[/quote]
,
is called charrua if there is a positive integer 
such that 
and 
are both square numbers; if there is no such number 
such that 
is charrua.
,
,
,
,
denote the set of positive integers. A function 
is said to be bonza if![\[
f(a)~~\text{divides}~~b^a-f(b)^{f(a)}
\]](http://latex.artofproblemsolving.com/7/b/b/7bb6b2b4fcde900ee6f7ef90b906525d61312796.png)
for all positive integers 
and 
.
for all bonza functions 
and all positive integers 
are defined by 
and ![\[a_{n + 1} = \frac {1 + a_n + a_nb_n}{b_n}, \quad b_{n + 1} = \frac {1 + b_n + a_nb_n}{a_n}.\]](http://latex.artofproblemsolving.com/e/4/9/e497e9872753206cebdf99b9a830e82821eb9bc3.png)
.
for positive integers 
across all ratings for which they are eligible.
be a parabola that passes through the vertices of a triangle 
.
that is tangent to 
such that![$$\left(a+b+c+d-K\right)^2+2\left[1-\left(3\sqrt2-4\right)K\right]abcd\geq\left(3\sqrt2-K\right)^2$$](http://latex.artofproblemsolving.com/3/5/3/35382fd52c29efa5185d1b9fbeef944a0fd2fee3.png)
for all 
satisfying
.
be its circumcircle, 
its orthocenter, and 
the foot of the altitude from 
.
be the midpoint of 
.
be the point on 
such that 
and let 
.
,
,
and 
are tangent to each other.
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
, for all 
are different and chosen from the set 
.![\[\frac{(V+U+Q+A+R)^2}{V-U-Q+A+R}=V^{{{U^Q}^A}^R}.\]](http://latex.artofproblemsolving.com/d/4/f/d4fd4785add9eec6ad20ad288503712c8e31686a.png)
and 
,
intersects 
,
lie on 
be the circumcentre of triangle 
.
meets 
and meets 
.
.
.
Y by Rounak_iitr
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
I have done some constructions, but still did not quite get to the answer, see diagram attached below
Attachments:
is midpoint of arc 
is perpendicular bisector of segment 
intersect 
and 
implies that 
.
collinear
on 
. so line 
.
implies that 
is midpoint of 
![[asy]
import graph;
import geometry;
size(9cm);
defaultpen(fontsize(10pt));
pair A = (0,2);
pair B = (-0.8,0);
pair C = (3,0);
pair M = (B+C)/2;
pair in = unit(B-A)+unit(C-A)+A;
pair ex = rotate(90,A)*in;
pair D = extension(A,ex,B,C);
path circ1 = circumcircle(A,B,C);
path circ2 = circumcircle(A,D,M);
pair F = intersectionpoints(A+3*(A-C)--A, circ2)[0];
pair E1 = intersectionpoints(B+3*(B-A)--A, circ2)[0];
pair N1 = (E1+F)/2;
pair P = intersectionpoints(circ1,circ2)[1];
pair Q = extension(F,E1,D,C);
path circ3 = circumcircle(Q,N1,M);
path circ4 = circumcircle(Q,B,E1);
draw(A--B--C--cycle, black);
draw(A--F, black);
draw(B--E1, black);
draw(D--B, black);
draw(E1--F, blue);
draw(D--P, orange);
draw(M--N1, royalblue);
draw(A--D, royalblue);
draw(circ1);
draw(circ2, red);
draw(circ3, deepgreen+dashed);
draw(circ4, deepgreen+dashed);
dot(A);
dot(B);
dot(C);
dot(D);
dot(M);
dot(F);
dot(E1);
dot(N1);
dot(P);
dot(Q);
label("$A$", A, N, black);
label("$B$", B, S+1.25W, black);
label("$C$", C, SE, black);
label("$D$", D, W, black);
label("$M$", M, NE, black);
label("$E$", E1, S, black);
label("$F$", F, N, black);
label("$N$", N1, SW, black);
label("$P$", P, SE, black);
label("$Q$", Q, NW, black);
[/asy]](http://latex.artofproblemsolving.com/8/8/f/88f1dd7b5e6ab5fbcc48f98610cee5e316da32bb.png)
because 
.
be an intersection of the line 
and 
and 
.
is a harmonic quadrilateral with diameter 
,
:
So, 
bisects 
.
,
.
and 
)
and 
meets 
.
.
.
and 
are concyclic.
,
meets 
again at 
.
.
,
.
We must have 
Therefore, 
as desired.
