[/quote]
,
?
Prove that
vertices, each edge has one of the colors 
,
,
.
,
groups such that any two vertices connected by an edge of color 
are in different groups, find the minimum possible value of 
.
such that, for all 
,
such that for all integer 
and a subset 
of ![$[n],$](http://latex.artofproblemsolving.com/d/2/7/d27d2715f4f95c24a28b7c6371a006d4c803b2cf.png)
satisfy 
for all 
then 
be the set of positive integers. Find all functions 
such that for all 
,![\[m^2f(m) + n^2f(n) + 3mn(m + n)\]](http://latex.artofproblemsolving.com/0/f/f/0ff783fa217b7be4bda5006e417f1471ea3a6fca.png)
is a perfect cube.
and 
.
![$\mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
be the set of all polynomials with real coefficients. Find all functions ![$f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$](http://latex.artofproblemsolving.com/0/3/3/033be3ab043bd47e537f01791c813739ad5c7da5.png)
satisfying the following conditions:
maps the zero polynomial to itself,![$P \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/f/7/c/f7c4c970908aa5570a9afcc45fc16c1d36cc3d5d.png)
,
,![$P, Q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
,
and 
have the same set of real roots.
be a square of center 
and let 
be the symmetric of the point 
be the intersection of 
and 
,
and 
.
is the angle bisector of 
.
be reals such that 
Show that
Let 
and 
represent the number of unit squares that have exactly 
edges on the boundary of 
such that the inequality 
holds for each polygon constructed with these conditions.
and 
Find the value range of 
and 
with real coefficients satisfying
Y by 606234, Math_olympics, justJen, samrocksnature, megarnie, CrazyInMath, Rounak_iitr, GeoKing, Funcshun840
Let 
, 
, 
, 
be four points such that no three are collinear and is not the orthocenter of 
. Let 
, 
, 
be the orthocenters of 
, 
, 
, respectively. Suppose that the lines 
, 
, 
are pairwise distinct and are concurrent. Show that the four points , , , lie on a circle.
Andrew Gu
, 
, 
, 
be four points such that no three are collinear and is not the orthocenter of 
. Let 
, 
, 
be the orthocenters of 
, 
, 
, respectively. Suppose that the lines 
, 
, 
are pairwise distinct and are concurrent. Show that the four points , , , lie on a circle.Andrew Gu
This post has been edited 2 times. Last edited by v_Enhance, Mar 1, 2021, 5:21 PM
be the rectangular circumhyperbola of 
are distinct. Let 
and 
with 
.
on 
and are concyclic iff 
.
.
,
then 
.
,![\[\begin{vmatrix}1&-a^2t^{-1}&a(1-t^{-1})\\1&-b^2t^{-1}&b(1-t^{-1})\\1&-c^2t^{-1}&c(1-t^{-1})\end{vmatrix}=\begin{vmatrix}1&ap&a+p\\1&bq&b+q\\1&cr&c+r\end{vmatrix}=0.\]](http://latex.artofproblemsolving.com/6/5/f/65f2e6e483df9e3a22465a417fa4d223acc12a54.png)
Then, 
are distinct roots of 
for some 
which are not all 
,
,
be the orthocenter of 
,
be a nondegenerate circumrectangular hyperbola of 
back onto 
lie on 
,
.
.
all equal the other point at infinity on 
concur.
:
,
![[asy]
unitsize(4cm);
pair O = (0, 0);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
draw(A--B--C--A);
dot("$A$", A, dir(90));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
pair D = 1.3*dir(10);
pair P = orthocenter(B, C, D);
pair Q = orthocenter(A, C, D);
pair R = orthocenter(A, B, D);
dot("$D$", D, dir(D));
dot("$P$", P, dir(270));
dot("$Q$", Q, dir(270));
dot("$R$", R, dir(270));
draw(D--A);
draw(D--B);
draw(D--C);
draw(P--D, deepgreen);
draw(C--foot(D, B, C), deepgreen);
draw(D--foot(Q, A, C), deepgreen);
draw(D--foot(D, A, B), deepgreen);
pair H = orthocenter(A, B, C);
pair K = (D + H)/2;
pair L = (A + P)/2;
pair M = (B + Q)/2;
pair N = (C + R)/2;
dot("$H$", H, dir(270));
dot("$K$", K, dir(90));
dot("$L$", L, dir(270));
dot("$M$", M, dir(270));
dot("$N$", N, dir(225));
draw(A--P, heavyblue);
draw(B--Q, heavyblue);
draw(C--R, heavyblue);
draw(H--D, heavyblue);
pair E = foot(D, B, C);
dot("$E$", E, dir(270));
pair Z = foot(A, B, C);
dot("$Z$", Z, dir(270));
draw(A--Z, deepgreen);
[/asy]](http://latex.artofproblemsolving.com/a/2/d/a2d8b4a039a707b09da11440233bf2a429c5a8c6.png)
,
,
be the midpoints of 
,
be the feet from 
Also, let 
and ![$S = 2[ABC]$](http://latex.artofproblemsolving.com/7/7/a/77a6e824aa87f63441d230ab2832479aac03b2e5.png)
.
