Isotomic Reciprocity

What Might This Be About?


Let $P$ and $Q$ be two points in $\Delta ABC$ and $P'$ and $Q'$ their respective isotomic conjugates.

Let $DEF$ be the cevian triangle of $Q'$ and $UVW$ that of $P',$ as shown below:

Isotomic Reciprocity, problem

Then $P$ lies on $EF$ iff $Q$ lies on $VW.$ In addition, if $M=BC\cap VW$ and $N=BC\cap EF,$ then $(EFNP)=(VWMQ).$


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



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


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


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.

Barycenter and Barycentric Coordinates

  1. 3D Quadrilateral - a Coffin Problem
  2. Barycentric Coordinates
  3. Barycentric Coordinates: a Tool
  4. Barycentric Coordinates and Geometric Probability
  5. Ceva's Theorem
  6. Determinants, Area, and Barycentric Coordinates
  7. Maxwell Theorem via the Center of Gravity
  8. Bimedians in a Quadrilateral
  9. Simultaneous Generalization of the Theorems of Ceva and Menelaus
  10. Three glasses puzzle
  11. Van Obel Theorem and Barycentric Coordinates
  12. 1961 IMO, Problem 4. An exercise in barycentric coordinates
  13. Centroids in Polygon
  14. Center of Gravity and Motion of Material Points
  15. Isotomic Reciprocity
  16. An Affine Property of Barycenter
  17. Problem in Direct Similarity
  18. Circles in Barycentric Coordinates
  19. Barycenter of Cevian Triangle
  20. Concurrent Chords in a Circle, Equally Inclined

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