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Minimize the square of the Euclidean distance, using minimizers under strictly increasing transformations. For $s=x+y>0$,For fixed $s$ the unique minimum has $x=y=s/2$. Put $x=y=\alpha >0$ and $J(\alpha )=2{\alpha }^2+{\ln }^2(2\alpha )$. Its derivative isLet $h(\alpha )=2{\alpha }^2+\ln (2\alpha )$. Since $h^{\prime }(\alpha )=4\alpha +1/\alpha >0$, it is strictly increasing, with limits $-\infty$ and $+\infty$ at the two ends of $(0,\infty )$. It therefore has exactly one zero. The sign of $J^{\prime }$ changes there from negative to positive, proving the unique global minimum. The nearest point is $P=(\alpha ,\alpha ,\ln (2\alpha ))$ at that zero.^[ref]

The Newton root-finding iteration isStarting at $0.4$ gives $0.376376476$, $0.376544572$, and $0.376544582$. Moreover $h(0.3765)<0<h(0.3766)$ certifies the rounding. Thus $\alpha =0.377$ to three significant figures; more accurately $P\approx (0.3765445825,0.3765445825,-0.2835716452)$.

Place $N$ objects into $m\ge 1$ boxes, assigning each object to exactly one box. Here $N$ is a nonnegative integer and $m$ is a positive integer. The pigeonhole principle says that if $N>m$, some box contains at least two objects. Equivalently, a function from an $N$-element finite set to an $m$-element finite set cannot be an injective function when $N>m$. The boxes may stand for categories rather than physical containers. ^[ref]

For thirteen objects and twelve boxes, a box containing at least two is unavoidable. Thinking of thirteen people and their birth months gives the same conclusion: at least two share a birth month. Twelve people could occupy the twelve months one apiece, so thirteen is the smallest number that forces a shared month. The generalized pigeonhole principle quantifies how many objects must share a box when there are still more objects.

The useful surprise is that the pigeonhole principle proves existence without needing to locate the objects involved. Choosing a good assignment rule is often the creative step: equal labels can mean equal remainders, nearby points, or repeated states of an algorithm. Once the rule is chosen, a short count can force a relationship among a much larger collection of objects. The pigeonhole principle itself does not name the crowded box. When the objects and their assignments are explicitly available, a simple algorithm finds one: scan the objects, keep the first object seen in each box, and stop when a later object enters an occupied box. At most $m+1$ scanned objects are needed when $N>m$.

A geometric example. Any five points in a square of side one contain two whose Euclidean distance is at most $1/\sqrt{2}$. Give each point Cartesian coordinates $(x,y)\in [0,1{]}^2$. Divide both ranges into the intervals $[0,1/2)$ and $[1/2,1]$, making four cells. Thus $x<1/2$ belongs to the left half and $x\ge 1/2$ to the right half; $y<1/2$ belongs to the lower half and $y\ge 1/2$ to the upper half. This convention assigns every boundary point to exactly one cell.

The pigeonhole principle puts two of the five points in the same cell. Their differences in Cartesian coordinates have magnitude at most $1/2$, so the Pythagorean theorem gives their Euclidean distance $d$ the boundFor the plotted points $(0.1,0.1)$, $(0.4,0.4)$, $(0.8,0.2)$, $(0.2,0.8)$ and $(0.8,0.8)$, the first two occupy the lower-left cell and have Euclidean distance $\sqrt{0.3^2+0.3^2}=\sqrt{0.18}\approx 0.424264$. These Cartesian coordinates exhibit the forced pair; the argument applies to every placement of five points.

An arithmetic example. Assign eight integers to their seven residue classes modulo the integer modulus $7$. The pigeonhole principle gives two with the same residue class, so $7$ is an integer divisor of their difference. In $\{2,5,9,12,16,20,24,27\}$, for example, $2$ and $9$ satisfy the integer congruence $9\equiv 2(\mathrm{mod}7)$, and $9-2=7$. Likewise, assigning fractional parts to short intervals gives the Dirichlet approximation theorem; its existing article explains the approximation bound.

In planar polar coordinates $(r,\theta )$, the polar radius $r\ge 0$ is the Euclidean distance to the origin. It satisfies $x=r\cos \theta$ and $y=r\sin \theta$. For a differentiable plane curve with parameters $r(t),\theta (t)$, the orthogonality of radial and angular directions gives squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$.

