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Translate one vertex of the tetrahedron to the origin and denote the other vertex vectors by $a,b,c$. Translation preserves edge vectors and their dot products. We may label the two known perpendicular pairs of opposite edges of a tetrahedron as the pairs represented by $a,c-b$ and $b,a-c$. Their orthogonality givesHence all three pairwise dot products have the same value $s$. The third pair satisfiesso the third pair of opposite edges is also perpendicular.

Expanding squared lengths using the dot product givesThus the sums of squared opposite-edge lengths are equal. These two conclusions are the basic identities for orthogonal opposite edges of a tetrahedron.

Set $w=(x-y)/2$. The sum condition then gives $x=a+w$ and $y=a-w$, and their dot product isTherefore $|w{|}^2=|a{|}^2-c$. Since $c<|a{|}^2$, this is a strictly positive squared radius. The complete solution isEvery choice of unit vector $n$ satisfies both conditions, and every solution has this form. The endpoints lie on the sphere with center $a$ and radius $R$, and their midpoint is $a$. They are therefore opposite ends of one diameter of that sphere.

To construct a potential for the exact differential, first integrate its $dx$ coefficient with respect to $x$:Matching the $dy$ coefficient gives $g_y=-2z^2$, hence $g=-2y{z}^2+h(z)$. The remaining partial derivative is thenMatching the $dz$ coefficient forces $h^{\prime }=0$. Therefore a potential, valid throughout ${\mathbb{R}}^3$, isDifferentiating this expression recovers all three coefficients, proving that the differential form really is an exact differential.

For an exact differential $Md\theta +Nd\varphi =dF$, equality of mixed partial derivatives requires $M_{\varphi }=N_{\theta }$. Here the two partial derivatives areThey are not identically equal, so the differential form is not a closed differential form and hence is not an exact differential. Equality along special lines such as $\theta =0$ does not supply a potential on a two-dimensional open neighborhood, because every such neighborhood contains points where the two partial derivatives differ.

Yes: every finite group of rotations about the origin in the plane is cyclic. A planar rotation is determined by an angle modulo $2\pi$, and composition adds those angles. If $G$ consists only of the identity, it is a cyclic group. Otherwise, finiteness allows us to choose the smallest positive representative angle $\alpha \in (0,2\pi )$ occurring in $G$.

Take any planar rotation in $G$ with representative angle $\theta \in [0,2\pi )$. Write $\theta =q\alpha +r$, where $q$ is an integer and $0\le r<\alpha$. The composition $R_{\theta }{R}_{\alpha }^{-q}$ belongs to $G$ and has angle $r$. Minimality of $\alpha$ forces $r=0$, so every element of $G$ is a power of $R_{\alpha }$.

Similarly, write $2\pi =m\alpha +r$ with $0\le r<\alpha$. If $r>0$, the planar rotation $R_{\alpha }^{-m}$ has representative angle $r$, a contradiction. Thus $\alpha =2\pi /m$ andThis proves finite groups of planar rotations are cyclic by an elementary angle argument.

A group $G$ is a cyclic group if some element $g\in G$ generates it:For a finite cyclic group, the order of a group element of a generator equals the number of elements of the group.

An example of a finite abelian group that is not a cyclic group is the Klein four-group, realized aswith componentwise addition modulo two. Addition commutes, so this is an abelian group of four elements. Each of its three nonzero elements satisfies $g+g=0$, so the cyclic subgroup it generates has just two elements; the identity generates only itself. No element generates all four elements.

