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Weitzenböck's InequalityThe applet below illustrates a geometric proof of Weitzenböck's Inequality. For any triangle ABC with sides a, b, c, and the area S,
This wonderful geometric proof is due to C. Alsina and R. B. Nelsen.
References
Copyright © 1996-2010 Alexander Bogomolny
TheoremFor any triangle ABC with sides a, b, c, and the area S,
ProofForm, on the sides of ΔABC, Napoleon triangles, ABC', BCA', CAB'. For a triangle with all angles less than 120°, the three lines AA', BB', CC' concur at Fermat's point of the triangle. The point solves the minimization problem: for all points X, Multiply (1) by √3/4. The inequality becomes
where Sa is the area of ΔBCA', etc.
In fact, a stronger result holds. Let F be interior to ΔABC, with distances x, y, z to A, B, C, respectively. Let Sk denote the area of an equilateral triangle with side k. Then
To see that (3) holds, observe that F splits ΔABC into three triangles each with angle 120° at F. This shows that the remaining angles in each triangle add up to 60°. It follows that three copies of each of the triangles fit into the corresponding equilateral triangle leaving equilateral triangles with sides When F coincides with, say, vertex C, then
Stronger than (1) is the Hadwiger-Finsler inequality: if a, b, c are the sides of a triangle with area S then
In terms of areas, the latter is equivalent to
In case F is internal to ΔABC, this inequality is a consequence of (3) because
To prove these, assume a ≥ b ≥ c. Then by reflecting the triangle with sides b, x, z in the segment z leads to the following configuration:
The triangle inequality (in the yellow triangle) then implies
In case where, say, angle C is 120° or more so that F coincides with C,
Now observe that a ≥ |b - c| and b ≥ |c - a| from which, again, (6) follows.
Copyright © 1996-2010 Alexander Bogomolny |
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