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Let the triangles have vertices $A_0,A_1,A_2$ and $B_0,B_1,B_2$. Work initially with nonzero-area triangles. A triangle-triangle intersection algorithm begins with the unnormalized normal vector $N_A=(A_1-A_0)\times (A_2-A_0)$ and the three signed plane evaluations $d_j=N_A\cdot (B_j-A_0)$. If all $d_j$ are strictly positive, or all strictly negative, the second triangle lies strictly on one side of the first one's plane, so the triangles are disjoint. Otherwise compute $N_B$ and $e_i=N_B\cdot (A_i-B_0)$ and apply the symmetric rejection test. Strict inequalities are important: contact on a vertex or edge counts as intersection.

If the planes are not parallel, put $L=N_A\times N_B$. Each triangle meets the other one's plane in an empty set, a point or a line segment. Its endpoints are obtained from zero-valued vertices and the edges whose endpoint plane evaluations have opposite signs. For example, an edge $B_j{B}_k$ contributesAll these points lie on the common line of the two planes. Choose a coordinate $r$ for which $L_r\ne 0$ and form the coordinate intervals $I_A=[a_{-},a_{+}]$, $I_B=[b_{-},b_{+}]$ of the two slices. This coordinate is one-to-one along the common line, so the exact final test isOnly that coordinate of each slice endpoint needs to be evaluated; no full three-dimensional line parameterization is necessary.

Parallel planes which have survived the strict plane tests are coplanar. Drop the coordinate corresponding to the largest component of $N_A$ and test the two projected planar triangles. For each edge of either projected triangle, project both vertex sets onto the perpendicular direction. A strictly disjoint pair of projection intervals certifies separation. If none of the six edge directions separates them, the planar convex sets overlap: a separating line for two disjoint convex polygons can be chosen parallel to an edge of one of them. This includes containment and boundary contact.

A zero normal vector signifies a degenerate triangle; reduce its convex hull to a point or its extreme line segment and use point containment or segment intersection instead. For floating-point input, reliable sign predicates and scale-aware error bounds are preferable to treating a tiny signed distance as an arbitrary exact zero. Plane rejection followed by interval overlap, with a planar test for coplanar input, determines all nondegenerate cases.

The supplied subdivision mask describes the regular, valence-six triangular grid. Its coefficients split into four parity classes after binary refinement: one class for old vertices and three classes for the three edge directions. For an old vertex $V$ with cyclic neighbors $V_1,\dots ,V_6$, the even stencil isFor an edge with endpoints $A,B$ and opposite vertices $C,D$ in its two incident triangles, the new odd vertex isRotating this edge stencil supplies the other two odd parity classes. All stencil weights sum to one, giving affine invariance; their first moments place an affine regular grid at the old positions and edge midpoints. Apply the same weights separately to $x,y,z$. Boundary edges and extraordinary vertices need additional rules, which are not specified by this uniform mask. The subsequent arguments concern the regular interior grid and compatible extruded boundary treatment.

Use the limit subdivision surface and limit subdivision curve, rather than treating their initial control meshes as the reflecting geometry. Build adaptive refinement hierarchies for both. Each surface cell supplies a limit-position evaluator, local tangent vectors and a normal vector; each curve interval supplies a limit-position evaluator and tangent. Regular regions may be evaluated as local splines, while extraordinary neighborhoods require the evaluation machinery of the specified subdivision rules. A piecewise smooth surface is handled patch by patch, with explicit boundary or crease joins.

A useful initial polygonal construction refines the surface to small triangles and the source curve to small line segments. For a candidate mirror triangle in a plane $n\cdot (X-X_0)=0$, reflect the eye across that plane:For a source edge $Y(t)=(1-t)Y_0+t{Y}_1$, a physical reflection point is the plane intersection of the line $E^{\ast }Y(t)$:Keep only $0\le t\le 1$, $0<\lambda <1$, points inside the mirror triangle, and eye/source points on the appropriate reflecting side. Unfolding the reflected path across the mirror proves the specular reflection condition. As $Y$ moves along a straight edge, $X$ lies on the intersection of the mirror plane with the plane through $E^{\ast },Y_0,Y_1$, so the retained locus is a straight segment, a point or the empty set. Plane and triangle clipping computes its endpoints. If these defining points are collinear, or a denominator vanishes, use the corresponding limiting geometric case rather than a generic plane formula. A bounding volume hierarchy accelerates candidate search where geometric bounds exclude pairs; refined facet segments provide seeds and approximate connectivity.

