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.
Theorem 1
The proof depends on the following
Lemma
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
(1) |
a/d_{a} = a/(PB·PC)·(2R) = a·PA·(2R)/(PA·PB·PC), b/d_{b} = b/(PC·PA)·(2R) = b·PB·(2R)/(PA·PB·PC), c/d_{c} = c/(PA·PB)·(2R) = c·PC·(2R)/(PA·PB·PC). |
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
(2) | a/d_{a} = b/d_{b} + c/d_{c}, |
as needed.
It follows from (1) that say, when a = b = c,
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.
Theorem 2
Given point P on the circumcircle of a cyclic convex polygon X_{1}X_{2}X_{3} ... X_{n-1}X_{n}
- a_{i j} = length of side X_{i}X_{j}
- d_{i j} = distance from P to sidelines X_{i}X_{j}
- R_{i j} = ratio a_{i j}/d_{i j}
Without loss of generality we may assume that P is on the arc X_{1}X_{2}. Then the set of ratios R_{12}, R_{23}, R_{34}, ..., R_{(n-1)n}, R_{n1} has an engaging property that, in a sense, extends Theorem 1:
(2) | R_{12} = R_{23} + R_{34} + ... + R_{(n-1) n} + R_{n1}. |
Proof
The proof is by induction where we start with
We proceed with the inductive step. Assume
X_{1}X_{3}X_{4} ... X_{k} is a convex
In ΔX_{1}X_{2}X_{3},
R_{12} | = R_{23} + R_{13} |
= R_{23} + R_{34} + R_{45} + ... + R_{k1} |
So the theorem is true for any
Bui Quang Tuan also came up with a direct proof that avoids inductive reasoning. Assuming P is on the arc X_{1}X_{2}, consider a triangulation of the polygon
R_{13} = R_{34} + R_{14},
...
R_{1k} = R_{k(k + 1)} + R_{1(k + 1)},
...
R_{1(n-2)} = R_{(n-2)(n-1)} + R_{1(n-1)},
R_{1(n-1)} = R_{(n-1)n} + R_{1n}.
Summing them up we again obtain the desired result:
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