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The chosen unit normal points toward the plane from the origin, so the plane equation isTo locate the circle centre, project $\mathbf{p}$ onto this plane. PutThen $\mathbf{n}\cdot \mathbf{h}=d$, so $\mathbf{h}$ lies in the plane. For any other point $\mathbf{r}$ of the plane, $\mathbf{r}-\mathbf{h}$ is perpendicular to $\mathbf{n}$. The Pythagorean theorem therefore givesRestricting the sphere equation to the plane consequently gives a circle centred at $\mathbf{h}$. Thus $\mathbf{m}=\mathbf{h}$, proving that the displacement of its centre is purely normal:The same Pythagorean theorem gives $q^2={\rho }^2+{\gamma }^2$, and henceThe sign of $\gamma$ is fixed by which side of the plane contains $\mathbf{p}$; $q$ and $\rho$ alone determine only its magnitude. Substituting the signed offset yields the sphere-plane intersection radiusA genuine circle requires $|d-\mathbf{n}\cdot \mathbf{p}|<q$; equality gives tangency and $\rho =0$, while a larger distance gives no intersection. The signed distance from a point to a plane in the direction $\mathbf{n}$ is $\mathbf{n}\cdot \mathbf{p}-d=-\gamma$; the ordinary distance from a point to a plane is its absolute value.

Interpret $Y_n$ as the length of a shortest Euclidean travelling salesman tour, and take the points to be independent random variables with the uniform distribution on the square. We assume $n\ge 2$; a closed tour on one point has length zero, so the printed lower bound would be false for $n=1$. For two points the closed tour traverses the joining segment twice.

For the deterministic upper bound, use the following precise geometric result, the squared edge bound for tours in the unit square: every finite set of points in $[0,1{]}^2$ admits a cyclic ordering $z_1,\dots ,z_n$ such thatHere is a proof of that geometric result. First consider a right triangle with hypotenuse endpoints $a,b$, hypotenuse length $c$, and right-angle vertex $v$. The quadratic path bound in a right triangle supplies a path from $a$ to $b$ through any finite set inside the triangle whose sum of squared edge lengths is at most $c^2$. For an empty set use the edge $ab$. For a single point $z$, use $a,z,b$: the triangle lies in the disk with diameter $ab$, so $\angle azb\ge \pi /2$ and the law of cosines gives $\Vert a-z{\Vert }^2+\Vert z-b{\Vert }^2\le c^2$.

For several points, draw the altitude from $v$ to the hypotenuse. It splits the triangle into two smaller right triangles with hypotenuses $av$ and $vb$. Their squared hypotenuse lengths sum to $c^2$ by the Pythagorean theorem. Repeated altitude subdivision eventually makes every cell small enough to contain at most one of the given distinct points: each child is similar to the original triangle and its diameter contracts by at most the fixed factor $\max (\Vert a-v\Vert ,\Vert b-v\Vert )/c<1$. Construct the base-case paths and combine them from the bottom of this finite subdivision tree. The two child paths concatenate at $v$. If $v$ is an auxiliary point, delete it by joining its two neighbours directly. Both neighbours lie in the right-angle sector of the parent triangle at $v$, so their angle at $v$ is at most $\pi /2$. The law of cosines now shows that this shortcut does not increase the sum of squared edge lengths. The combined path still has endpoints $a,b$ and squared cost at most $c^2$.

One can first put the given points in generic positions, avoiding subdivision boundaries, and then pass to a limit. There are only finitely many possible visiting orders, so a subsequence keeps the same order and its squared cost converges; this also handles coincident locations. No assumption about a minimum separation remains in the result.

Split the square along a diagonal $ab$. Apply the triangle result in each half, following one path from $a$ to $b$ and the other back to $a$. Each half contributes at most $\Vert a-b{\Vert }^2=2$. Remove either diagonal endpoint if it was auxiliary. At a square corner all neighbours lie in a sector of angle $\pi /2$, so the same squared-cost shortcut applies. This gives the asserted cyclic ordering with total squared cost at most four. It does not assert that the tour minimizing ordinary length itself has this squared-edge property. The Cauchy-Schwarz inequality applied to the tour supplied by the theorem provesThus the upper bound is deterministic, independently of the sampling law.