To obtain the reflection curve of the actual limits, correct these seeds. In a local surface chart write $S(u,v)$, and on a source interval write $C(t)$. SetThe exact limit reflection locus is $F=0$ together with the front-side conditions. For a two-sided mirror, orient $n$ locally so that $b\cdot n>0$ and require $a\cdot n>0$; for a one-sided mirror use its prescribed orientation and discard the wrong side. At grazing incidence this orientation condition ceases to be strict and the case needs separate handling. Unlike the eye-facing assumption in the parametric question, no global orientation condition is supplied here.

Apply a continuation method: at a rank-two root of $DF$, predict in a nullspace tangent direction and correct by the Newton method on a transverse section. Map corrected parameters to $S(u,v)$, crossing patch and curve-interval boundaries through their hierarchy adjacency. Trace all branches, recognizing closed components and endpoints. Supplement facet seeds by recursive searches of the surface-cell/source-interval product, with interval arithmetic or other certified enclosures of $F$; otherwise a small loop absent from the initial polygonal approximation could be missed. Split singular neighborhoods where the Jacobian matrix loses rank, and test the open source-to-mirror and mirror-to-eye segments for occlusion when selecting actually visible components.

Allocate a geometric error budget to limit evaluation, root correction and polygonal chords. For example, require combined position-evaluation and corrected-root error at most $ϵ/2$ and, on every regular arc, require the chord error bound $M(\Delta s{)}^2/8\le ϵ/2$. Use subdivision position and normal bounds for evaluations: a tiny position error alone is insufficient, since a wrong normal vector changes the reflected ray. Away from singularities an inverse transverse Jacobian matrix bound converts reflection residual into parameter error and a bound on $DS$ converts this to space error. Refine until these bounds hold, or use validated enclosures of entire arcs and their distances to the chords.

Near a caustic the inverse bound can become large, so the hierarchy must refine more strongly or isolate the singular branch geometry. Mesh flatness alone does not certify a reflection tolerance. With regular convergent evaluators and finite resolvable components, the resulting ordered vertices and chords satisfyThe problem supplies no particular subdivision rules or smoothness guarantees, so this construction specifies what their evaluators and bounds must provide; an arbitrary nonsmooth or degenerate limit need not have an ordinary finite reflection curve. Refine for seeds, enforce reflection on the actual limit geometry, then refine the traced arcs to the required spatial tolerance.

We construct a family by seven-triangle Dirichlet transplantation. Start with a scalene triangle $\tau$ sufficiently close to an equilateral triangle. Its three side lengths are the three independent parameters; restrict to an open set where its angles $\alpha ,\beta ,\gamma$ are distinct and lie between $\pi /4$ and $\pi /2$. Label the sides $a,b,c$ opposite the corresponding vertices $A,B,C$.

For each of two planar domains, begin with tile $7$ and attach the remaining congruent copies by Euclidean reflection across the indicated sides. The attachment tables areFor example, $(6,7)$ in the first $a$ row means to reflect tile $7$ across its $a$ side to obtain tile $6$. Both attachment graphs are trees: there is a central tile with three arms, each consisting of two tiles. Every side not used in the table is an exterior boundary side.

At the equilateral triangle, these seven reflected tile interiors are disjoint; at each central vertex four sectors of angle $\pi /3$ meet, leaving an exterior gap $2\pi /3$. Other boundary vertices have angles $\pi /3$ or $2\pi /3$, and distinct nonincident pieces are separated. The positive angular gaps and separations persist under sufficiently small perturbations of the three side lengths. Thus this construction really gives connected polygonal planar domains on an open three-dimensional parameter set. Their only reentrant vertices are the central vertices $A,B,C$, with interior angles $4\alpha ,4\beta ,4\gamma$; all other boundary angles are one or two triangle angles and are less than $\pi$.

