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.
+2 w
be a circle and 
a given point in the plane. Each line through which intersects determines a chord of . Show that the midpoints of these chords lie on a circle.
a number from a set 
is written --- in the first row numbers 
, in the second 
and so on. Exactly 
of them have been chosen, no two from the same row or the same column. Let us denote by 
a number chosen from row number 
. Show that:![\[ \frac{1^2}{a_1}+\frac{2^2}{a_2}+\ldots +\frac{n^2}{a_n}\geq \frac{n+2}{2}-\frac{1}{n^2+1}\]](http://latex.artofproblemsolving.com/a/d/5/ad5ae6861da79a08d32093f63827016868856a3a.png)
lying outside the circle 
, two tangents are drawn touching at points 
and 
. A point 
is chosen on the segment 
. Let points and 
be the midpoints of segments 
and 
respectively. The circumcircle of triangle 
intersects again at point 
(
). Prove that the line 
passes through the centroid of triangle 
.
has circumcircle 
, drop the perpendicular line from 
to 
and meet at point 
, similarly, altitude from 
to 
meets at 
. Prove that if 
be the circumcircle of a triangle 
. Denote by and the midpoints of the sides 
and , respectively, and denote by 
the midpoint of the arc of not containing . The circumcircles of the triangles 
and 
intersect the perpendicular bisectors of and at points 
and 
, respectively; assume that and lie inside the triangle . The lines and 
intersect at . Prove that 
.
be the set of positive rational numbers. Construct a function 
such that![\[ f(xf(y)) = \frac {f(x)}{y}
\]](http://latex.artofproblemsolving.com/5/9/1/59172e31cf6b7d885295f6e1daf6891a71739247.png)
, 
in .
be an acute-angled triangle 
with 
and incenter 
. Let 
be the mixtilinear circle at vertex 
, i.e. the circle internally tangent to the circumcircle of 
and also tangent to lines 
and 
. A circle 
passes through points 
and 
and is tangent to at point 
, with 
and being inside 
. Prove that:
and 
with 
, and announces them to the deputy leader and a contestant. The leader then secretly tells the deputy leader an -digit binary string, and the deputy leader writes down all -digit binary strings which differ from the leader’s in exactly positions. (For example, if 
and 
, and if the leader chooses 
, the deputy leader would write down 
and 
.) The contestant is allowed to look at the strings written by the deputy leader and guess the leader’s string. What is the minimum number of guesses (in terms of and ) needed to guarantee the correct answer? be an acute scalene triangle. A point 
varies on its side . The points 
and 
are the midpoints of the arcs 
and 
(not containing ) of the circumcircles of triangles 
and 
, respectively. Prove that the circumcircle of triangle 
passes through a fixed point, independent of the choice of on .
are positive integer.
touches the sides , 
and at , and 
respectively. Let denote the other point of intersection of 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
bisects 
. Denote 
and 
.![\[(PQ; EF) = -1.\]](http://latex.artofproblemsolving.com/a/7/7/a778b533bd051f336bd505b11b4a0b07dafde6e2.png)
Combining this with 
, we use the Apollonius Circle Lemma to deduce the desired.![[asy]
size(225); defaultpen(linewidth(0.4)+fontsize(10));
pair A, B, C, H, D, E, F, P, Q;
A = dir(110);
B = dir(220);
C = dir(320);
D = foot(A, B, C);
H = .5A + .5D;
E = extension(A, C, B, H);
F = extension(A, B, C, H);
P = extension(E, F, B, C);
Q = extension(E, F, A, D);
filldraw(E--P--D--cycle, palegreen);
draw(A--B--C--A--D);
draw(F--D--E--P--B);
draw(B--E^^C--F);
dot("$A$", A, N);
dot("$B$", B, S);
dot("$C$", C, S);
dot("$P$", P, SW);
dot("$Q$", Q, dir(60)*1.5);
dot("$H$", H, dir(0));
dot("$E$", E, NE);
dot("$F$", F, NW);
dot("$D$", D, S);
[/asy]](http://latex.artofproblemsolving.com/0/1/b/01b15151e074158d0cfed459dc83a90462df7134.png)
and 
. Using the Ceva-Menelaus Configuration, it then follows that 
. Projecting through , we thus find that
Since 
and 
, the Angle Bisector Configuration then implies that 
bisects 
, so we therefore conclude that 
, as desired. 
and 
. We first use a lemma:
be collinear points in that order and let be any point not on this line. Then any of the following two conditions imply the third:
is harmonic;
;
bisects 
.
we have 
, and further that 
. Hence 
bisects , and we are done. 
and let 
. Then by right angles and bisectors lemma it suffices to show that 
. However, projecting we find,
as desired.
but this is true since 
are concurrent cevians in 
and 
. Then by EGMO 
, we have 
. And then by 
we have 
harmonic and 
, so 
bisects .
, 
, 
, 
, 
.
and 
.
and 
are opposites so 
so , as desired.
.
as the image of a point , and 
as the image of a line 
. lies on , 
and 
are parallel lines.
, so 
and 
are perpendicular to 
.
is a rectangle.
.
as desired.
. If 
the statement is obviously true.
to meet at ; it's well-known that 
. Then, we cite EGMO lemma 9.18. Because and 
, then we have that bisects , as desired. 
and 
. From EGMO 9.11,![\[(X,D;B,C) \overset{A}{=} (X,Y;F,E) = -1.\]](http://latex.artofproblemsolving.com/f/5/5/f55404d99ef4c0e5b595c7f9dc36d391a1aa8808.png)
, EGMO 9.18 finishes. 
which finishes from right angles and bisectors
be any interior point on 
,
at 
respectively. Prove that 
and by ceva's theorem, 
which gives 
and since both are acute, 
.
,
,
are Cevians of 
.
perpendicular to 
.
.
,
,
,
.
,
,
.
and 
,
respectively. This gives 
and 
.
,
to get 
.
,
,
.
,
.
,
.
,
and 
.
and 
,
until 
. Can someone please explain to me.
.
.
.
being a pencil we actually get that 
.
and 
,
or just 
Therefore by right angles and bisectors we conclude 
and 
Its clear that (perspectivity at 
From this, 
bisects 
(since 
and follows from law of sines), as desired.