Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3

The integral is finite because a continuous function on a compact closed interval is bounded. Nonnegativity is immediate, and absolute homogeneity of a norm follows from $|cf(x)|=|c||f(x)|$. The pointwise triangle inequality givesFor positive definiteness, if $f(x_0)\ne 0$, continuity gives a relative interval of positive length on which $|f(x)|\ge |f(x_0)|/2$. Its integral is positive, including when $x_0$ is an endpoint. Thus $\Vert f\Vert =0$ implies $f=0$, and this is a norm on the given vector space.

For the given sequence,The last bound tends to zero as both indices tend to infinity, proving the Cauchy sequence property. Moreover, the sequence converges to the zero function in this norm:There is no conflict with the nonzero endpoint values: convergence in this norm does not imply pointwise convergence.

Write the cone as $z=r\cot \alpha$, with $r$ the distance from its axis and $0<\alpha <\pi /2$. Its Euclidean metric givesConsequently a differentiable path expressed as $r=r(\theta )$ has arc lengthFor $L(r,r^{\prime })=\sqrt{r^2+{\csc }^2\alpha (r^{\prime }{)}^2}$, the Euler-Lagrange equation has the Beltrami identity because $L$ has no explicit $\theta$ dependence:For a nonradial minimizing path, $C>0$. Rearranging and integrating givesThe endpoint conditions determine $C$ and ${\theta }_0$ on the selected angular branch.

To verify that this stationary path really minimizes arc length, use the local isometry from a circular cone to the plane. Set $s=r/\sin \alpha$ and $\varphi =\theta \sin \alpha$; then $d{s}^2+s^{2}d{\varphi }^2$ is the plane metric. The displayed path is a straight line $s\cos (\varphi -{\varphi }_0)=C/\sin \alpha$, so its segment realizes the plane distance. Choose endpoint angular lifts with $|\Delta \theta |\le \pi$; then $|\Delta \varphi |<\pi$, and the joining segment avoids the apex. Its length isOther lifts with developed angular difference less than $\pi$ give at least this distance; an angular change of at least $\pi$ has infimum at least $s_1+s_2$, attained only if passage through the apex is allowed. This proves the global choice above. Endpoints on a common generator have the radial segment as shortest path, which is the limiting case omitted by the $r(\theta )$ parametrization. If an endpoint is the apex, its shortest path is a generator. Thus shortest paths are straight segments after developing the cone.

Fix any $z_0\in \mathbb{C}$. The Cauchy estimate on the circle $|z-z_0|=R$ applies for every $R>0$ because $f$ is an entire function. It givesTherefore $f^{\prime }(z_0)=0$. Since $z_0$ was arbitrary and $\mathbb{C}$ is connected, $f$ is constant. This is the sublinear case of the polynomial growth theorem for entire functions.

The composition $g(z)=e^{-f(z)}$ is an entire function, andBy Liouville's theorem, $g$ is constant. Since the exponential function never vanishes,Thus $f$ is constant. Notice that constancy of an exponential alone need not be inverted by choosing a global logarithm; differentiating supplies the conclusion directly.

Let $T$ be the isometry. Since $T(0)=0$, it preserves norms, and the polarization identity applied to distances gives $⟨T(x),T(y)⟩=⟨x,y⟩$. For an orthonormal basis $e_1,e_2,e_3$, the vectors $T(e_i)$ are again an orthonormal basis, andExpanding in that basis proves $T(x)={\sum }_i⟨x,e_i⟩T(e_i)$. Hence $T$ is a linear map represented by an orthogonal matrix $Q$.

A reflection in a hyperplane with nonzero normal $v$ is $H_v(x)=x-2⟨x,v⟩v/⟨v,v⟩$. If $Q{e}_1\ne e_1$, choose $v=Q{e}_1-e_1$. Because $\Vert Q{e}_1\Vert =\Vert e_1\Vert$, substitution shows $H_{v}Q{e}_1=e_1$. If they agree, no reflection is needed. The resulting orthogonal transformation fixes $e_1$ and preserves its orthogonal complement. Apply the same construction there to fix $e_2$; the second normal is perpendicular to $e_1$, so the first vector remains fixed. Finally the restriction to the remaining line is either the identity or negation; in the latter case a third reflection, normal to $e_3$, fixes it. Reversing these steps writes $Q$ as a product of at most three reflections, each in a plane through the origin. This proves the three-dimensional case of the Cartan–Dieudonné theorem.

