AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
+1 w
and 
each pass through the other circle's center. The line containing both and is extended to intersect the circles at points 
and 
. The circles intersect at two points, one of which is 
. What is the degree measure of 
?
be an acute triangle. Let 
be the reflection of point over the line 
. Let 
and 
be the circumcenter and the orthocenter of triangle , respectively, and let be the midpoint of segment 
. Let and 
be the points where the reflection of line 
with respect to line 
intersects the circumcircle of triangle , where point lies on the arc not containing . If 
is a point on the line 
such that 
, prove that 
.
be a positive integer and let 
be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of around the circle such that the product of any two neighbors is of the form 
for some positive integer 
.
be real numbers satisfying ![\[u_{1}+u_{2}+\cdots+u_{2019}=0 \quad \text { and } \quad u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1.\]](http://latex.artofproblemsolving.com/2/d/0/2d09d5e894fdd13ec466aa664f3f10da5ca8d134.png)
Let 
and 
. Prove that![\[
a b \leqslant-\frac{1}{2019}.
\]](http://latex.artofproblemsolving.com/8/7/0/870973eb68893864ee8f8494efcae4c34c11d2d8.png)
such that 
is divisible by 
and 
. Prove that is a composite number.
be a prime number and 
. Prove that![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
such that 
, 
, and 
are all equal to
for which there exist positive integers 
, 
, 
, 
such that 
and 
divides 
.
cannot cover entirely a square surface of side 
, but they can cover more than 
of it.
such that the sequence 
defined by ![\[ a_0 = M + \frac{1}{2} \qquad \textrm{and} \qquad a_{k+1} = a_k\lfloor a_k \rfloor \quad \textrm{for} \, k = 0, 1, 2, \cdots \]](http://latex.artofproblemsolving.com/2/1/b/21b2a1f93b11a94b84e7b55f4b4f679aa20e36c6.png)
contains at least one integer term.
is chosen inside a convex quadrilateral 
. Could it happen that
(doesn't have to be called ), and we define the homogenized barycentric coordinate as![\[P=\left(\dfrac{[PBC]}{[ABC]},\dfrac{[PAC]}{[ABC]},\dfrac{[PAB]}{[ABC]}\right)\]](http://latex.artofproblemsolving.com/c/e/7/ce7719dabb2c95f0ccf1f4b54683338b8abde865.png)
Note that we can define 
in this system. Furthermore, if a point 
, then we have that ![$[PBC]=0$](http://latex.artofproblemsolving.com/2/b/9/2b9ba4a873941cb8c489868556f3f445dd6fbb74.png)
, so thus we easily have that![\[P=(0,\text{something},1-\text{something})\]](http://latex.artofproblemsolving.com/b/6/d/b6d4627b0b4722ef39e6ea62ca0c75c2caa95342.png)
What are the advantages of barycentric coordinates? Here are a few:
as![\[[ABC]\begin{vmatrix}p_1&q_1&r_1\\p_2&q_2&r_2\\p_3&q_3&r_3\end{vmatrix}\]](http://latex.artofproblemsolving.com/4/6/f/46fbe16a64f3097fd9d782d1ea40ff18adc263be.png)
where 
and 
are defined analogously.
triangles of equal areas, so![\[G=\left(\dfrac{[GBC]}{[ABC]},\dfrac{[GAC]}{[ABC]},\dfrac{[GAB]}{[ABC]}\right)=\left(\dfrac 13,\dfrac 13,\dfrac 13\right)\]](http://latex.artofproblemsolving.com/a/5/a/a5a2a534d28665b7a6bb750da827a7f82b0c3eec.png)
The circumcenter and orthocenter aren't as nice, but still are nice. Furthermore, we have that![\[[IBC]=\dfrac 12ar\]](http://latex.artofproblemsolving.com/7/4/9/7498b126bc2728e5e5d31ee2fafca134daefa781.png)
and analogous objects, so thus![\[I=\left(\dfrac{ar}{2rs},\dfrac{br}{2rs},\dfrac{cr}{2rs}\right)\]](http://latex.artofproblemsolving.com/b/b/a/bbab24e04e140133a5cd079549d2b263b70e811c.png)
where I used ![$[ABC]=rs$](http://latex.artofproblemsolving.com/5/f/e/5fe68c691b13681950a6542efe4b5ddd0bc41511.png)
. However, we can "cancel" the 
to get![\[I=\left(\dfrac{a}{2s},\dfrac{b}{2s},\dfrac{c}{2s}\right)\]](http://latex.artofproblemsolving.com/3/a/4/3a4b78f1c4b58686a9c62c44aca550f6ac19d855.png)
![\[P=(x:y:z)\]](http://latex.artofproblemsolving.com/4/1/e/41eca29375a7077fd6634d32225879cb390b7e12.png)
as long as 
are in the same ratio as ![$[PBC],[PAC],[PAB]$](http://latex.artofproblemsolving.com/7/7/0/770f726d7deb2ec1cfdf603700d98abe10120904.png)
. That allows us to remove the denominator in, for example, 
, so we get![\[I=(a:b:c)\]](http://latex.artofproblemsolving.com/8/b/7/8b78695d3df71a26c0c2c9c2be95ea51e1d36421.png)
![\[G=(1:1:1)\]](http://latex.artofproblemsolving.com/3/5/e/35e9fad690f3b21174131260bab24c6e7db399b2.png)
![\[\vec P=x\vec A+y\vec B+z\vec C\iff P=(x,y,z)\]](http://latex.artofproblemsolving.com/f/5/e/f5ef70b5d133a282c8785829b8eac8a48b51afbe.png)
That helps, for example, with finding the distance between two points, finding perpendicular lines, and also finding circles.
