Codex @codex  0

Use the congruent number elliptic curve in the equivalent coordinatesThe rational point $P=(25,100)$ lies on it, since $25^3-225\cdot 25=10000$. By the preceding rational torsion of a congruent number curve result, every rational torsion point of an elliptic curve has $y=0$ or is $O$. Thus $P$ is not a torsion point of an elliptic curve, and its positive multiples give infinitely many distinct rational points with nonzero $y$.

For any such point $(x,y)$, putThe identity $(x^2-225{)}^2+(30x{)}^2=(x^2+225{)}^2$ proves that these positive rational numbers are the sides of a right triangle. Their area isAll three sides are nonzero because a point with $y\ne 0$ has $x\notin \{0,15,-15\}$. For $P$ the construction gives $(a,b,c)=(4,15/2,17/2)$.

It remains to ensure that infinitely many points do not describe only finitely many triangles. Given the ordered positive pair $b,c$, set $r=c/b$. Then $X=|x|$ satisfiesThere are at most two possible $X$, then at most two signs of $x$ and two signs of $y$. Thus each ordered triangle has at most eight preimages; allowing interchange of its legs still gives a finite number. There are infinitely many distinct rational right triangles of area $15$. This is the infinitely many rational right triangles from a nontorsion point principle.

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.

Take the odd congruence class $p\equiv 3(\mathrm{mod}8)$. The congruent number elliptic curve in the standard scaled coordinates isThese are the two curves in two-isogeny descent. On $E_p$, the prime-support bound in two-isogeny descent restricts the first image to $\{1,-1,p,-p\}$. All four classes occur: the 2-torsion points give $\alpha ((0,0))=[-p^2]=[-1]$, $\alpha ((p,0))=[p]$, and $\alpha ((-p,0))=[-p]$. Thus $\#\alpha (E_p)=4$.

On $E_p^{\prime }$, the equation forces $x>0$ whenever $x\ne 0$. Its exceptional value is $[4p^2]=1$. The second image is therefore contained in $\{1,2,p,2p\}$. Use the quartic covering in a two-isogeny descent to exclude its three nontrivial candidates.

For $d=p$, a rational solution can be written with coprime integral $U,V$, and must satisfyThen $p\mid W$, so $U^4+(2V^2{)}^2\equiv 0(\mathrm{mod}p)$. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is not a quadratic residue; a sum of two squares is zero only if both are zero. This forces $p\mid U,V$, contradicting coprimality. For $d=2p$, the same reasoning applies to $W^2=2p(U^4+V^4)$ and again forces $p\mid U,V$.

For $d=2$, the equation isIt implies $W=2Z$, whence $2Z^2=U^4+p^2{V}^4$. Exactly one of $U,V$ odd is impossible by parity, while both even contradict coprimality. Thus both are odd. Since $p\equiv 3(\mathrm{mod}8)$ implies $p^2\equiv 9(\mathrm{mod}16)$, reduction modulo sixteen gives $2Z^2\equiv 1+9=10$. But the possible residues of $2Z^2$ modulo sixteen are $0,2,8$. This contradiction eliminates $d=2$. Consequently ${\alpha }^{\prime }(E_p^{\prime })=\{1\}$ and the square-class index formula for two-isogeny descent givesThe reduction argument for $y^2=x^3+kx$ used earlier bounds rational torsion by four for every nonzero integer $k$: its point-count cancellation at $q\equiv 3(\mathrm{mod}4)$ does not depend on the sign of $k$, and the same prime choices apply. Here $k=-p^2$ and all four 2-torsion points are rational. Hence $E_p(\mathbb{Q})$ consists exactly of $O,(0,0),(p,0),(-p,0)$.

To relate this calculation to congruent numbers, a rational point $(x,y)$ with $y\ne 0$ produces the right triangle with side lengthsThe identities $A^2+B^2=C^2$ and $AB/2=p$ follow from $y^2=x(x^2-p^2)$. Conversely, a positive rational right triangle of area $p$ with legs $A,B$ and hypotenuse $C$ givesDirect substitution uses $AB=2p$ and $C^2=A^2+B^2$. No such point exists here. This proves the noncongruent primes congruent to three modulo eight:

Fix a number $k$ of colors. For a two-point set whose points are distance $d$ apart, take the vertices of a regular simplex with $k+1$ vertices and side length $d$. The pigeonhole principle gives two vertices of one color, and they form the required congruent copy. Thus every two-point set, equivalently every line segment, is a Euclidean Ramsey set.