Three reflections are sometimes necessary. The map $-I$ is the product of the three coordinate-plane reflections. Its determinant is $-1$, excluding zero or two reflections. A single plane reflection fixes a two-dimensional vector subspace, whereas $-I$ fixes only zero, excluding one. Thus the bound is sharp.

Player A maximizes the payoff, while B minimizes it. The fourth row is strictly worse for A than the first in every column, so remove $A_4$ by dominated strategy elimination. On the remaining rows, column $B_3$ is strictly better for B than $B_2$, so remove $B_2$. Now $A_3$ is strictly worse than $A_2$ in the two remaining columns. The reduced zero-sum game has payoff matrixwith rows $A_1,A_2$ and columns $B_1,B_3$.

If A uses mixed strategy $(p,1-p)$, the expected payoffs against these columns are $-2+6p$ and $3-8p$. The first increases and the second decreases, so their minimum is maximized where they are equal: $14p=5$. ThusFor an optimality certificate, B uses mixed strategy $(q,0,1-q)$, making A's first two row payoffs $-5+9q$ and $3-5q$. Equality gives $q=4/7$. Against this B strategy the four row payoffs are $(1/7,1/7,-6/7,-6/7)$, while against the displayed A strategy the three column payoffs are $(1/7,13/7,1/7)$. Therefore A guarantees $1/7$ and B holds A to $1/7$, verifying the value of a zero-sum game and the original, unreduced mixed strategy.

The Lagrange basis polynomials satisfy ${\ell }_i(x_j)={\delta }_{ij}$: for $j\ne i$, the product has a zero factor, whereas for $j=i$ every factor is one. The polynomialhas degree of a polynomial at most $n$ and has $n+1$ distinct roots of a polynomial. The root bound for a polynomial implies $q=0$. Hence the interpolation identity holds for every polynomial of degree at most $n$.

At a simple root of a polynomial $x_i$ of $\omega$, differentiating the product gives ${\omega }^{\prime }(x_i)={\prod }_{k\ne i}(x_i-x_k)\ne 0$. Consequently the Lagrange basis polynomials can be writtenDivide the polynomial interpolation identity from part (a) by $\omega (x)$, at points outside the nodes. This gives the partial fraction decompositionEvery denominator of $\omega$ is simple, and the rational function is proper because $\mathrm{deg}p<\mathrm{deg}\omega$, so no polynomial part is missing.

For $x\in \ker (\alpha -\lambda I)$, commutation givesThus each eigenspace of $\alpha$ is an invariant subspace for $\beta$.

If $\alpha ={\beta }^2$, the two endomorphisms commute. The $n$ distinct real eigenvalues of $\alpha$ have one-dimensional real eigenspaces, since their direct sum is the $n$-dimensional vector space. Take a nonzero real eigenvector $x_i$. Its eigenspace is invariant under $\beta$, so $\beta x_i=c_i{x}_i$ for a real scalar $c_i$. Squaring givesThe simple-spectrum hypothesis is essential: the real $90$-degree rotation matrix squares to $-I$ in dimension two, whose negative eigenvalue has a two-dimensional eigenspace.

With polar unit vectors ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, the second term is $\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$. It is tangent to circles, independent of $z$, and has zero divergence and zero vorticity away from the axis. Its singular axis carries a line vortex; adding the constant velocity field gives a uniform flow plus that line vortex.

The counterclockwise circulation around a positively oriented loop is $\oint \mathbf{u}\cdot d\mathbf{r}$. For a radius-$R$ circle, the uniform flow contributes zero and the line vortex contributesAny loop winding once around the axis has the same circulation, by Stokes theorem in the nonsingular intervening region.

