Skip to content

Cevians

Creative Commons License
This work is licensed under a Creative Commons Attribution 4.0 International License.

Section 3, session 18. Discussed together with Superposed Filters.

The problem

The syllabus gives only the name. This is the editors' statement of the puzzle it most probably means; see the note at the end.

A cevian is a straight segment from a corner of a triangle to a point on the opposite side (a median is one example).

Draw a large triangle ABC. Mark D one-third of the way from B to C, E one-third of the way from C to A, and F one-third of the way from A to B. Draw the cevians AD, BE and CF. They enclose a small triangle in the middle.

  1. Before measuring, write down a guess: what fraction of the area of ABC is the central triangle?
  2. Measure it and compare.
  3. Explain the number, then find a second, independent explanation.
  4. Does the answer depend on the shape of the triangle?
  5. What if the marks sit at one-quarter of each side, or at different fractions on different sides?

Triangle ABC with each side divided into thirds by tick marks; segments from A to D, B to E and C to F cross to enclose a small shaded triangle in the middle

Drawn for this site (CC BY 4.0).

Why it is in the course

Section 3 is "Observations and Questions", and here a measurement overrules a confident first impression. The figure is built entirely from thirds. When James Randi set the puzzle in early 2001, he reported that an instinctive answer "might be one-ninth, which many readers gave us." A careful drawing disagrees, and the honest next step is a question: why that number?

The syllabus says the puzzles exist "to slow you down for a few minutes so you can examine the working of your own mind". Writing the guess down first is how you catch your mind at work. As with session 12's Sums of Integers, "like a jig-saw puzzle of cross-checks", the answer can be reached several ways that check one another.

Where it comes from

Robert Potts's 1859 school edition of Euclid poses the figure for an equilateral triangle (Problem 59). A later exercise (Problem 100) asks the reader to show what share of the area each of its two triangles has: one joining the marks, and one enclosed by the cevians. (His numbers are in the Resolution.)

The general formula, for cevians cutting the sides in any ratios, was set in the Cambridge Tripos of January 1878 and solved in Glaisher's 1879 volume of solutions. Edward John Routh, himself a Tripos coach then, printed it in a footnote to his Treatise on Analytical Statics (first edition 1891; p. 82 of the 1896 second edition), noting that "the author has not met with these expressions". It is called Routh's theorem all the same.

The puzzle became famous through Hugo Steinhaus's Mathematical Snapshots, which gives a proof with no algebra. Martin Gardner sent that proof to Randi's column in 2001. A 2004 Mathematical Gazette note, "Feynman's triangle", reportedly tells how Richard Feynman was set the puzzle in a dinner conversation and worked it out himself. The editors have not read it.

Hints
  • Draw it big on graph paper and count squares before reasoning.
  • Start with a convenient triangle: (0,0), (3,0), (0,3). Then ask whether an affine map, which takes any triangle to any other and scales every area by the same factor, could change the fraction.
  • Find how each cevian is cut by the other two. The ratio is simple.
  • No algebra: through each corner of the central triangle, draw a line parallel to the side opposite that corner. Add parallels in the same three directions through A, B and C. Compare the pieces outside ABC with the gaps inside.
Resolution

The central triangle is exactly one-seventh of ABC, for any triangle.

Coordinates. With A = (0,0), B = (3,0), C = (0,3) (area 9/2), D = (2,1), E = (0,2), F = (1,0). The cevians cross at P = (6/7, 3/7) (AD and CF), Q = (12/7, 6/7) (AD and BE) and R = (3/7, 12/7) (BE and CF). The shoelace formula gives triangle PQR an area of 9/14, and (9/14)/(9/2) = 1/7. Affine maps preserve area ratios, so this holds for every triangle.