For the lower bound, put $R_i={\min }_{j\ne i}\Vert X_i-X_j\Vert$, the nearest neighbour distance at $X_i$. Both edges incident to $X_i$ in any closed tour have length at least $R_i$. Summing over vertices and then dividing by two gives $Y_n\ge {\sum }_i{R}_i$. Conditional on $X_i=x$, the union bound and the uniform distribution giveBoundary clipping only makes the ball smaller. Apply the layer cake representation of the expected value, restricting the integral to $0\le t\le 1/(2\sqrt{n})$:Linearity of the expected value now yieldsThe unspecified sampling phrase in the paper must therefore be understood as independent uniform sampling: an arbitrary distribution concentrated in a tiny region cannot satisfy this lower bound.

Give $X\times Y$ the uniform probability measure and let $f$ be the indicator function of the edge set. For partitions $\mathcal{P}=\{X_i\}$ and $\mathcal{Q}=\{Y_j\}$, define the energyBecause conditional expectation is an orthogonal projection in L2 space, $0\le \mathcal{E}\le \Vert f{\Vert }_2^2\le 1$.

Suppose the pair of partitions is not $\epsilon$-regular. For each irregular $(X_i,Y_j)$ choose witnesses $A_{ij}\subseteq X_i$ and $B_{ij}\subseteq Y_j$ withRefine each $X_i$ by all sets $A_{ij}$ and each $Y_j$ by all sets $B_{ij}$. If the old partitions have $r,s$ cells, the refined ones have at most $r{2}^s,s{2}^r$ cells.

The Pythagorean theorem for the two nested conditional-expectation projections givesInside an irregular $X_i\times Y_j$, the witness rectangle occupies at least an ${\epsilon }^2$ proportion and the mean of the displayed difference over it has magnitude greater than $\epsilon$. The Cauchy-Schwarz inequality therefore gives an energy gain greater thanfrom that pair. Since irregular pairs have total weight greater than $\epsilon$, the complete refinement raises the energy by more than ${\epsilon }^5$.

Energy is at most one, so after at most $\lceil {\epsilon }^{-5}\rceil$ refinements the process stops at an $\epsilon$-regular pair of partitions. Iterating the cell-count bounds $r\mapsto r{2}^s$ and $s\mapsto s{2}^r$ from $r=s=1$ a bounded number of times produces a finite $K(\epsilon )$ independent of $G$. This proves the Bipartite Szemerédi regularity lemma.

To first order, the radial and transverse comoving separations arewhereThe local spatial geometry is that of Euclidean space, so the Pythagorean theorem givesConsequently the requested function is

With $b\cdot c=0$, substitute $a=\alpha b+\beta c$ into the plane constraints. They become $\alpha |b{|}^2=\lambda$ and $\beta |c{|}^2=\mu$, givingThe Pythagorean theorem for the two orthogonal terms in $a$ then gives the closest distanceZero values of either scalar are allowed, including zero distance if both planes contain the origin.

Let $b_0=b(0)$ in centimetres and take $t$ in seconds. The constant rate gives $b(t)=b_0+4t$. By the Pythagorean theorem, the diagonal length isfor times when the side length is nonnegative. No initial value of $b$ is specified, so it must remain as a parameter. Choosing the time origin when $b=0$ gives the special convention $d(t)=\sqrt{100+16t^2}\mathrm{cm}$.

Interpret the stated side lengths as all eight edges of the square pyramid. The equal lateral edges put its apex above the base centre. The base-centre-to-vertex distance is half the diagonal, $1/\sqrt{2}\mathrm{m}$. The Pythagorean theorem therefore givesUsing the pyramid volume formula with a $1{\mathrm{m}}^2$ base,

A space diagonal of the unit cube has displacement vector $(1,1,1)$. Its Euclidean norm, or two successive applications of the Pythagorean theorem, givesThe requested diagonal joins opposite vertices; $\sqrt{2}$ would be the diagonal of a face.

The two equal legs meet at the right angle, so the Pythagorean theorem gives $A{B}^2+A{C}^2=B{C}^2=4$, with $AB=AC$. Thus $AC=\sqrt{2}$. The midpoint condition then yields

The triangle $ADB$ is right-angled at $A$, with $AB=\sqrt{2}$ and $AD=1/\sqrt{2}$. Another application of the Pythagorean theorem gives $DB=\sqrt{2+1/2}=\sqrt{5/2}$. At the angle with vertex $D$, the adjacent side is $AD$ and the hypotenuse is $DB$. Hence

Let the semicircle have radius $R$, and write $A=|\mathbf{a}|$, $B=|\mathbf{b}|$. The right triangle has hypotenuse $2R$, so the Pythagorean theorem gives $A^2+B^2=4R^2$. Its area is $AB/2$, whereas half the area of the semicircle is $\pi R^2/4$. Equating these and eliminating $R$ givesEquivalently, the relation between the two norms can be writtenThe two possible ratios are reciprocals, corresponding to interchanging the two sides.

