One-Seventh Triangle and Routh’s Theorem

Certain geometric facts have an appealing simplicity that makes one hope for an equally simple picture-proof. The one-seventh triangle is a good example. Start with a triangle ABC . Choose points D,E,F on BC,CA,AB respectively, so that each chosen point divides its side in the ratio 1:2.

\displaystyle BD=\frac13BC,\quad CE=\frac13CA,\quad AF=\frac13AB.

The three cevians AD,BE,CF form a smaller central triangle. Its area is exactly one seventh of the area of ABC .

It is natural to search for a clever dissection or an auxiliary construction that makes the fraction 1/7 visually obvious. One can certainly draw useful parallels and use similar triangles, but the number 1/7 does not simply leap from the picture. The reason is that the problem belongs to a more general three-parameter family: if the three side divisions are changed, the central area changes according to a rational formula. This is Routh’s theorem. The one-seventh configuration is the particularly symmetric case in which all three ratios are equal. This very elusiveness often hints that the general case, possesses an algebraic richness that may not readily yield to elementary geometric intuition alone, guiding us toward more structured approaches.

The problem is affine in nature. An affine transformation preserves lines, division ratios along lines, and ratios of areas. Thus the specific angles and side lengths of ABC are irrelevant. We could send ABC to any convenient reference triangle, but barycentric coordinates are even better: they retain the triangle itself while encoding the side-division data directly.

Barycentric Coordinates

Barycentric coordinates are especially well suited to this problem because they encode both affine position and side ratios directly. Use ABC as the reference triangle. Its vertices have normalized barycentric coordinates

\displaystyle A=(1,0,0),\qquad B=(0,1,0),\qquad C=(0,0,1).

A point X=(\alpha,\beta,\gamma) , with \alpha+\beta+\gamma=1 , means that \displaystyle X=\alpha A+\beta B+\gamma C. The three coordinates measure the affine weights of the vertices. In particular, a point lies on BC exactly when its first coordinate is zero, and similarly for the other sides.

Let points D, E, F be on sides BC, CA, AB respectively, such that:

\displaystyle \frac{BD}{BC} = x, \quad \frac{CE}{CA} = y, \quad \frac{AF}{AB} = z

where 0<x,y,z<1 . Thus x,y,z describe where the three chosen points lie along their respective sides. Their barycentric coordinates are

\displaystyle D = (1-x)B + xC = (0, 1-x, x)
\displaystyle E = (1-y)C + yA = (y, 0, 1-y)
\displaystyle F = (1-z)A + zB = (1-z, z, 0)

The cevians are AD, BE, CF . Let the inner triangle be PQR , where P = AD \cap BE , Q = BE \cap CF , and R = CF \cap AD . Our goal is to compute the area of PQR relative to the area of ABC .

Coordinates of Inner Vertices

Let us derive P=AD\cap BE in full detail. A point on the cevian AD can be written as \displaystyle (1-t)A+tD=(1-t,t(1-x),tx). Likewise, a point on the cevian BE can be written as \displaystyle (1-u)B+uE=(uy,1-u,u(1-y)). At their intersection P , these coordinates are equal: \displaystyle 1-t = uy, \quad t(1-x) = 1-u, \quad tx = u(1-y). The middle equation gives \displaystyle u=1-t(1-x). Substitute this into the first equation: \displaystyle 1-t=\big(1-t(1-x)\big)y=y-ty(1-x). After rearranging, \displaystyle 1-y=t(1-y+xy), so we get

\displaystyle t=\frac{1-y}{1-y+xy}.

Substituting this value into the parametrization of AD , we obtain

\displaystyle P= \frac{1}{1-y+xy} \big(xy,(1-x)(1-y),x(1-y)\big).

The same calculation, performed cyclically, gives the other two vertices:

\displaystyle Q=\frac{1}{1-z+yz} \big(y(1-z),yz,(1-y)(1-z)\big),

\displaystyle R= \frac{1}{1-x+zx} \big((1-z)(1-x),z(1-x),zx\big).

For three normalized barycentric points, the determinant of their coordinate matrix gives the signed ratio of their triangle’s area to the area of the reference triangle:

Area Ratio Calculation

The ratio of the area of \displaystyle \triangle PQR to \displaystyle \triangle ABC is given by the determinant:

\displaystyle \mathcal{A}_{\text{ratio}} = \frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)}=\begin{vmatrix} P_A & P_B & P_C \\ Q_A & Q_B & Q_C \\ R_A & R_B & R_C \end{vmatrix}

One way to see this is to apply an affine map sending A,B,C to (0,0),(1,0),(0,1) . Under this map, a barycentric point (\alpha,\beta,\gamma) becomes the ordinary point (\beta,\gamma) . Since every normalized barycentric row has sum 1 , the barycentric determinant becomes the usual planar determinant for the signed area.

Substituting the coordinates of P,Q,R , we get

\displaystyle \mathcal{A}_{\text{ratio}} =\frac{\Delta}{(1-y+xy)(1-z+yz)(1-x+zx)},

where \displaystyle \Delta= \det \begin{pmatrix} xy&(1-x)(1-y)&x(1-y)\\ y(1-z)&yz&(1-y)(1-z)\\ (1-z)(1-x)&z(1-x)&zx \end{pmatrix}.

