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Since $K\ge 0$, Gauss-Bonnet gives $4\pi (1-g)\ge 0$, hence $g\le 1$. If $g=1$, the integral of the continuous nonnegative function $K$ is zero, so $K$ vanishes everywhere.

That is impossible for a compact regular surface in ${\mathbb{R}}^3$. Choose $p\in S$ maximizing the Euclidean distance from the origin, after translating the origin off the surface if necessary. The sphere centered at the origin through $p$ is a supporting sphere. The supporting-sphere curvature bound shows that both principal curvatures at $p$ have the same sign and nonzero magnitude, so $K(p)>0$. Therefore $g\ne 1$, and $g=0$. The classification theorem for surfaces now shows that $S$ is diffeomorphic to the sphere.

Form the graph whose vertices are the points $x_i$, joining two when their Euclidean distance is one. Two distinct points have at most two common unit-distance neighbours, because two unit circles intersect in at most two points. The graph is therefore $K_{3,2}$-free. Part (c) bounds its unordered edges by $2n^{3/2}$, so the number of ordered unit-distance pairs is at most

Let $X=\{v_1,\dots ,v_m\}$ be the vertices of a regular polygon, indexed cyclically so that $v_i\mapsto v_{i+1}$ is an isometry. We first prove the product lemma needed for the cyclic-symmetry argument.

Suppose $A\subseteq X$ has the property that, for every number of colors, a finite witness forces a copy of $X$ on which the vertices corresponding to $A$ are monochromatic. We claim that for every $N,k$ there is a finite witness forcing an $A$-invariant copy of $X^N$. Proceed by mathematical induction on $N$. The case $N=1$ is the assumption. For the step, choose a finite set $S$ that forces an $A$-invariant $X^{N-1}$ for a coloring with $k^{|X|-|A|+1}$ colors, and choose $T$ that forces an $A$-monochromatic $X$ for a coloring with $k^{|S|}$ colors.

Given a $k$-coloring $c$ of $S\times T$, color $t\in T$ by its complete vector $(c(s,t){)}_{s\in S}$. In the resulting copy of $X$ in $T$, that vector is the same for every member of $A$. Each $s\in S$ can therefore be colored by the vector consisting of $c(s,a)$ for one $a\in A$ and the colors $c(s,x)$ for $x\in X\setminus A$. Applying the choice of $S$ gives an $A$-invariant copy of $X^{N-1}$, and its product with the chosen copy of $X$ in $T$ is the desired $A$-invariant copy of $X^N$. This proves the lemma rather than assuming it.

We now prove by induction on $r$ that a finite witness forces a copy of $X$ whose first $r$ vertices have one color. The case $r=1$ is trivial. For $r=2$, use that a line segment is a Euclidean Ramsey set; each occurrence of a segment congruent to $v_1{v}_2$ can be extended to a copy of the regular polygon, and only finitely many extensions are needed for a finite witness.

Assume the assertion for $r$, and put $A=\{v_1,\dots ,v_r\}$. The product lemma lets us work inside an $A$-invariantly colored copy of $X^N$, with $N$ as large as required. For an increasing $(m-1)$-element subset $H=\{a_1<\cdots <a_{m-1}\}$ of $[N]$ and $1\le i\le r$, define a word $w_i(H)$ by putting $v_{i+j}$ in coordinate $a_j$, with subscripts read modulo $m$, and putting $v_1$ in every other coordinate. Color $H$ by the $r$-tupleBy the Finite Ramsey theorem, for sufficiently large $N$ there are coordinates $b_1<\cdots <b_m$ on which all $(m-1)$-subsets have the same tuple color.

For $1\le i\le r+1$, let $z_i$ be the word whose entries in coordinates $b_1,\dots ,b_m$ arecyclically, and whose other entries are $v_1$. For $i\le r$, the word $z_i$ differs from $w_i(\{b_2,\dots ,b_m\})$ only in coordinate $b_1$, where the two entries $v_i,v_1$ lie in $A$. Similarly $z_{i+1}$ differs from $w_i(\{b_1,\dots ,b_{m-1}\})$ only in coordinate $b_m$, again by two members of $A$. The $A$-invariance and the homogeneous choice of the $b_j$ therefore giveHence $z_1,\dots ,z_{r+1}$ have one color.

