You can for instance radially divide a disk of radius 2 into lots of approximately triangular wedges, divide each wedge into four roughly triangular pieces, and rearrange the pieces to form four approximate disks of radius 1.

an excellent observation. here's a small related puzzle. it's clear that you can cut a unit disk into two pieces that each contain a disk of radius 1/2. let phi = (sqrt5 - 1) / 2. let epsilon be a small positive number. how can you cut a unit disk into four pieces that can be rearranged in a non-overlapping fashion to form two regions that each contain a disk of radius phi - epsilon? (no piece covers more than one disk) no idea whether this is optimal, but it's the best i can do on an hour's notice and 6 hours sleep. erich

I came across an interesting formula in an optics book for the solid angle of a spherical triangle.  Let a, b, c, be the unit vectors from the center to the vertices of a spherical triangle on the unit sphere.  Then the solid angle Omega, which equals the area of the triangle, is given by tan(Omega/2) = [a,b,c] / (1 + b.c + a.c + a.b), where a.b etc. are dot products, and [a,b,c] is the scalar triple product.  This paper is the cited reference. http://www.maa.org/sites/default/files/Eriksson14108673.pdf   --  Gene

For the length, theta, of an arc on the circle, tan(theta/2) = sin(theta) / (1 + cos(theta)) = |a x b| / (1 + a.b) But the obvious generalization fails in dimension 4 or higher.  Remembering that the volume of the unit n-ball is (pi^(n/2) / (n/2)!), the expression for its total solid angle is a rational number times pi^2 or higher power of pi.  On the other hand, consider an orthant, for which the unit vectors are orthogonal.  The volume determinant is 1, and all the dot products are 0, so we have tan(Omega/2) = 1 and Omega = pi/2. Back in 3 dimensions, here's a nice relation between the scalar triple product and the sides of the triangle.  Let V be the matrix whose rows are the components of the vectors a, b, c, so that [a,b,c] = det V.  Then [a,b,c]^2 = det(V V^T) = det [[1, a.b, a.c], [b.a, 1, b.c], [c.a, c.b, 1]] The sides of a spherical triangle are measured as angles, and a.b = cos c etc.  Expanding the determinant, we get [a,b,c]^2 = 1 - (cos a)^2 - (cos b)^2 - (cos c)^2 + 2 cos a cos b cos c.   --  Gene From: James Propp <jamespropp@gmail.com> To: Eugene Salamin <gene_salamin@yahoo.com>; math-fun <math-fun@mailman.xmission.com> Sent: Saturday, February 20, 2016 11:17 AM Subject: Re: [math-fun] Solid angle of a spherical triangle I like it! Is there a generalization to n dimensions? Presumably dot products and triple products give way to determinantal expressions. Jim Propp On Saturday, February 20, 2016, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote: I came across an interesting formula in an optics book for the solid angle of a spherical triangle.  Let a, b, c, be the unit vectors from the center to the vertices of a spherical triangle on the unit sphere.  Then the solid angle Omega, which equals the area of the triangle, is given by tan(Omega/2) = [a,b,c] / (1 + b.c + a.c + a.b), where a.b etc. are dot products, and [a,b,c] is the scalar triple product.  This paper is the cited reference. http://www.maa.org/sites/default/files/Eriksson14108673.pdf   --  Gene