.
is 
,
and then this means 
But note that 
,
has the same direction as 
because 
.
.
.
,
.
.
,
,
But we have 
,
So then using 
,
Then 
,
,
concur if and only if 
,
.
,
(where in this product the LHS evaluates an expression cyclically through 
and the right hand side uses 
,
,
we get 
on the LHS, and its counterpart on the RHS. The second term expands into 
The first term is symmetric in 
and 
,
,
The first term is symmetric in 
,
if and only if two of the variables are equal -- clearly these are roots of the equation, and clearing denominators gets an equation of total degree 
if and only if ![[asy]
unitsize(4cm);
pair O = (0, 0);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
draw(A--B--C--A);
dot("$A$", A, dir(90));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
pair D = 0.6*A + 0.4*foot(A, B, C);
pair P = orthocenter(B, C, D);
pair Q = orthocenter(A, C, D);
pair R = orthocenter(A, B, D);
dot("$D$", D, dir(45));
dot("$P$", P, dir(45));
dot("$Q$", Q, dir(270));
dot("$R$", R, dir(270));
draw(D--A);
draw(D--B);
draw(D--C);
//draw(P--D, deepgreen);
//draw(C--foot(D, B, C), deepgreen);
//draw(Q--foot(Q, A, C), deepgreen);
//draw(D--foot(D, A, B), deepgreen);
pair H = orthocenter(A, B, C);
pair K = (D + H)/2;
pair L = (A + P)/2;
pair M = (B + Q)/2;
pair N = (C + R)/2;
dot("$H$", H, dir(270));
dot("$K$", K, dir(225));
dot("$L$", L, dir(0));
dot("$M$", M, dir(270));
dot("$N$", N, dir(270));
draw(A--P, heavyblue);
draw(B--Q, heavyblue);
draw(C--R, heavyblue);
//draw(H--D, heavyblue);
pair E = foot(D, B, C);
//dot("$E$", E, dir(270));
pair Z = foot(A, B, C);
dot("$Z$", Z, dir(270));
draw(A--Z, deepgreen);
[/asy]](http://latex.artofproblemsolving.com/3/6/5/365cc207001471a4a9283d549ffcfb31779bb61a.png)
and 
are line 
be the concurrency point. We have 
because both lines are perpendicular to 
.
centered at 
.
,
.
.
and 
,
by symmetry. Hence if 
,
which are reflections of the previous four points about 
be the projections of the corresponding point onto line 
is the midpoint of 
,
is the midpoint of 
)
.
is the foot of the altitude from 
so multiplying the similar terms, we get that 
where 
are defined similarly to 
are collinear by menalauas, and we are done by simson lines.
.
as both are perpendicular to 
,
.
.
and 
to get![\[\frac{AO}{OP}=\frac{QO}{OB}=\frac{CO}{OQ}=\frac{OP}{AO},\]](http://latex.artofproblemsolving.com/1/9/2/192cbd9baff8ee04a3eeaef5170890fc5fa49399.png)
.
,
.
and 
are both the diameter of 
,
.
and 
are either supplementary or equal. A quick case check on whether 
and 
are on the same side of 
and 
,
.
.
and 
.
, and so on, so 
,
collinear with 
.
,
lie on opposite sides of 
and 
denote the circumradii of 
and 
respectively. Let 
denote the acute angle 
.
,
.
,
.
,
.
.
be the isogonal conjugate of 
,
,
;
is the pedal triangle of 
or 
,
is 
reflected across some point. But, as they all lie on a rectangular hyperbola, and 
is now a parallelogram, the intersection point is the center of the hyperbola and thus, is the Poncelet point of 
be the circumcenters of triangles 
respectively. If two of them coincide, we are done. Suppose for the sake of contradiction that 
.
