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.
has only postive integers in forms
where 
![$f:[1,b]\rightarrow \mathbb{R},\ f(x)=x+\dfrac{bc}{x}$](http://latex.artofproblemsolving.com/5/6/d/56d1aa809cfad6667896f3ec4e2a1951b13a394c.png)
. As 
, we infer that ![$f(x)\ge f(b)=b+c,\ \forall x\in [1,b]$](http://latex.artofproblemsolving.com/3/9/6/396ded9b7014efff81a94e281d94b95dc71b62ab.png)
; in particular, 
.
, hence 
. It is easy to see that for 
, if 
, then 
. If 
, 
, so 
, which is in contradiction with the fact that 
. Thereby, 
.
. If 
, contradiction; so 
and 
, or equivalently 
and 
is odd, we get that 
. Furthermore, 
, hence 
. Now 
which together with the fact that 
, we get that 
, so the family of the solutions 
is described by the set
, then using 
, and 
, we get 
, which is clearly impossible. Thus, 
must hold.
and implies 
. Keeping this in mind, we now observe that, 
, implies 
. Thus, 
. Now, note also that, if 
, then 
, and thus, the largest power of 
dividing either 
or 
is exactly . Clearly, 
, and thus, 
. Moreover, if 
, then we again have a contradiction, as 
. Thus, 
. This yields, 
, which brings us, 
. Now, using 
, we immediately obtain 
. This immediately establishes (since is odd), that .
It follows that 
and 
we get 
Now 
and is an odd multiple of 2, we have that 4 must not divide either one of them. It follows that 
or 
But 
So 
Q.E.D.
with odd positive integers 
such that 
and 
. It is easy to see that 
. Further,
Of course, numbers 
are even. If both 
and 
are divisible by 
, then 
are not divisible by , so 
, that's impossible. Similar things happen when both 
and 
are divisible by , so we have two cases:
. So 
, hence 
. Further, 
, therefore 
, and we finally get 
.
. Here 
, hence . Also we have that 
, therefore and again.
and thus . Let
where 
are odd integers. Then, the condition becomes![\[2^{2k-2} - x^2 = 2^{2m-2} - y^2 \Longrightarrow 2^{2k-2} - 2^{2m-2} = x^2-y^2\]](http://latex.artofproblemsolving.com/4/d/d/4ddafb7c4a2ed27635fd75217fca16fcc439da28.png)
.
Note that![\[k\geq v_2(x^2-y^2) = v_2(2^{2k-2} - 2^{2m-2}) = 2m-2\]](http://latex.artofproblemsolving.com/2/2/a/22aa55b0d3cbecc2c70081b403a9f8361021af84.png)
where the first inequality is true because 
odd implies that
, and 
implies that 
. 
with equality when 
.
Note that![\[2^{2k-2} - 2^{2m-2} = x^2-y^2 \leq (2^{k-1}-1)^2 - 1^2,\]](http://latex.artofproblemsolving.com/f/8/b/f8b20fd1b76a3e6c7b65073e836c05410fbc908d.png)
thus 
, so . , and so the 
equality case must hold and we're done. 
.
:, we have: 
Therefore 
or . Note that: 
Note that 
are odd integers.
have a difference of 
which is an odd multiple of , we must have either: or 
. Notice that: 
which implies . Notice that: 
which implies . Also, we have: 
. Thus: 
Since is odd integer, and we are done.
. So, 
. Note that 
, and 
.
, 
, a contradiction as 
. So, 
and 
. Since 
, 
and 
. Hence, 
, and 
.
, consider a prime 
such that 
. 
, a contradiction, so 
.
Prove that
Where 
and 
Prove that
satisfying
Prove
be odd integers such that 
and 
and 
for some integers 
and 
,
or 
, contradiction )