of the triangle 
which doesn't coincide with 
or 
intersectes the altitudes from 
and 
at 
and 
respectively. Let 
be the reflection of 
be the reflection of 
Prove that the circles 
and 
are tangent.
be positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
such that for any positive integer 
there exists a positive integer 
such that the last 
are equal to 
such that for any positive real numbers 
,![\[
\frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C.
\]](http://latex.artofproblemsolving.com/9/1/c/91c8a4af2bd2f3107c8921d2e06f8a09dedd0164.png)
such that 
.
Where 
with legs 
and 
is drawn with a semicircle inscribed into the triangle. What is the smallest possible radius of the semi-circle and the largest possible radius?
be non-negative real numbers such that 
.![\[
P = \frac{a}{a^2 + 2} + \frac{b}{b^2 + 2} + \frac{c}{c^2 + 2}.
\]](http://latex.artofproblemsolving.com/d/b/e/dbea7db769c4996fc081ee75837e177b298cffcb.png)
be a permutation of 
such that there are exactly 
ordered pairs of integers 
satisfying 
and 
.
students, 
students play cricket, 
students play hockey and 
students play basketball. 
students play both basketball and hockey, 
students play both cricket and basketball, 
students play both cricket and hockey, and 
students play all three: basketball, hockey, and cricket. Find the number of students who do not play any game.
be reals such that 
. Prove that
Let 
. Prove that
,
,
is not greater than 
.
Write only the answers to the questions given.
to be more exact:
in years 2002-08 time was 90' for part A and 120' for part B
white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...
unit. The area of the hexagon is ...
and 
.
is ...
so that 
and 
is ...
,
,
,
has an average of 
and a median of 
.
are integers greater than 
which are four consecutive quarters of an arithmetic row with 
.
and 
are squares of two consecutive natural numbers, then the smallest value of 
is ...
,
and 
.
and 
lies on the line segment 
.
and 
.
is ...
meet 
for each natural number 
is ....
provided that the difference of any two numbers at least 
is ...
which satisfies 
are as many as ...
and 
.
.
is a point on the side extension 
so that 
is perpendicular to 
and 
.
is ...
consists of 
with 
positive reals is ....
with 
is ....
square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of 
coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of 
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Hello fellow AoPSers (and Mathlinkers),
Over the past couple weeks, Mewto55555 and I have put together an article to introduce a new powerful, computational technique in olympiad geometry: barycentric coordinates! We hope that this method proves to just as powerful, if not more so, than the already well-known complex numbers and Cartesian coordinates.
Attached here are two versions of the article. One of them takes the time to develop the entire theory, step-by-step, before leading in to the examples; the second version omits the derivation in favor of a quick "formula sheet" (which is Appendix B in the first), with the emphasis placed squarely on the examples.
As there are bound to be typos, please post any mistakes you have in-thread so that I can correct them in this post.
Happy bashing!
EDIT: Just finished up version 2, linked below. These links update dynamically and will always reflect the most recent version.
Barycentric Coordinates in Olympiad Geometry, Version 2
Long version
Brief version
Over the past couple weeks, Mewto55555 and I have put together an article to introduce a new powerful, computational technique in olympiad geometry: barycentric coordinates! We hope that this method proves to just as powerful, if not more so, than the already well-known complex numbers and Cartesian coordinates.
Attached here are two versions of the article. One of them takes the time to develop the entire theory, step-by-step, before leading in to the examples; the second version omits the derivation in favor of a quick "formula sheet" (which is Appendix B in the first), with the emphasis placed squarely on the examples.
As there are bound to be typos, please post any mistakes you have in-thread so that I can correct them in this post.
Happy bashing!
EDIT: Just finished up version 2, linked below. These links update dynamically and will always reflect the most recent version.
