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.
be an isosceles trapezoid with 
. The inscribed circle 
of triangle 
meets 
at 
. Let 
be a point on the (internal) angle bisector of 
such that 
. Let the circumscribed circle of triangle 
meet line at 
and 
. Prove that the triangle 
is isosceles.
such that, for all positive integers 
and 
,
be real numbers so that 
and 
Show that
inscribed in a circle 
with center 
. Let 
the incenter of triangle and 
the contact points of the incircle with sides 
of triangle respectively . Let also 
the foot of the perpendicular line from 
to the line 
.Prove that line 
passes from the antidiametric point 
of 
in the circle .( 
is a diametre of the circle ).
be a positive integer. Two players and 
play a game in which they take turns choosing positive integers 
. The rules of the game are: takes the first turn. Determine the outcome of the game, assuming that both players use optimal strategies.
and 
, of equal radius intersect at different points 
and 
. Consider a circle externally tangent to at 
and internally tangent to at point 
. Prove that lines 
and 
intersect at a point lying on .
and be positive integers. In an 
grid, two cells are considered neighboring if they share a common edge. Kritesh performs the following actions:
in any cell of the grid.
. and .
be non-negative real numbers such that 
Prove that
be circumscribed about circle 
, and let 
be the orthocenter of 
. The circle touches line 
at . The tangent to the circle 
at meets at . Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
such that![\[ f(x)f\left(x + 4f(y)\right) = xf\left(x + 3y\right) + f(x)f(y) \]](http://latex.artofproblemsolving.com/d/b/1/db14b64aa71cb3c8dbf53707f719f5d9a43bd7e4.png)
for any positive real numbers 
.
+1 w
and 
be coprime integers. Call a set 
balanced if the numbers 
form an arithmetic progression with difference 
, and the numbers 
form an arithmetic progression with difference 
.
and 
, determine the maximum size of a collection of balanced sets such that every two of them have a non-empty intersection.
be an integer. The sequence 
is defined via 
, 
and![\[ a_{n+2} = ka_{n+1} + a_n \]](http://latex.artofproblemsolving.com/b/5/e/b5e3e6f9f0b0709113498266e41b372299a2986f.png)
for any positive integer . Prove that there are finitely many pairs 
of positive integers such that 
.
be a circumscribed quadrilateral with incircle 
and no two opposite angles equal. Let 
be an arbitrary point on the diagonal 
, which is inside . The segments 
and 
intersect at 
and 
. The tangents to at and intersect at . Prove that lies on the line .
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.
.
.
.