Proper motion describes the change in a celestial object's direction on the sky due to its motion relative to the observer. Its angular velocity differs from the annual stellar parallax signal caused by the observer's orbit.

Thus proper motion is an angular velocity, rather than a speed in kilometres per second. It describes the relative transverse motion of source and observer in a specified reference frame. A star moving towards us can have little proper motion, while a nearer star can show larger proper motion than a faster but more distant one. In the example above, the observed coordinate actually decreases over the first half-year, although $\mu$ is positive: the annual stellar parallax excursion temporarily outweighs the steady drift. A short observed displacement alone therefore need not estimate proper motion well.

Real catalogues give two proper motion components. For right ascension $\alpha$ and declination $\delta$, the angular velocities in perpendicular directions on the local sky areThe cosine factor accounts for the shorter east-west arc associated with a given right ascension increment away from the celestial equator. ESA's Gaia DR3 fields pmra and pmdec already contain these two components; pm is their total proper motion. Do not multiply pmra by another cosine. Components of $60$ and $80\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$, for example, give total proper motion $100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$. ^[ref].

For nearby stars in the ordinary nonrelativistic approximation, Euclidean distance $d$ converts total proper motion into transverse speed $v_{\perp }$:The small-angle approximation gives transverse speed as Euclidean distance times angular velocity in radians per unit time. With arcseconds and parsecs, the parsec convention turns their product into astronomical units per year. One astronomical unit per Julian year of $365.25$ days is approximately $4.74047\mathrm{km}{\mathrm{s}}^{-1}$. The coefficient is a unit conversion, rather than a property of a particular star.

Taking the model's $100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}=0.100\mathrm{arcsec}\mathrm{y}{\mathrm{r}}^{-1}$ as the total proper motion, with the other component zero, gives $v_{\perp }=4.74047\times 0.100\times 20\simeq 9.48\mathrm{km}{\mathrm{s}}^{-1}$. At the same total proper motion, Euclidean distances of $10$, $20$ and $100$ parsecs instead give transverse speeds of $4.74$, $9.48$ and $47.40\mathrm{km}{\mathrm{s}}^{-1}$, respectively. Their stellar parallax values are $100$, $50$ and $10\mathrm{mas}$. The same annual angular displacement can therefore describe quite different journeys through space.

The transverse speed still omits the line-of-sight velocity, which can be measured through the Doppler effect in spectral lines. Combining these measurements permits a three-dimensional velocity estimate in a consistent reference frame. ^[ref].

Historically, Edmond Halley's 1718 paper compared contemporary and ancient stellar directions and argued that some supposedly fixed stars had moved. Its title, "Considerations on the change of the latitudes of some of the principal fixt stars", captures the conceptual surprise: long-separated measurements could reveal motion invisible in an ordinary night's watching. Modern astrometry retains that insight, using repeated precise measurements to turn the apparently static sky into a record of changing positions. ^[ref]. ^[ref].

Take $N=4p-1$, $|A|=|B|=2p-1$ and sign vectors $u_A=21_A-1$. The Frobenius inner product of two outer products is $(u_A\cdot u_B{)}^2$, proving the formula. The squared Euclidean distance is largest precisely when $|A\cap B|=p-1$, where the inner product is minus one. The common diagonal can be deleted, leaving an isometric representation in dimension $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ if off-diagonal coordinates are multiplied by $\sqrt{2}$. The Frankl-Wilson theorem then bounds each smaller-diameter class.

The radius of a circle or sphere is the Euclidean distance from its centre to its boundary. The diameter is twice the radius.

The vertex set of a triangle with three distinct side lengths has trivial setwise stabilizer subgroup in the plane isometry group. Each vertex is characterized by the unordered pair of lengths of its incident edges, so a setwise-preserving Euclidean isometry fixes every vertex. An isometry fixing three noncollinear plane points is the identity: squared Euclidean distances to them determine every other point, since subtracting two of the Euclidean distance equations gives two independent linear equations. For example, $(0,0),(1,0),(0,2)$ have side lengths $1,2,\sqrt{5}$.