Barycentric Coordinates in Olympiad Geometry, Version 2
Long version
Brief version
This post has been edited 1 time. Last edited by v_Enhance, Aug 28, 2015, 3:23 PM
Reason: Fix links
Reason: Fix links
and similar. now we use the general form of a circle; that is:![\[a^2yz+b^2xz+c^2xy=px+qy+zr\]](http://latex.artofproblemsolving.com/8/e/7/8e717da6a3fa7e8093051e135aed4e761203aa42.png)
is normalized. Plugging in 
gives 
.
give 
and 
.
with 
solved gives:![\[b^2\cdot\frac{a}{a+c}\cdot\frac{c}{a+c}=\frac{c^2b}{a+b}\cdot\frac{a}{a+c}+\frac{a^2b}{b+c}\cdot\frac{c}{a+c}\]](http://latex.artofproblemsolving.com/3/6/d/36dea5d044d9de1c60be6a644f3c675f71b4d664.png)
![\[\Leftrightarrow abc[\frac{b}{a+c}-\frac{a}{b+c}-\frac{c}{a+b}]=0\]](http://latex.artofproblemsolving.com/4/0/f/40fdc5c41aac10b2f10d508c0eee7860c2c1042a.png)
.![\[
m=BD=(1-k)a
\]](http://latex.artofproblemsolving.com/d/6/5/d6567775e5257ed3644968467b0f7e326f25858d.png)
![\[
n=DC=ka
\]](http://latex.artofproblemsolving.com/c/0/6/c0635f2d1e02899ebaf9505ee1dc5d754a2f2e57.png)
![\[
d^2=|AD|^2=-(k(1-k)a^2-(1-k)b^2-kc^2)=k(k-1)a^2+(1-k)b^2+kc^2
\]](http://latex.artofproblemsolving.com/2/d/5/2d52453ec1032c67952cf3f599de3a75979f230a.png)
![\[
man+dad=k(1-k)a^3+k(k-1)a^3+(1-k)ab^2+kac^2=(1-k)ab^2+kac^2
\]](http://latex.artofproblemsolving.com/3/f/b/3fbaae305d53ad657380d11e36613a0b9abcb67d.png)
![\[
bmb+cnc=(1-k)ab^2+kac^2
\]](http://latex.artofproblemsolving.com/c/1/c/c1ce94da5bf15893e6e6533bbc4e703bc8ec671f.png)
,
,
,
,
,
.
,
,
.
,
,
,
,![\[
(\frac{yz}{1+yz+z}, \frac{1}{1+yz+z}, \frac{z}{1+yz+z})
\]](http://latex.artofproblemsolving.com/9/a/4/9a4c0e8978a409ce348e9c1aa4f04480f05c93a3.png)
and 
are just cyclic variations of this. Then![\[
[PQR]=[ABC]\prod_{cyc} \frac{1}{1+yz+z}\det\begin{bmatrix} yz & 1 & z \\ y & xy & 1 \\ 1 & x & xz \end{bmatrix}
\]](http://latex.artofproblemsolving.com/8/a/6/8a6616b51ad8040db863c231e41f7292e332f897.png)
![\[
x^2y^2z^2+xyz+1-xyz-xyz-xyz=(xyz)^2-2xyz+1=(xyz-1)^2
\]](http://latex.artofproblemsolving.com/9/f/e/9feac7f04e9365f7160b6862cead2d53d5688996.png)
![\[
[PAB]+[QBC]+[RCA]+[PQR]=[ABC]
\]](http://latex.artofproblemsolving.com/3/a/2/3a2874eaede3a10112e0c64c4fd278ac1fd4b37c.png)
![\[
[PQR]=[ABC](1-\frac{z}{1+yz+z}-\frac{y}{1+xy+y}-\frac{x}{1+xz+x})
\]](http://latex.artofproblemsolving.com/d/3/b/d3bc5417dda1da80b876fd0b0a45c1bef7b7728a.png)
with 
.
,
,
.![\[
[PAF]=[ABC](\det\begin{bmatrix} 1 & 0 & 0 \\ x & y & z \\ \frac { x }{ x+y } & \frac { y }{ x+y } & 0 \end{bmatrix})
\]](http://latex.artofproblemsolving.com/a/5/b/a5b28ae46c7fea1ee4968b99f134d9277afdf7e4.png)
.![\[
\sum_{cyc} \frac{yz}{1-z}=\sum_{cyc} \frac{yz}{x+y}=\frac{1}{2}
\]](http://latex.artofproblemsolving.com/a/1/d/a1dff99d6d8eeefbd302d3d49ac06abd98b1bf5f.png)
![\[
\sum \frac{yz-y+y}{1-z}=\sum -y+\frac{y}{1-z}=-1+\sum\frac{y}{x+y}
\]](http://latex.artofproblemsolving.com/4/c/9/4c90d6463ce6e6f584e57ca4a9ef10e92e668602.png)
![\[
\sum \frac{y}{x+y}=\frac{3}{2}
\]](http://latex.artofproblemsolving.com/d/f/b/dfbde49534134dba7713335236e520e116636738.png)
![\[
2(2x^2y+x^2z+xy^2+3xyz+2xz^2+2y^2z+yz^2)=3(x+y)(y+z)(x+z)
\]](http://latex.artofproblemsolving.com/4/b/4/4b4d0d1f7554aba8ca13ce3d45bd656e531c92df.png)
![\[
0=3(x+y)(y+z)(x+z)-2(2x^2y+x^2z+xy^2+3xyz+2xz^2+2y^2z+yz^2)=x^2z+xy^2+yz^2-x^2y-y^2z-z^2x=(x-y)(y-z)(z-x)
\]](http://latex.artofproblemsolving.com/0/f/c/0fcd5564baa9da14e0c4812b11a31d607f20de1f.png)
must be equal, giving the result.
