[/quote]
,
so that:
with all real numbers 
satisfies 
be a positive integer. Do there exist distinct 
and 
,
or 
,
divides 
?
be the midpoints of 
respectively. Let 
be points in plane such that 
.
pass through a fixed point.
such that:
such that:
such that for any 
,
such that 
.
of the acute-angled triangle 
is tangent to its side 
at a point 
.
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
is not a perfect cube of an integer.
over 
is linear
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
be a cyclic pentagon with 
and 
.
be points on 
and 
,
is cyclic. Let 
.
,
,
concur.
be real numbers such that 
and 
and determine the equality cases.
be the set of real numbers. Determine all functions ![\[f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)\]](http://latex.artofproblemsolving.com/6/b/2/6b2acc0a1d76866e461fdf5d96126806a9ead8dd.png)
for all real numbers 
with incenter 
.
and 
intersects 
at 
,
be the midpoints of 
meets the external bisector of angle 
at 
and line 
meets the external bisector of angle 
at 
.
are concylic.
Y by Adventure10, Mango247
Let 
be an acute,non-isosceles triangle with 
.Let 
be the midpoints of 
respectively and segments 
are altitudes.In the extension of 
we take a point 
such that the parallel from to 
crosses the extensions of 
at 
respectively (we extend 
to 
-side and 
to -side and 
to 
-side).If the circumcirle 
of 
crosses the line 
at 
and the circumcirle 
of 
crosses the line 
at 
prove that 
.
be an acute,non-isosceles triangle with 
.Let 
be the midpoints of 
respectively and segments 
are altitudes.In the extension of 
we take a point 
such that the parallel from to 
crosses the extensions of 
at 
respectively (we extend 
to 
-side and 
to -side and 
to 
-side).If the circumcirle 
of 
crosses the line 
at 
and the circumcirle 
of 
crosses the line 
at 
prove that 
.
are the midpoints of 
we have 
and 
.
,
are concyclic. Similarly, we have 
,
are concyclic. Thus, by POP, we have 
and 
,
.
and 
,
and 
.
,
.
and 
,
by AA Similarity. Thus we have 
.
.
,
and consequently we get 
is cyclic and because 
,
is also cyclic and 
which implies that 
is also cyclic. Similarly, 
is cyclic. Also because 
and 
,
and 
.
is an isosceles trapezoid and 
.![[asy]
/* Free script by liberator */
unitsize(0.7cm);
defaultpen(fontsize(11pt));
pen c = rgb(0.9,0,0);
pen d = rgb(0.4,0.6,0.8);
/* Initialize objects */
pair A = (3,6);
pair B = origin;
pair C = (10,0);
pair D = midpoint(B--C);
pair E = midpoint(C--A);
pair Z = midpoint(A--B);
pair K = foot(B,C,A);
pair L = foot(C,A,B);
pair M = (-2,6);
pair T1 = rotate(180,A)*B;
pair T = intersectionpoint(Z--T1, circumcircle(B,D,M));
pair S1 = rotate(180,A)*C;
pair S = intersectionpoint(M--T, C--S1);
pair N1 = rotate(180,T)*M;
pair N2 = rotate(180,N1)*M;
pair N = intersectionpoint(T--N2, circumcircle(C,D,S));
pair R = intersectionpoint(D--N, circumcircle(B,D,M));
pair P = intersectionpoint(D--Z, circumcircle(C,D,S));
/* Draw objects */
draw(A--B--C--cycle, d+linewidth(1));
draw(D--E--Z--cycle, d);
draw(K--L, d);
draw(M--N, d);
draw(M--Z, d);
draw(N--E, d);
draw(A--S, d);
draw(A--T, d);
draw(P--R, d);
draw(T--R, d);
draw(S--P, d);
draw(circumcircle(B,D,M), c);
draw(circumcircle(C,D,S), c);
/* Place dots on and label each point */
dot(A); label("$A$", A, dir(100));
dot(B); label("$B$", B, dir(200));
dot(C); label("$C$", C, dir(340));
dot(D); label("$D$", D, dir(-90));
dot(E); label("$E$", E, 1.5*dir(100));
dot(Z); label("$Z$", Z, dir(180));
dot(K); label("$K$", K, dir(40));
dot(L); label("$L$", L, dir(-60));
dot(M); label("$M$", M, dir(180));
dot(N); label("$N$", N, dir(90));
dot(S); label("$S$", S, dir(120));
dot(T); label("$T$", T, dir(90));
dot(P); label("$P$", P, dir(-120));
dot(R); label("$R$", R, dir(0));
[/asy]](http://latex.artofproblemsolving.com/3/7/f/37f9d51c9dbaac2690d79a57a5f46dc7bedb9eae.png)
,
,
.
.
is an isosceles trapezoid, and 
.
