[/quote]
,
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
and has exactly 
then
.
on 
and 
on 
lies on the circumcircle of triangle 
be the circumcircle of triangle 
is the midpoint of 
be the second intersection of 
and 
be a transversal of 
be intersections of 
with 
be a tangent line of 
is cyclic
be positive integers such that 
and 
.
.
and 
Prove that
at 
,
perpendicular to 
be the intersections of 
with 
,
be the projections of 
onto line 
be the second intersections of 
with the incircle 
be the intersection of 
and 
.
bisects segment 
.
.
,
.
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 center of 
.
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 
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 
Y by Mathandski, EpicBird08, KevinYang2.71, Alex-131, aidan0626, Pengu14, eg4334, arfekete, Yiyj1, megarnie, krithikrokcs, OronSH, MathRook7817, sixoneeight, Math4Life2020, blueprimes, vincentwant, mathfan2020, elasticwealth, cowstalker, StressedPineapple, lpieleanu, ehuseyinyigit
1. Let 
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 .
2. Let
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.
3. Alice the architect and Bob the builder play a game. First, Alice chooses two points
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:
Note:
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
, 
to 
, and 
to 
.
4. Let
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 
.
5. Determine, with proof, all positive integers such that ![\[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]](//latex.artofproblemsolving.com/3/c/5/3c5984d0003c1ca2df2e68841dd9e93ed6c2d27c.png)
is an integer for every positive integer .
6. Let
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 
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 .2. Let
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.3. Alice the architect and Bob the builder play a game. First, Alice chooses two points
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:
For every city distinct from 
and 
, there exists 
such
distinct from 
and 
, there exists 
such
that 
is directly similar to either 
or 
.
Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.
is directly similar to either 
or 
.
Note:
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
, 
to 
, and 
to 
.4. Let
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 
.![[asy]
size(300);
defaultpen(linewidth(0.6)+fontsize(11));
pair A = (0,6), B = (-3,0), C = (5,0), H = orthocenter(A,B,C), D = foot(A,B,C), P = 2*D - H, F = foot(C,A,B), Xx = 3*C-2*B;
path circ = circumcircle(A,F,P);
pair[] oops = intersectionpoints(circ, Xx--B);
pair X = oops[0], Y = oops[1];
draw(A--B--C--cycle, rgb(0.8,0.1,0.1));
draw(C--F^^A--P, rgb(0.9,0.4,0.1));
draw(circ,rgb(0.1,0.5,0.1));
draw(circumcircle(A,B,C),rgb(0.1,0.1,0.7));
draw(F--X--P--C--Y,rgb(0.8,0.1,0.8));
dot("$A$",A,N,linewidth(3.3));
dot("$B$",B,SW,linewidth(3.3));
dot("$C$",C,SE,linewidth(3.3));
dot("$D$",D,NE,linewidth(3.3));
dot("$F$",F,NW,linewidth(3.3));
dot("$X$",X,SW,linewidth(3.3));
dot("$Y$",Y,SE,linewidth(3.3));
dot("$P$",P,S,linewidth(3.3));
dot("$H$",H,NE,linewidth(3.3));
[/asy]](http://latex.artofproblemsolving.com/5/4/b/54b45b4ebef7f8e82abfa7c1daaeee177d6ff3dd.png)
5. Determine, with proof, all positive integers
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 .6. Let
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 .
alternate solutions (if they exist) and appropriate point values. This only barely does that for P4, with nothing for the remaining 5 problems.
at least using angles shows you are on the right path (but the ironic part is that's pretty much the only thing you can do once you identify the circular region LOL)
