An Extension of van Schooten's Theorem

van Schooten's Theorem asserts that, for a point P on the circumcircle of an equilateral ΔABC, the longest of the segments PA, PB, PC is the sum of the shorter two.

Bui Quang Tuan has shown how van Schooten theorem can be extended to triangles that are not equilateral.

In ΔABC, h = b·c/(2R), where h is the altitude from A and R is the circumradius. (For the proof see Metric Relations in a Triangle.)

To continue with the theorem, for ΔPBC, lemma gives

and similarly for triangles PCA and PAB. Next we obtain

ΔABC divides its circumcircle into three arcs, say (BC), (CA), (AB). If, for instance, P is on arc (BC) then by Ptolemy's theorem,

Therefore

It follows from (1) that say, when a = b = c, PA = PB + PC is equivalent to a/d_a = b/d_b + c/d_c. However, while the latter holds for all triangles, the former holds only for the equilateral ones.

Observe that, as it follows from the proof, the ratio in the left-hand side of (2) (i.e., the largest ratio possible) corresponds to the arc that contains P.

Now, as Bui Quang Tuan has shown, (2) admits a further extension to polygons with more than three vertices.

Without loss of generality we may assume that P is on the arc X₁X₂. Then the set of ratios R₁₂, R₂₃, R₃₄, ..., R_(n-1)n, R_n1 has an engaging property that, in a sense, extends Theorem 1:

Proof

The proof is by induction where we start with n = 3, the triangle case. Theorem 1 may serve as the base of the induction.

We proceed with the inductive step. Assume k > 3 that our theorem has been shown for n = k-1. Consider convex k-gon X₁X₂ ... X_k. This polygon divides the circumcircle into k arcs. We shall assume that P is on the arc X₁X₂. Consider ΔX₁X₂X₃. This triangle divides circumcircle into three arcs and, by our assumption, P is on the arc X₁X₂.

X₁X₃X₄ ... X_k is a convex (k-1)-gon and P is on the arc X₁X₃. (This is because the original k-gon is convex.) Therefore, by the inductive assumption,

In ΔX₁X₂X₃,

So the theorem is true for any n ≥ 3.

Bui Quang Tuan also came up with a direct proof that avoids inductive reasoning. Assuming P is on the arc X₁X₂, consider a triangulation of the polygon X₁X₂ ... X_n by diagonals emanating from X₁. We get triangles X₁X₂X₃, X₁X₃X₄, ..., X₁X_kX_(k+1), ..., X₁X_(n-1)X_n. Since then P can be considered to lie in arc X₁X₂, but also in arcs X₁X₃, X₁X₄, and so on, Theorem 1, when applied to these triangles, implies the following sequence of identities:

Summing them up we again obtain the desired result: