Olympiad Geometry IV: Barycentric & Complex Coordinates

Complex bash works because rotations and reflections are multiplications and conjugations. Its sweet spot: problems on ONE circle with points defined by intersections of chords and tangents, where every point has a short formula. Its failure mode: two circles or an incircle, where the formulas explode. Decide in the first five minutes; a half-finished bash scores nothing.

Each condition is a single equation to verify, and with unit-circle points the conjugates are $1/a$, so everything becomes a rational identity. Clear denominators and compare — the algebra is mechanical, and that is the point. Write the target condition FIRST, then compute only what it needs.

Barycentrics shine on cevians, ratios along sides, and anything defined by areas or by the incircle/excircles — precisely where complex numbers struggle. The distance formula is $|PQ|^2 = -a^2yz - b^2zx - c^2xy$ for the displacement $(x, y, z)$ with $x + y + z = 0$, and the circumcircle is $a^2yz + b^2zx + c^2xy = 0$. Every named center has a short coordinate; memorize the six that recur (centroid, incenter, circumcenter, orthocenter, symmedian point, Nagel point).

The standard olympiad use: a point defined by two ratios on the sides, a claim that some third line passes through it. Compute the point in unnormalized coordinates, plug into the line equation, and the claim reduces to an identity in $a, b, c$. Keep everything homogeneous and only normalize (divide by $u + v + w$) when distances are needed.

A bash is a proof because every step is an identity in the free parameters: nothing depends on a picture, so configuration issues vanish, but only if the setup is fully general. Losing generality is the classic failure — placing a triangle with a right angle, or on a circle of radius $1$, is allowed only when the problem's statement is invariant under the similarity you used. State that invariance in one sentence and the grader has nothing to object to. Then do the algebra you would do anyway, and stop the moment the target equation is an identity.

Step 1 — barycentrics: $D = (0 : 2 : 1)$, $E = (1 : 0 : 2)$, $F = (2 : 1 : 0)$ (the weight on $B$ is $DC$, on $C$ is $BD$, and so on).

Step 2 — the cevian $AD$ is the set of points with $y : z = 2 : 1$, the line $y = 2z$; similarly $BE$ is $z = 2x$ and $CF$ is $x = 2y$.

Step 3 — intersections: $AD \cap BE$ has $z = 2x$, $y = 4x$, the point $\tfrac17(1, 4, 2)$; $BE \cap CF$ is $\tfrac17(2, 1, 4)$; $CF \cap AD$ is $\tfrac17(4, 2, 1)$.

Step 4 — the area of a triangle with normalized barycentric vertices is $[ABC]$ times the determinant of their coordinates:

How to find it. Ratios along sides and an area: that is what barycentrics were built for. Each cevian is a one-term equation, so the whole problem is three tiny linear systems and one determinant.

Step 1 — setup: complex numbers, $ABC$ labeled counterclockwise, $\zeta = e^{i\pi/3}$, so $\zeta^2 = \zeta - 1$ and $\bar\zeta = 1 - \zeta$. The outward apex on $BC$ is on the right of the direction $B \to C$, obtained by rotating $C - B$ clockwise by $60^\circ$: $A_1 = B + \bar\zeta(C - B) = C - \zeta(C - B)$.

Step 2 — the center $X$ of the triangle on $BC$ is its centroid: $3X = B + C + A_1 = \zeta B + B + (2 - \zeta)C$. Cycling the letters, the centers $Y$ on $CA$ and $Z$ on $AB$ are $3Y = \zeta C + C + (2 - \zeta)A$ and $3Z = \zeta A + A + (2 - \zeta)B$.

Step 3 — target: $Z - X = \zeta(Y - X)$, which says $XZ$ is $XY$ turned by $60^\circ$ — an equilateral triangle. Compute

Multiplying the first by $\zeta$ and using $\zeta^2 = \zeta - 1$: $\zeta(2 - \zeta) = \zeta + 1$, $\;-\zeta(1 + \zeta) = 1 - 2\zeta$, $\;\zeta(2\zeta - 1) = \zeta - 2$. These are exactly the coefficients of the second. $\blacksquare$

How to find it. Equilateral triangles are rotations by $60^\circ$, and rotation is multiplication. Reduce everything with $\zeta^2 = \zeta - 1$ and the identity becomes a comparison of three coefficients.

Step 1 — setup: axes along $BC$ and the altitude from $A$, so $A = (0, a)$, $B = (b, 0)$, $C = (c, 0)$ with $a > 0$ and $b < 0 < c$. Every acute triangle can be placed this way.

Step 2 — the altitude from $B$ is perpendicular to $AC$, whose direction is $(c, -a)$, so it is $c(x - b) = ay$; at $x = 0$ this gives $H = \left(0, -\tfrac{bc}{a}\right)$. Also $M = \left(\tfrac{b + c}{2}, 0\right)$, so $\overrightarrow{HM} = \left(\tfrac{b + c}{2}, \tfrac{bc}{a}\right)$ and $\mathbf v = \left(\tfrac{bc}{a}, -\tfrac{b + c}{2}\right)$ is perpendicular to it.

Step 3 — the line is $H + t\mathbf v$. Substituting into $AB$, $\tfrac xb + \tfrac ya = 1$:

The line $AC$ is the same equation with $b$ and $c$ swapped, so $t_Q = \dfrac{2(a^2 + bc)}{a(b - c)} = -t_P$.

Step 4 — $P = H + t_P\mathbf v$ and $Q = H - t_P\mathbf v$, so $H$ is the midpoint of $PQ$. $\blacksquare$

How to find it. The statement is about $H$ and a perpendicular, so put an axis along an altitude: $H$ then has a one-term coordinate, and the symmetry $b \leftrightarrow c$ between the two sides does the final step for you.