On $C_R$, $\mathbf{n}={\mathbf{e}}_r$, $\mathbf{u}\cdot \mathbf{n}=U\cos \theta$, and $dl=Rd\theta$. ThusIn fact this equality holds for every $R>0$ for the stated field, since the remaining vector integral is $(0,\pi ,0)$.

This is the advective part of the momentum flux. To obtain the force, include pressure. For steady irrotational flow, the Bernoulli equation givesThe constant terms integrate to zero against $\mathbf{n}$; the remaining pressure contribution is $\oint p\mathbf{n}dl=\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The total outward momentum flux plus pressure term is therefore $\rho \mathbf{\Gamma }\times \mathbf{U}$, the force of the obstacle on the fluid. The opposite force, of fluid on the obstacle per unit span, isThis is the Kutta–Joukowski theorem, with positive $\Gamma$ defined counterclockwise: positive $U$ and $\Gamma$ give force in the negative $y$ direction. For an actual two-dimensional aerofoil, faster-decaying far-field terms do not change this limit. The displayed advective integral supplies half the lift; pressure supplies the other half.

The discriminant of a binary quadratic form is $d=b^2-4ac$; it is unchanged by a unimodular matrix change of integral variables and satisfies $d\equiv 0$ or $1(\mathrm{mod}4)$. For such a $d$, the discriminant criterion for prime representation by a binary quadratic form says that some integral binary quadratic form of discriminant of a binary quadratic form $d$ represents an odd prime $p$ exactly when $d$ is a square modulo $p$, including zero. Equivalently,These versions agree: a square root modulo $p$ can be chosen even when $d\equiv 0(\mathrm{mod}4)$ and odd when $d\equiv 1(\mathrm{mod}4)$, by adding $p$ if needed, and then the Chinese remainder theorem applies.

For completeness, if $f(x,y)=p$, then $(x,y)$ is primitive, since any common divisor would have its square divide $p$. Complete $(x,y)$ to a basis of ${\mathbb{Z}}^2$ with determinant one using Bézout's identity. The transformed binary quadratic form has leading coefficient $p$ and discriminant of a binary quadratic form $B^2-4pC=d$. Conversely, a suitable $B$ makes $[p,B,(B^2-d)/(4p)]$ an integral binary quadratic form representing $p$ at $(1,0)$.

For $x^2+3y^2$, the discriminant of a binary quadratic form is $-12$. The prime $3$ is represented. If $p\ne 3$ is represented, reducing modulo $3$ gives $p\equiv x^2\equiv 1(\mathrm{mod}3)$, which also excludes $p=2$. To prove sufficiency for $p\equiv 1(\mathrm{mod}3)$, quadratic reciprocity givesThe criterion constructs a positive definite binary quadratic form $[p,B,C]$ of discriminant of a binary quadratic form $-12$. It is primitive: a common divisor of its coefficients would divide $p$, and if that divisor were $p$, then $p^2$ would divide $12$, impossible for $p\ne 2,3$.

We must still establish that this form is equivalent to $x^2+3y^2$, rather than merely some form of the same discriminant of a binary quadratic form. In a positive definite integral binary quadratic form, replace $x$ by $x+ky$ to arrange $|b|\le a$. If $c<a$, apply $(x,y)\mapsto (-y,x)$ and repeat. Each such exchange strictly decreases the positive integer leading coefficient, so this reduction algorithm for a positive definite binary quadratic form terminates with $|b|\le a\le c$. Then $12=4ac-b^2\ge 3a^2$, so $a\le 2$. With $a=1$, parity forces $b=0$, giving $c=3$. With $a=2$, the only integral choices are $b=\pm 2,c=2$, and these are not primitive. Thus every primitive positive definite binary quadratic form of discriminant of a binary quadratic form $-12$ is equivalent to $[1,0,3]$, and preserves its represented integers. Therefore

A photon of wavelength $\lambda$ has energy $E=h\nu =hc/\lambda$ and momentum $\mathbf{p}=(h/\lambda )\hat{\mathbf{n}}$ along its propagation direction.

