[/quote]
,
and 
be the incircle and circumcircle of the acute triangle 
so that all of its sides are tangent to 
and 
,
,
and 
intersects on 
be real numbers such that 
Prove that if 
is a positive constant, then
is not always true
.
in which 
and 
.
be a point on the side 
and 
.
.
satisfying the equation![\[
(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/2/5/e/25eed911f303ed7cc4e03765a2f943ad4641c451.png)
Prove that
Let 
Prove that
,
,
and 
there exists infinitely many positive integers 
for which 
and 
are not relatively primes.
or 
tile.
be an integer. Dominoes are placed on an 
board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some 
such that each row and each column has a value of 
.
meet at 
intersects 
and 
,
and 
.
and 
Y by ChimkinGang, lpieleanu, zhenghua, Rounak_iitr
Let 
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
be the reflection of 
be the foot from 
to 
be the circumcenter of 
.
is right, by homothety by a factor of 
at 
is right. If we apply a homothety by a factor of 
at 
and 
map to the perpendicular bisectors of 
and 
,
maps to 
,
.
be center of 
.
be the point so that 
is a rectangle; it suffices to show 
.
Now checking 
we get 
.
.
so 
.
with 
.![\[
-a^2S_CS_B-b^2S_Bp-c^2pS_C=0.\tag{1}
\]](http://latex.artofproblemsolving.com/1/1/6/1161ffe16824cb3b00142d28a4efa40a707dace5.png)
Note that 
.
so 
.
gives 
.
.
to intersect with 
and![\[
-a^2yz+S_By-S_Cz=0
\]](http://latex.artofproblemsolving.com/9/6/4/9642d2d3f74e0b48f987465ca82ddea330aa1cc1.png)
so![\[
-a^2y(1-y)+S_By-S_C(1-y)=0.
\]](http://latex.artofproblemsolving.com/1/0/f/10f9d162db73b06ae53a918cc9dbdf0456660aab.png)
Solving yields 
so 
and 
.
,
.
note that 
Let the line through 
at point 
Then 
by the parallel lines, so 
is a diameter in 
Thus if 
then 
Since 
by midlines we get 
and this implies, once again by midlines, that 
Hence 
and together with 
this implies 
as desired.
.
.
is an isosceles trapezoid, so the perpendicular bisector of 
,
,
.
,
,
and we are done.
?
Let 
.
.
in terms of 
gives: 
Simplifying 
gives ![\[a(d-b) = b^2+bc.\]](http://latex.artofproblemsolving.com/d/7/8/d788c34f9d0515a0f15efdc7d847e72fe28259ce.png)
Solving 
for 
.
,![\[d^2 - bd - bc-c^2=(d-b-c)(d+c) = 0.\]](http://latex.artofproblemsolving.com/d/c/2/dc25cd031e8c1a1dd008b9a4e268010d380574b6.png)
Since 
,
as desired.
,
,
.
intersect on the line 
,
.
and 
,
is the line 
,
has slope 
.
.
,
,
.
,
.
,![\[\begin{cases}
y=\frac{bx}{a} \\
y=-\frac{ax}{b}+\frac{ac}{b}
\end{cases} \implies \frac{x(a^2+b^2)}{ab}=\frac{ac}{b} \implies x=\frac{a^2c}{a^2+b^2},\]](http://latex.artofproblemsolving.com/7/f/c/7fc201fc320294bbf76988ed102f0365ef813bba.png)
and![\[y=\frac{bx}{a} \implies y=\frac{abc}{a^2+b^2}.\]](http://latex.artofproblemsolving.com/d/b/1/db177eeb8d3cf0c4c7a85ea3972887ef4e9d63d2.png)
Thus,![\[F=\left(\frac{a^2c}{a^2+b^2},\frac{abc}{a^2+b^2}\right).\]](http://latex.artofproblemsolving.com/8/5/d/85d5b65be4dc7ab740364ecfe55a38f3337f1f38.png)
The midpoint of ![\[\left(\frac{1}{2}\left(a+\frac{a^2c}{a^2+b^2}\right),\frac{1}{2}\left(b+\frac{abc}{a^2+b^2}\right)\right)=\left(\frac{a^3+a^2c+ab^2}{2(a^2+b^2)},\frac{a^2b+abc+b^3}{2(a^2+b^2)}\right).\]](http://latex.artofproblemsolving.com/2/9/4/29480f23081b3d4f152fbdba1d72ca0ca86d32d3.png)
Since 
,
is 
,![\[y=-\frac{a}{b}\left(x-\frac{a^3+a^2c+ab^2}{2(a^2+b^2)}\right)+\frac{a^2b+abc+b^3}{2(a^2+b^2)}.\]](http://latex.artofproblemsolving.com/9/8/3/98361d93bd1ab6da7e65c6b825be56b719e894d8.png)
At 
-
The midpoint of ![\[\left(\frac{1}{2}\left(a+\frac{a^2c}{a^2+b^2}\right),\frac{1}{2}\left(\frac{a^2-ac}{b}+\frac{abc}{a^2+b^2}\right)\right)=\left(\frac{a^3+a^2c+ab^2}{2(a^2+b^2)},\frac{a^4+a^2b^2-a^3c}{2b(a^2+b^2)}\right).\]](http://latex.artofproblemsolving.com/1/6/d/16d8b38820d8e74de949540dba40ead29beb3d4a.png)
The slope of ![\[\frac{\frac{abc}{a^2+b^2}-\frac{a^2-ac}{b}}{\frac{a^2c}{a^2+b^2}-a}=\frac{\frac{2ab^2c+a^3c-a^4-a^2b^2}{b(a^2+b^2)}}{\frac{a^2c-a^3-ab^2}{a^2+b^2}}=\frac{2b^2c+a^2c-a^3-ab^2}{abc-a^2b-b^3}.\]](http://latex.artofproblemsolving.com/c/c/3/cc3bfbaef81e1ce30aefb09ee42ad1b06b09837d.png)
The slope of the perpendicular bisector of ![\[y=\frac{b^3+a^2b-abc}{2b^2c+a^2c-a^3-ab^2}\left(x-\frac{a^3+a^2c+ab^2}{2(a^2+b^2)}\right)+\frac{a^4+a^2b^2-a^3c}{2b(a^2+b^2)}.\]](http://latex.artofproblemsolving.com/e/a/8/ea8752d7b641762523f4f5b32447db58b0e619ce.png)
At 
Hence, the perpendicular bisectors of 
is a parallelogram and hence a rectangle. Since 
,
.