The cyclic words $z_1,\dots ,z_m$ form an isometric copy of $\sqrt{m}X$: in each of the $m$ varying coordinates the cyclic shift is an isometry, so every squared distance is multiplied by $m$. Rescaling the finite witness by $1/\sqrt{m}$ gives a copy of $X$. The induction reaches $r=m$, proving that every regular polygon is a Euclidean Ramsey set.

For three consecutive vertices $p_0,p_1,p_2$ of a regular $M$-gon scaled to have side length one,Since the polygon is Euclidean Ramsey, choosing $M$ large enough proves that $\{0,1,2\}$ is an approximately Euclidean Ramsey set.

In fact every finite $X\subseteq {\mathbb{R}}^d$ is approximately Ramsey. First perturb each coordinate of each point onto a sufficiently fine lattice $\delta \mathbb{Z}$, changing all pairwise distances by less than $\epsilon /2$. For each coordinate, map the finitely many required integers to consecutive vertices of a very large regular polygon, scaled so that one angular step has arc length $\delta$. If the step angle is $\theta$, the chord replacing a difference of $q$ lattice steps has lengthas $\theta \to 0$, uniformly over the finitely many differences involved. Taking the orthogonal Cartesian product of $d$ such polygons therefore embeds the perturbed points with a further distance error below $\epsilon /2$. Each polygon is Euclidean Ramsey, and their product is Euclidean Ramsey by the product theorem for Euclidean Ramsey sets. A monochromatic copy of that product contains the corresponding approximate copy of $X$, completing the proof.

Assume $\pi \in C$ andfor every $z\in C$. Expanding the squared Euclidean distance givesHence $\pi$ minimizes the distance from $x$ over $C$, so $\pi ={\pi }_C(x)$. This proves the sufficient direction of the variational characterization of convex projection.

Use the corrected decompositionThe hint's minus sign before $({\mu }_j-w_j)$ is a typographical error. Squaring, summing, and using ${\sum }_i(y_{ij}-{\mu }_j)=0$ for each coordinate gives the centroid sum-of-squares identitySince $m>0$ and $w\ne \mu$, the final term is strictly positive. HenceThus the centroid uniquely minimizes the sum of squared Euclidean distances to the observations.

Write $x_i\in {\mathbb{R}}^p$ for row $i$ of the data matrix. The k-means clustering objective is the within-cluster sum of squared Euclidean distances:The nonempty $C_k$ form a partition of a set of observation indices. For fixed clusters, part (a) shows that the best $m_k$ is their centroid.

Initialize distinct centres, for example by choosing $K\le n$ distinct observations. Then repeat the following steps:

The empty-cluster repair is possible for $K\le n$ and cannot increase $J$: the transferred observation now has squared distance zero, while all other squared distances stay unchanged before the centre update.

Assignment cannot increase $J$, and any changed assignment strictly decreases it under the tie rule. Updating centres cannot increase $J$ by the centroid sum-of-squares identity. After an update, the centres are determined by the partition of a set. There are only finitely many possible assignments. Unless this is the initial round, any nonterminal round starts with an assignment that strictly reduces $J$; repairing empty clusters and updating cannot undo that decrease. Thus a partition cannot recur, proving finite termination of K-means clustering. The final result need not be a global minimum; different initial centres can lead to different final clusterings.

The polynomial factors as $p(x)=(3x-2{)}^2(3x+1)$, so $\alpha =2/3$ is a multiple root of multiplicity two. The Newton root-finding iteration isFor $x\ne 0,2/3$, cancelling a factor givesThis expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this closed interval,Thus $g$ is increasing, andBy the mean value theorem, $|g(x)-g(y)|\le (16/27)|x-y|$ for all $x,y\in I$. The contraction mapping theorem states that a contraction mapping from a nonempty complete metric space into itself has a unique fixed point, and every fixed-point iteration converges to it. The closed interval $I$ with ordinary Euclidean distance is complete, and $g(2/3)=2/3$, soIn fact $|x_k-2/3|\le (16/27{)}^k|x_0-2/3|$. The Newton iteration at a multiple root here has linear convergence, since $g^{\prime }(2/3)=1/2$. Thus the convergence assertion is true for the simplified fixed-point iteration, and for the usual Newton root-finding iteration that stops upon reaching an exact root of a polynomial. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p^{\prime }$ vanish there; that literal version of an infinite sequence requires the stated stopping convention or continuous extension.