(both perpendicular to 
)
hence 
be the midpoints of 
.
Lemma, we obtain 
and similar relations. Cyclically we have 
The first equality implies 
and since 
and 
we obtain 
.
hence 
.
be the point of concurrence, and let 
and 
,
.
is a parellelogram. Therefore 
is the reflection of 
and so on 
be the unique rectangular hyperbola passing through 
also lie on 
.
and 
,
be the intersection of 
and 
,
be the intersection of 
.
and 
,
,
,
concur. But they are all perpendicular to 
,
,
.
,
are not concurrent) we have that this can only happen if they are all the same line. But this indicates that ![[asy] pair A = dir(130); pair B = dir(210); pair C = dir(330); pair D = dir(97); pair P = B+C+D; pair Q = C+A+D; pair R = A+B+D;
filldraw(R--B--D--cycle, invisible, palecyan); filldraw(Q--A--C--cycle, invisible, palered); filldraw(C--B--D--cycle, invisible, palegreen);
draw(R--foot(R,B,D), lightgrey+dashed); draw(B--foot(B,D,R), lightgrey+dashed); draw(D--foot(D,R,B), lightgrey+dashed);
draw(A--foot(A,C,Q), lightgrey+dashed); draw(C--foot(C,Q,A), lightgrey+dashed); draw(Q--foot(Q,A,C), lightgrey+dashed);
draw(C--foot(C,B,D), lightgrey+dashed); draw(B--foot(B,D,C), lightgrey+dashed); draw(D--foot(D,C,B), lightgrey+dashed);
draw(A--foot(A,B,C), lightgrey+dashed); draw(B--foot(B,C,A), lightgrey+dashed); draw(C--foot(C,A,B), lightgrey+dashed);
pair T = midpoint(A--P); pair S = (A+B+C+D)/4; pair O = 2*S-T; /* A' = foot A B C R270 T' = foot T B C R270 D' = foot D B C R300 O' = foot O B C R300 S' = foot S B C R270 T--Tp lightgrey dashed S--Sp lightgrey dashed O--Op lightgrey dashed */
draw(T--O, lightblue);
filldraw(A--B--C--cycle, invisible, lightgrey); draw(A--C, palered);
draw(A--Q, red+1); draw(B--P, red+1);
draw(A--P, blue); draw(B--Q, blue); draw(C--R, blue); pair H = A+B+C; draw(D--H, lightblue);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(80)); dot("$P$", P, dir(285)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$T$", T, 1.8*dir(90)); dot("$S$", S, 1.8*dir(75)); dot("$O$", O, dir(315)); dot("$H$", H, dir(45));
/* TSQ Source:
A = dir 130 B = dir 210 C = dir 330 D = dir 97 R80 P = B+C+D R285 Q = C+A+D R = A+B+D
R--B--D--cycle 0.1 lightcyan / palecyan Q--A--C--cycle 0.1 lightred / palered C--B--D--cycle 0.1 lightgreen / palegreen
R--foot(R,B,D) lightgrey dashed B--foot(B,D,R) lightgrey dashed D--foot(D,R,B) lightgrey dashed
A--foot(A,C,Q) lightgrey dashed C--foot(C,Q,A) lightgrey dashed Q--foot(Q,A,C) lightgrey dashed
C--foot(C,B,D) lightgrey dashed B--foot(B,D,C) lightgrey dashed D--foot(D,C,B) lightgrey dashed
A--foot(A,B,C) lightgrey dashed B--foot(B,C,A) lightgrey dashed C--foot(C,A,B) lightgrey dashed
T = midpoint A--P 1.8R90 S = (A+B+C+D)/4 1.8R75 O = 2*S-T R315
T--O lightblue
A--B--C--cycle 0.1 yellow / lightgrey A--C palered
A--Q red+1 B--P red+1
A--P blue B--Q blue C--R blue H = A+B+C R45 D--H lightblue
*/ [/asy]](http://latex.artofproblemsolving.com/1/1/7/117b0d130be6bf2beea101a6d36c7f7bcbf42f09.png)
,
,
,
,
,
.
is a hexagon with opposite sides parallel and concurrent diagonals as ![\[ \frac{AT}{TP} = \frac{TQ}{BT} = \frac{CT}{TR} = \frac{TP}{AT} \]](http://latex.artofproblemsolving.com/a/6/0/a600f74ca17efb306b1dd698319591bc60ccd1fc.png)
so all the ratios are equal to 
.
