
dot("$C_2$", C2, dir(X + Y));
draw(B--C--C1--B2--cycle, heavycyan);
draw(C--A--A1--C2--cycle, heavycyan);
draw(A--B--B1--A2--cycle, heavycyan);
draw(B--C1, heavycyan);
draw(C--B2, heavycyan);
draw(C--A1, heavycyan);
draw(A--C2, heavycyan);
draw(A--B1, heavycyan);
draw(B--A2, heavycyan);
draw(Circle(X, abs(X - P)), heavyred);
draw(Circle(Y, abs(Y - P)), heavyred);
draw(Circle(Z, abs(Z - P)), heavyred);
draw(C1--A2, deepgreen+dotted);
draw(B1--C2, deepgreen+dotted);
draw(A1--B2, deepgreen+dotted);
[/asy]](http://latex.artofproblemsolving.com/c/1/c/c1cf59f1e59b8b783f92b9420e2a6cfd246c1032.png)
,
be the centers of the rectangles, and let 
,
,
be their circumcircles.![\[\measuredangle CPB = \measuredangle CPA + \measuredangle APB = \measuredangle CA_1A + \measuredangle AB_1B = -\measuredangle BC_1C = \measuredangle CC_1B\]](http://latex.artofproblemsolving.com/7/1/8/718ad7e1663f5f33aa6ad5ad4f52ed8597d05433.png)
by the given condition (the rectangles all being outside means the orientation is correct). So 
as well.
over 
,
over 
,
over 
.
be the centers of the rectangles (so 
let it take 
to 
.
so 
and 
Angle and length chase to show the reflections of 
and 
coincide, likewise with 
This first reflection lies on 
are reflections of 
so by symmetry all three lines coincide at this "reflection" point
and similar for other three we are done that the lines concur at 
.
and ![\[(\angle BOC,\angle COA)=(180^\circ-\angle BC_1C,180^\circ-\angle CA_1A),\]](http://latex.artofproblemsolving.com/1/4/d/14dd39c22d248f7360f7a9e9bbab1c6331deed78.png)
whence the angle condition gives 
.
as claimed.
and so 
and so on. In all, 
;
too. It follows that 
,
and cyclic permutations.
,
,
are diameters of their respective circles, and both circles go through 
.
The circumcircles of 
and 
concur.
and 
Notice that
so 
Also, 
so 
implying that 
are collinear. By similar logic, we find 
more triples of collinear points, concluding the problem.
,
and 
.
,
and 
,
.
is cyclic, 
.
is cyclic, 
.
is cyclic 
,
is a diameter in 
is a diameter in 
,
is a diameter in 
is a diameter in 
,
is a diameter in 
is a diameter in 
,![\[\angle B_1KX_2 = \angle B_1KA + \angle AKC_2 = \angle AA_2B_1 + \angle AA_1C_2 = 90 + 90 = 180,\]](http://latex.artofproblemsolving.com/c/a/f/cafc47d5287018d77bc05c194f461906ffc84975.png)
.
so 
is cyclic and the three circumcircles of the rectangles concur. Also, 
are collinear and cyclic variations hold by angle chasing: 
so done.
.
.
and 
which forces 
and therefore 
and 
.
passes through point 
and 
meet at 
.
and 
.
,
are rectangles, we have 
are the centers of the respective circumcircles.
.
,
,
are in the same direction as each other as the rectangles are all erected outside 
,![\[
\angle B C_1 C + \angle C A_1 A + \angle A B_1 B = 180^\circ
\]](http://latex.artofproblemsolving.com/c/1/1/c11522a99ea3088f9204dd0dca7b68a99139e612.png)
![\[
\Rightarrow \measuredangle B C_1 C + \measuredangle C A_1 A + \measuredangle A B_1 B = 0
\]](http://latex.artofproblemsolving.com/c/e/3/ce357e88ea2086cec8fb4467a6f6fda37f063b3c.png)
![\[
\Rightarrow \measuredangle B P C + \angle C P A + \measuredangle A B_1 B = 0
\]](http://latex.artofproblemsolving.com/c/e/2/ce270befb4bb5ab96197ef493fd9f81bdd4b0ed9.png)
![\[
\Rightarrow \measuredangle B P A + \measuredangle A B_1 B = 0
\]](http://latex.artofproblemsolving.com/d/8/7/d874962f3561045d8f9cedd7a57a8779eeb66ca8.png)
![\[
\Rightarrow \measuredangle B P A = \measuredangle B B_1 A \Rightarrow (A P B B_1) \text{ cyclic.}
\]](http://latex.artofproblemsolving.com/0/d/7/0d791617d258d12520e11d42fe6ff9e139b3efec.png)
as the other two follow analogously.
of 
satisfies 
since 
and 
.
of scale 
suffices.
.
Therefore the 
drawn circles concur at 
.
,
at the midpoint of 
,
,
.
We have that 
so all 3 circles concur at 