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Put $U=B_0-A_0$, $T=D_0-C_0$, $D=A_0-C_0$ and $v=V-W$. A point on each moving segment has the form $A_0+sU+tV$ and $C_0+rT+tW$, with $s,r\in [0,1]$. Contact is therefore equivalent toThe contact time of translating line segments is the minimum feasible $t$. This reduces the entire continuous-motion problem to three linear equations and linear inequalities.

In three-dimensional space form the matrix $H=[U,-T,v]$. If it is nonsingular, solveusing a stable linear solver. A candidate with $s,r$ in the unit interval and $t\ge 0$ is the only contact event and hence the earliest one; otherwise no contact occurs. Check initial intersection explicitly, so an already touching pair returns time zero.

If $H$ is singular, use a rank-revealing factorization to test consistency. Write every solution as $x=x_0+Nz$, with columns of $N$ spanning the null space, and minimize its time coordinate subject to the box and nonnegative-time constraints. This is a small linear programming problem. It handles parallel segments, collinear overlap, a continuum of contact times and segments degenerated to points. Infeasibility means no future contact. When $v=0$, the relative geometry is stationary: initially disjoint segments never meet. The first contact is the least feasible nonnegative time, not just an intersection of the two supporting lines.

Begin with two points at distance one. A ruler draws the line through two known points; a compass draws a circle with a known centre and a constructible radius. New points are intersections of such lines and circles. A real number is constructible when it occurs as a coordinate of a point obtained by finitely many such operations; a complex point is constructible when both coordinates are.

If current coordinates lie in a field $K\subseteq \mathbb{R}$, a line-line intersection solves linear equations over $K$, while line-circle or circle-circle intersections reduce to a quadratic equation over $K$. Thus each step requires at most adjoining a square root. Consequently every constructible number lies in a towerConversely field sums and differences are obtained by laying off segments, products and quotients by similar triangles, and a positive square root by the geometric-mean construction in a semicircle. Hence every real number in such a tower is constructible. This proves the quadratic-tower criterion, not merely its necessity.

In particular a constructible algebraic number has degree a power of two. Degree alone is not sufficient in general; the exact criterion is containment in a quadratic tower. Equivalently the Galois group of its normal closure is a finite two-group: a tower's normal closure remains a two-extension, and a two-group admits a chain of subgroups with successive index two, yielding the converse by Galois correspondence.

The standard impossibility results follow. Doubling a unit-volume cube requires $\sqrt[3]{2}$, whose polynomial $X^3-2$ is irreducible by Eisenstein and has degree three. Trisecting an arbitrary angle would in particular trisect $60$ degrees, requiring $2\cos 20^{\circ }$, a root of $X^3-3X-1$. This cubic has no rational root and is irreducible, so this construction is impossible, although some particular angles can be trisected. Squaring the unit circle requires side length $\sqrt{\pi }$, impossible because $\pi$ is transcendental whereas constructible numbers are algebraic.

For a regular $n$-gon, constructibility is equivalent to that of $e^{2\pi i/n}$. The cyclotomic extension is Galois of degree $\phi (n)$, so it is constructible exactly when this degree is a power of two. Writing the prime factorization of $n$ shows the equivalent conditionIndeed odd prime exponents must be one, and each $p_i-1$ must be a power of two. If $2^m+1$ is prime, $m$ must itself be a power of two by factoring when $m$ has an odd divisor. Conversely this factorization makes the cyclotomic Galois group a two-group and provides a quadratic tower. This gives both positive constructions, such as the regular seventeen-gon, and obstructions, such as the regular seven-gon.

Writing $X=\left(\begin{array}{cc}x& y\\ z& w\end{array}\right)$ and using the Frobenius inner product, the three linear equations areThe first two differ by twice the third; equivalently $A-B=2C$. Thus there are only two independent constraints. Substituting $z=-w$ gives $y=2w-x$, so every element of the vector subspace has the unique expressionThe two coefficient matrices satisfy all three constraints. Their linear independence follows by looking first at entry $(1,1)$ and then at entry $(2,2)$. Therefore a basis for this orthogonal complement is

The instantaneous kinematic viscosity law is $\bar{\nu }(r,\Sigma )$, with no explicit constitutive memory. Expand its transported quantity at the possibly evolving background:The viscous transport response exponent satisfies ${\partial }_{\Sigma }(\bar{\nu }\Sigma ){|}_0=q{\bar{\nu }}_0$. Subtract the background equation and retain only linear terms to obtainThis remains a linear equation with space- and time-dependent background coefficients; neither a steady background nor constant $q$ is required for this step. The sign of the response coefficient is the negative-diffusion criterion for viscous disk instability.

For $\bar{\nu }=C{r}^a{\Sigma }^b$, the response is $q=b+1$. A nonaccreting background has ${\bar{\nu }}_0{\Sigma }_0\propto r^{-1/2}$, while ${\Sigma }_0\propto r^{-p}$, soConsequently ${\bar{\nu }}_0\propto r^{p-1/2}$. On this steady background, introduce the square-root-radius diffusion transformSince $q$ is constant and ${\partial }_r=(2x{)}^{-1}{\partial }_x$, the linear equation becomesThus $A(x)\propto x^{2p-3}$, and the required choice isFor a Fourier mode $g\propto e^{\lambda t+i{k}_{x}x}$, $\lambda =-A{k}_x^2$. Positive kinematic viscosity makes $A$ have the sign of $q$, so $q<0$ gives growing modes whose rate increases with $k_x^2$. The formal equation is a backward heat equation and predicts arbitrarily rapid small-scale amplification. Physically the thin disc diffusion closure applies only to wavelengths sufficiently larger than the thickness and stress-relaxation scales; this formal limit identifies the need for a cutoff, rather than a finite fastest wavelength absent from the model. The boundary case $q=0$ has vanishing linear transport response.