Let $A=(Y+Z)/2$, $B=(Z+X)/2$, and $C=(X+Y)/2$, interpreting these expressions as position vectors. The resulting medial triangle is nondegenerate: for example,Thus $BC$ is parallel to $YZ$, and the same calculation holds cyclically. The altitude of a triangle through $A$ is consequently the line through the midpoint of $YZ$ that is orthogonal to $YZ$, namely its perpendicular bisector. Each remaining altitude of a triangle similarly gives one of the other two perpendicular bisectors.

By iii, the medial triangle has a unique orthocenter, so the three perpendicular bisectors meet in one point. Every point on the perpendicular bisector of $YZ$ is equally far from $Y$ and $Z$: expanding the squares of the two Euclidean distances gives $2\mathbf{q}\cdot (\mathbf{z}-\mathbf{y})=\Vert \mathbf{z}{\Vert }^2-\Vert \mathbf{y}{\Vert }^2$. Applying this to the three sides shows that the common point is equally far from $X,Y,Z$. It is the circumcenter.

For observations ${\mathbf{x}}_i\in {\mathbb{R}}^p$, K-means clustering seeks nonempty clusters in cluster analysis $C_1,\dots ,C_K$ and centroids ${\mathbf{m}}_k$ minimizingAt fixed centroids, assigning each point to a nearest centroid minimizes its squared Euclidean distance. At fixed assignments, the centroid sum-of-squares identity shows that the sample mean uniquely minimizes each group's contribution. These alternating steps cannot increase $W$.

For finite termination of K-means clustering, more than a lower bound on $W$ is needed. Keep an assignment unchanged when nearest centroids tie, and stop when assignments no longer change. Every change then strictly decreases $W$. After updating the centroids, $W$ is determined by the partition of a set of $n$ indices; there are only finitely many choices for that partition of a set, so no strictly improving partition of a set can recur. Empty clusters in cluster analysis require a convention, such as moving a point from a nonsingleton cluster in cluster analysis to become its own centroid, which cannot increase $W$. With such a consistent convention the algorithm terminates, although it may reach only a local minimum.

Every Euclidean distance obeys the triangle inequality. The three recorded entries $d_{14}=0.67$, $d_{16}=2.88$ and $d_{46}=4.37$ violate it:At least one of $0.67,2.88,4.37$ must be incorrect. Two are in the final row as required. The gap is far larger than any possible rounding of entries to two decimal places.

It is single-linkage clustering. Its intercluster distance is the smallest cross-pair Euclidean distance. Successive nearby points can join around the ring and within the dense central cloud before any edge crosses the gap between them. Equivalently, at a threshold below the between-group gap but above the lengths needed to connect each group, the single-linkage clustering components are precisely the central group and the ring.

Complete-linkage clustering uses the largest cross-pair Euclidean distance. Opposite points on the ring are far apart, so an intact ring has a large diameter. That rule favors compact pieces and can attach a ring segment to the center before all ring segments join. The visible gap and short links within each group explain why single-linkage clustering recovers the two intended clusters in cluster analysis in this example; it is not a universal guarantee for arbitrary noisy rings.

With two distinct centroids, K-means clustering assigns points on opposite sides of their nearest-centroid decision hyperplane, which in two dimensions is a straight line. This constraint explains the typical outcomes in the three source plots; exact assignments depend on initialization and the realized points.