we would be done, as we would have 
by parallelogram and 
with 
.
simple terms are equivalent.
.
.
![[asy]
size(250);
defaultpen(linewidth(0.6)+fontsize(11));
pair A = (0,9), B = (-3,0), C = (7,0), H = orthocenter(A,B,C), D = foot(A,B,C), P = 2*D - H, F = foot(C,A,B), Xx = 3*C-2*B;
path circ = circumcircle(A,F,P);
pair[] oops = intersectionpoints(circ, Xx--B);
pair X = oops[0], Y = oops[1];
draw(A--B--C--cycle, rgb(0,0,0));
draw(C--F^^A--P, rgb(0,0,0));
draw(circ,rgb(0,0,0));
draw(circumcircle(A,B,C),rgb(0,0,0));
draw(F--X--P--C--Y,rgb(0,0,0));
dot("$A$",A,N,linewidth(3.3));
dot("$B$",B,SW,linewidth(3.3));
dot("$C$",C,SE,linewidth(3.3));
dot("$D$",D,NE,linewidth(3.3));
dot("$F$",F,NW,linewidth(3.3));
dot("$X$",X,SW,linewidth(3.3));
dot("$Y$",Y,SE,linewidth(3.3));
dot("$P$",P,S,linewidth(3.3));
dot("$H$",H,NE,linewidth(3.3));
[/asy]](http://latex.artofproblemsolving.com/d/c/7/dc76772b5e5fb7e4644c430af07ce29523a372c4.png)
Since 
,
.
is cyclic, we compute 
But, if above we multiply by 
the first factor and by 
the second while in the denominator we multiply by 
the first, by 
the second and by 
where we replace 
by what we get at the equation of line and get 
At this point, I got really frustrated, wondering how in the world I would compute the value of 
.
,
,
,
and 
:
And, making many cancellations(which took me about 30 minutes and 3 failed attempts), we find 
and hence 
,
be the two intersection of line 
.![[asy]
import geometry;
size(9cm);
defaultpen(fontsize(11pt));
pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair H = orthocenter(A,B,C);
pair F = extension(C,H,B,A);
pair D = extension(A,H,B,C);
pair P = 2*D - H;
pair Q = 2*C - H;
pair X = intersectionpoints(line(D,C), circumcircle(A,F,P))[0];
pair Y = intersectionpoints(line(D,C), circumcircle(A,F,P))[1];
draw(A--B--C--A, fuchsia);
draw(circumcircle(A,B,C), red);
draw(A--P, fuchsia);
draw(F--Q, fuchsia);
draw(P--Q, fuchsia);
draw(C--Y, fuchsia);
draw(C--P, fuchsia);
draw(circumcircle(F,A,P), red);
markscalefactor = 0.01;
draw(rightanglemark(A,D,C), deepgreen);
draw(rightanglemark(A,P,Q), deepgreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(273)*1.5);
dot("$H$", H, dir(210));
dot("$F$", F, dir(F));
dot("$D$", D, dir(D));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
[/asy]](http://latex.artofproblemsolving.com/1/b/5/1b5a1ba4cbefc17a0251dea99deb4a38aebdc22a.png)
.![\[AD \cdot HD = FH \cdot HC \implies AD \cdot 2HD = FH \cdot 2CH \implies AH \cdot HP = FH \cdot HQ\]](http://latex.artofproblemsolving.com/7/a/8/7a80afd263cf98dd67bdf2903cf52d1354aefecd.png)
which proves the claim. Since ![\[\overline{CP} = \overline{CH} = \overline{CQ}\]](http://latex.artofproblemsolving.com/9/2/5/92574cd31cb489d626cab182192c9b2bffef0d1c.png)
which means 
in 
at 
,now notice two spiral similar triangles
,now we have here: 
and 
so so we would have also 
and we are done:
be the feet of the 
-
.
.
with 
;
and 
respectively. Then it suffices to show that 
is a harmonic quadrilateral. Projecting from 
is harmonic. But this is easy as 
and 
is a cyclic isosceles trapezoid it follows that the midpoint of 
lies on the perpendicular bisectors of 
,
,
.
:
:
:
.
:
:
Now clearly 
since 
,
,
,
Q
with 
with the way you labeled your coordinates, so idt this is right
,
,
and that whether 
