and 
,
as the greatest exponent 
such that 
divides 
.
and infinitely many 
such that 
.
and 
.
with circumcircle 
is given. Let 
be the midpoint of 
,
be the foot of altitude from 
to 
,
be the intersection point of 
and 
.
lies on the arc 
.
.
be a trapezium inscribed in a circle 
with diameter 
. The circle with center 
and radius 
meets 
and 
(where 
)
at 
.
is perpendicular to 
.
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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)
,![\[
a c (-a^3-a^2 b+a b^2+b^3-a^2 c-a b c+b^2 c-a c^2-b c^2-c^3)=0
\]](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 
.
,
(they turned into lines) and the other leftover sub-triangle 
![$$P=\left(\frac{[PBC]}{[ABC]},\frac{[PAC]}{[ABC]},\frac{[PAB]}{[ABC]}\right)=(x,y,z)$$](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.
