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The two labelled regions in the original diagram are complementary circular sectors. Their areas add to $\pi r^2$ and satisfy $A=2B$, so $B=\pi r^2/3$. The sector with central angle $\theta$ has area $r^2\theta /2$. Hence

The two radii enclose the angle $2\pi /3$, so the triangle has area $T=\frac{1}{2}{r}^2\sin (2\pi /3)=\sqrt{3}{r}^2/4$. From the sector calculation, $r^2=3B/\pi$. Consequently

Choose two distinct space-diagonal directions $\mathbf{u}=(1,1,1)$ and $\mathbf{v}=(-1,1,1)$. Their dot product is one and their norms are both $\sqrt{3}$. Thus the acute angle between their unoriented lines isAny two distinct space diagonals have an absolute dot product of one when represented with components $\pm 1$, so they give the same acute angle.

Take $r>0$ and select a complex argument $\theta$ for $z$. To specify the complex exponentiation, use $z^{1+i}=\exp ((1+i)(\ln r+i\theta ))$. Expanding its exponent givesIts complex logarithm therefore has valuesThe continuous logarithm obtained by retaining the displayed exponent uses $k=0$. If the outer logarithm is the principal logarithm, choose $k$ to put the imaginary part in $(-\pi ,\pi ]$ instead; the real part is unchanged. In particular one cannot generally assert $\mathrm{Log}(e^w)=w$ without checking its imaginary part.

The condition on the real part gives $\ln r=\theta$, so the required logarithmic spiral has polar coordinatesAs $\theta$ increases, it runs from the negative real axis near radius $e^{-\pi }$ through $-i{e}^{-\pi /2}$, $1$ and $i{e}^{\pi /2}$ to the negative real axis at radius $e^{\pi }$. The angle increases by one complete turn. Using the conventional principal argument $-\pi <\theta \le \pi$, the inner negative-axis endpoint is excluded and the outer endpoint is included. The convention $-\pi \le \theta <\pi$ reverses those endpoint choices. If both limiting arguments are admitted by the paper's wording, both limiting endpoints are drawn; the interior curve is the same. The origin is never included because the complex logarithm is undefined there.

The triangle $ADB$ is right-angled at $A$, with $AB=\sqrt{2}$ and $AD=1/\sqrt{2}$. Another application of the Pythagorean theorem gives $DB=\sqrt{2+1/2}=\sqrt{5/2}$. At the angle with vertex $D$, the adjacent side is $AD$ and the hypotenuse is $DB$. Hence

Powers of the imaginary unit repeat in blocks $i,-1,-i,1$, each with sum zero. The first $44$ terms comprise eleven complete blocks; the last is $i^{45}=i$. Applying Euler's formula to the resulting complex exponential gives $e^i=\cos 1+i\sin 1$. HenceThe angles here are in radians.

The volume element is the absolute Jacobian determinant times the coordinate differentials. In the Cartesian coordinate system, cylindrical coordinates $(r,\theta ,z)$ and spherical coordinates $(\rho ,\vartheta ,\phi )$, respectively,Here $\rho$ is distance from the origin, $\vartheta$ is the polar angle from the positive $z$ axis, and $\phi$ is the azimuthal angle. The spherical coordinates convention fixes which angle enters the sine factor.

The Pythagorean trigonometric identity makes the second factor $1+{\cos }^2\theta$, which is strictly positive. It is therefore enough to solve $\sin \theta -\cos \theta \le 0$. The sine addition formula givesThe shifted angle lies between $\pi /4$ and $5\pi /4$. On this interval the sine is nonpositive precisely from $\pi$ onwards. Restoring the original angle and retaining its strict upper bound gives

The two directed segments have vectors $\mathbf{u}=(1,0,-1)$ and $\mathbf{v}=(0,1,1)$. Their dot product is $-1$ and both norms equal $\sqrt{2}$. HenceThe common starting point fixes the obtuse angle between these segments.

Since $z^2=(x^2-y^2)+2ixy$, equating its real part and imaginary part gives $x^2-y^2=2xy$, orThe quadratic equation in $y$ factors as $[y-(-1+\sqrt{2})x][y-(-1-\sqrt{2})x]=0$. Thus the set is the union of two straight lines through the origin:Their slopes multiply to $-1$, so they are perpendicular. Their angles to the real axis are $\pi /8$ and $-3\pi /8$, modulo $\pi$. The origin is included.

