Geometric inequalities

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Let AA be a finite subset of the unit disc in the plane, and let N(A,r)N(A,r) be the set of points at distance ≤r\leq r from AA, where 0<r<10<r<1. Show that the length of the boundary N(A,r)N(A,r) is, at most, C/rC/r for some constant CC independent of AA.

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If a convex curve of circumference ll and of area aa can roll unimpeded in the interior of another convex curve of circumference LL and area AA, then we have

lL≤2π(a+A). lL \leq 2\pi (a+A) .
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Remark.
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G. Pólya, problem: Arch. Math. Phys. Ser. 3, 27, p. 162 (1918).

Denote the volume, the surface area and the integral of the mean curvature of two convex surfaces by vv, ss, mm and VV, SS, MM respectively. If the first body can roll unimpeded in the interior of the second, then we have the inequalities

mS+sM≤12π(V+v),mS−sM≤4π(V−v). mS + sM \leq 12\pi (V+v) , \qquad mS - sM \leq 4\pi (V-v) .
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Remark.
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G. Pólya, problem: Arch. Math. Phys. Series 3, 27, p. 162 (1918).

Let aa, bb, cc be the sides, and α\alpha, β\beta, γ\gamma the angles of a triangle in radian measure, aa lying opposite to α\alpha, bb to β\beta and cc to γ\gamma. Then we have

π3≤aα+bβ+cγa+b+c≤π2. \frac{\pi }{3} \leq \frac{a\alpha +b\beta +c\gamma }{a+b+c} \leq \frac{\pi }{2} .

Equality holds in the first inequality if and only if the triangle is equilateral, and in the second inequality if and only if the triangle degenerates into a straight line counted twice.

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Remark.
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Following J. Kürschák.

Let k1,k2,…,k6k_1, k_2, \ldots , k_6 be the lengths of the six edges of a tetrahedron and α1,α2,…,α6\alpha_1, \alpha_2, \ldots , \alpha_6 the radian measures of the angles formed by the faces of the tetrahedron that meet at the respective edges. Then we have

π3<k1α1+k2α2+⋯+k6α6k1+k2+⋯+k6<π2. \frac{\pi }{3} < \frac{k_1\alpha _1+k_2\alpha _2+\cdots +k_6\alpha _6}{k_1+k_2+\cdots +k_6} < \frac{\pi }{2} .

The given bounds are the best possible.

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Remark.
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G. Pólya, problem: Arch. Math. Phys. Ser. 3, 27, p. 162 (1918).

A lampshade has the shape of a frustum of a right circular cone. Its perimeter is PP at the bottom, pp at the top, its slant height is ss, and the inclination ii of the slant height to the bottom plane is ≥60°\geq 60°; P≥pP \geq p. Show that such a lampshade can be cut out in one piece from a rectangular sheet of paper of dimensions PP and s+[p(P−p)]/8ss + \left[p(P-p)\right]/8s. There will even be paper left over for a flap to glue the ends together, except in the extreme case where P=pP=p, i=π/2i=\pi /2, and no paper will be wasted.

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Remark.
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Cf. G. Pólya, problem: Amer. Math. Monthly, 54, p. 340 (1947). Solution by Joseph Rosenbaum: Amer. Math. Monthly, 55, pp. 162--164 (1948).

Consider a triangle. Let aa, bb and cc denote its sides and AA its area. Prove that

A≤34(abc)2/3 A \leq \frac{\sqrt3}{4} (abc)^{2/3}

with equality only for equilateral triangles.

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(continued). If p>0p > 0

A≤34(ap+bp+cp3)2p, A \leq \frac{\sqrt3}{4} \left(\frac{a^p+b^p+c^p}{3}\right)^{\frac{2}{p}} ,

with equality only for equilateral triangles.

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(continued). Let rr and RR denote the radii of the inscribed and the circumscribed circle of a triangle, respectively. Prove that

2r≤R, 2r \leq R ,

with equality only for equilateral triangles.

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Formulate a proposition that has an analogous relation to 17.1 as 17 has to 16.

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(continued). Try to find simple particular cases for which the proposition formulated above can be conveniently tested.

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References

[deSi04] de Souza P.N., Silva J. Berkeley Problems in Mathematics. 3rd ed., Springer, 2004.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.