The specified Compton scattering formula requires the electron to be initially at rest in the frame where the wavelengths and angle are measured. This initial-rest assumption is implicit in the printed request; it is not valid for arbitrary initial electron momentum. Use four-momentum with Minkowski metric signature $(+---)$: initially $P=(mc,0)$, while $K=(E/c,(E/c)\hat{\mathbf{n}})$ and $K^{\prime }=(E^{\prime }/c,(E^{\prime }/c){\hat{\mathbf{n}}}^{\prime })$ are null. Conservation gives $P^{\prime }=P+K-K^{\prime }$, and the outgoing electron satisfies the same mass-shell condition $P^{\prime 2}=P^2=m^2{c}^2$. HenceAfter dividing by $E{E}^{\prime }$ and substituting the photon energy relation,The scale $h/(mc)$ is the electron Compton wavelength. For a moving initial electron, the earlier invariant identity remains valid, but $P\cdot K$ includes its initial velocity and the wavelength shift is different.

The contraction mapping theorem states that a contraction mapping $f$ of a nonempty complete metric space into itself has a unique fixed point. More explicitly, if $d(f(x),f(y))\le qd(x,y)$ with $0\le q<1$, iteration from any $x_0$ converges to that fixed point.

To prove this, put $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$ the triangle inequality givesThus $(x_n)$ is a Cauchy sequence, and completeness supplies a limit $x_{\ast }$. A contraction mapping is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing to the limit in the bound also gives the useful error estimate $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

For the function space, take $X$ nonempty; otherwise the supremum in the printed definition is undefined without a convention. Boundedness of $X$ makes the uniform metric $\rho (f,g)$ finite. Nonnegativity and symmetry follow from $d$; $\rho (f,g)=0$ means $f(x)=g(x)$ at every $x$. Taking suprema in $d(f(x),h(x))\le d(f(x),g(x))+d(g(x),h(x))$ proves the triangle inequality. Hence $\rho$ is a metric.

Suppose $(f_n)$ is a Cauchy sequence for this uniform metric. At each $x$, $d(f_n(x),f_m(x))\le \rho (f_n,f_m)$, so completeness of $X$ supplies a value $f(x)={\lim }_n{f}_n(x)$. Given $\epsilon >0$, choose $N$ with $\rho (f_n,f_m)<\epsilon /2$ for all $n,m\ge N$. Fix $n\ge N$ and let $m\to \infty$ pointwise; then $d(f_n(x),f(x))\le \epsilon /2$ for every $x$. Thus $\rho (f_n,f)\le \epsilon /2<\epsilon$. The uniform limit theorem makes this limit continuous, so $f\in F$. This proves the completeness of the continuous-map space in the uniform metric.

Finally fix $f\in C$ with contraction constant $q_f<1$, and write $x_f=\theta (f)$ and $x_g=\theta (g)$. Using the fixed point equations,Therefore the fixed-point stability in the uniform metric estimate isChoosing $\rho (f,g)<(1-q_f)\epsilon$ proves continuity of $\theta$ at every $f$. The estimate uses the constant of the fixed map $f$ only; it does not require a common contraction constant for all nearby $g$.

Assume the impulse point is in the interior, $0<a<l$. Expand the displacement in the fixed-end Fourier sine seriesOrthogonality gives the initial velocity coefficient by applying the Dirac delta to the sine function:Substitution in the wave equation yields $T_n^{\prime \prime }+2k{T}_n^{\prime }+{\alpha }_n^2{T}_n=0$. For $k<{\alpha }_n$, its roots are $-k\pm i{\omega }_n$, where ${\omega }_n=\sqrt{{\alpha }_n^2-k^2}$. The two initial conditions determineHence the normalized velocity-impulse Green function for a damped string isFor a mode with critical damping, use the limit $e^{-kt}t$ in place of the time factor. For $k>{\alpha }_n$, put ${\eta }_n=\sqrt{k^2-{\alpha }_n^2}$ and use $e^{-kt}\sinh ({\eta }_{n}t)/{\eta }_n$. These choices solve the same modal ordinary differential equation and initial conditions, so the formula covers arbitrary $k>0$. The impulse initial velocity is a distribution, so initial differentiation of the series is interpreted distributionally.

