[/quote]
,
have the Dirichlet propriety, if 
,
.
with 
subset of 
and have the Dirichlet propriety.
?
?
denote the set of positive integers. Determine all 
satisfying the following equation:
and prime 
.
,
)
and there exists no index 
with 
such that![\[a_m > \frac{i_{j+1} - m}{i_{j+1} - i_j} \cdot a_{i_j} + \frac{m - i_j}{i_{j+1} - i_j} \cdot a_{i_{j+1}};\]](http://latex.artofproblemsolving.com/c/a/b/cab180c10974a06ae24c73c6f31fc071cc061dbc.png)
- A lower subsequence if 
and there exists no index ![\[a_m > \frac{i_{j+1} - m}{i_{j+1} - i_j} \cdot a_{i_j} + \frac{m - i_j}{i_{j+1} - i_j} \cdot a_{i_{j+1}}.\]](http://latex.artofproblemsolving.com/e/b/4/eb4126427c43773d8e11a43cb0f2c281eb809913.png)
Find the smallest positive integer 
such that in any sequence of 
or a lower subsequence of length 
.
is inscribed in circle 
.
intersects line 
at point 
.
intersects line 
at point 
.
be the circumcenter of 
.
meets it at point 
.
and 
are either equal or supplementary.
with the following property: For any six non-negative real numbers 
summing to 
,
summing to 
,![\[(a_i + a_j + a_k)(b_i + b_j + b_k) \leq M.\]](http://latex.artofproblemsolving.com/3/e/a/3ea74e39bd863244cb937144fdf0c2d329361248.png)
Proposed by Wu Zhuo and Xu Wenchang
satisfies that for any positive integer 
,![\[f(f(x)) + f(x) \geq 301 + x.\]](http://latex.artofproblemsolving.com/e/e/7/ee7d6ea65f3b290774f02f4be80570711552ecf3.png)
Find the minimum possible value of 
.
,
such that no matter how each integer point in the coordinate plane 
is colored with one of 
of the same color such that the area of 
is at most 
denote the sum of the digits of the positive integer 
in base 
,
such that for any infinite subset 
and 
in 
for each 
.
with 
-
,
is 
be 
,
lies on arc 
,
intersects the circumcircle of 
.
and 
,
intersects 
at 
respectively, prove that 
are concurrent.
and circle 
are externally tangent at point 
.
on circle 
and 
.
is extended to intersect circle 
.
is taken on the arc 
of circle 
is extended to intersect the angle bisector of 
at point 
.
and the circle passing through points 
intersect at another point 
.
.
be a set of points in space such that no four points are coplanar. Some pairs of points are connected by line segments, and let 
of segments in 
of real numbers.
grid 
.
in every square 
.
in the 
-
.
.
for all 
,
for all 
.
denote the set of positive integers. Let 
be a function satisfying the following property: for 
,![\[
f(mn)^2 = f(m^2)f(f(n))f(mf(n))
\]](http://latex.artofproblemsolving.com/7/1/8/71854a10658c426000e2ff925d2f8891d762d911.png)
holds if and only if 
and 
.
is called a "good number" if there exists a sequence of integers 
such that for any positive integer 
,![\(a_{n+1} = [\alpha a_n]\)](http://latex.artofproblemsolving.com/b/5/a/b5a81fcae0d4bedc310fe29408fdd439634ce6fc.png)
,
is even if and only if 
and 
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 
be an intersection of 
and 
.
is a harmonic quadrilateral with diameter 
,
:
So, 
bisects 
.
,
.
and 
are concyclic (since 
)
and 
meets 
.
.
.
and 
are concyclic.
,
meets 
again at 
.
.
,
.
We must have 
Therefore, 
as desired.