Assume $r_0$ is an interior point and choose $\epsilon$ small enough that its closed ball lies inside $V$. Apply Green's second identity to the punctured region $V_{\epsilon }=V\setminus \overline{B_{\epsilon }(r_0)}$, where the potential $\psi =1/|r-r_0|$ is smooth and satisfies Laplace equation. Since ${\nabla }^2\varphi =-\rho$, this givesThe crucial orientation is the outward unit normal vector of the punctured region: on the inner sphere, it is $n_{\epsilon }=-e_r$, pointing toward $r_0$. Consequently $\psi =1/\epsilon$ and ${\partial }_{n_{\epsilon }}\psi =1/{\epsilon }^2$. The inner boundary term becomesWriting $r=r_0+\epsilon \omega$ and $dS={\epsilon }^{2}d\Omega$, its first part tends to ${\int }_{|\omega |=1}\varphi (r_0)d\Omega =4\pi \varphi (r_0)$ by continuity. The second part is $O(\epsilon )$ because the gradient of $\varphi$ is bounded near $r_0$, and hence tends to zero.

The omitted volume term also tends to zero: for bounded $\rho$ near $r_0$, its absolute value is bounded by a constant times $4\pi {\int }_0^{\epsilon }rdr=O({\epsilon }^2)$. Thus the limiting integral exists, and rearranging the outer boundary terms givesThis is Green's third identity for the Poisson equation with the stated sign convention. It retains both boundary terms; no homogeneous boundary condition has been assumed. The punctured-domain calculation accounts for the singular source without incorrectly applying the regular divergence theorem through $r_0$.

Use the product rule for the divergence of the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two dot products cancel. Apply the divergence theorem with outward unit normal vector $n$ to obtain Green's second identity:Smoothness of the functions and boundary supplies the regularity needed for the divergence theorem.

In Cartesian coordinates, put $r=(x_1^2+x_2^2+x_3^2{)}^{1/2}$. Away from the origin the partial derivatives of $\psi =1/r$ areSumming the three second partial derivatives yieldsThus $1/r$ is a harmonic function on the punctured space and solves Laplace equation there. The origin is excluded: the singularity cannot be treated as an ordinary harmonic point.

The 36 ordered outcomes are equiprobable. For $2\le s\le 12$, let $c_s=6-|s-7|$ count outcomes with sum $s$. Then $\mathbb{P}(A_s)=c_s/36$ and $\mathbb{P}(B_i)=1/6$ for $1\le i\le 6$. Their joint probability is $1/36$ if $1\le s-i\le 6$, and zero otherwise.

If the joint probability is zero, independence is impossible in these ranges because both marginal probabilities are positive. If it is $1/36$, the independent events condition requires $1/36=c_s/216$, or $c_s=6$. This happens only for $s=7$, and then all six values of $i$ allow the partner $7-i$. Therefore within the possible sums and facesIf arbitrary integer labels outside these ranges are included, an impossible sum or face defines an empty event, which is trivially independent of every event.

Two independent events satisfy $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$. If $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning that $B$ occurred does not change the probability of $A$. The product definition also covers zero-probability events without dividing by zero.

Every urn contains $n-1$ balls. Let $k=n-r$ be its number of blue balls; choosing an urn uniformly makes $k$ uniform on $0,\dots ,n-1$. Conditional on that urn, the probability of a blue first draw is $k/(n-1)$, so the law of total probability givesDrawing twice without replacement requires $n\ge 3$. Given $k$, the probability that both draws are blue is $k(k-1)/((n-1)(n-2))$, and henceTaking the ratio defining conditional probability gives the second requested probabilityAlthough a blue first draw removes a blue ball from its urn, it also makes an urn rich in blue balls more likely. Averaging with this changed conditional distribution is essential.

For $\mathbb{P}(B)>0$, the conditional probability is $\mathbb{P}(A\mid B)=\mathbb{P}(A\cap B)/\mathbb{P}(B)$. Since the events $B_i$ partition the sample space, their intersections with $A$ are disjoint and exhaust $A$. The law of total probability therefore gives $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Substituting this and $\mathbb{P}(A\cap B_i)=\mathbb{P}(A\mid B_i)\mathbb{P}(B_i)$ into the definition proves Bayes' theorem:The denominator is positive by the given hypothesis on $A$.