Dataset A: one cluster in cluster analysis contains most of the horizontal cloud and the bottom of the vertical cloud; the other contains the upper part of the vertical cloud. The two visually elongated groups are poorly matched to squared Euclidean distance from their centroids. Points at the lower end of the vertical group can be closer to the horizontal group's centroid, so the two original shapes need not be recovered intact.

Dataset B: a roughly straight cut divides the ring and generally the central cloud between the two clusters in cluster analysis. A Voronoi cell for either centroid is a half-plane; it cannot select a surrounding ring while excluding its center. The cut's direction can vary because the geometry is approximately rotationally symmetric.

Dataset C: the two dense lower clouds typically receive different centroids. The much smaller upper cloud is absorbed into one or divided between both clusters in cluster analysis, according to the nearly vertical separating line. With $K=2$, the objective usually benefits more from separating the two large clouds than from spending one centroid on the sparse third group. These are typical qualitative outcomes, rather than guarantees about every initialization.

With normal vector $n=(-1,-2,2)$, a plane through $A=(2,-1,1)$ has equation $n\cdot (r-A)=0$. As $n\cdot A=2$, its plane equation isThe distance from a point to a plane is the absolute signed offset divided by the Euclidean norm of its normal vector. Hence the Euclidean distance from the origin is $\boxed{2/3}$, since $|n|=3$. The closest point is $2n/9=(-2,-4,4)/9$, whose Euclidean distance from the origin is $2/3$ and which satisfies the plane equation.

Substituting $y=x/2$ into $xy=1$ gives $x^2=2$. The intersection points are $(\sqrt{2},1/\sqrt{2})$ and $(-\sqrt{2},-1/\sqrt{2})$. Their Euclidean distance is

The squared Euclidean distance along the line isIts minimum occurs at $t=0$ exactly when its derivative there vanishes, so $a\cdot (b\times c)=0$. Since $b,c$ are independent, their linear span is the two-dimensional orthogonal complement of $b\times c$. Consequentlyfor unique real $\alpha ,\beta$. This is the closest point on the intersection of two planes; the positive quadratic coefficient $|d{|}^2$ proves that the stationary distance is a strict global minimum.

The Euclidean distance is $D(x,y)=\sqrt{x^2+y^2}$. Along the parametric curve,This is at least $a^2$, with equality exactly when $t=0$. Therefore

Let $\mathbf{d}=\mathbf{f}-\mathbf{e}$. A point on the line has position vector $\mathbf{r}=\mathbf{e}+t\mathbf{d}$. By bilinearity and antisymmetry of the cross product,For a point with position vector $\mathbf{c}$, the area of the parallelogram spanned by $\mathbf{c}-\mathbf{e}$ and $\mathbf{d}$ equals base length times perpendicular Euclidean distance. Hence the shortest Euclidean distance to the line isFor $\mathbf{c}=(0,0,1)$, these quantities areThus $\boxed{d=\sqrt{6}/4}$. The perpendicular meets $EF$ at $(1/4,1/4,1/2)$, whose displacement from $C$ has precisely this length.

The evaluated position vectors are ${\mathbf{r}}_1(1)=(2,1,0)$ and ${\mathbf{r}}_2(-1)=(-1,-1,0)$. Their difference is $(3,2,0)$, so the Euclidean distance is

Use only the dot product to minimize the squared Euclidean distance from the origin:Completing the square givesThe square is minimized at $t_0=-({\mathbf{a}}_2\cdot {\mathbf{n}}_2)/\Vert {\mathbf{n}}_2{\Vert }^2$, yieldingGeometrically this subtracts the squared length of the orthogonal projection parallel to the line. At the minimizing point, $({\mathbf{a}}_2+t_0{\mathbf{n}}_2)\cdot {\mathbf{n}}_2=0$, as required for perpendicular distance.

For a point $\mathbf{p}={\mathbf{r}}_2(t)$, the magnitude of $(\mathbf{p}-{\mathbf{a}}_1)\times {\mathbf{n}}_1$ is the area of a parallelogram. Dividing by its base length $\Vert {\mathbf{n}}_1\Vert$ gives the perpendicular Euclidean distance to the first line. Explicitly,HenceThis use of the cross product removes any component parallel to the first line automatically.