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 a quadrilateral inscribed in a circle with center 
and 
be the intersection of segments 
and 
. Let 
be the circumcircle of 
and 
be the circumcircle of 
. The tangent to at 
and the tangent to at 
meet at 
. The tangent to at 
and the tangent to at 
meet at 
. Show that 
.
, let 
denote the number of 
s in the binary representation of . Prove that for any positive integer 
,![\[\sum_{i=1}^{n}(-1)^{s(3i)} > 0.\]](http://latex.artofproblemsolving.com/d/e/9/de975cc2632685d7a3f71c023115535d0aeb5dbb.png)
Holden Mui
+2 w
grid (
), we color some of its cells black.A coloring is called balanced if each row and each cell contains exactly black cells.Detemine the number of balanced colorings.
inscribed in 
with 
being the midpoint of 
. The tangents at 
of intersect at . Let 
be the projection of onto 
. On the perpendicular bisector of , take a point 
that is not on and different from M. Circle 
intersects 
at 
. Lines 
and 
intersect at . Prove that 
is an isosceles triangle.
for which there exist real numbers 
satisfying 
, 
and
for 
.
be an integer. Find the smallest integer 
with the property that there exists a set of distinct real numbers such that each of its elements can be written as a sum of other distinct elements of the set.
be a regular 
-gon. Assign integers between and to the vertices of such that each integer appears exactly once. (If two assignments coincide under rotation, treat them as the same. ) An operation is a swap of the integers assigned to a pair of adjacent vertices of . Find the smallest integer such that one can achieve every other assignment from a given one with no more than operations., the point is the midpoint of and is a point on the side such that 
. The cotangents of the angles 
, 
, and 
are positive integers 
.
is also an integer and equals 
.
, some of which consisting of Fibonacci numbers.
such that for all positive real numbers 
and 
:
and be two isogonal conjugate with respect to triangle . Let 
and 
be two points lying on the circle 
such that 
and 
are perpendicular and parallel to bisector of , respectively. Prove that 
and 
bisect two arcs containing and not containing , respectively, of 
.
be a triangle such that 
and 
. The tangent at to the circumcircle 
of triangle meets the line at . Let be the reflection of in the line , let 
be the foot of the perpendicular from to 
, and let 
be the midpoint of the segment 
. Let the line 
intersect the circle again at 
. is tangent to the circumcircle of triangle 
.![\[\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\]](http://latex.artofproblemsolving.com/f/a/6/fa6c46c12e7e6bf0b04f6f9a910df6c2b452fe6a.png)
for all real numbers , , 
, each different from , and satisfying 
., , , each different from , and satisfying . there exists an -digit number divisible by 
all of whose digits are odd.
be a strictly increasing sequence of positive integers such that for all positive integers 
![\[d(2025^{a_n}-1)|a_{n+1}.\]](http://latex.artofproblemsolving.com/8/f/d/8fda531266d39c7f11c58d3c1077d2cab3357bea.png)
Show that for any positive real number 
there is a positive integers 
such that 
for all 
.
denotes the number of positive divisors of the positive integer 
.
is a polynomial of real coefficients that has exactly real zeros (counting multiplicity), then 
has exactly 
real zeros (counting multiplicity). To solve the original problem, we need an additional result: be a polynomial of real coefficients. Assume that there exists 
such that 
and 
. Then, has a non-real zero. is monic. Let be the degree of . Assume the contrary that has real zeros (counting multiplicity), let them be 
, and therefore
Since , we have 
for all 
. For all 
, we have
Take the derivative w. r. t , we have
But this yield contradiction, since for 
, the L. H. S equals 
while the R. H. S is strictly less than . 
be the degree of and 
be the coefficients of 
in , for 
. Assume the contrary that there exists a smallest 
such that 
(note that 
since 
, so we can not have 
). By the minimality of , we must have 
. Let 
, namely, the 
-th order derivative of . Since has exactly real zeros (counting multiplicity), 
must have exactly 
real zeros (counting multiplicity). And since is the degree of , we can conclude that all zeros of are real. On the other hand, it is clear that 
. However, we also have 
since . According to Lemma 1, must exist a non-real zero (contradiction).
be a polynomial with real coefficients such that all its roots are real and nonzero. Prove that there are no 2 consecutive coefficients both zeros.