,
,
,
,
be the projections of 
,
gives that 
.
and we're done. 
and let 
,
are colinear. Similarly, by Pascal’s Theorem on 
are collinear. Thus, 
are collinear. Hence, since these points are the orthogonal projections of 
(because 
are the ortocenter of 
,
through the points 
,
.
,
,
.
,
,
concur at 
,
,
.
is a parallelogram: indeed,
and 
are both perpendicular to 
,
.
and 
,
by Desargue's theorem on 
and 
.
is the midpoint of 
be the image of 
,
.
of 
.
and 
,
are collinear. Furthermore, from 
is the degenerate pedal triangle of 
with respect to 
where 
(they all lie on the line at infinity). By Brocard's Theorem we know that the polar of 
.
and 
.
as both are perpendicular to 
are collinear. But by the previously derived parallellisms, the latter two are points at infinity, hence 
.
is a parallelogram, since 
is clear as they are both perpendicular to 
is actually the midpoint of each of the three segments.![\[CD\cot{\angle DBC} = BP = AQ = CD\cot{\angle DAC}\]](http://latex.artofproblemsolving.com/9/0/d/90d6cc6b480ca98b3d4047e58afa65b0f3307ec4.png)
hence 
so 
are both perpendicular to 
meet at the line at infinity, so they are also parallel. Thus 
,
,
is a parallelogram. Let 
.
,
.
has midpoint 
,
,
because 
is a parallelogram. The centroid of 
,
Now focus on 
which yields 
and everything else follows. 
be the coordinate for 
.
is a parallelogram, we see that 
.
.
![[asy] size(8cm); pen prim = blue; pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = dir(70) * 3/2; pair E = foot(D, B, C); pair F = foot(C, B, D); pair G = foot(D, A, C); pair H = foot(C, A, D); pair P = extension(D, E, C, F); pair Q = extension(D, G, C, H); draw(A--B--C--cycle); draw(CP( (C+D)/2, C)); draw(D--E); draw(C--F); draw(D--G);draw(C--H); draw(A--D); draw(D--B); draw(G--E, prim); draw(F--H, prim); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, S); dot("$F$", F, dir(20)); dot("$G$", G, dir(220)); dot("$H$", H, dir(H-Q)); dot("$P$", P, dir(290)*2); dot("$Q$", Q, dir(350)); draw(B--Q, prim); draw(A--P, prim); [/asy]](http://latex.artofproblemsolving.com/3/8/6/386ba371f49215633fe0f75c4b24c0cd14a9cdcf.png)
be the feet from 
and 
be the foot from 
lies on 
is cyclic with diameter 
Now, we apply Pascals on 
to get 
and 
are collinear. Therefore, 
are concurrent. 
then 
all pass through 
which implies 
is collinear. By converse of Simson line 
![[asy]
size(250);
pair A = (-3.6488936772554, 4.7629376368042), B = (4.064712684455, 4.4134012725823), C = (4.709553074972, -3.7175408315205), D = (0.4691154449004, -5.9816327787115);
pair P = orthocenter(B, C, D), Q = orthocenter(A, C, D), R = orthocenter(A, B, D);
pair X = extension(B, Q, A, P);
draw(A--B--C--D--cycle, blue);
draw(circumcircle(A, B, C), lightblue);
draw(A--C, green); draw(B--D, green);
draw(A--P, orange); draw(B--Q, orange); draw(C--R, orange); draw(B--R--Q--C, red+dashed); draw(A--R, magenta+dashed); draw(C--P, magenta+dashed);
dot("$A$", A, NW);
dot("$B$", B, NE);
dot("$C$", C, SE);
dot("$D$", D, S);
dot("$Q$", Q, SE);
dot("$P$", P, SE);
dot("$R$", R, N);
dot("$X$", X, E);
[/asy]](http://latex.artofproblemsolving.com/a/c/b/acb5cca5e548a9720008b4d0920b895e41b4c029.png)
is a parallelogram.
,
by definition.
,
.
implies 
.
However, because 
and 
,
are the vertices of a parallelogram.
,
.
.
which is already known, we have 
,
,
so 
Similarly, 
so 
But because 
due to parallelogram 
,
.
,
or 
is obtuse, it actually follows that 
,
,
.
.
so 
.
