[/quote]
,
be a positive integer. Determine all sequences 
of positive integers such that
for all positive integers 
.
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and 
such that 
is an obtuse angle. Let the circumcircle of 
intersect 
and 
.
intersect 
; let the line 
intersect 
.
are concyclic.
such that 
.
.
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of 
.
be the circle with diameter 
.
be the intersection of 
and 
.
be the orthocenter of 
be the intersection of 
and 
be the lines through 
tangent to 
respectively. Let 
and 
.
be the intersection of 
and 
.
be the line through 
and let 
be the reflection of 
is tangent to 
.
where 
are integers in 
are each labeled with a positive integer less than or equal to 
such that every label is used exactly 
and 
such that the following property holds: For any two lattice points 
and 
that are both labeled 
,
and 
are both divisible by 
be an odd prime. Define 
over the positive integers as follows:
where 
such that 
.
such that 
and 
is a perfect square.
denote the set of all positive integers less than 
that are relatively prime to 
.
reducible over the rational numbers, given that it has integer coefficients?
students, each pair of students have watched a movie together. Let the set of movies watched by a student be his movie collection. If every student has watched every movie at most once, at least how many different movie collections can these students have?
be non-negative real numbers such that 
Prove that
and 
.
and 
be real numbers. Let 
be the matrix with entries ![\[a_{ij} = \begin{cases}1,&\text{if }x_i + y_j\geq 0;\\0,&\text{if }x_i + y_j < 0.\end{cases}\]](http://latex.artofproblemsolving.com/c/0/8/c0856fe7b2b113180659952c4d216a01d8849f64.png)
Suppose that 
matrix with entries 
,
such that the sum of the elements in each row and each column of 
.
.
When does equality hold?
Y by HWenslawski, Adventure10, Mango247, DEKT
Let 
be the incenter of triangle 
, and let 
, 
, and 
be the circumcenters of triangles 
, 
, and 
, respectively. Prove that the circumcircles of triangles and 
are concentric.
be the incenter of triangle 
, and let 
, 
, and 
be the circumcenters of triangles 
, 
, and 
, respectively. Prove that the circumcircles of triangles and 
are concentric.This post has been edited 2 times. Last edited by djmathman, May 27, 2018, 5:18 PM
Reason: official wording!
Reason: official wording!
is a triangle with incenter 
lie on a circle whose center is the circumcenter of 
with the circumcircle of 
and the circumcircle of 
![[asy]
import olympiad;
unitsize(50);
pair A,B,C,O,I;
A=origin; B=2*right; C=1.5*dir(70);
I=incenter(A,B,C); // olympiad - incenter
draw(A--B--C--cycle);
dot(I);
draw(circumcircle(A,B,C)); // olympiad - circumcircle
draw(incircle(A,B,C)); // olympiad - incircle
label("$I$",I,S);
label("$A$",B,E);
label("$B$",A,W);
label("$C$",C,N);
[/asy]](http://latex.artofproblemsolving.com/9/a/c/9ac368ff9f71989b350bddd4e16b1d095ec92e04.png)
Now, let 
and 
Thus, 
Similarly, 
be circumcentres of triangle 
respectively.
is cyclic quad.
all lie on circumcircle of 
and triangle 
lie on the circumcircle of 
,
is also known as the circum-incentral triangle of 
and 
.
,
and 
hit 
,
respectively. By the Incenter-Excenter Lemma, 
,
,
and 
.
and 
denote the circumcircle of 
.
and 
so 
are concyclic, implying the result.
even with the diagram in front of me. 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(8cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.403215344845714, xmax = 31.496302855665586, ymin = -11.226497779920303, ymax = 5.8723246089452745; /* image dimensions */
pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqffff = rgb(0.,1.,1.);
pair A = (8.82,1.78), B = (5.34,-5.36), C = (15.26,-5.36), I = (9.463831335028141,-2.7823620420270165), D = (9.812433245429258,-5.36);
draw(A--B--C--cycle, linewidth(0.4) + blue);
draw(circle(I, 2.60110375308607), linewidth(0.8) + ffxfqq);
/* draw figures */
draw(A--B, linewidth(0.4) + blue);
draw(B--C, linewidth(0.4) + blue);
draw(C--A, linewidth(0.4) + blue);
draw(circle((10.3,-3.3594117647058823), 5.348266381473275), linewidth(0.8) + qqffff);
draw(C--(5.492370646566023,-1.0161974579901623), linewidth(0.8) + red);
/* dots and labels */
dot(A,dotstyle);
label("$A$", (8.891419393101584,1.9880487969547336), NE * labelscalefactor);
dot(B,dotstyle);
label("$B$", (5.427606426738809,-5.159819578718582), NE * labelscalefactor);
dot(C,dotstyle);
label("$C$", (15.338516359395303,-5.159819578718582), NE * labelscalefactor);
dot(I,linewidth(4.pt) + dotstyle);
label("$I$", (9.55214672194535,-2.6170204646835367), NE * labelscalefactor);
dot(D,dotstyle);
label("$D$", (9.89252140650123,-5.159819578718582), NE * labelscalefactor);
dot((10.3,-8.707678146179159),linewidth(4.pt) + dotstyle);
label("$A'$", (10.473160574273024,-9.1041615115131), NE * labelscalefactor);
dot((14.271460688462117,0.22269003273053659),linewidth(4.pt) + dotstyle);
label("$B'$", (14.417502507067628,0.36626353524734273), NE * labelscalefactor);
dot((5.492370646566023,-1.0161974579901623),linewidth(4.pt) + dotstyle);
label("$C'$", (5.287452144862858,-0.7149266392242511), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/d/3/9/d39cc7ea1142641c582923c7b2820302db6dd860.png)
![\[C'I=C'A=C'B\]](http://latex.artofproblemsolving.com/b/d/1/bd128cb2433e6db2191e15b7b595d95529a814df.png)
and 
and also it the circumcenter of 
be the center of 
and 
the 
Since 
and there is only one such point, 
must be the midpoint of 
and therefore on 
and 
are on 
.
,
,
.
finishing the problem 
and 
.
,
is the excenter of the 
,
is equidistant from 
are on 
so they have the same center
as 
respectively.
,
is midpoint of 
,
coincides with that of 
![[asy]
size(200);
pair A, B, C, I, OA, OB, OC;
A = dir(100);
B = dir(200);
C = dir(340);
I = incenter(A,B,C);
OA = circumcenter(I,B,C);
OB = circumcenter(I,C,A);
OC = circumcenter(I,A,B);
dot("$A$", A, A);
dot("$B$", B, B);
dot("$C$", C, C);
dot("$M_A$", OA, OA);
dot("$M_B$", OB, OB);
dot("$M_C$", OC, OC);
dot("$I$", I, dir(I));
draw(unitcircle);
draw(A--B--C--cycle);
[/asy]](http://latex.artofproblemsolving.com/7/2/5/72565c6d2bf4c68d3645ebbc38fb712a7d1aebd2.png)
be the midpoints of arcs (on the circumcircle of 
respectively. We know that 
because they are the angle bisectors. It suffices to prove that 
We have 
and 
This means 
by triangle angle sum and since 
we have 
A similar argument proves that 
and so 
is the circumcenter of 
or 
Repeating this for 
yields the desired result.
,
,
are the midpoints of arcs 
and 
are cocentric, as desired.
