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 
of a circle intersect at 
, which is the midpoint of the chord 
. The points 
and 
are the intersections of the segments 
and , respectively, and 
and , respectively. Show that is the midpoint of 
.
such that 
is a perfect square.
. Points 
and 
lie on segments and 
, respectively. Lines 
and 
intersect at point 
. The circumcircles of triangles 
and 
intersect at a second point 
. The circumcircles of triangles and 
intersect at a second point 
. Point 
lies on segment 
such that 
. Prove that triangles 
and 
are similar.
the set of all positive real numbers.
which have the property:![\[f(x)f(y)=2f(x+yf(x))\]](http://latex.artofproblemsolving.com/4/f/f/4ffbde35ff082585a188c1a4368d39338d1b07a4.png)
and 
.
such that :
for which there exists a sequence 
of positive integers satisfying the following conditions:[list]
for all integers 
for all integers 
the sequence 
is a permutation of
[/list]
such that 
is a positive integer.
be positive real numbers such that 
. Prove that
.
questions, each with 
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
, 
, and 
of a hexagon 
inscribed in a circle 
intersect at a point 
, and the acute angle between any two of them is 
. Let 
be the radius of the circle tangent to segments 
and 
and internally tangent to ; the radii 
, 
, 
, 
, and 
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
such that 
for every 
.
and be polynomials with integer coefficients. Suppose that the integers 
and 
are roots of , and that 
. Prove that the equation 
has no integer solutions.
. Find .
![$A=[1,0,0]$](http://latex.artofproblemsolving.com/5/e/d/5edd90e5880a3e676c65b04ec622149ac1c9e6fd.png)
, ![$B=[0,1,0]$](http://latex.artofproblemsolving.com/8/0/8/808aee5270277243ef8023672d85c99f1f4c7c8d.png)
, ![$C=[0,0,1]$](http://latex.artofproblemsolving.com/5/3/5/5350fac5fb4b9fa81700a91f7f11ea10503deffb.png)
, 
, 
, 
, we get:
, we have 
and 
, so 
and , that is, is tangent to 
![[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A = origin, B = (2,0);
path El = ellipse((1,0), sqrt(2),1);
pair C = relpoint(El, 0.36);
pair M = (A+B)/2, P = foot(A,C,B);
real r = abs(B-C), s = abs(A-C);
pair T = C * (r+s)/s, N = (A+T)/2, D = s/(r+s)*B;
draw(A--B--C--cycle,rgb(0.1,0.1,0.8));
draw(C--N,rgb(0.1,0.5,0.1));
draw(C--D^^N--M,rgb(0.9,0.1,0.1));
draw(circumcircle(B,M,P),orange+linetype("3 3"));
draw(A--P^^rightanglemark(A,P,C,2),rgb(0.6,0.1,0.6));
clip((-2,-1)--(-2,1.6)--(2.2,1.6)--(2.2,-1)--cycle);
dot("$A$",A,SW);
dot("$B$",B,SE);
dot("$C$",C,NW);
dot("$D$",D,S);
dot("$P$",P,dir(A--P));
dot("$M$",M,S);
dot("$N$",N,NW);
[/asy]](http://latex.artofproblemsolving.com/8/2/8/82837c16b10c60f37621b4f091dcc4d3898f00bb.png)
be the point on ray for which 
(not pictured above for space reasons), and let 
be the midpoint of 
. Then the given length condition implies![\[
AN = \frac{AC+CB}2 = \frac{AB}{\sqrt 2},
\]](http://latex.artofproblemsolving.com/4/3/c/43cdd9f20d046d60f127129dc4e2e911b9bcc8b8.png)
or 
. This means that the circumcircle of 
is tangent to at , implying 
.
is cyclic. To prove this, let be the foot of the angle bisector from 
. Observe that is the reflection of 
across the external angle bisector of 
, so 
. Thus, a (simple?) angle chase yields
Thus quadrilateral 
is cyclic, as desired.
is tangent to at , which is exactly what we wanted.
, let be the midpoint of and the foot of the altitude to . Prove that if 
, then the circumcircle of triangle 
is tangent to .
and prove F,M,B,P is clycic,as shown in the diagram.
be a point on such that 
, We claim that it is the tangency point
tangents 
tangents 
. and and tangents at so 
are concyclic and we are done.
be a point on rey behind such that
, similar argument to the first case finishes.
Trivial
and from B be 
.
and
and 
.