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On Some Geometrical Inequalities

Published online by Cambridge University Press:  03 November 2016

Extract

In § 1 of this Note we first investigate some geometrical inequalities connecting a set of n unordered points in a plane. An ellipse is associated with the set of points, and use is made of the invariants of this ellipse under rotation of the coordinate axes. Some inequalities, first obtained by Weitzenböck, are derived and extended. A new geometrical entity, which we call the mutual area, is then defined for a set of n and a set of m points with the same centroid. Using the mutual invariants of two ellipses with a common centre under rotations of the coordinate axes, we obtain some inequalities connecting two sets of points with a common centroid.

Type
Research Article
Copyright
Copyright © Mathematical Association 1942

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References

page 202 of note * Weitzenböck, , Math. Zeitschrift, 5 (1919), (137146)CrossRefGoogle Scholar.

page 202 of note † Loc. cit.

page 203 of note * Pedoe, , Math. Gazette, 25 (1941), 224 CrossRefGoogle Scholar.

page 203 of note † Cf. Weitzenböck, loc. cit.; and Pedoe, loc.cit.

page 204 of note * Orthogonal projection shows clearly that a triangle is not uniquely determined by the unique ellipse which touches the sides of the triangle at their midpoints. This ellipse has its centre at the centroid of the triangle and the semi-diameters parallel to the sides of the triangle are respectively proportional to those sides. The ellipse is therefore similar to the ellipse (3). We deduce that the shape of a triangle is not determined by that of the central ellipse for unit particles imagined at the vertices, that is, by the shape of the ellipse (3). Similarity of the central ellipses of two triads does not therefore involve the similarity of the triads.

page 206 of note * Loc. cit.

page 206 of note † Loc. cit.

page 206 of note ‡ Let O be the centroid of the four non-coplanar points ABGD. We show that if the products of inertia ∑yz, ∑zx, ∑xy of unit particles at the vertices about any three mutually perpendicular planes through O vanish, then ABCD is equilateral. The product of inertia about the plane BCA and a plane through BC perpendicular to this plane equals the product of inertia of a unit particle at D with respect these planes. It also equals the product of inertia of four unit particles at O with respect to these planes. It follows that DO is parallel to the plane through BC perpendicular to ABC. Continuing thus, DO must be perpendicular to ABC. Similarly AO is perpendicular to DBC. Therefore BC is perpendicular to the intersection of DOA and DBC; i.e. DB = DC. Proceeding thus, the result follows.

page 208 of note * The associated ellipsoids are then similar and similarly situated.