One-Seventh Triangle and Routh’s Theorem

Certain mathematical gems sparkle with an apparent simplicity, teasing us with the promise of an equally simple, “aha!” proof. The one-seventh area–triangle problem exemplifies this, inviting us to think about a direct, first-principles argument.

Recall the setup: in any triangle \displaystyle ABC , points \displaystyle D, E, F are chosen on sides \displaystyle BC, CA, AB respectively, such that they divide the sides in a \displaystyle 1:2 ratio (e.g., \displaystyle BD = \frac{1}{3}BC , \displaystyle CE = \frac{1}{3}CA , \displaystyle AF = \frac{1}{3}AB ). The cevians \displaystyle AD, BE, CF then delineate a central triangle. Its area? Precisely \displaystyle 1/7 th that of \displaystyle \triangle ABC .

Many of us have undoubtedly pursued that elusive, purely synthetic demonstration—a clever dissection, perhaps, or an inspired choice of auxiliary lines that would render the \displaystyle 1/7 th ratio visually manifest. Even leveraging affine invariance by simplifying to an equilateral triangle doesn’t typically make the specific fraction “leap out” without algebraic engagement with the ratios. This very elusiveness often hints that the general case, known as Routh’s Theorem, possesses an algebraic richness that may not readily yield to elementary geometric intuition alone, guiding us toward more structured approaches.

The Algebraic Engine: Barycentric Coordinates

When direct geometric insight for the general case stalls, a robust strategy is to “let the symbols do the talking.” Barycentric coordinates offer a natural and powerful language for problems steeped in ratios and affine properties.

Let \displaystyle \triangle ABC be the reference triangle. Its vertices in normalized barycentric coordinates are \displaystyle A=(1,0,0) , \displaystyle B=(0,1,0) , and \displaystyle C=(0,0,1) . Let points \displaystyle D, E, F be on sides \displaystyle BC, CA, AB respectively, such that: \displaystyle BD/BC = x_s , \displaystyle CE/CA = y_s and \displaystyle AF/AB = z_s

The barycentric coordinates of \displaystyle D, E, F are:

\displaystyle D = (1-x_s)B + x_sC = (0, 1-x_s, x_s)
\displaystyle E = (1-y_s)C + y_sA = (y_s, 0, 1-y_s)
\displaystyle F = (1-z_s)A + z_sB = (1-z_s, z_s, 0)

The cevians are \displaystyle AD, BE, CF . Let the inner triangle be \displaystyle PQR , where \displaystyle P = AD \cap BE , \displaystyle Q = BE \cap CF , and \displaystyle R = CF \cap AD .

Coordinates of Inner Vertices

Let’s derive the coordinates for \displaystyle P = AD \cap BE . A point on line \displaystyle AD can be written as \displaystyle (1-t)A + tD = (1-t)(1,0,0) + t(0, 1-x_s, x_s) = (1-t, t(1-x_s), tx_s) . A point on line \displaystyle BE can be written as \displaystyle (1-u)B + uE = (1-u)(0,1,0) + u(y_s, 0, 1-y_s) = (uy_s, 1-u, u(1-y_s)) . At their intersection \displaystyle P , these coordinates are equal:

  1. \displaystyle 1-t = uy_s
  2. \displaystyle t(1-x_s) = 1-u
  3. \displaystyle tx_s = u(1-y_s)

From (2), \displaystyle u = 1 - t(1-x_s) . Substituting into (1): \displaystyle 1-t = (1 - t(1-x_s))y_s = y_s - ty_s(1-x_s) . \displaystyle 1-y_s = t - ty_s(1-x_s) = t(1 - y_s + x_s y_s) . So, \displaystyle t = \frac{1-y_s}{1 - y_s + x_s y_s} .

The coordinates of \displaystyle P=(P_A, P_B, P_C) are:

\displaystyle P_A = 1-t = 1 - \frac{1-y_s}{1 - y_s + x_s y_s} = \frac{x_s y_s}{1 - y_s + x_s y_s}
\displaystyle P_B = t(1-x_s) = \frac{(1-y_s)(1-x_s)}{1 - y_s + x_s y_s}
\displaystyle P_C = tx_s = \frac{(1-y_s)x_s}{1 - y_s + x_s y_s}

Let \displaystyle k_P = 1 - y_s + x_s y_s . So we get \displaystyle P = \frac{1}{k_P} (x_s y_s, (1-x_s)(1-y_s), x_s(1-y_s)) .

By cyclic permutation of variables (\displaystyle x_s \to y_s \to z_s \to x_s ) and corresponding coordinate positions, we find \displaystyle Q and \displaystyle R :

Let \displaystyle k_Q = 1 - z_s + y_s z_s . Then \displaystyle Q = \frac{1}{k_Q} (y_s(1-z_s), y_s z_s, (1-y_s)(1-z_s)) .
Let \displaystyle k_R = 1 - x_s + z_s x_s . Then \displaystyle R = \frac{1}{k_R} ((1-z_s)(1-x_s), z_s(1-x_s), z_s x_s) .

