## Sunday, March 19, 2006

### Barycentric Coordinates. Topic: Analytic Geometry. Level: Olympiad.

Definition: We define barycentric coordinates (mass points) as follows. Given a triangle $ABC$, we have $A(1,0,0)$, $B(0,1,0)$, and $C(0,0,1)$.

Let $d(P,l)$ denote the signed distance from a point $P$ to a line $l$.

A point $Q(x,y,z)$ in the plane is represented by

$\left(\frac{d(Q,BC)}{d(A,BC)}, \frac{d(Q,CA)}{d(B,CA)}, \frac{d(Q,AB)}{d(C,AB)}\right)$.

It's easy to show that all points $Q(x,y,z)$ satisfy $x+y+z = 1$.

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Theorem: Three points $P_i(x_i,y_i,z_i)$ for $i = 1,2,3$ are collinear iff

$\left|\begin{array}{ccc} x_1 \quad y_1 \quad z_1 \\ x_2 \quad y_2 \quad z_2 \\ x_3 \quad y_3 \quad z_3 \end{array} \right|$ = 0.

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Comment: The proof of this is pretty standard. It works the same way as with Cartesian coordinates (you can use vectors and cross products).

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Problem: (WOOT Message Board) Let $D, E, F$ lie on sides $BC, CA, AB$, respectively, of triangle $ABC$. Suppose that $D, E, F$ are collinear. Prove that the midpoints of $AD, BE, CF$ are collinear.

Solution: We use barycentric coordinates with triangle $ABC$.

Since $D$ lies on $BC$, we know $d(D, BC) = 0$ so the $x$-coordinate will be zero. Hence the coordinates of $D$ have the form $(0,t_1,1-t_1)$ for some real $t_1$.

Similarly, $E$ is of the form $(1-t_2, 0, t_2)$ and $F$ is $(t_3, 1-t_3, 0)$ for real $t_2, t_3$.

Since $D, E, F$ are collinear, we know the determinant of the matrix with the three points is zero. This comes out to be

$t_1t_2t_3-(1-t_1)(1-t_2)(1-t_3) = 0$. (1)

We can find the midpoints by simply averaging the points (work this out rigorously if you want). These are

$\left(\frac{1}{2},\frac{t_1}{2},\frac{1-t_1}{2}\right)$, $\left(\frac{1-t_2}{2},\frac{1}{2},\frac{t_2}{2}\right)$, and $\left(\frac{t_3}{2},\frac{1-t_3}{2},\frac{1}{2}\right)$.

If we take the determinant of the matrix with these three points, we find that it is equivalent to

$\frac{t_1t_2t_3-(1-t_1)(1-t_2)(1-t_3)}{4}$

after simplification. But by (1) we know that it zero as well, which means the midpoints are indeed collinear. QED.

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Comment: Barycentric coordinates are very powerful, and give easy proofs to some results which would be very difficult without them.

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Practice Problem: (Menelaus' Theorem) Let $D, E, F$ be points on sides $BC , CA, AB$ (possibly extended), respectively, of triangle $ABC$. Prove that $D,E,F$ are collinear iff $BD \cdot CE \cdot AF = -DC \cdot AE \cdot BF$ with signed distances.