AoPS Academy
: 
, 
on 
, 
, 
on 
and 
, 
on 
, such that they are the vertices of a convex hexagon 
with equal side lengths.
, 
and 
are concurrent.
, let 
be the result.
, where 
is the Euler's totient theorem, which calculates the number of integers less than relatively prime to . again because why not, let the result be 
.
for 
. Let 
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)
.
, and 
, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance 
away. What is the least number of moves that Carina can make in order for triangle to have area 2021?
in the coordinate plane where and 
are both integers, not necessarily positive.)
be an integer. We say that an arrangement of the numbers , 
, 
, 
in a 
table is row-valid if the numbers in each row can be permuted to form an arithmetic progression, and column-valid if the numbers in each column can be permuted to form an arithmetic progression. For what values of is it possible to transform any row-valid arrangement into a column-valid arrangement by permuting the numbers in each row?
be a convex polygon with 
sides, 
. Any set of 
diagonals of that do not intersect in the interior of the polygon determine a triangulation of into 
triangles. If is regular and there is a triangulation of consisting of only isosceles triangles, find all the possible values of .
![$[n] = \{1, 2, . . . , n\}$](http://latex.artofproblemsolving.com/4/f/f/4ff58e6e89a225ffa8db0ab50038fdadd9d4e2bc.png)
is a sequence 
such that each element of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
appears precisely one time as a term of the sequence. For example, 
is a permutation of ![$[5]$](http://latex.artofproblemsolving.com/c/4/2/c4216c69e041fca0904e8c5366d4480d1b2ca8b8.png)
. Let 
be the number of permutations of for which 
is a perfect square for all 
. Find with proof the smallest such that is a multiple of 
.
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 .
and 
. Use directed angles modulo 
. It is easy to show that 
. Also, 
in cyclic quadrilateral 
. But 
as 
is an angle bisector and ![\[ \measuredangle SBA = \measuredangle EBA = \measuredangle EDA = \measuredangle LDA \]](http://latex.artofproblemsolving.com/7/1/5/715ba8f0a53b77b270a5f249a38b4a3655930545.png)
in cyclic quadrilateral . So 
and therefore 
. Thus ![\[ \measuredangle BAE = \measuredangle CDE = 90^{\circ} - \measuredangle LCD = 90^{\circ}-\measuredangle ACB. \]](http://latex.artofproblemsolving.com/c/c/c/ccc61519c685d24f5be570f833129fc3bc1c79a8.png)
So 
which implies that 
are collinear. Done. 
.
, let 
be a point in such that is the angle bisector of 
. Let 
be the point of intersection of the circumcircle of triangle 
with the line perpendicular to drawn through . Let 
be the circumcenter of triangle . Prove that 
, , and are collinear.
. It's easy to observe that this equality holds in reverse we have on the other side of line than point , which contradicts collinearity 
, because foot of on segment is a center of this segment.
be the foot of the perpendicular from to then 
for some . Using EFFT we see:
Then solving for circle 
we get 
. Letting 
we can solve for and see:
And as 
it is then obvious are colinear.
unless of your age can be written as 
.
.
,
.
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 
,
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 intersection of 
and the line through 
,
.
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 
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 
,
at 
.
.
,
by circumcenter angles. But 
,
,
,
.
.
,
are concyclic.
.
.
be equal to 
so 
.
as 
on the circumcircle. Thus, it suffices to prove that 
.
and 
.
and by inscribed angles, we have 
.
.
.
so 
are collinear. 
.
.
,
.
.
.
it is enough to show that the angle between 
![[asy]
/* Argentina Cono Sur TST 2014 Problem 5, free script by liberator, 22 September 2014 */
unitsize(3cm);
pointpen=black;
pen dsp = rgb(0.4,0.6,0.8);
pen rcp = rgb(0.9,0,0);
pen gcp = rgb(0,0.6,0);
/* Initialize objects */
pair O = origin;
pair A = dir(110);
pair B = dir(200);
pair C = dir(340);
pair Ap = dir(-A);
pair D1 = bisectorpoint(B,A,C);
pair D = IntersectionPoint(Line(A,D1, 2014,2014), B--C);
pair E = IntersectionPoint(Line(A,O,-0.01, 2014), circumcircle(A,B,D));
/* Draw objects */
draw(A--B--C--cycle, dsp+linewidth(1));
draw(A--Ap, dsp);
draw(D--E, dsp);
draw(B--E, dsp);
draw(A--D, dsp);
draw(unitcircle, gcp);
draw(circumcircle(A,B,D), rcp);
/* Place dots on and label each point */
Drawing("A", A, dir(A));
Drawing("B", B, dir(B));
Drawing("C", C, dir(C));
Drawing("D", D, dir(-80));
Drawing("E", E, dir(0));
Drawing("O", O, dir(60));
[/asy]](http://latex.artofproblemsolving.com/8/3/7/837df7b1b82ef748c85c4332eb5a5f936f2892ad.png)
is a triangle but doesn't assume that 
and 
?
but then my solution wouldn't work.
the orthocenter of 
collinear on the perpendicular from 
in 
,
.
is the intersection of 
;
.
,
.
all lie on the line 
,
collinear.
be the intersection of 
.
.
.
.
and hence the result follows.
By angle chasing we get that 
is perpendicular to 
is parallel to 
=