Because these three reentrant angles are distinct, an isometry of the two planar domains would send each labelled central vertex to its counterpart. A planar isometry is the restriction of a Euclidean rigid motion: its differential preserves the Euclidean connection, hence is constant on the connected domain. After aligning the central triangles, such a motion fixes three noncollinear points and is the identity. But the domains are not identical. On the arm attached to side $a$, the point construction attaches the last tile across side $b$, whereas the line construction attaches it across side $c$. The former side is therefore interior in one domain and exterior in the other. This proves the two domains are nonisometric throughout this three-parameter family.

We now prove equality of the Dirichlet Laplacian spectra, including multiplicities. Index $1,\dots ,7$ by the nonzero binary vectors $x=(x_1,x_2,x_3)$, with integer label $x_1+2x_2+4x_3$. The transformationsare involutions over ${\mathbb{F}}_2$. Their point permutations $P_s$ are precisely the transpositions in the left attachment table, with the remaining points fixed. On covectors the inverse-transpose transformations give permutations $Q_s$ with exactly the right table. Let $B$ be the point-line incidence matrix of a set system of the Fano plane:The incidence relation is invariant under transforming $x$ and its covector $y$ contragrediently, so $B{P}_s=Q_{s}B$ for all three sides. Each point lies on three lines, and two distinct points lie on exactly one common line. Consequentlywhere $J$ is the all-ones matrix. Its eigenvalues are $9$ once and $2$ six times, so $B$ is invertible.

The Dirichlet boundary condition requires a sign change at an unglued side. Give each tile the sign $(-1{)}^{\text{distance from tile }7}$ and putThe signs alternate across each glued side in both domains. The signed side matrices are thereforeThey have entry $+1$ across a glued pair and entry $-1$ on a fixed exterior tile. The invertible seven-triangle Dirichlet transplantation matrix $C=DBD$ satisfiesThis is an actual Dirichlet boundary condition intertwiner, not merely the unsigned incidence relation.

To justify the spectral conclusion even at the reentrant corners, use the closed Dirichlet energy form. Pull a function on either domain back to its seven reference triangles and write its components as a vector $F$. The form domain consists of $H^1$ components whose side traces agree on glued pairs and vanish on exterior sides; equivalently $F=S_{s}F$ on each side for the point domain, and $G=T_{s}G$ for the line domain. These trace conditions exactly characterize $H_0^1$ of the glued polygon. Both its $L^2$ norm and its Dirichlet energy are the sums of their seven reference-tile integrals.

Since the signed side matrices are symmetric, the intertwining identities also give $S_s{C}^{T}C=C^{T}C{S}_s$. Thus $(C^{T}C{)}^{-1/2}$ commutes with every $S_s$, and the polar factoris an orthogonal matrix with $U{S}_s=T_{s}U$. Multiplication of the component vector by this constant matrix preserves the $L^2$ inner product and the integrated squared gradients, and maps the two form domains bijectively. The self-adjoint operators associated with these forms are consequently unitarily equivalent, so their eigenvalues and multiplicities agree. The eigenfunctions are smooth in the interiors; no unwarranted smoothness at polygonal corners is required. We have proved

The precise external concentration result we use is Talagrand's convex distance inequality: for a finite product of independent standard probability spaces, a measurable event $B$ of positive probability, and $u\ge 0$,Our coordinates are the independent uniformly sampled points of the triangle, so the hypothesis holds. Part (ii) therefore gives, for integers $a>b\ge 0$,Indeed every configuration in the first event has at least the distance from part (ii) to the second event. If one event is empty, the displayed product bound is automatic. This is the only use of the product-space theorem; the geometric separation was proved directly in part (ii). A precise source for that theorem is Section 4.1 of Talagrand's original product-space concentration paper.