The determinant has an unexpectedly clean factorization. Expanding along the first row, the third cofactor vanishes because its two monomials are identical. The two remaining cofactors give

\displaystyle \begin{aligned} \Delta &= x^2y^2z^2 -2xyz(1-x)(1-y)(1-z)\\ &\quad+(1-x)^2(1-y)^2(1-z)^2. \end{aligned}

Thus we have \displaystyle \Delta= \big(xyz-(1-x)(1-y)(1-z)\big)^2. We have therefore obtained the general formula

\displaystyle \frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = \frac{ \big(xyz-(1-x)(1-y)(1-z)\big)^2}{(1-y+xy)(1-z+yz)(1-x+zx)}.

This is already a complete answer in the variables x,y,z , which record the fractions of the three sides traversed from B,C,A respectively.

For the original configuration, x=y=z=\frac13.
The numerator becomes ((\frac13)^3-(\frac23)^3)^2 =(\frac1{27}-\frac8{27})^2 =\frac{49}{729}.
Each factor in the denominator is 1-\frac13+\frac13\cdot\frac13 =\frac23+\frac19 =\frac79.
Therefore (1-y+xy)(1-z+yz)(1-x+zx) =(\frac79)^3=\frac{343}{729}.
Hence we have

\displaystyle \frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} =\frac{49/729}{343/729} =\frac17.

Thus the central triangle has exactly one seventh of the area of ABC .

Routh’s Theorem

The traditional form of Routh’s theorem uses the ratios into which the points divide the sides:

\displaystyle p:=\frac{BD}{DC},\qquad q:=\frac{CE}{EA},\qquad r:=\frac{AF}{FB}.

These are related to the earlier variables by

\displaystyle x=\frac{p}{p+1},\qquad y=\frac{q}{q+1},\qquad z=\frac{r}{r+1}.

and the preceding formula becomes

\displaystyle \frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = \dfrac{(pqr-1)^2} {(pq+p+1)(qr+q+1)(rp+r+1)}.

This is Routh’s theorem. For the one-seventh triangle, the side divisions are all 1:2 , so p=q=r=\frac12. and we get \frac17, as before.

exactly as before.

Vanishing Area: The Condition of Ceva

The central triangle has area zero precisely when the numerator vanishes:

\displaystyle pqr=1.

But this is exactly Ceva’s theorem:

\displaystyle \frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1

is the necessary and sufficient condition for the three cevians AD,BE,CF to be concurrent. Thus Ceva’s theorem appears as the degenerate case of Routh’s theorem: when the cevians meet at one point, the inner triangle collapses.

The fact that the numerator is a square also has a geometric meaning. The quantity pqr-1 measures the failure of the Ceva condition. As the three cevians pass through the concurrent configuration, the signed orientation of the small triangle can change, but its ordinary area cannot become negative. Hence the unsigned area naturally contains the square (pqr-1)^2 .

This is more than a convenient consistency check. Routh’s formula does not merely coexist with Ceva’s theorem; it contains Ceva’s concurrency criterion as the precise boundary case in which the inner triangle loses all area. The geometric event of three lines meeting at one point becomes the algebraic condition pqr-1=0 , and the area is governed by its square. This is a recurring pattern in geometry: an incidence condition is encoded by the vanishing of a polynomial, while a quantitative invariant such as area measures how far the configuration lies from that degenerate case. Here the factor pqr-1 measures the failure of concurrency, and Routh’s theorem shows exactly how that failure controls the size of the central triangle.

The Synthetic Approach

Coordinates give a direct and systematic calculation, but there is also a classical synthetic route. One proves, using parallels, similar triangles, or Menelaus’s theorem, that the three corner triangles satisfy

\displaystyle \frac{\text{Area}(\triangle ABP)}{\text{Area}(\triangle ABC)} =\frac{p}{pq+p+1},

\displaystyle \frac{\text{Area}(\triangle BCQ)}{\text{Area}(\triangle ABC)}= \frac{q}{qr+q+1},

\displaystyle \frac{\text{Area}(\triangle CAR)}{\text{Area}(\triangle ABC)} =\frac{r}{rp+r+1}.

The three corner triangles and the central triangle fill ABC . Therefore

\displaystyle \frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} =1-\frac{p}{pq+p+1}-\frac{q}{qr+q+1}-\frac{r}{rp+r+1}.

Putting these fractions over a common denominator gives

\displaystyle \dfrac{(pqr-1)^2} {(pq+p+1)(qr+q+1)(rp+r+1)}.

This synthetic proof is a testament to classical geometric reasoning, yet the final algebraic step underscores that the path to the explicit formula, even via synthesis, often involves significant algebraic manipulation.

What begins as a simple geometric curiosity—the appearance of the fraction 1/7—ultimately reflects a deeper mathematical structure. Through barycentric coordinates and determinants, the seemingly mysterious area ratio emerges naturally from the algebra governing cevians in a triangle. Routh’s theorem reveals that classical geometry, affine transformations, and area ratios are all connected within a single elegant framework. Even Ceva’s theorem appears as a limiting case, showing how concurrency arises when the inner triangle collapses. In the end, the 1/7 triangle is not just a puzzle, but a window into the rich interplay between geometry and algebra.

Leave a comment