,![\[\frac{RT}{TC} = \frac{RT}{TB}\cdot \frac{TB}{TP} \cdot \frac{TP}{TC} = \frac{CT}{CQ}\cdot \frac{TQ}{TA}\cdot \frac{TA}{TR} = \frac{CT}{TR}.\]](http://latex.artofproblemsolving.com/a/d/4/ad48f27e511291890ca0141fd14ca7ee59533495.png)
Thus 
,
,
,
,
.
,![\[\frac{AM}{MQ} = \frac{MP}{BM}, \frac{AM}{MR} = \frac{MP}{CM}\Rightarrow \frac{MR}{MQ} = \frac{CM}{BM}\]](http://latex.artofproblemsolving.com/1/a/d/1ada1f84bb07eea2024306e869b20441beed7757.png)
However 
so 
.
.
,
,
.
.
.
different line segments on the hyperbola, this means 
as both are perpendicular to 
.
as both are perpendicular to 
.
as both are perpendicular to 
is a parallelogram!
and 
is parallelogram.
is perpendicular to 
,
using the parallelogram 
is perpendicular to 
,
is perpendicular to 
.
,
to 
with scale factor 
gets mapped to orthocenter of 
.
has midpoint X due to the homothety at 
is cyclic quadrilateral.
which is 
.
(Note 
to 
,
,
be the orthocenters of 
,
.
,
,
all concur at a point 
of 
obeys 
,
,
,
,
concur at ![\[ \frac{AX}{XD} = \frac{BX}{XE} = \frac{CX}{XF} = \frac{DX} {XA} \implies AX = DX\]](http://latex.artofproblemsolving.com/0/e/d/0ed73a39e712cdc4a28aa0939fd5abfc17a20618.png)
So, 
,
,
and 
is rotationally symmetric around 
are reflections of each other over 
.
and 
be reflections of 
,
,
are all cyclic. Since 
have distinct circumcircles. Since 
and 
intersect on the radical axis of 
.
.
and 
due to orthocentres, so 
and the rest follow by symmetry.
is a parallelogram
so 
giving 
by symmetry so 
so 
so 
and similar for all the other 
kinda distances. We'll show that 
and 
which will then solve the problem.
Notice that 
and 
Thus 
so we get that 
which implies that 
parallelogram, which implies that 
So now, 
and 
which ends the problem.
has opposite sides parallel.
and 
are both perpendicular to 
,
.
.
,
.
so the ratios are all 
as
Since 
,
,
We obtain cyclic variations for 
.
,
absurd. Thus 
,
.
we get 
Similarly, 
thus all ratios must be 
in directed lengths as clearly they cannot be 
Then 
It must then pass through 
Now if its center is some point 
the reflection of hexagon 
Since there is at most one conic through six points, this implies that the hyperbola is invariant under some translation, which is impossible.
so the reflection of 
But this point is 
and 
are perspective and 
.
.
and 
.
.
which is also called the anticenter. This is also a motivation for considering such a point, by knowing a priori the converse solution.
and 
.
is a paralellogram.
lies on 
and 
.![\[ \frac{XG_B}{G_BY} = \frac{XB}{YE} = \frac{XC}{YF} = \frac{XG_C}{YG_C} \]](http://latex.artofproblemsolving.com/f/3/4/f348c6e1b9ba284dc1a146fd327d588a79e8ee39.png)
so 
.
lies on 
,
which suffices. 
is fixed under reflection about 
passing through the orthocentre 
for the concurrence to hold. This combined with varying 
,
works trivially as each of 
,
passing through 
all projective, hence lines 
--- points 
lie on the 
.
which is the same for other points. Therefore 
are concurrent.
.
.
are concurrent there is an involution 
which swaps 
.
.
and suppose 
and define 
accordingly. Since converse of the problem is true we have 
concurrent so analogously 
.
are on a conic. 
are collinear so this conic must be union of 2 lines which is impossible unless 
(since certainly 
)
.
which is a contradiction since 
,
,
are all points at infinity, so they lie on the line at infinity. Hence, by the converse of Pascals theorem, the points 
This implies that 
and 
.
,
,
.
be the line through 
and 
similarly. Then the lines 
,
.
.
,
are concurrent. (possibly on the point at infinity) Since both of the lines 
and 
is perpendicular to 
.
to be collinear. Similarly, by Pascal on 
,
to be collinear. Therefore, 
are collinear. Since the pedal triangle of 