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}} = \begin{vmatrix} P_A & P_B & P_C \\ Q_A & Q_B & Q_C \\ R_A & R_B & R_C \end{vmatrix}

Substituting the coordinates:

\displaystyle \mathcal{A}_{\text{ratio}} = \frac{1}{k_P k_Q k_R} \begin{vmatrix} x_s y_s & (1-x_s)(1-y_s) & x_s(1-y_s) \\ y_s(1-z_s) & y_s z_s & (1-y_s)(1-z_s) \\ (1-z_s)(1-x_s) & z_s(1-x_s) & z_s x_s \end{vmatrix}

Let \displaystyle X=x_s, Y=y_s, Z=z_s , and \displaystyle X_c=1-X, Y_c=1-Y, Z_c=1-Z . The determinant of the numerators, Det’, is:

\displaystyle Det' = XY( (YZ)(ZX) - (Y_cZ_c)(ZX_c) ) - X_cY_c( (YZ_c)(ZX) - (Y_cZ_c)(Z_cX_c) ) + XY_c( (YZ_c)(ZX_c) - (YZ)(Z_cX_c) )

The third term’s parenthesis \displaystyle (YZ_c)(ZX_c) - (YZ)(Z_cX_c) = YZX_cZ_c - YZX_cZ_c = 0 .

So, \displaystyle Det' = XY( XYZ^2 - X_c Y_c Z_c Z ) - X_cY_c( XYZ Z_c - X_c Y_c Z_c^2 )
\displaystyle = X^2Y^2Z^2 - XYZ(X_cY_cZ_c) - XYZ(X_cY_cZ_c) + (X_cY_cZ_c)^2 \displaystyle = (XYZ - X_cY_cZ_c)^2

Substituting back \displaystyle X=x_s, Y=y_s, Z=z_s and \displaystyle X_c=1-x_s, Y_c=1-y_s, Z_c=1-z_s :

Det’ \displaystyle = (x_s y_s z_s - (1-x_s)(1-y_s)(1-z_s))^2 .

The product of the denominators \displaystyle k_P k_Q k_R equal to

((1-x_s)+x_s y_s)((1-y_s)+y_s z_s)((1-z_s)+z_s x_s)

after appropriate matching of terms (the set of factors is the same due to cyclic symmetry).

Thus,

\displaystyle \mathcal{A}_{\text{ratio}} = \frac{(x_s y_s z_s - (1-x_s)(1-y_s)(1-z_s))^2}{((1-x_s)+x_s y_s)((1-y_s)+y_s z_s)((1-z_s)+z_s x_s)}

For the \displaystyle 1/7 th case, \displaystyle x_s=y_s=z_s=1/3 . Numerator is \displaystyle ((1/3)^3 - (2/3)^3)^2 = (1/27 - 8/27)^2 = (-7/27)^2 = 49/729 .
Denominator terms are each \displaystyle (1-1/3)+(1/3)(1/3) = 2/3 + 1/9 = 7/9 . So the denominator is \displaystyle (7/9)^3 = 343/729 . And the total ratio is \displaystyle (49/729) / (343/729) = 49/343 = 1/7 . The algebra delivers!.

Routh’s Theorem

This formula is a specific instance of Routh’s Theorem. The more traditional parameterization uses ratios \displaystyle k_1 = BD/DC , \displaystyle k_2 = CE/EA , and \displaystyle k_3 = AF/FB . The substitution \displaystyle x_s = k_1/(k_1+1) (and similarly for \displaystyle y_s, z_s ) transforms the barycentric formula into the familiar Routh’s Theorem:

\displaystyle \mathcal{A}_{\text{ratio}} = \frac{(k_1 k_2 k_3 - 1)^2}{(k_1k_2+k_1+1)(k_2k_3+k_2+1)(k_3k_1+k_3+1)}

For the \displaystyle 1/7 th triangle, \displaystyle k_1=k_2=k_3=1/2 , which again yields \displaystyle 1/7 .

Vanishing Area: The Condition of Ceva

The inner triangle’s area vanishes if and only if the numerator \displaystyle k_1 k_2 k_3 - 1 = 0 , i.e., \displaystyle k_1 k_2 k_3 = 1 . This is precisely Ceva’s Theorem for the concurrency of \displaystyle AD, BE, CF . This confluence, where a fundamental theorem emerges as a limiting case, is a hallmark of deep algebraic structure — Geometric properties captured in terms of polynomials equalities (inequalties).

The Synthetic Approach

While coordinates give a direct computational route, one can also prove Routh’s Theorem purely by drawing parallels and invoking similar triangles.

Let ratios be \displaystyle x = BD/DC , \displaystyle y = CE/EA , \displaystyle z = AF/FB (these are \displaystyle k_1,k_2,k_3 in the common form of Routh’s Theorem).

  1. Area of a Corner Triangle: Using similar triangles (often by constructing parallels) or Menelaus’s Theorem, they establish, for instance: \displaystyle \frac{\text{Area}(\triangle APB)}{\text{Area}(\triangle ABC)} = \frac{x}{xy+x+1}
  2. Summation: By symmetry, the areas of \displaystyle \triangle BQC and \displaystyle \triangle CRA are found. The inner triangle’s area is then: \displaystyle \frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = 1 - \left( \frac{x}{xy+x+1} + \frac{y}{yz+y+1} + \frac{z}{zx+z+1} \right)
  3. Algebraic Simplification: This expression simplifies (requiring, as they note, “a bit of algebraic jugglery”) to the standard Routh’s formula: \displaystyle \frac{(xyz - 1)^2}{(xy+x+1)(yz+y+1)(zx+z+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.

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