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 
,
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 
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.
and 
.
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 
,
,
is the Euler's totient theorem, which calculates the number of integers less than 
for 
.
be the remainder 
leaves when you divide it by 
.
be your predicted USA(J)MO score.![$\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$](http://latex.artofproblemsolving.com/c/d/3/cd3f7d917b0ec944ec3017e1f737c538e1edd10e.png)
.
Y by USA, samrocksnature, RedFlame2112, icematrix2, player01, HWenslawski, megarnie, mathematicsy, jhu08, mathlearner2357, JVAJVA, suvamkonar, ImSh95, son7, mainstain, mathmax12, jmiao, Adventure10, Mango247, Aopsauser9999, FrenchFry99, AlexWin0806, ItsBesi, Kempu33334
Points 
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects 
.
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects 
.
and Z is outside 
Though it's not competition-oriented, I believe it's around the same audience.
be the midpoint of 
is cyclic. Let 
Then, 
so 
colinear.
is a isosceles trapezoid, we wish to show 
.
which implies 
is a cyclic quadrilateral. Thus,
as desired.
.
,
,
By Power of a Point and cyclic quadrilaterals, 
So 
are on the same circle, which implies our desired conclusion.
is similar to 
,
is similar to 
.
.
is a median
be the point at infinity along lines 
and 
,
.![\[-1 = (A,C ;B , D) \stackrel{E}{=} (A, C; M, P_{\infty}) \]](http://latex.artofproblemsolving.com/3/b/a/3bac455ec07839e6b78f06a8fd19884b103e48b3.png)
so 
is the midpoint of 
,
.
cyclic. Now let 
.
and clearly 
where 
,
and thus 
.
is perpendicular to 
is cyclic.
and the tangent criterion, we have ![\[\angle BMP = \angle BMD = \angle DBP = \angle BOP\]](http://latex.artofproblemsolving.com/a/3/6/a363b3afbc14fcf16302abdb30b3f845fadecc28.png)
as desired.
,
and 
let the center of the circle be 
and let 
so we have 
we have 
from which it follows that 
so we are done. 
and 
and 
respectively. Then, by the formula for the intersection of tangents, we get 
Then, since 
and 
must have the same measures, so it follows that 
Then, since 
we have
Therefore, the midpoint of 
It now suffices to show that 
and 
are collinear. By the complex number collinearity condition, this is equivalent to
concluding the proof. 
,
.
are all concyclic.![\[ \angle{BMP} = \angle{MED} = \angle{BDP}, \]](http://latex.artofproblemsolving.com/2/d/d/2dddfc10aac77de9281ee88e4f3cf7a47c46cf03.png)
where the last part comes from the so-called Tangent Lemma. In particular, 
is a cyclic quadrilateral, so 
.
is clearly cyclic 
)
tells us that 
,
is cyclic with 
,
implies 
,
.
,
is cyclic. But since 
has diameter 
,
and hence 
.
;
is isosceles.
is cyclic. We can do so by proving 
:![\[\angle{BXO} = \angle{BED} = \frac{\overarc{BD}}{2} = \frac{\angle{BOD}}{2} = \angle{BOP}\]](http://latex.artofproblemsolving.com/f/a/7/fa7332189b642ee5ad9f46b558a4b8e5aae1b7d4.png)
,
(this can be shown by proving 
,
,
(as 
by the tangent condition). Since 
and then project through 
onto 
is cyclic and by the 
criterion, 
.
is cyclic. This means 
,
and we're done. 