Evaluation of the probability generating function at zero selects the constant coefficient, soTo identify this limit with ultimate extinction, let $E_n=\{X_n=0\}$. The absorbing-zero property of the branching process gives $E_n\subseteq E_{n+1}$. Ultimate extinction means that some finite generation is empty, namely $E={\bigcup }_{n\ge 0}{E}_n$. Continuity of probability on increasing events therefore givesThis set argument is essential: it uses absorption, rather than inferring extinction merely from a numerical limit of generation probabilities.

The binomial series givesThus $p_0=1-\alpha >0$ and, for $k\ge 1$,with the factors after $\beta$ omitted for $k=1$. The signs follow from $0<\beta <1$. As $s\uparrow 1$, the series has nonnegative terms and its sum increases to $G(1)=1$, so monotone convergence gives ${\sum }_{k\ge 0}{p}_k=1$. These are therefore genuine probabilities on the nonnegative integers, proving that $G$ is a probability generating function. Its expected value is infinite, since $G^{\prime }(s)=\alpha \beta (1-s{)}^{\beta -1}\to \infty$; finiteness of the mean is not required for this branching process.

Write $G_n(s)=1-A_n(1-s{)}^{{\beta }^n}$. The composition formula gives $A_{n+1}=A_n{\alpha }^{{\beta }^n}$, with $A_0=1$. Consequently $A_n={\alpha }^{1+\beta +\cdots +{\beta }^{n-1}}$, and summing the geometric series yields the explicit generation generating functionThis is the fractional-power offspring generating function and its iterates.

Given $X_n=k$, the next size is a sum of $k$ independent reproduction counts. Multiplication of their probability generating functions gives $\mathbb{E}[s^{X_{n+1}}\mid X_n=k]=G(s{)}^k$. Averaging this conditional expectation provesWith one initial individual, $G_0(s)=s$ and $G_1(s)=G(s)$, so $G_n$ is the $n$-fold composition of $G$ with itself.

A branching process models reproduction through successive generations. In the Galton-Watson process used here, every individual independently produces a nonnegative integer number of offspring with a common offspring distribution; these reproduction counts are independent across individuals and generations. If $X_n$ is the generation size, thenwhere the ${\xi }_{n,j}$ have that distribution and the empty sum is zero. Thus zero is absorbing: once a generation is empty, every subsequent generation is empty.

Another convolution of probability densities gives, for $0\le s\le 1$,Only this part of the density is needed because the fourth length is at most one.

Four positive lengths $a,b,c,d$ can form a nondegenerate quadrilateral if and only if every length is less than the sum of the other three. Necessity follows by following the other three sides between the endpoints of the longest side and applying the triangle inequality. For sufficiency, select a diagonal length $r$ in the intersectionBoth intervals are nonempty, and their intersection is nonempty exactly when $|a-b|<c+d$ and $|c-d|<a+b$. These are precisely the four strict length inequalities. Construct triangles with sides $(a,b,r)$ and $(c,d,r)$ on opposite sides of their common diagonal. Their union has the required simple, nondegenerate quadrilateral as boundary.

The failure events, in which one length is at least the sum of the others, are disjoint except on sets of probability zero. For the specified length $W$, independence and the density above giveThere are four symmetric choices of the overly long side, so the probability isThis is the four-sided case of the polygon probability for independent uniform rods.

Three positive lengths form a nondegenerate triangle precisely when each is less than the sum of the other two. Necessity is the triangle inequality. For sufficiency, two circles centered at the ends of one side intersect off that side whenever the remaining lengths satisfy these strict inequalities, giving the third vertex.

Failure thus occurs when $X\ge Y+Z$, $Y\ge X+Z$, or $Z\ge X+Y$. Two such inequalities cannot hold together for strictly positive lengths. Zero lengths and equality cases have probability zero because the variables have continuous probability density functions. By symmetry each failure event has the probability $1/6$ just calculated. Hence the probability of forming a triangle is

Conditioning on $U=X+Y$, independence of $Z$ gives $\mathbb{P}(Z>u)=1-u$ for $0\le u\le 1$ and zero for $u>1$. Using the preceding probability density function,