Ratios. Each cevian is cut 3 : 3 : 1 counting from its corner: P and Q sit 3/7 and 6/7 of the way from A to D. Triangle AFC is one-third of ABC, because AF is one-third of AB. Triangle AFP has the same apex A and a base FP one-seventh as long as FC, so it is 1/21 of ABC. The same argument makes BDQ and CER 1/21 each. Triangle ABD is also one-third (7/21) of ABC, and it holds AFP, BDQ and one quadrilateral, so each quadrilateral is 5/21. The centre gets what is left: 21 − 3 − 15 = 3 twenty-firsts, or 1/7.

Steinhaus. Through each corner of the central triangle, draw a line parallel to the opposite side. Each passes through one of the unused one-third marks. Parallels in the same directions through A, B and C complete six copies of the central triangle around it. The parts of the copies that stick out beyond the sides of ABC match the gaps left inside it, so ABC has the area of seven central triangles.

Routh's theorem. If CD/BD = x, AE/CE = y, BF/AF = z, the central fraction is (xyz − 1)² / [(xy + y + 1)(yz + z + 1)(zx + x + 1)]. With all ratios equal to n it is (n − 1)² / (*n*² + n + 1): 0 for medians, 1/7 for thirds, 4/13 for quarter points. Potts's Problem 100 gives "one-third and one-seventh" for his two triangles. For more proofs see Coxeter's Introduction to Geometry and Klamkin and Liu (1981).

Sources

  • Robert Potts, Euclid's Elements of Geometry, fifth school edition, Problems 59 and 100, pp. 78 and 80 (1859) — Internet Archive 🔓
  • J. W. L. Glaisher, G. H. Prior and N. M. Ferrers, Solutions of the Cambridge Senate-House Problems and Riders for the Year 1878, rider (vii), p. 33 (1879) — Internet Archive 🔓
  • Edward John Routh, A Treatise on Analytical Statics, with Numerous Examples, vol. I, 2nd ed., footnote on p. 82 (1896) — Internet Archive 🔓
  • Hugo Steinhaus, Mathematical Snapshots, revised English edition (1960) — Internet Archive 🔓 (borrow)
  • H. S. M. Coxeter, Introduction to Geometry, 2nd ed. (1969) — Internet Archive 🔓 (borrow)
  • M. S. Klamkin and A. Liu, "Three more proofs of Routh's theorem", Crux Mathematicorum 7, 199–203 (1981) — CMS scan of the issue 🔓
  • James Randi, Commentary of 9 February 2001 (the puzzle's answer and the Steinhaus proof sent by Martin Gardner) — Wayback Machine 🔓
  • R. J. Cook and G. V. Wood, "88.46 Feynman's triangle", The Mathematical Gazette 88(512), 299–302 (2004) — doi:10.1017/S002555720017514X 🔒 (not read by the editors)
  • Wikipedia contributors, "One-seventh area triangle" — Wikipedia 🔓
  • Wikipedia contributors, "Routh's theorem" — Wikipedia 🔓
  • Eric W. Weisstein, "Routh's Theorem", MathWorld — MathWorld 🔓
  • Arthur T. Winfree, The Art of Scientific Discovery: original course syllabus — PDF 🔓

How sure are we that this is Winfree's problem?

Probable, not certain. The syllabus lists only "Cevians" in session 18, beside Superposed Filters, and no surviving Winfree document describes it. This puzzle is the standard one told in terms of cevians, needs no apparatus, has a surprising measurable answer and cross-checking proofs, and was in Randi's column in 2001 (the syllabus says many handouts came "from current periodicals"). Two weaker readings remain:

  • Counting regions made by several cevians from each corner, a pattern hunt; but the disk-slicing problem, n dots on a circle, is listed separately in session 19.
  • Ceva's theorem: discover when three cevians meet in one point, as medians do. The name comes from it, but it lacks a one-number surprise.

Back to Section 3 · All problems · The schedule

Machine-readable: this page as Markdown · raw source on GitHub · llms.txt · llms-full.txt (whole site). See For AI agents.