[/quote]
,
,
is chosen on the side 
.
and 
,
and 
are drawn, respectively. The point 
is the circumcenter of 
.
and 
lies on the line 
.
be a point in the plane of triangle 
such that the segments 
,
,
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to 
is acute.
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray 
and ray 
.
.
be the intersection of lines 
and 
be the intersection of line 
is tangent to 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of 
,
denote the incenter and 
.
perpendicular to 
intersects the tangent from 
.
and 
.
.
such that for any positive integer number 
and any arrangement of the numbers from 1 to 
on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than 
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be the binary representation of 
.
,
,
,
equals 
.
such that![\[
4k^3 + 4k + 1
\]](http://latex.artofproblemsolving.com/f/3/5/f355e456cd47ecfdb13597d52fec3d848b67c594.png)
numbers, none of which appears more than 
times. Prove that they can be divided into 
groups of 
be a cyclic quadrilateral. 
be the projections of 
,
is a kite.
.
and 
,
,
.
such moves the blackboard will contain a single number. What is its maximum possible value? (Reminder that 
is the maximum integer not exceeding 
.
Y by anantmudgal09, Davi-8191, tenplusten, Combi12345, Combi1234, mathleticguyyy, megarnie, Adventure10, Mango247
Let 
points be given inside a rectangle 
such that no two of them lie on a line parallel to one of the sides of . The rectangle is to be dissected into smaller rectangles with sides parallel to the sides of in such a way that none of these rectangles contains any of the given points in its interior. Prove that we have to dissect into at least 
smaller rectangles.
Proposed by Serbia
points be given inside a rectangle 
such that no two of them lie on a line parallel to one of the sides of . The rectangle is to be dissected into smaller rectangles with sides parallel to the sides of in such a way that none of these rectangles contains any of the given points in its interior. Prove that we have to dissect into at least 
smaller rectangles.Proposed by Serbia
This post has been edited 1 time. Last edited by hajimbrak, Jul 23, 2015, 10:42 AM
Reason: Added proposer
Reason: Added proposer
usable corners), we deduce the inequality 
,
.
smaller rectangles. The induction solution uses the fact that if we start with any 
plus the four corners on the big rectangle for a total of 
intersecting lines parallel to either the length or breadth of 
is easy. Suposse is true for 
.
.
the number of segments above the leg and ending at the leg, and 
the number of segments below the leg and ending at the leg (do not count the segments at the endpoints of the leg, only those touching the leg in its interior, also note that this numbers could be cero). Then this leg has al least 
rectangles above it and at least 
rectangles below it, a total of 
rectangles. Remove the leg (and extend the segments ending at the leg until they end in another leg). Now you have exactly 
legs, and in the space occupied by the 
rectangles. Since 
the number of rectangles before the leg was removed is 
rectangles, then now we have at least 
rectangles, done. This also proofs that the number of rectangles is equal to the number of legs (with our definition of leg).
,
to be the longest line segment that runs through 
angles. It cannot form only one right angle because then we would have a rectangle with a 270 degree angle...not possible. If we have four right angles formed, then that doesn't work out because then that isn't the true endpoint of the segment. Therefore, each endpoint contributes 
.
.
where we add on an extra 360 from the right angles of 
.
and observe the trivial base cases. Suppose the claim holds till 
,
cases.
(since we also lose the top-edge of 
(since we also lose at least 
and we are done. 
.
as desired. 
)
lines. Therefore, it suffices to show that 
line segments in a dissection ensure 
rectangles.
T-intersections (no two line segments can have a common endpoint). Also, there are 
L-intersections; the ones at the corners of 
X-intesections.
,
.
,
is the number of smaller rectangles in 
or
which finishes the proof.
segments; in total, the rectangles cover 
endpoints (excluding corners of the large rectangle). On the other hand, each endpoint is the intersection of at least two segment, and there are two endpoints on any of the m segments, hence we have 
.
coordinate).
corners and there are at least 
corners from maximal segments and another 
rectangles, as desired.
.
or 
.
passing through each point 
by hypothesis, each such segment contributes 
vertices (including the four corners), thus![\[F = E+2-V \geq \frac 12 V \geq n+2\]](http://latex.artofproblemsolving.com/e/8/5/e8596a34e5c3a97a9d586f0537051fe46803f225.png)
which means there are at least 
being true.
![[asy]
size(2cm);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
dot((0.5,0.5));
draw((0.5,0)--(0.5,1));
[/asy]](http://latex.artofproblemsolving.com/1/b/4/1b4e869d24bb52573c0baeaa26c3ed23011bc95e.png)
we cannot have any less than 2 partions. So 
this is clear by simply drawing a diagram and showing that with 
,
rectangles.
rectangles, we show we cannot possibly form such a dissection.
or 
segments attached be 
.
,
pairs. Counting by each vertex gives 
pairs, so we can write 
.
,
given 
,
,
.
which is at most 
,
,
lines. Eliminating the 
,
and 
.
and 
respectively, and set 
and 
.
,
be the rectangle bounded by 
,
,
,
be the rectangle bounded by 
,
,
rectangles.
rectangles (equality holds most of the time; if 
are part of the same line for some 
(which is very inefficient) we get at most 
rectangles). We finish by induction hypothesis. 
The base case is trivial, so assume that the proposition holds for 
Then if we have 
,
or 
,
.![\[360R \ge 360n + 360 \implies R \ge n+1. \quad \blacksquare\]](http://latex.artofproblemsolving.com/1/7/e/17efc663f93adfa9bf2a7a792e028f011fc870c7.png)