Unroll a circular cylinder of radius $R>0$ using the coordinate $u=R\varphi$. Its line element becomes $d{s}^2=d{u}^2+d{z}^2$, so a geodesic lifts to a straight line in the plane. An endpoint has infinitely many lifts differing by $2\pi Rk$. Minimize the ordinary Euclidean distance over those lifts to obtain the unrestricted shortest path. A fixed lift instead specifies a winding class and need not give the global shortest path.

For angular difference $0\le \delta \le 2\pi$, put $m=\min (\delta ,2\pi -\delta )$. Then $d=\sqrt{R^2{m}^2+(z_1-z_0{)}^2}$. At $\delta =\pi$, two different geodesics have the same shortest length. This is the source of the corner in the distance-versus-angle graph at the opposite generator of the circular cylinder.

A sphere is the set of points at one fixed positive Euclidean distance from a center.

With fixed positive scale estimates $s_j$, this is Euclidean distance in standardized coordinates. It removes marginal units but ignores correlations. Small estimated variances can greatly amplify measurement noise; a zero scale must be handled separately.

For an injective function that does not increase distances, one stretch-factor convention measures the largest compression that must be undone:If the sharp Lipschitz constant of $f$ is one, this equals bi-Lipschitz distortion. Fixing the scale matters: uniform compression $f(x)=x/2$ has stretch factor two in this convention, but bi-Lipschitz distortion one. In a geometric spanner, the corresponding stretch factor is the largest graph-route distance divided by direct Euclidean distance; the graph-to-Euclidean identification does not increase distances. ^[ref].

Take the four-point setthe vertices of a polygon of a rectangle whose side lengths are four and two. Three choices of structure give three different symmetry groups on these same points.

As an unstructured set, $R$ has all $4!=24$ permutations as symmetries, forming the symmetric group $S_4$. If Euclidean distance is part of the data, the allowed maps are only the identity, a half-turn rotation, and the reflections across the two coordinate axes. They form a Klein four-group of group order four. To see that nothing is missing, fix the image of one vertex of a polygon: its three other distances, $2$, $4$, and $\sqrt{20}$, are distinct, so they uniquely determine the images of the remaining vertices.

Preserving affine structure permits more. The affine mapcycles all four points of $R$. It sends a horizontal side of length four to a vertical side of length two, so it is not an isometry. Yet it preserves the rectangle's vertex set and all affine combinations among those points.

Let $A=\mathrm{diag}(2,1)$ and let $r(x,y)=(-y,x)$ be the square's quarter-turn rotation. ThenConjugating every square symmetry by $A$ gives eight affine maps preserving $R$. This group isomorphism makes the affine symmetry group a dihedral group, although some of its maps do not preserve the original Euclidean distances. There are no further affine symmetries: an affine map preserves the common midpoint of the diagonals, so it must preserve opposite pairs. The image of one vertex has four choices and the image of an adjacent vertex has two; those choices determine the map through three noncollinear points.

ThusAn arbitrary permutation need not be an affine map. For example, exchanging the two top vertices while fixing the two bottom vertices sends the common diagonal midpoint to two different points, which no single affine map can do. The phrase "the symmetry of these points" becomes unambiguous only after stating which structure matters.

A symmetry group collects bijections of an object to itself that preserve a specified structure in both directions, with function composition as the operation. For a geometric object one may require Euclidean distances to be preserved, giving an isometry group; for a graph, one preserves adjacency, giving a graph automorphism group. For an unstructured finite set, every permutation is allowed. The structure to be preserved is part of the definition.

For a square with its usual geometric structure, the symmetry group consists of four rotations and four reflections. These form the dihedral group of group order eight. The existing group action example shows how its eight elements move the vertices of a polygon; changing the object to its edges or a marked configuration changes the group action being studied.^[ref]

Specifying the preserved data makes it possible to test a proposed symmetry exactly. After the preserved data are declared, function composition, the identity map, and inverse maps automatically preserve them too. Symmetry transformations therefore form a group, and questions about equivalent configurations become questions about group orbits, stabilizer subgroups, and group homomorphisms. This supplies reusable mathematics for comparing many different kinds of objects.

For a finite set of at least two points in the plane, some point determines the displayed number of different positive Euclidean distances. The circular-arc graph for distinct distances has quadratically many edges. The rich-bisector deletion bound removes high-multiplicity edges while retaining that order of growth. The crossing lemma for multigraphs then compares its lower crossing bound with the quadratic count of pairs of circles. This strengthens the corresponding bound for the whole distinct-distance set. The original result is Theorem11 of Székely's 1997 paper.

