be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
.
intersects line 
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 
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 
.
.
states are taking mathcounts state and 
territories are taking mathcounts state 
,
national qualifiers each).
perfect square dates this century, 
this year.
Y by ihatemath123, lpieleanu, RoyalPrince, LostDreams, Countmath1
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 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 .
.
bubbles. Remove that set, and repeatedly remove any set with 
.
such that 
have a matching, and no person in 
matches any bubble in 
delicious for person 
if the sum of 
into disjoint counterclockwise-oriented arcs 
such that
is the shortest arc starting at some cupcake 
,
is the shortest arc starting at the first cupcake counterclockwise of 
that is delicious for someone.
now. For each arc 
to consist of all the people for which 
of the sets 
with
;
(indexed without loss of generality) to the sets in 
such that 
.
,
we can remove them and induct down (which is false) otherwise (this case doesn’t even exist??) there exists a bubble not matched with anyone which we can induct down on preserving score. I missed the immediate match and win case and I think my second case is a subset of my first which is skullers. Like I’m pretty sure I have a solve if in my first case instead of matching and winning I match and consider the complement which eventually reduces to something like my second case.
people who you are about to delete settled. For the people who aren't settled, you need to make sure they lose exactly 
.
unsettled people value 
,
in value).
people and 
people with a bubble and give 
such that if the set of its matches is 
,
.
and 
.
,
still. Additionally, note that no person in 
.
,
is trivial.
between the people and the arcs for Pip.
of people, who are compatible with fewer than 
of the arcs. Let's imagine deleting 
for the remaining graph. We again delete 
is possible in the remaining graph, i.e.\ there are no more bad sets (and then terminate once that occurs).
.![[asy]usepackage("amssymb");
size(8cm); dotfactor *= 1.3; real w = 3; real eps = 0.4; label("People", (-w,10)); label("Arcs of Pip", (w,10)); filldraw(box((-w-eps, 9+eps), (-w+eps,7-eps)), invisible, red+1.2); filldraw(box((-w-eps, 6+eps), (-w+eps,5-eps)), invisible, orange+1.2); filldraw(box((-w-eps, 4+eps), (-w+eps,2-eps)), invisible, brown+1.2);
filldraw(box((w-eps, 9+eps), (w+eps,8-eps)), invisible, red+1.2); filldraw(box((w-eps, 7+eps), (w+eps,7-eps)), invisible, orange+1.2); filldraw(box((w-eps, 6+eps), (w+eps,5-eps)), invisible, brown+1.2);
draw((-w+eps, 9+eps)--(w-eps, 9+eps), red+dashed); draw((-w+eps, 7-eps)--(w-eps, 8-eps), red+dashed); draw((-w+eps, 6+eps)--(w-eps, 9+eps), orange+dashed); draw((-w+eps, 5-eps)--(w-eps, 7-eps), orange+dashed); draw((-w+eps, 4+eps)--(w-eps, 9+eps), brown+dashed); draw((-w+eps, 2-eps)--(w-eps, 5-eps), brown+dashed);
label((-w-eps, 8), "Bad set $\mathcal{B}_1$", dir(180), black); label((-w-eps, 5.5), "Bad set $\mathcal{B}_2$", dir(180), black); label((-w-eps, 3), "Bad set $\mathcal{B}_3$", dir(180), black);
draw((-w,1)--(w,1), deepgreen+1.3); draw((-w,0)--(w,0), deepgreen+1.3); label((0, 0.5), "Final perfect matching $\mathfrak{M}$", deepgreen);
for (int i=0; i<10; ++i) { dot((w,i), blue); dot((-w,i), blue); } label("Pip", (-w,0), dir(180), blue);
[/asy]](http://latex.artofproblemsolving.com/5/7/a/57a351245aaaaf5518f8ececbcbb58b8dc5c2576.png)
of ![\[ (\ge 1) + (\ge 1) - (\le 1) \qquad \ge \qquad 1 \]](http://latex.artofproblemsolving.com/0/5/8/058f38c90ef6a220bd70fe2840c7e84cf768f5e9.png)
meaning the induction is OK. See below for a cartoon of the deletion, where Pip's arcs are drawn in blue while 
)![[asy]
size(13cm); usepackage("amsmath"); pair O = (0,0); picture before; picture after;
real r = 1; real s = 0.9; real t = 0.65;
draw(before, arc(O, s, -20, 80), red+1.3); draw(before, arc(O, s, 100, 200), red+1.3); draw(before, arc(O, s, 220, 320), red+1.3);
draw(before, "Pip arc to delete", arc(O, r, 40, 140), blue+1.3); draw(before, rotate(-60)*"Pip arc", arc(O, r, 160, 260), blue+1.3); draw(before, rotate( 60)*"Pip arc", arc(O, r, 280, 380), blue+1.3);
label(before, "$\boxed{0.2}$", t*dir(120), red); label(before, "$\boxed{0.3}$", t*dir( 60), red); label(before, "$\boxed{0.8}$", t*dir(180), red); label(before, "$\boxed{0.43}$", t*dir(240), red); label(before, "$\boxed{0.57}$", t*dir(300), red); label(before, "$\boxed{0.7}$", t*dir( 0), red); label(before, "$Q$'s values", O, red);
draw(after, arc(O, s, -20, 200), red+1.3); draw(after, arc(O, s, 220, 320), red+1.3);
draw(after, rotate(-60)*"Pip arc", arc(O, r, 160, 260), blue+1.3); draw(after, rotate( 60)*"Pip arc", arc(O, r, 280, 380), blue+1.3);
label(after, "$\boxed{0.8}$", t*dir(180), red); label(after, "$\boxed{0.43}$", t*dir(240), red); label(after, "$\boxed{0.57}$", t*dir(300), red); label(after, "$\boxed{0.7}$", t*dir( 0), red); label(after, "$Q$'s values", O, red);
add(before); add(shift(3.2,0)*after);
[/asy]](http://latex.artofproblemsolving.com/b/8/d/b8dbd78a98423d5c16be57fce0a886df36a1824f.png)
)
then we will now perform induction on 
can be considered), thus we are done 
.