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

Let $h_i={\mathbb{E}}_i{T}_6$. Then $h_6=0$, and first-step analysis on the closed class givesSolving these linear equations gives

The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r_1<|z|<r_2$. The Itô formula therefore makes $\log |B_{t\wedge \tau }|$ a bounded martingale. The optional sampling theorem for a supermartingale giveswhere $p=\mathbb{P}(|B_{\tau }|=r_1)$. Solving this linear equation yields the planar Brownian annulus hitting probability

Seek a similarity solution in the dimensionally completed formwhere the fixed scales absorb $A,D$, and $Q_0$. Its inlet volume flux scales asso the prescribed source requiresThe time derivative, horizontal lubrication theory term, and pressure-dependent Darcy drainage term scale respectively asEquating the three exponents givesSolving these linear equations together with the inlet condition yields the similarity exponents of a draining gravity currentEquivalently,and, up to constant dimensional scales,

An elementary row operation is one of three reversible changes: interchange two rows; multiply one row by a nonzero scalar; or add a scalar multiple of one row to a different row. Their inverses are, respectively, the same interchange, multiplication by the reciprocal, and addition of the negative scalar multiple.

Applied to an augmented matrix, these changes interchange linear equations, rescale an equation by a nonzero factor, or add a multiple of one equation to another. Every old solution satisfies the new equations. Applying the inverse elementary row operation gives the converse, so the solution set is unchanged at every step. Equivalently, all the operations together givewhere $E$ is an invertible matrix, a product of elementary matrices.

Writing the unknown vector as $(x,y,z{)}^T$, the first two linear equations give $x=z-1$ and $y=z-\beta$. Substitution into the third givesThus, when $\alpha \ne 2$, put $t=-\beta /(\alpha -2)$. There is exactly one solution:When $\alpha =2$, the remaining equation is $0=-\beta$. HenceThe second case is an affine line in a vector space, so it contains infinitely many solutions. These cases exhaust all real numbers $\alpha ,\beta$.

First find the fixed axis of rotation by solving $(M-I)\mathbf{n}=0$. After removing a nonzero common factor, the linear equations areThey give $n_2=n_3=-n_1$. Choose the following unit vectors:Their dot products show that they form an orthonormal basis; hence $P=({\mathbf{p}}_1{\mathbf{p}}_2{\mathbf{p}}_3)$ is an orthogonal matrix. Since $M$ is an orthogonal matrix fixing ${\mathbf{p}}_1$, it preserves the orthogonal complement of ${\mathbf{p}}_1$. Its restriction there has determinant $1$, so it is a planar rotation.

The trace gives $1+2\cos \theta =\mathrm{tr}M=9/7$, so $\cos \theta =1/7$. To determine the sign of the sine, rather than just its magnitude, calculateThus a suitable angle and change of basis areIn fact direct matrix multiplication gives

Multiplication of a matrix by a column vector forms the linear combination of its columns with coefficients from that vector. Hereso $A\mathbf{r}=x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3$. Equating this expression with $\mathbf{b}$ reproduces the original three linear equations.

Set $X=x^2$ and $Y=y^2$ in the two conic sections. A real intersection requires $X,Y\ge 0$. Solving the resulting linear equations givesFor $a>0$ the denominator and $X$ are positive, so intersections exist exactly when $a\ge 1$. At $a=1$ they are $(\pm 1,0)$; for $a>1$ both choices of sign for each square root give intersections. Consequently

For $a\ne 1,5/2$, the matrix is invertible, so there is exactly one solution. To exhibit it, subtract the first linear equation from the third to obtain $(a-1)x=2$, then subtract twice the second from the first to obtain $-x+(5-2a)z=-3$. Back-substitution givesAt $a=1$, the first and third linear equations have identical left-hand sides but right-hand sides one and three, so there is no solution.

At $a=5/2$, subtracting the first from the third forces $x=4/3$, while the first minus twice the second forces $x=3$. These contradict each other, so there is again no solution. Thus there is one solution for $a\ne 1,5/2$, no solution for either exceptional value, and no parameter value with infinitely many solutions.

All three linear equations are multiples of the first, leaving the single constraint $x+y+z=0$. Choose two real numbers $s,t$ freely:A basis for this two-dimensional linear subspace is $(1,0,-1{)}^T,(0,1,-1{)}^T$.

Here $\alpha \ne 3$, so the same elimination forces $z=0$. The first two linear equations would then require $x-y=1$ and $x-y=-1$ simultaneously. Thus

When $\alpha =3$, the third linear equation is $-3$ times the second, so it adds no constraint. Take $z=t$ freely. The remaining linear equations are $x-y=-1-3t$ and $x+6y=1-\beta t$. Solving gives the straight line of solutionsThis exists for every real $\beta$.

Adding three times the second linear equation to the third gives $3(3-\alpha )z=0$. Since $\alpha \ne 3$, it follows that $z=0$. The second linear equation then gives $y=x+1$, and the first becomes $(1+2\alpha )x=1-2\alpha$. Thus the unique solution isThe assumption $\alpha \ne -1/2$ makes the denominator nonzero; $\beta$ is unrestricted.

Put $D_n=(-1{)}^n{e}^{\pi }-1$. The two linear equations from parts (i) and (ii) give $I_n=D_n+n{J}_n$ and $J_n=-n{I}_n$, so $(1+n^2)I_n=D_n$. ThereforeThese formulas include $n=0$, when $I_0=e^{\pi }-1$ and $J_0=0$.