by 
grid of squares coloured in the same way as a standard checkerboard. Find the total number of ways to place 
counters on white squares so that each square contains at most one counter and no two counters are in diagonally adjacent white squares.
are roots of 
,
be a positive integer . Prove that there exists infinitely many pairs of positive integers 
such that 
and :
matrix whose entries come from the set 
is called a silver matrix if, for each 
,
-
.
;
.
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
,
are all greater than 
.
,
.
and 
to be the remainder when 
is divided by 
.
and 
is odd for 
.
or an equivalent formulation. To prove this construction works, we need to show that
is always an integer between 
and 
inclusive.
is divided by 
.
,
since it telescopes.
,![\[\frac{r_{i+1} n - r_i}{2^i} > \frac{r_{i+1} n}{2^i} - 1 \geq \frac{n}{2^i} - 1 \geq \frac{n}{2^k} - 1.\]](http://latex.artofproblemsolving.com/4/2/0/420d26b6099ca264846885e7eef23b2c9c7b2dd0.png)
Thus, each digit is lower bounded by 
,
becomes arbitrarily large.
fails.
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
,
such that 
has no nonreal roots. We can also assume 
case is trivial as the constant term must be nonzero, so fix 
.
.
.
which has the 
for all integers 
.
be the coefficient of the 
term of 
for all 
,
for which 
.
and 
.
and 
.
is the coefficient of the 
term in 
and 
.
,
.
,
)
has 
distinct real roots as well, along with two consecutive nonleading coefficients equal to zero, but one degree lower, by the Power Rule. By doing this repeatedly, we eventually end up with a polynomial where the two consecutive coefficients have degree 
,
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
distinct from 
and 
,
such[/center]
is directly similar to either 
or 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
,
to 
,
to 
.
to be the set of points strictly outside the circle with diameter 
.
is always acute. Suppose two roads 
and 
cross. Then, 
is a convex quadrilateral. Since 
.
is not acute but there does not exist an 
such that 
are connected by some path. This is trivially true for 
,
are also connected.
is obtuse, we have 
.
,
,
is not mentioned. Do not reward points if the induction argument is incomplete or incorrect.
be the orthocenter of acute triangle 
be the foot of the altitude from 
.
intersects line 
.
be the center of 
.
be the intersection of 
and the line through 
,
.
be the foot of the altitude from 
and 
,
is a midline and thus 
.
is a midline of 
,
and thus 
.
(by definition), by either HL congruence or the Pythagorean theorem, we must have 
.
.
and 
to attempt to identify 
.
or 
,
.
.![\[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]](http://latex.artofproblemsolving.com/3/c/5/3c5984d0003c1ca2df2e68841dd9e93ed6c2d27c.png)
is an integer for every positive integer 
is divisible by 
,
be a prime power dividing 
.
.
,
if 
and 
otherwise, so for each term, either both the numerator are divisible by 
,
,
,
.
has an inverse modulo 
.![\[\binom ni \equiv \pm 1 \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor} \pmod{p^a}.\]](http://latex.artofproblemsolving.com/0/1/c/01c00beb6efc07ed028fd0273f262b8bfb965d33.png)
modulo 
.
,![\[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (\pm 1)^k \equiv p \equiv 0 \pmod p.\]](http://latex.artofproblemsolving.com/2/e/c/2ececb6afa8a10e116056cdc57a6c948298d0c29.png)
where 
,
satisfies the induction hypothesis for the prime 
.![\[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (\pm 1)^k \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}^k \equiv p\sum_{i=0}^d \binom di^k \pmod {p^a}\]](http://latex.artofproblemsolving.com/6/c/9/6c9a4b6aac62b1befef1511451c9fa30993d2cf7.png)
By the induction hypothesis, 
divides the inner binomial sum, so since we are multiplying it by 
.
.
is incorrect, reward up to 
points total.
.
between the set of people except Evan 
if the total score of the cupcakes in that bubble with respect to 
.
denotes the set of neighbors of 
people.
.
from the graph, noting that no person in 
elements are removed from the set 
,
.
-
.
which are simultaneously 
-
-
-
satisfying ![\[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]](http://latex.artofproblemsolving.com/2/6/e/26edbd87f49eb68ec9b662b602db712e5151dc9c.png)
,
.
and 
intersect at point 
,
.
lie on a circle 
and point 
and 
are tangent to 
are collinear, and (iii) 
.
bisects 
Y by ehuseyinyigit
Let 
be an acute scalene triangle inscribed in a circle 
. The angle bisector of 
intersects 
at 
and at 
. The point 
is the midpoint of 
. Let 
be the altitude in 
, and the circumcircle of 
intersects again at 
. Let 
be the midpoint of , and let 
be the reflection of 
with respect to . Prove that the triangles 
and 
are similar.
be an acute scalene triangle inscribed in a circle 
. The angle bisector of 
intersects 
at 
and at 
. The point 
is the midpoint of 
. Let 
be the altitude in 
, and the circumcircle of 
intersects again at 
. Let 
be the midpoint of , and let 
be the reflection of 
with respect to . Prove that the triangles 
and 
are similar.This post has been edited 1 time. Last edited by Assassino9931, Mar 30, 2025, 1:09 PM
midpoint of arc 
on 
,
isosceles trapezoid, let 
diameter of 
point on 
is tangent to 
which by ratio Lemma it happens that 
is symedian.
are colinear.
colinear and thus stacking ratio lemmas:![\[ \frac{BP}{PC} \cdot \frac{BA_1}{A_1C}=\frac{BD}{DC} \cdot \left(\frac{CA'}{A'B} \right)^2 \cdot \frac{BK}{KC}=\left( \frac{BA}{AC} \right)^2=\frac{BT}{TC} \]](http://latex.artofproblemsolving.com/0/6/2/06228fd0f202dc193b4e9a5c79817b23b3f08d93.png)
Happens to finish (notice 
colinear from reflecting was used).
is a rectangle so 
but also using Claim 1 and projecting cross ratios:![\[ -1=(A, E; P, A_1) \overset{S'}{=} (A, L; S'P \cap AL, \infty_{AL}) \implies P,M,S' \; \text{colinear!} \]](http://latex.artofproblemsolving.com/4/b/a/4baf385de77dd5b8efbecc35f835fe7aab99355b.png)
and from that we get 
thus we are done 
.
.
.
,
is cyclic. From here, we find 
.
intersect 
.
on 
.
,
intersects 
at its midpoint 
(from the midsegment in 
since 
is a diameter of 
due to the parallelism of 
and 
.
is cyclic, leading to 
,
.
be the second point of intersection of the circumcircles of 
.
lies on 
cross the line 
at 
.
lie on the line 
by Power of a point, which simplifies to 
or 
or 
or 
so 
.
so 
which implies that 
is cyclic. Now 
and we conclude by showing that the two angles in 
