Isotomic Reciprocity

The main tool we'll use for solving the problem is the barycentricc coordinates. Much of what we'll need for the proof can be found in Paul Yiu's online book Introduction to the Geometry of the Triangle.

Let point $S$ have homogeneous barycentric coordinates $(x:y:z)$ with respect to $\Delta ABC.$ The cevian $AS$ hits side $BC$ at point $(0:y:z),$ etc. The isotomic conjugate $S'$ has coordinates $\displaystyle (\frac{1}{x}:\frac{1}{y}:\frac{1}{z}).$

A line through two points $S_{i}=(x_{i}:y_{i}:z_{i}),$ $i=1,2,$ is defined by the equation

$\left|\begin{array}{ccc} x & y & z\\ x_{1} & y_{1} & z_{1}\\ x_{2} & y_{2} & z_{2} \end{array}\right|=0.$

Assume $Q=(x:y:z)$ and $P=(u:v:w).$ Then $Q'=\displaystyle (\frac{1}{x}:\frac{1}{y}:\frac{1}{z})$ and $P'=\displaystyle (\frac{1}{u}:\frac{1}{v}:\frac{1}{w}).$ Thus $E=\displaystyle (\frac{1}{x}:0:\frac{1}{z})$ while $F=\displaystyle (\frac{1}{x}:\frac{1}{y}:0).$ Similarly, $V=\displaystyle (\frac{1}{u}:0:\frac{1}{w})$ and $W=\displaystyle (\frac{1}{u}:\frac{1}{v}:0).$

Point $Q=(x:y:z)$ lies on $VW$ only if

$\left|\begin{array}{ccc} x & y & z\\ \displaystyle\frac{1}{u} & 0 & \displaystyle\frac{1}{w}\\ \displaystyle\frac{1}{u} & \displaystyle\frac{1}{v} & 0\\ \end{array}\right|=0.$

The determinant equation evaluates to

$\displaystyle -\frac{x}{vw}+\frac{y}{uw}+\frac{z}{uv}=0,$

which is equivalent to

(*)

$-xu+yv+zw=0.$

This same equation is the condition for $P$ to lie on $EF.$

Now, to evaluate the cross-ratio $(VWMQ)=\displaystyle\frac{VM}{WM}:\frac{VQ}{WQ}$ we'll need to find those ratios. To this end, let's find $\alpha$ and $\beta$ such that $Q=\alpha V+\beta W.$ It will then follow that $\displaystyle\frac{VQ}{WQ}=\frac{\beta}{\alpha}.$ I rewrite $V=(w(v+u):0:u(v+u))$ and $W=(v(w+u):u(w+u):0)$ which insures that the pair of the homogeneous coordinates refers to the same frame of reference. $Q=\alpha V+\beta W$ then gives three equations:

$\begin{align} x&=\alpha w(v+u)+\beta v(w+u),\\ y&=\beta u(w+u),\\ z&=\alpha u(v+u). \end{align}$

From the last two $\displaystyle\alpha=\frac{z}{u(v+u)}$ and $\displaystyle\beta=\frac{y}{u(w+u)}.$ We need to check that these also satisfy the first equation:

$\begin{align}\displaystyle \alpha w(v+u)+\beta v(w+u)&=\frac{z}{u(v+u)}w(v+u)+\frac{y}{u(w+u)}v(w+u)\\ &=\frac{zw}{u}+\frac{yv}{u}\\ &=\frac{yv+zw}{u}\\ &=\frac{xu}{u}\\ &=x, \end{align}$

where we made use if the condition (*). Thus

$\displaystyle\frac{VQ}{WQ}=\frac{\beta}{\alpha}=\frac{y}{u(w+u)}\frac{u(v+u)}{z}=\frac{y(v+u)}{z(w+u)}.$

Point $M$ lies at the intersection of $VW$ and $BC$ whose equation is $x=0.$ Thus, from the determinant equation above $M=\displaystyle (0:\frac{1}{v}:-\frac{1}{w})=(0:w:-v).$ Like $Q,$ $M$ is a linear combination of $V$ and $W,$ say $M=\gamma V+\delta W.$ This again leads to three equations:

$\begin{align} 0&=\gamma w(v+u)+\delta v(w+u),\\ w&=\delta u(w+u),\\ -v&=\gamma u(v+u). \end{align}$

From the last two equations, $\displaystyle\gamma=-\frac{v}{u(v+u)}$ and $\displaystyle\delta=\frac{w}{u(w+u)}.$ The first equation is automatically satisfied. Thus we have

$\displaystyle\frac{VM}{WM}=-\frac{w(v+u)}{v(w+u)}.$

We conclude that

$\begin{align}\displaystyle (VWMQ)&=\frac{VM}{WM}:\frac{VQ}{WQ}\\ &=-\frac{w(v+u)}{v(w+u)}\cdot\frac{z(w+u)}{y(v+u)}\\ &=-\frac{wz}{vy}. \end{align}$

Due to the apparent symmetry $(EFNP)$ has the same value.

Note that $M$ and $U$ are harmonic conjugates with respect to the pair $B,C$ as are $N$ and $D.$

The above statement has been posted by Francisco Javier García Capitán at the CutTheKnotMath facebook page.