Van Schooten's and Pompeiu's Theorems
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A Mathematical Droodle

3 November 2014, Created with GeoGebra


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Copyright © 1996-2017 Alexander Bogomolny


Van Schooten's and Pompeiu's Theorems

The applet illustrates two theorems discovered three hundred years apart. Let $P$ be a point in the plane of an equilateral $\Delta ABC.$ Both theorems are concerned with the relative lengths of the segments $AP,$ $BP,$ and $CP,$ joining $P$ the vertices of $\Delta ABC.$

Theorem (F. van Schooten, 1615 - 1660)

For $P$ on the circumcircle of $\Delta ABC.$ The longest of the three segments equals the sum of the shorter two.

Theorem (D. Pompeiu, 1873 - 1954)

For $P$ not on the circumcircle of $\Delta ABC,$ there exists a triangle with sides $PA,$ $PB,$ and $PC.$

(The triangle in question is known as Pompeiu's triangle.)

Pompeiu's theorem

Ptolemy's theorem offers one approach to proving the theorems. A proof below is more elementary.

For a proof, rotate the plane $60^{\circ}$ around $B$ so as to make vertex $C$ overlap vertex $A.$ Let $A'$ and $P'$ be the images of $A$ and $P,$ respectively. I'll show that $APP'$ is Pompeiu's triangle.

By the construction, $\Delta BPP'$ is equilateral, so that $PP' = BP.$ In addition, $AP'$ is the image under the above rotation of $CP,$ so that $AP' = CP.$ The remaining side in $\Delta APP'$ is just $AP,$ and we are done with Pompeiu's theorem.

Assuming $A,$ $P,$ and $P'$ are collinear (i.e., when Pompeiu's triangle is degenerate), $\angle AP'B$ is either $60^{\circ}$ or $120^{\circ}.$ (This is because it is either equal to $\angle BPP',$ which is $60^{\circ},$ or is supplementary to it.) But then the quadrilateral $ABCP$ is cyclic, proving van Schooten's theorem and its converse.

Remark 1

Van Schooten's configuration appears in the following simple lemma:

Assume in $\Delta ABC$ $\angle BAC=120^{\circ}.$ Let $D$ be the foot of the angle bisector from $A.$ Then $\displaystyle AD=\frac{AB\cdot AC}{AB+AC}.$

For a proof, consider the circumcircle $(ABC)$ and let $E$ be the point where $AD$ crosses $(ABC).$ $E$ is the midpoint of the arc $BC$ opposite $A.$ This makes $\Delta BCE$ equilateral and the required proportion could be written as $\displaystyle AD=\frac{AB\cdot AC}{AE}$ (or, in the notations used for Pompeiu's theorem, $\displaystyle DP'=\frac{AP'\cdot BP'}{CP'},$ where $D$ is the intersection of $CP'$ and $AB.)$

corollary to van Schooten's theorem

Triangles $CDE$ and $ADB$ are similar, so that $\displaystyle\frac{AD}{CD}=\frac{AB}{CE}.$ But $CE=BC,$ hence,

$\displaystyle AD=AB\frac{CD}{BC}.$

By a property of angle bisectors, $\displaystyle\frac{AB}{AC}=\frac{BD}{CD},$ implying


Combining the two ratios yields $\displaystyle AD=AB\frac{AC}{AB+AC}=\frac{AB\cdot AC}{AB+AC}.$

Remark 2

van Schooten's theorem admits a non-trivial extension for triangles that are not equilateral!

(There is another proof.)

There is also a converse construction of finding an eqilateral triangle.


  1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, p. 4.

Ptolemy's Theorem

  1. Ptolemy's Theorem
  2. Sine, Cosine, and Ptolemy's Theorem
  3. Useful Identities Among Complex Numbers
  4. Ptolemy on Hinges
  5. Thébault's Problem III
  6. Van Schooten's and Pompeiu's Theorems
  7. Ptolemy by Inversion
  8. Brahmagupta-Mahavira Identities
  9. Casey's Theorem
  10. Three Points Casey's Theorem
  11. Ptolemy via Cross-Ratio
  12. Ptolemy Theorem - Proof Without Word
  13. Carnot's Theorem from Ptolemy's Theorem

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Copyright © 1996-2017 Alexander Bogomolny


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