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 acute-angled scalene triangle with circumcenter 
. Denote by 
, 
, and 
the midpoints of sides 
, 
, and 
, respectively. Let 
be the circle passing through 
and tangent to 
at . The circle intersects and 
at points 
and 
, respectively (where and are distinct from ). Let 
be the midpoint of segment 
, and let 
be the intersection of lines and 
. Prove that 
and that triangle 
is isosceles.
chessboard has a real number written in it in such a way that each number is equal to the geometric mean of all the numbers a knight's move away from it.
is inscribed in circle 
. The interior angle bisector of angle 
intersects side 
and at 
and 
(other than ), respectively. Let 
be the midpoint of side . The circumcircle of triangle 
intersects sides 
and 
again at 
and 
(other than ), respectively. Let 
be the midpoint of segment 
, and let 
be the foot of the perpendicular from to line 
. Prove that line 
is tangent to the circumcircle of triangle 
.
is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
and 
be different points on a circle such that 
is not a diameter. Let 
be the tangent line to at . Point 
is such that is the midpoint of the line segment 
. Point 
is chosen on the shorter arc of so that the circumcircle 
of triangle 
intersects at two distinct points. Let be the common point of and that is closer to . Line 
meets again at 
. Prove that the line 
is tangent to . be a right angled triangle with 
. Point is the midpoint of side And be an arbitrary point on 
. The reflection of 
over intersects lines and at and , respectively. The circumcircles of 
and intersect again at 
. Prove that the center of the circumcircle of 
lies on 
.
be a trapezoid with parallel sides 
. Points 
and 
lie on the line segments 
and 
, respectively, so that 
. Suppose that there are points 
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points , , 
and 
are concyclic.
be a positive integer. A board with a format 
is divided in equal squares.Determine all integers ≥3 such that the board can be covered in 
(or 
) pieces so that there is exactly one empty square in each row and each column.
positive integers 
such that 
for any 
, 
such that 
?
![$A \in \mathbb Q[x]$](http://latex.artofproblemsolving.com/d/1/5/d158f344cf30f8beb5f2199da53272bd8e304704.png)
with ![$\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$](http://latex.artofproblemsolving.com/1/1/9/119a7b617e09ba16bb266b44ab3c85acc529f88f.png)
and 
.
. For 
, 
is the smallest number different from 
which doesn't divide 
. Proof that every number large enough appears in the sequence.
, every two opposite sides are equal. Let 
denote the point of intersection of with the perpendicular bisector of 
. Define 
and 
similarly. Prove that , , and are collinear.
is the midpoint of and 
are the orthogonal projections of 
onto 
From 
we get 
On the other hand, from 
and 
we get
into 
yields 
we conclude that 
are collinear.
be the midpoint of and let the perpendiculars from 
and 
to the line meet it at points 
, respectively. Notice that 
since we are taking projections along , the 
factor cancels out. Notice that 
and 
. Dividing the two relations yields that 
. Hence, multiplying similar relations gives that 
and by the Menelaus' theorem, the points are collinear. and radius 
and analogous ones for 
are coaxial. Any proof along these lines? 
Prove that
Where 
and 
Prove that
satisfying
Prove
