AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
be positive integers. If 
is divisible by 
for every positive integer 
, prove that 
.
with 
and 
. Lines 
and 
intersect at 
, and lines 
and 
intersect at 
. Let 
be the midpoints of sides 
, respectively. Let 
and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to 
if and only if 
divides into two triangles with equal area.
kawaii if it satisfies 
(
is the number of positive factors of , while 
is the number of integers not more than that are relatively prime with ). Find all that is kawaii.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
, 
, 
be nonzero real numbers such that 
and 
. Find the value of
.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be a function satisfying
for all integers and . Show that there exist positive integers 
and such that 
for all integers .
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
of positive integers with 
prime, 
, and 
, such that 
is a multiple of 
.
a point in the space and 
the centroid of a tetrahedron . Prove that:
be a logarithmic spiral centered at the origin (ie curve satisfying for any point on it, line 
makes a fixed angle with the tangent to at ). Let 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
on the spiral, the polar of wrt. is tangent to the spiral.
be a monic polynomial so that there exists another real coefficients 
that satisfy![\[P(x^2-2)=P(x)Q(x)\]](http://latex.artofproblemsolving.com/6/9/7/697faa929e4fda7a6e9b1cd97849bd42ffc14306.png)
Determine all complex roots that are possible from that satisfies (a), determine whether should have real coefficients or not.
outside triangle 
, 
intersects 
at 
, we easily have 
are collinear and 
are concyclic, this leads to 
.
, now we only have to prove that 
. Notice that 
are tangents of at 
, so we have 
is a harmonic quadrilateral and 
is the symmedian of triangle . We have 
and after several angle calculations, we have 
, hence proved
; additionally, construct the equilateral triangle 
, 
and on different sides of .
rotation about 
will map 
to and 
to , hence 
(Torricelli problem).
(s.a.s.), and 
, hence 
, done.
be the circumcenter of . Is is obvious that must lie on the circumcircle of 
meets this circle again at . Then 
is equilateral. If we choose a point 
on 
so that 
then 
is also equilateral. But then it is easy to see that 
, hence 
. Thus 
and to complete the proof it remains to show that 
are in fact tangents to the circumcircle of and is the A-symmedian, then 
. By sinus law in 
: 
and in triangle 
: 
. Indeed, q.e.d. and median 
is 
, 
, 
, 
. By Stewart's Theorem we have ![\[AM^2=(2d)^2=2b^2+2c^2-a^2=2b^2+2c^2-(b^2+c^2-2bc\cos 60^{\circ})=b^2+c^2+bc.\]](http://latex.artofproblemsolving.com/5/5/a/55a8f3d968a5ae50821ead4aa06e96ba17b6068a.png)
Hence it suffices to show 
. Now Cosine Rule on 
, 
, 
yields
Now 
gives 
. Denote this as 
, then 
gives 
. It suffices to show this is equivalent to 
.
to 
such that 
and extend 
to 
such that 
. Then 
. Therefore 
is cyclic, so by Power of a Point 
, as desired. Hence done.
is the circumcenter of 
implying that is an equilateral triangle!
is the circumcenter of implying that is an equilateral triangle! is not the circumcenter of so is not equilateral
is the circumcenter of implying that is an equilateral triangle! is not the circumcenter of so is not equilateral
(120 = 2*60)
is the circumcenter of implying that is an equilateral triangle! is not the circumcenter of so is not equilateral (120 = 2*60) lies on circumcircle of 
not circumcenter
is not the circumcenter of so is not equilateral (120 = 2*60) lies on circumcircle of not circumcenter and happen to be one and the same as per the given inputs in the problem using central angle theorem on Point and vertex .
(120 = 2*60) lies on circumcircle of not circumcenter and happen to be one and the same as per the given inputs in the problem using central angle theorem on Point and vertex .[/quote]
be the point on such that 
is equilateral triangle. Let 
be the reflection of over .
. Hence, we need 
. Reflect diagram over , note that goes to 
, the midpoint of arc 
and 
goes to 
, where 
is the midpoint of arc as 
. Thus, 
. Also lies on the perpendicular bisector of 
as it is center of 
. We conclude that .![[asy]import olympiad;
size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta;
pair A,B,C,M,a,I,x,X,T,N,K,O;
A=dir(120);B=dir(210);C=dir(330);M=midpoint(B--C);a=2M-A;path w=circumcircle(a,B,C);I=incenter(A,B,C);x=foot(a,A,I);X=2x-a;T=intersectionpoints(A--X,w)[0];N=2M-X;K=extension(X,N,A,I);O=(0,0);
draw(A--B--C--cycle,deep);draw(w,deep);draw(A--X,med);draw(A--a,med);draw(B--X--C,deep);draw(B--T,med);draw(C--T,med);draw(circumcircle(A,B,C),deep);draw(A--N,deep);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$M$",M,dir(M));
dot("$A'$",a,dir(a));
dot("$X$",X,dir(X));
dot("$T$",T,dir(T));
dot("$N$",N,dir(N));dot("$K$",K,dir(K));dot("$O$",O,N);
[/asy]](http://latex.artofproblemsolving.com/5/f/e/5fee5ca11507b9b3d6fa0bba02f0d96954eb854b.png)
be the point in circumcircle of equilateral triangle . Let's assume lies on minor arc , Then we have 
.
and 
outside of the . Let 
. Let 
. It's easy to conclude that 
. Since 
is a isosceles trapezoid, we have that 
. But 
, by our lemma and we are done.
.
.
.