Choose an integer median $m=m_n$, so $\mathbb{P}(L_n\le m)\ge 1/2$ and $\mathbb{P}(L_n\ge m)\ge 1/2$. First set $a=m+t$, $b=m$ in the product bound, and then set $a=m$, $b=m-t$. For positive integers $t$ this yields the convex-chain median concentration estimatesThe lower event is empty if $m-t<1$ in our sampling model; the upper event is empty if $m+t>n$. Thus the stated bounds remain valid for all positive integers $t$. Also $m\ge 1$ for $n\ge 1$.

By part (i), $m\le C{n}^{1/3}$ with $C=4e$. Let $h_n=\omega (n)n^{1/6}/2$, take $t_n=\lfloor h_n\rfloor +1$, and use the deterministic interval $I_n=[m-h_n,m+h_n]$. Its length is exactly $\omega (n)n^{1/6}$. As $L_n$ and $m$ are integers, falling outside this interval implies one of the two tail events with $t=t_n$. ThereforeFor positive $m,t$,Since $t_n\ge h_n$ and $m\le C{n}^{1/3}$, this exponent is at leastConsequentlyThus $L_n$ lies with high probability in the required interval. This proof works for every $\omega (n)\to \infty$, including growth faster than $n^{1/6}$; no extra restriction on $\omega$ is needed. The scale comes from the square root of the $O(n^{1/3})$ certificate size at the median.

Choose a star centre $a$ of the open star-shaped set $D$ and define the candidate antiderivative by a straight-segment contour integral:For each fixed $z\in D$ and all sufficiently small $h$, the filled triangle with vertices $a,z,z+h$ lies in $D$. Indeed, the compact segment $[a,z]$ has a neighbourhood of some positive radius contained in the open set $D$, and every point of this triangle lies within $|h|$ of that segment. The zero triangle integral consequently givesBy continuity, the last integral tends to $f(z)$ as $h\to 0$. HenceIn particular, this proves the existence of the antiderivative without incorrectly assuming the star-shaped set is convex.

On a general domain, the conclusion can fail. Take $D=\mathbb{C}\setminus \{0\}$ and $f(z)=1/z$. Every filled triangle contained in $D$ has zero boundary integral by Cauchy integral theorem, but the unit circle has integral $2\pi i$. An antiderivative would make every closed contour integral zero. This is the period obstruction to a holomorphic antiderivative: the local conclusion of Morera's theorem does not remove the global obstruction.

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

For this stellar population, let $m=M/M_{\odot }$ denote initial mass in units of the solar mass, and let $\tau =t/\mathrm{Gyr}$ denote present stellar age. Constant formation of equal numbers of stars per unit time makes the stellar age a uniform distribution on $[0,10]$ Gyr, so $Y=\tau /10$ is uniform on $[0,1]$. The normalized initial mass function has probability density function $f(m)=0.1m^{-2}$ for $m>0.1$, since ${\int }_{0.1}^{\infty }{m}^{-2}dm=10$. ConsequentlyThese formulae have the stated age and mass domains; outside them the relevant cumulative fractions saturate at zero or one. The probability integral transform makes $X$ uniform on $[0,1]$: $m=0.1/X$ gives $f(m)|dm/dX|=1$. The time-independent initial mass function and constant number formation rate give independence of $X$ and $Y$. All fractions here count objects, including white dwarfs, using the stipulated stellar evolution law.

A red giant hasThus the mass boundaries are $m=9/\tau$ and $m=10/\tau$ for positive $\tau$. Equivalently, for $0.9<m<1$ the red giant region runs from $9/m$ to the age cap $10$, and for $m\ge 1$ it runs from $9/m$ to $10/m$. There are no red giants with $m\le 0.9$. Boundaries have zero probability and their endpoint convention does not affect the fractions. In the uniform $(X,Y)$ square the red giant region is $9X<Y<10X$, with triangle vertices $(0,0)$, $(1/10,1)$ and $(1/9,1)$. The white dwarf region is the triangle $0<X<Y/10$.