The printed formula requires correction: the factor depending on $n$ must lie inside the sum, and sine-mode normalization requires $2/l$, rather than $2$. The printed slash after the final sine is also stray. Direct differentiation at $t=0$ verifies the corrected normalization: the initial sine coefficients are exactly those of $\delta (x-a)$.

A mode has dimensionless damping ratio $k/{\alpha }_n$, and characteristic rootsIf $k/{\alpha }_n\ll 1$, the normal mode oscillates at ${\omega }_n={\alpha }_n+O(k^2/{\alpha }_n)$ with exponentially decaying envelope $e^{-kt}$. As $k\to 0$, the solution approaches the undamped wave equation response.

If $k/{\alpha }_n\gg 1$, the roots are real and negative, withThe mode has a fast transient and a slow nonoscillatory relaxation. After the fast transient, neglecting $y_{tt}$ gives the diffusion equation $y_t\simeq c^2{y}_{xx}/(2k)$ for these long-wavelength modes. At $k={\alpha }_n$, the repeated root produces $t{e}^{-kt}$: this is critical damping.

Thus the critical parameter is $k/{\alpha }_n=1$ for each mode, and the fundamental threshold isWhen $k<\pi c/l$ every mode is oscillatory. For any fixed, finite $k$, sufficiently large $n$ still gives oscillatory modes, so large damping of the lowest modes does not mean that every mode of the full string is overdamped.

The complex Taylor series theorem states that if $f$ is holomorphic in $|z-z_0|<R$, thenThe power series converges absolutely there and uniformly on each smaller closed disc; its coefficients are unique. For $|z-z_0|<r<R$, the Cauchy integral formula also gives $f^{(n)}(z_0)/n!=(2\pi i{)}^{-1}{\int }_{|\zeta -z_0|=r}f(\zeta )(\zeta -z_0{)}^{-n-1}d\zeta$. This is the analytic, rather than finite real-variable, form of Taylor theorem relevant here.

Let $h=f-g$, whose power series coefficients are $c_n=a_n-b_n$. Suppose some coefficient is nonzero, and let $N$ be the least such index. Inside the common convergence disc,The parenthesized analytic function tends to $c_N\ne 0$ as $z\to z_0$, so it does not vanish in some neighborhood. Hence $h$ has no zeros sufficiently near $z_0$ except possibly $z_0$ itself. This contradicts $h(z_k)=0$ with $z_k\ne z_0$ and $z_k\to z_0$. Thus all coefficients agree:This also proves the local isolation of zeros of a nonzero analytic function.

Put $h=f-g$. Choose a disc about $z_0$ contained in the domain $D$. Part (a) gives a convergent Taylor series there, and the accumulating zeros together with part (b) force all its coefficients to vanish. Thus $h=0$ near $z_0$.

To propagate this to the entire domain, defineThe set is nonempty by the preceding local argument. Each derivative is continuous, so $S$ is closed relative to $D$ as the intersection of their zero sets. If $z\in S$, its Taylor series is zero on a disc about $z$, and all derivatives vanish throughout that disc; hence $S$ is open relative to $D$. Connectedness forces $S=D$. Therefore $f=g$ throughout $D$, proving the identity theorem for holomorphic functions rather than merely invoking it.

Take the analytic functionIt vanishes at $z=1/n$ for every positive integer $n$, but $f(2)=1$, so it is not identically zero. The identity theorem does not apply at the accumulation point $0$, because that point lies outside the domain.

Let $f$ have the stated zeros but not be identically zero. The origin is an isolated singularity. If it were a removable singularity, the extended holomorphic function would have zeros accumulating at an interior point. The proof in part (c) would force the extension, and then $f$ on its domain, to be identically zero.

If the origin were a pole of order $m$, then $g(z)=z^{m}f(z)$ would have a holomorphic extension with $g(0)\ne 0$. Continuity would make $g$ nonzero on a small disc, so $f$ would have no zeros in its punctured disc, again a contradiction. The classification of isolated singularities leaves only an essential singularity at the origin. This argument applies to every function in part (d), not just the particular sine example.