For qubit density operators, the trace distance has a particularly simple geometry. Write their Bloch vectors as $\mathbf{r},\mathbf{s}$ in the Bloch ball:where $I$ is the identity operator and $\mathbf{\sigma }=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ denotes the Pauli matrices. This Bloch vector representation includes both pure states on the Bloch sphere and mixed states in the interior of the Bloch ball. ^[ref]

The Pauli matrix multiplication law gives $[(\mathbf{r}-\mathbf{s})\cdot \mathbf{\sigma }{]}^2=|\mathbf{r}-\mathbf{s}{|}^{2}I$. Hence the two eigenvalues of $\Delta$ are $\pm |\mathbf{r}-\mathbf{s}|/2$, and their absolute sum givesThe trace distance is half the straight-line Euclidean distance between the Bloch vectors, not the angular or surface distance along the Bloch sphere. Opposite Bloch vectors have trace distance one; the centre $I/2$ is at trace distance $1/2$ from every pure state.

For the pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, the Bloch vectors are $\mathbf{r}=(0,0,1)$ and $\mathbf{s}=(1,0,0)$. Their straight-line separation is $\sqrt{2}$, soThese Bloch vectors are perpendicular, but the pure states are not orthogonal: $|⟨0|+⟩|=1/\sqrt{2}$. The Helstrom measurement for the zero and plus states measures along the difference direction $\mathbf{n}=(\mathbf{r}-\mathbf{s})/\sqrt{2}=(-1,0,1)/\sqrt{2}$, rather than along either input Bloch vector.

Given finitely many locations and pairwise travel costs, find a cyclic visiting order minimizing the sum of its costs. For Euclidean distance this becomes the Euclidean travelling salesman tour problem.

A metric satisfies $d(x,z)\le d(x,y)+d(y,z)$: visiting an intermediate point cannot beat the direct endpoint distance. A norm satisfies the corresponding inequality $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert$.

Take $O=(0,0)$, $Q=(3,0)$, and $P=(3,4)$. For Euclidean distance, the triangle inequality reads $5\le 3+4=7$. The direct line segment has metric path length $5$, while the route through $Q$ has metric path length $7$. For Manhattan distance, the same comparison becomes $7=3+4$: both routes are shortest. The line segment and the bent route both become metric geodesics when progress is measured by metric path length. For Chebyshev distance, the direct line segment has metric path length $4$, while the same bent route still costs $7$.

Repeated application of the triangle inequality bounds the endpoint distance by the sum along any finite chain of intermediate points. This explains why metric path length is always at least the distance between endpoints. Equality reveals a shortest route, but its shape and uniqueness depend on the chosen metric.

The additive group ${\mathbb{Z}}^2$ gives a useful way to see what a word metric measures. Let $e_1=(1,0)$ and $e_2=(0,1)$, and choose the four-move generating set of a group $S=\{\pm e_1,\pm e_2\}$. Its Cayley graph is the square lattice. Each move changes exactly one coordinate by one, so the word length is the L1 norm, and the word metric is Manhattan distance:At least this many moves are necessary, and moving horizontally then vertically attains the bound. Bridson gives this square lattice as a basic Cayley graph example.^[ref]

Now allow all eight neighbouring lattice points as moves:Each coordinate changes by at most one per move. Use diagonal moves until the smaller displacement is exhausted, then use axis moves: the resulting word length is the maximum norm, and the word metric is Chebyshev distance. From $(0,0)$ to $(2,3)$, an $S$-route has five steps, while the $T$-route $(0,0),(1,1),(2,2),(2,3)$ has three.

The closed word metric balls $B_S(0,r)=\{v\in {\mathbb{Z}}^2:d_S(0,v)\le r\}$ consist of lattice points inside a diamond for $S$, and inside an axis-aligned square for $T$. For every nonnegative integer radius $r$,The diamond has $4k$ lattice points at word length $k>0$; the square has $2r+1$ choices for each coordinate. At radius three the counts are $25$ and $49$. These are discrete closed balls for Manhattan distance and Chebyshev distance; an Euclidean distance bound would instead select lattice points inside a Euclidean disk.