Take the dumbbell axis to be the $z$ axis, with its midpoint at $z=0$. The right-hand sphere has centre $z=\alpha a$ and radius $R=\beta a$, while its junction with the cylinder is at $z=\gamma a$. Put $d=(\alpha -\gamma )a$, so $0<d<R$. The Pythagorean theorem in a meridional cross-section gives the handle radius $H$ from $H^2+d^2=R^2$. Thus

The Pythagorean theorem gives $c^2=a^2+b^2$. Taking a total differential gives $2cdc=2ada+2bdb$, soThe area of the right triangle is $A=ab/2$, so the product rule gives $dA=(bda+adb)/2$ andThe prescribed small changes therefore give $dA/A\approx 0.01-0.02=-0.01$: the area decreases by approximately $1\%$. For comparison, retaining the product of the two finite percentage changes gives $(1.01)(0.98)-1=-0.0102$, an exact decrease of $1.02\%$; the total differential retains only the first-order terms.

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.

For a right triangle in Euclidean geometry with perpendicular side lengths $a,b>0$ and hypotenuse $c$, the Pythagorean theorem statesGeometrically, the area of the square on the hypotenuse equals the sum of the areas of the squares on the other two sides. Euclid's Elements, Book I, Proposition 47 states this area equality; Proposition 48 states its converse: a triangle satisfying the equality has a right angle opposite side $c$.^[ref][ref]

For $a=3$ and $b=4$, we compute $c=\sqrt{9+16}=5$. The triangle has area $ab/2=6$, and its side lengths form a Pythagorean triple. Conversely, a triangle measured to have sides $3,4,5$ must have a right angle between the sides $3$ and $4$. This makes the equality both a calculation tool and a test for perpendicularity.

The three squares in the next figure have areas $3^2=9$, $4^2=16$ and $5^2=25$ in squared length units. These are different from the central right triangle's area $3\cdot 4/2=6$. The Pythagorean theorem adds the areas of the attached squares, not their side lengths and not the area of the triangle. The square on the hypotenuse is tilted, but a rotation preserves its area. This makes the formula visible without suggesting that the hypotenuse has to align with an axis.

Historically, Euclid's Elements records an organized proof of the Pythagorean theorem as an equality of areas, rather than the modern equation between numbers representing lengths. Its placement in Book I makes it available before the treatment of similar triangles in Book VI. This is evidence of how the result was established within that text; the name Pythagorean theorem alone does not establish who first discovered or proved it. The precise historical anchor here is the surviving proposition, rather than an unsupported story of a single inventor.^[ref]

The dissection below makes the general identity tangible. Both outer squares have side $a+b$ and contain the same four congruent right triangles. In the first arrangement, their hypotenuses bound a square of side $c$: adjacent acute angles together make a right angle. In the second, the uncovered regions are squares of sides $a$ and $b$. Subtracting the same total triangle area from equal outer areas givesFor the pictured dimensions, the shared uncovered area is $49-24=25=9+16$.

The same mechanism survives in inner product spaces: for orthogonal vectors $u,v$, expanding the inner product gives $\Vert u+v{\Vert }^2=\Vert u{\Vert }^2+\Vert v{\Vert }^2$. This is why orthogonal projection separates a squared distance into independent contributions. The familiar right triangle is the simplest picture of a much more reusable principle.

For every real angle $\theta$,It is the equation of the unit circle written in terms of its sine and cosine coordinates. For an acute angle, dividing the Pythagorean theorem by the square of the hypotenuse gives the same relation in a right triangle. The circle interpretation explains why it remains true outside that first-quadrant triangle.

For example, if $\sin \theta =3/5$, the identity gives ${\cos }^2\theta =16/25$, so $\cos \theta$ is either $4/5$ or $-4/5$. The sign needs the angle's location: it is positive in the first quadrant and negative in the second. Thus the identity determines the magnitude of the missing coordinate without determining the direction.

The relation is useful for eliminating one squared trigonometric function, checking that a coordinate pair has unit length, and checking that the columns of a rotation matrix are unit vectors. It also holds for complex arguments under the complex exponential definitions, although then the squares are algebraic squares rather than squared moduli.