Work first on the unit sphere. A spherical lune of angle $\alpha$ has area $2\alpha$: its fraction of sphere area is $\alpha /(2\pi )$. The three great circles extending the sides of a nondegenerate spherical triangle divide the sphere into four antipodal pairs of triangular regions. Call the desired region $T$, of area $A$, and its three adjacent regions $T_1,T_2,T_3$, of areas $A_1,A_2,A_3$. These four regions contain one region from each antipodal pair, so $A+A_1+A_2+A_3=2\pi$. The lune at each vertex consists of $T$ and one of these adjacent regions. Therefore, if the three interior angles are $\alpha ,\beta ,\gamma$,which proves the spherical excess formula, the spherical-triangle form of the Gauss-Bonnet theorem:For a sphere of radius $s$, multiply this area by $s^2$.

Triangulate a convex spherical $n$-gon by diagonals from a vertex. There are $n-2$ spherical triangles, and their angles add to the polygon's interior angle sum. Thus its unit-sphere area isFor a nondegenerate convex regular polygon in an open hemisphere, area is positive and each angle is less than $\pi$, giving $(n-2)\pi /n<\alpha <\pi$. Every angle in this interval occurs. Indeed place $n$ vertices at equal longitude spacings and colatitude $t\in (0,\pi /2)$ about a pole, and join consecutive vertices by shorter great circle arcs. The polygon is convex and regular. Split a central isosceles spherical triangle along the perpendicular to its edge. The resulting right spherical triangle has angles $\pi /n$, $\alpha /2$, $\pi /2$, and hypotenuse $t$. The angular spherical law of cosines yieldsso $\cot (\alpha /2)=\cos t\tan (\pi /n)$. As $t$ increases from $0$ to $\pi /2$, $\alpha$ increases continuously through exactlyThe endpoints are degenerate limits, not strict convex polygons.

For the spherical triangle reflection group, positive area requires $1/p+1/q+1/r>1$. Each angle is less than $\pi$, so the integer denominators are at least two. Arrange $p\le q\le r$. If $p\ge 3$ the sum is at most one, hence $p=2$. If $q\ge 4$ the sum is again at most one; thus either $q=2$ and any $r\ge 2$ works, or $q=3$ and $r=3,4,5$. These triangles exist, for example as the reflection chambers of a regular dihedral configuration, tetrahedron, cube, and dodecahedron, respectively.

To count group elements using the assumed tessellation, first label the original triangle's three vertices by their types and propagate these labels through side reflections. Around a type-$p$ vertex there are $2p$ sectors; the associated alternating reflections make a full turn and return the labels unchanged. The same holds at the other types. Every closed path in the dual tessellation is generated by these circuits around vertices, since the sphere is simply connected. Thus the labels are globally consistent, including when two angle denominators coincide. Every generated isometry preserves these labels. An element stabilizing a triangle therefore fixes its three vertices and is the identity. Consequently group elements correspond bijectively to triangles, and area givesThe possibilities, including permutations of the entries, are

Here is an explicit dodecahedron construction for the last case, avoiding an assumption about which vertices of the tessellation to join. Put $\phi =(1+\sqrt{5})/2$, and take the convex hull of the twenty pointswith independent signs. They all have squared radius three. The supporting plane $\phi y+z={\phi }^2$ contains the five verticesin cyclic order. Consecutive distances are $2/\phi$, and nonconsecutive distances are $2$, so this is a regular pentagon. Changing the two signs in its normal $(0,\phi ,1)$ and cyclically permuting coordinates gives twelve supporting planes, all with a congruent pentagon. Each listed vertex lies on three of these faces; the sixty face-edge incidences pair into thirty edges. The faces form a closed convex polyhedral surface with $20-30+12=2$, hence exhaust the boundary of the convex hull. This constructs a regular dodecahedron.