,
,
.![\[
u=\frac{-(b c (a^3+a^2 b-a b^2-b^3+a^2 c-a b c-b^2 c-a c^2-b c^2-c^3))}{2 (a+b) (a+c) (b+c)}
\]](http://latex.artofproblemsolving.com/0/7/7/077784e74123867b45b4044a623164be6e4d9441.png)
lies on the circumcircle iff 
,![\[
a c (-a^3-a^2 b+a b^2+b^3-a^2 c-a b c+b^2 c-a c^2-b c^2-c^3)=0
\]](http://latex.artofproblemsolving.com/c/9/c/c9cf0f3c1669d8f932a1572d3e8586391cbc9932.png)
![\[
b(a+b)(b+c)=ab^2+abc+b^3+b^2c=a^3+a^2 b+a^2 c+2 a b c+a c^2+b c^2+c^3=(a+c)(b+c)c+(a+c)(a+b)a
\]](http://latex.artofproblemsolving.com/0/e/7/0e7280915705496f542eec11fd125fdd8f109f37.png)
gives the result.
and 
,
iff 
with 
.
,
is in fact the power of the point w.r.t. the circle. I don't see a way forward from there though.
in place of 
for non-normalized coordinates, in order make it clear which triples are normalized and which are not. Of course, this probably means that some new typos have appeared.
need not be in the directions of the x,y axes. It might be helpful for you to look at some sources for a basic understanding of vectors before trying to learn bary
,
and 
?
work?
,![$[CPB]$](http://latex.artofproblemsolving.com/2/7/2/2723e38a779dabba9d6d1a163b967d0641bf0a39.png)
is equal to ![$x[ABC]$](http://latex.artofproblemsolving.com/f/0/d/f0daf521c6206e933974e044ea620bdc0abb8b45.png)
,![$$P = \frac{1}{[ABC]}([PBC], [PCA], [PAB])$$](http://latex.artofproblemsolving.com/b/3/a/b3acafcf62b9e45ff9ea7d2c4d110998674b1803.png)
to one of the vertices, only one sub-triangle remains, which is equal to the area of the reference triangle. Thus, as the coordinates are normalized, the area of both the reference triangle and the sub-triangle is 
,
![$$P=\left(\frac{[PBC]}{[ABC]},\frac{[PAC]}{[ABC]},\frac{[PAB]}{[ABC]}\right)=(x,y,z)$$](http://latex.artofproblemsolving.com/4/a/7/4a793dbc47040c67dae6a9d467c734b0d9a6bd6e.png)
? Also, how do you define 
?
.
and 
have "turned into" lines, and their area is 
is equivalent to triangle ![$$A = P = \frac{1}{[ABC]}([PBC], [PCA], [PAB]) = (\frac{1}{1}, \frac{0}{1}, \frac{0}{1}) = (1, 0, 0)$$](http://latex.artofproblemsolving.com/2/3/e/23eb1b28866a266b35f74a463b8f43ff2e24ee72.png)
. And this can be done similarly for points 
?
I don't understand what they do.
or 
for any point 
.
its symmetric form 
![\[\vec P=x\vec A+y\vec B+z\vec C\iff P=(x,y,z)\]](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.
in 
which means that you can translate the circumcenter 
,
when 
is the zero vector. I have explained this very badly, but the article by Evan Chen explains everything clearly. To be honest though, you can probably just read the demonstrations given in the article, and then try the problems using the formula sheet.