At fixed $Y=y$, define the conditional probabilities of a red giant, white dwarf and main sequence star by $g(y)$, $w(y)$ and $s(y)$. Their interval widths areIntegration over the uniform distribution of age gives the individual-star fractionsThe systems form a coeval binary population: the two binary star components have the same age. Their masses are independent, so $(X_1,X_2,Y)$ has uniform probability density function on $[0,1{]}^3$ and their states have conditional independence given $Y$. Unconditional independence of their states would be incorrect: older systems make both evolved states more likely. The law of total probability now givesThe subtraction removes the double counting of systems containing two red giants.

Sperner's lemma says that if a triangle is triangulated, its vertices are labelled $1,2,3$, and label $i$ is forbidden on the edge opposite vertex $i$, then an odd number of small triangles have all three labels.

To prove it, count incidences with edges whose endpoint labels are $1$ and $2$. Along the outer $12$-edge, the labels begin at $1$ and end at $2$, so the number of $12$ transitions is odd; no other boundary edge contributes. An interior edge contributes twice. A small triangle contributes an odd number precisely when its three labels are $1,2,3$: a triangle using only $1,2$ contributes two, and every other non-tricoloured triangle contributes zero. The number of tricoloured triangles is therefore odd.

Now suppose the closed sets $A_i$ in the question existed. Let $d_i(x)$ be the distance from a point to a closed set $A_i$. Empty triple intersection makes $d_1+d_2+d_3>0$, sois a continuous function from the large triangle to itself, where $v_i$ is the vertex opposite ${\alpha }_i$. Since the $A_i$ cover the triangle, at least one $d_i(x)$ vanishes, so $r(x)$ always lies on the boundary. On ${\alpha }_i\subseteq A_i$, its $i$th barycentric coordinate vanishes, hence $r({\alpha }_i)\subseteq {\alpha }_i$. Thus $r$ is face-preserving on the boundary; its boundary restriction is homotopic there to the identity by the straight-line homotopy. This would give a map from the triangle into its boundary whose boundary degree is one, contradicting the no-retraction theorem, the standard topological consequence of Sperner's lemma. Therefore the three closed sets must have a common point.

Let $S$ be a finite Ramsey witness for $X$ under $k$ colors. Choose a finite witness $T$ for $Y$ under $k^{|S|}$ colors. Given a coloring of $S\times T$, color each $t\in T$ by the complete vectorThere is a copy $Y^{\prime }\subseteq T$ on which this vector is constant. Thus, for each $s$, the color $c(s,t)$ is independent of $t\in Y^{\prime }$. These values define a $k$-coloring of $S$, which has a monochromatic copy $X^{\prime }$. Then $X^{\prime }\times Y^{\prime }$ is a monochromatic isometric copy of $X\times Y$. This proves the product theorem for Euclidean Ramsey sets.

Every nondegenerate triangle and every line segment is a Euclidean Ramsey set. If $T$ is the given acute triangle and $I$ is a segment of length $t$ in a new orthogonal coordinate, then $T\times I$ is exactly the vertex set of the triangular prism with base $T$ and height $t$. The product theorem therefore makes it Euclidean Ramsey.

A planar graph admits at least one graph drawing in which edges meet only at common endpoints. This is a property of the abstract graph, not a requirement on every graph drawing. A plane graph specifies one such drawing.^[ref]

For example, the complete graph $K_4$ joins every pair of four vertices, giving six edges. Put its vertices at the corners of a square and its diagonals cross. Instead put $A,B,C$ at the corners of a triangle and $D$ inside it: the same six edges $AB,AC,BC,AD,BD,CD$ now meet only at their endpoints. The first graph drawing obscures planarity; the second proves it. In particular, its crossing number is zero.

The distinction matters whenever connections must fit on one surface without intersecting. A failed graph drawing may only need rearrangement, whereas nonplanarity by edge counts proves that every rearrangement fails.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.

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.