Choose the spherical directions of a face center $F=(0,\phi ,1)$, the midpoint $M=(0,\phi ,0)$ of its edge between $(\pm {\phi }^{-1},\phi ,0)$, and that edge's endpoint $V=({\phi }^{-1},\phi ,0)$. Normalize these to the unit sphere. The three side planes of triangle $FMV$ are $x=0$, $z=0$, and $\mathbf{n}\cdot \mathbf{x}=0$ with $\mathbf{n}=(-\phi ,{\phi }^{-1},-1)$. Their reflections are coordinate sign changes in the first two cases andin the third, since $\Vert \mathbf{n}{\Vert }^2=4$. Substitution in the twenty listed vertices, using ${\phi }^2=\phi +1$, shows that all three reflections permute the vertex set. The side-plane angles give interior angles $\pi /2$ at $M$, $\pi /3$ at $V$, and $\pi /5$ at $F$ because $|n_z|/\Vert \mathbf{n}\Vert =1/2$ and $|n_x|/\Vert \mathbf{n}\Vert =\phi /2=\cos (\pi /5)$. Thus this triangle is congruent to $\Delta$, and its side-generated group is $G$, conjugated by the aligning isometry.

It already has $120$ elements. Conversely, a symmetry of the dodecahedron is determined by the image of one face, one vertex on that face, and the orientation along its boundary: there are at most $12\cdot 5\cdot 2=120$ choices. An isometry fixing these choices fixes that face pointwise and must preserve the polyhedron's interior side, hence is the identity. Therefore the complete symmetry group of the constructed dodecahedron is $G$, including orientation-reversing symmetries.

A balanced transportation problem chooses nonnegative shipments $x_{ij}$ minimizing ${\sum }_{ij}{c}_{ij}{x}_{ij}$, subject to ${\sum }_j{x}_{ij}=s_i$ and ${\sum }_i{x}_{ij}=d_j$, where the total supply equals total demand. For multipliers $u_i,v_j$, its Lagrangian isTaking its infimum over $x_{ij}\ge 0$ produces the linear programming duality problem of maximizing ${\sum }_i{u}_i{s}_i+{\sum }_j{v}_j{d}_j$ subject to $u_i+v_j\le c_{ij}$. These are the transportation dual potentials.

The transportation simplex algorithm starts with a feasible transportation spanning tree of $m+n-1$ basic entries, obtainable by filling the north-west available cell with the smaller remaining supply or demand. Set $u_i+v_j=c_{ij}$ on basic cells; this determines all potentials up to adding a constant to all $u_i$ and subtracting it from all $v_j$. If all nonbasic reduced costs $r_{ij}=c_{ij}-u_i-v_j$ are nonnegative, feasibility and the dual bound prove optimality. Otherwise let a negative-cost cell enter. It closes a unique cycle in the basic tree. Alternate additions and subtractions around that cycle and choose the largest amount preserving nonnegativity, namely the minimum shipment on the subtraction cells. A cell that reaches zero leaves the basis. The cost changes by the entering reduced cost times the pivot amount. Zero basic cells and an anti-cycling rule can handle degeneracy; all pivots below are nondegenerate.

The north-west initial shipment matrix isWith $u_1=0$, its transportation dual potentials are $u=(0,1,0)$ and $v=(5,9,5)$. The most negative reduced cost is $r_{32}=-7$. The cycle is $+(3,2),-(3,3),+(2,3),-(2,2)$; the pivot amount is $\min (40,46)=40$. HenceNow $u=(0,1,-7)$ and $v=(5,9,5)$. Enter $(1,3)$, of reduced cost $-4$, on cycle $+(1,3),-(1,2),+(2,2),-(2,3)$. The amount is $\min (22,78)=22$, givingThe new potentials are $u=(0,5,-3)$ and $v=(5,5,1)$. Enter $(2,1)$, of reduced cost $-7$, on cycle $+(2,1),-(1,1),+(1,3),-(2,3)$. The amount is $\min (14,56)=14$. ThusTo certify optimality in the original transportation problem, take $u=(0,5,-3)$ and $v=(-2,5,1)$. Their entire reduced-cost matrix isThey satisfy dual feasibility, and their value is $5\cdot 84-3\cdot 40-2\cdot 14+5\cdot 68+78=690$, equal to the shipment cost. The minimum transportation cost is therefore $690$, with no unproved optimality assertion from the iterations alone.