AoPS Academy
index this yr (118.5 on amc10+ 9 on AIME) i’m in 7th rn btw first year comp math also. I will grind so hard until like 30 hrs/week. I’m ok at proofs. made mc nats
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
, the digits in the base-
representation of 
are all greater than .
and be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
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 
, there exists 
such[/center]
is directly similar to either 
or 
.[/center]
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
, 
to 
, and 
to 
.
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from to , and let be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points and . Prove that is the midpoint of 
. such that ![\[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]](http://latex.artofproblemsolving.com/3/c/5/3c5984d0003c1ca2df2e68841dd9e93ed6c2d27c.png)
is an integer for every positive integer .
and be positive integers with 
. There are cupcakes of different flavors arranged around a circle and people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person , it is possible to partition the circle of cupcakes into groups of consecutive cupcakes so that the sum of 's scores of the cupcakes in each group is at least 
. Prove that it is possible to distribute the cupcakes to the people so that each person receives cupcakes of total score at least with respect to .
+4 w
where 
and 
are real numbers with 
, Xiaoli rounded up by a small amount, rounded down by a small amount, then subtracted her rounded values. Which of the following is necessarily correct?
Her estimate is larger than 
Her estimate is smaller than 
Her estimate equals 
Her estimate equals 
Her estimate is 
.
,
.
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 
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 
,
fails.
,
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 
,
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 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 
,
.
.
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 
and 
,
and 
with 
is
?
,
,
.
.
?
students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 
.
.
times their difference. What is the ratio of the larger number to the smaller?
,
,
?
more than the radius of the first. What is the relationship between the heights of the two cylinders?
.
for some constant 
.
lines. How many different values of 
and 
intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area 
.
?
a true statement about the list of 
?
as a decimal?
,
,
,
,
,
.
times. Let 
,
.
Anyway, I'm pretty certain this is not the first time it appeared in the circle phrasing, but finding tons of questions isomorphic to "how many binary strings don't have consecutive ones" is not exactly hard.
are integers. What is the sum of all possible values of 
have only integer roots for 
as the constant term, but somehow I can't find it right now.
,
.
are possible?
and 
are not congruent but have the same area and the same perimeter. The sides of 
,
passes through both foci of, and exactly four points on, the ellipse with equation 
.
.
be the number of sequences of length 
is divided by 
"
is 
,
.
?
that can be written as fractions 
where 
.![\[(\cos(a\pi)+i\sin(b\pi))^4\]](http://latex.artofproblemsolving.com/e/c/8/ec8719603f64f25dacf3bda313a411e348d627c8.png)
is a real number?
consists of two circles of radii 
and 
that are externally tangent. For 
,
are ordered according to their points of tangency with the 
consists of the 
circles constructed in this way. Let 
,
its radius. What is ![\[\sum_{C\in S}\dfrac1{\sqrt{r(C)}}?\]](http://latex.artofproblemsolving.com/5/b/e/5be96aa0120717982c09c426fc9c7f557c109a84.png)
![[asy]
import olympiad;
size(350);
defaultpen(linewidth(0.7));
// define a bunch of arrays and starting points
pair[] coord = new pair[65];
int[] trav = {32,16,8,4,2,1};
coord[0] = (0,73^2); coord[64] = (2*73*70,70^2);
// draw the big circles and the bottom line
path arc1 = arc(coord[0],coord[0].y,260,360);
path arc2 = arc(coord[64],coord[64].y,175,280);
fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75));
fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75));
draw(arc1^^arc2);
draw((-930,0)--(70^2+73^2+850,0));
// We now apply the findCenter function 63 times to get
// the location of the centers of all 63 constructed circles.
// The complicated array setup ensures that all the circles
// will be taken in the right order
for(int i = 0;i<=5;i=i+1)
{
int skip = trav[i];
for(int k=skip;k<=64 - skip; k = k + 2*skip)
{
pair cent1 = coord[k-skip], cent2 = coord[k+skip];
real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2);
real shiftx = cent1.x + sqrt(4*r1*rn);
coord[k] = (shiftx,rn);
}
// Draw the remaining 63 circles
}
for(int i=1;i<=63;i=i+1)
{
filldraw(circle(coord[i],coord[i].y),gray(0.75));
}[/asy]](http://latex.artofproblemsolving.com/1/1/9/11999bae3eb2df3369375c128acf2c1551a21a07.png)
![\[\frac{1}{\sqrt{r_{\text{middle}}}}=\frac{1}{\sqrt{r_{\text{left}}}}+\frac{1}{\sqrt{r_{\text{right}}}}\]](http://latex.artofproblemsolving.com/0/f/7/0f7b36eeb55e61f9aefd4846adacd43ff8488ba4.png)