At each height, the removed circular sector has the same angle and $4/5$ of the original radius. Its area and hence its volume are therefore multiplied by $(4/5{)}^2=16/25$. Equating the removed volume to that of the reformed solid sphere givesConsequently

At height $z$, the circular sector has radius $r(z)=R{e}^{-z/H}$ and angle $\Theta (z)=2\pi -\pi z/H$. Its area is $\Theta (z)r(z{)}^2/2$, so slicing by horizontal cross-sections gives the volume integralUse the substitution rule $t=z/H$. ThenAn integration by parts gives the antiderivative $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore

The acute angles follow from part (ii) and the Snell's law relation:The path lengths below and above the interface are respectivelyHence the total distance and minimum travel time areFor a sketch to scale, choose the interface as $y=0$, with $A=(0,-\sqrt{3})$ and $B=(2,1)$. The crossing is $C=(1,0)$ because $d_1\tan \theta =d_2\tan \varphi =1$. The figure uses equal axis scales and shows the two normal-relative angles.

The angles in the PDF are measured from the vertical normal to the interface. The horizontal displacements of the two path segments are $d_1\tan \theta$ and $d_2\tan \varphi$, while their lengths are $d_1\sec \theta$ and $d_2\sec \varphi$. Thus the geometric constraint and travel time areThe Lagrange multiplier equations $\nabla T=\lambda \nabla g$ giveCancelling the positive geometric factors gives $\lambda =2\sin \theta /c=\sqrt{2}\sin \varphi /c$, henceThis is Snell's law, $\sin \theta /(c/2)=\sin \varphi /(c/\sqrt{2})$. The relation depends only on the ratio of the two speeds, so changing $D,d_1,d_2$ leaves the relation unchanged; it changes which angles satisfy it together with the geometric constraint.

For a minimum check, let $u$ be the horizontal distance to the crossing point. ThenandThus the travel time is a strictly convex function of $u$, and its stationary point is the unique global minimum. For positive $D,d_1,d_2$, $T^{\prime }(0)<0<T^{\prime }(D)$, placing the crossing between the two endpoint projections.

From the same multiplication, the imaginary part isHere the planar vectors are embedded in three dimensions with zero third coordinate. Its magnitude is the area of their parallelogram; its sign is positive when the oriented angle from ${\mathbf{v}}_2$ to ${\mathbf{v}}_1$ has positive sine. The reversed order matters: $\Im (z_1{z}_2^{\ast })$ is the negative of the usual signed area for the order $({\mathbf{v}}_1,{\mathbf{v}}_2)$.

Use spherical coordinates $(r,\theta ,\varphi )$ centred on the right-hand sphere, with polar axis pointing outwards along $s$. The Jacobian determinant is $r^2\sin \theta$, and the retained region is $r\le R$ with $r\cos \theta \ge -d$.

For $0\le r\le d$, every polar angle $0\le \theta \le \pi$ is allowed. For $d<r\le R$, the cut requires $0\le \theta \le \arccos (-d/r)$. Splitting the volume integral at $r=d$ therefore givesSubstitution of $R=\beta a$ and $d=(\alpha -\gamma )a$ reproduces the mass in part (c). The radial split avoids integrating over a polar angle range independent of $r$, which would incorrectly include part of the removed spherical cap.

The dot product is $1\cdot 0+2(-1)+(-1)\cdot 1=-3$, while the Euclidean norms are $\sqrt{6}$ and $\sqrt{2}$. Therefore the angle $\theta \in [0,\pi ]$ satisfies

Join the centre of the unit circle to adjacent vertices of the regular polygon. Each of the $n$ resulting triangles has two sides of length $1$ and included angle $2\pi /n$. Bisecting one gives a right triangle, so each polygon side, a chord of a circle, has length $2\sin (\pi /n)$. Thus the perimeter is $2n\sin (\pi /n)$.

Each of these triangles has area $\frac{1}{2}\sin (2\pi /n)$, so the polygon's area is $\frac{n}{2}\sin (2\pi /n)$. Comparing the two expressions gives

Since $\mathbf{v}=-\mathbf{u}$ and $\mathbf{u}\ne 0$, the vectors point in opposite directions. More explicitly, their dot product givesHence their angle is $\boxed{\theta =\pi \text{ radians}=180^{\circ }}$.

Use the bead angles ${\theta }_1,{\theta }_2,{\theta }_3$ as generalized coordinates. The straight-line spring length between two beads follows from the law of cosines:The hoop is horizontal, so the gravitational potential energy is constant and can be omitted. The kinetic energy is $(m/2){\sum }_i{\dot{\theta }}_i^2$, giving the LagrangianAt ${\theta }_i^0=\Theta +2\pi (i-1)/3$, every spring has its natural length, so the potential energy is zero and its first derivatives vanish. This is an equilibrium for every constant $\Theta$. It is stable against relative displacement because each spring energy is nonnegative; the overall rotation is a neutral direction.