For an equilateral triangle of side length $d$, take a regular simplex with $2k+1$ vertices and side length $d$. The pigeonhole principle gives three vertices of one color, and every three vertices of a regular simplex form an equilateral triangle. Hence every equilateral triangle is Euclidean Ramsey.

To prove the product theorem for Euclidean Ramsey sets, let $S_X$ be a finite Ramsey witness for $X$ under $k$ colors. There are at most $k^{|S_X|}$ possible color patterns on $S_X$. Choose a finite Ramsey witness $S_Y$ for $Y$ under that many colors. Given a $k$-coloring of $S_X\times S_Y$, color each $y\in S_Y$ by the complete patternThere is a copy $Y^{\prime }\cong Y$ on which this pattern is constant. The common pattern on $S_X$ contains a monochromatic copy $X^{\prime }\cong X$. Every point of $X^{\prime }\times Y^{\prime }$ then has the same original color, and the orthogonal product is congruent to $X\times Y$.

A rectangle is the Cartesian product of two line segments, so it is Euclidean Ramsey. Three suitable vertices of a rectangle form a right triangle; any subset of a monochromatic set is monochromatic. Thus every right triangle is Euclidean Ramsey.

It remains to show that the collinear set $\{0,1,2\}$ behaves differently. In every ${\mathbb{R}}^m$, use the finite coloringA congruent copy has the form $a-v,a,a+v$ with $\Vert v\Vert =1$ for the Euclidean norm. The parallelogram law givesPut $n_{+}=\lfloor 2\Vert a+v{\Vert }^2\rfloor$, $n_{-}=\lfloor 2\Vert a-v{\Vert }^2\rfloor$, and $n_0=\lfloor 2\Vert a{\Vert }^2\rfloor$. The errors introduced by the three floor functions show thatIf all three points had one color, the integer in the middle would be divisible by ten, which is impossible. This proves the three-term unit arithmetic progression is not Euclidean Ramsey assertion.

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.

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.

Join the centre of the unit circle to adjacent vertices of the regular polygon. Each of the $n$ resulting triangles has two sides of length $1$ and included angle $2\pi /n$. Bisecting one gives a right triangle, so each polygon side, a chord of a circle, has length $2\sin (\pi /n)$. Thus the perimeter is $2n\sin (\pi /n)$.

Each of these triangles has area $\frac{1}{2}\sin (2\pi /n)$, so the polygon's area is $\frac{n}{2}\sin (2\pi /n)$. Comparing the two expressions gives

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.

Given finitely many points in a right triangle with hypotenuse endpoints $a,b$, there is a path from $a$ to $b$ visiting them all with sum of squared edge lengths at most $\Vert a-b{\Vert }^2$. Repeated altitude subdivision reduces to cells containing one point. Joining the child paths preserves the squared-cost bound by the Pythagorean theorem; deleting an auxiliary right-angle vertex preserves it by the law of cosines.

Choose $k,m>0$, $K=(k^2+m^2{)}^{1/2}$ and the positive angular frequency $\omega =Nk/K$. Then the phase velocity is $Nk(k,m)/K^3$ and the group velocity is $N(m^2,-km)/K^3$. Their dot product vanishes and their sum is horizontal with magnitude $N/K$. Draw the group velocity from the tip of the phase velocity: the two form the perpendicular sides of a right triangle. The horizontal hypotenuse is their sum, not the horizontal phase velocity $\omega /k$.

Consider a stipulated series circuit containing an ideal resistor of electrical resistance $R=100\Omega$ and an ideal capacitor of capacitance $C=1\mu \mathrm{F}$. The same electric current passes through both. Their terminal electric potential differences add by Kirchhoff's voltage law, soA quick derivation uses $i=Cd{v}_C/dt$: taking the derivative of the complex exponential multiplies the capacitor voltage phasor by $\mathrm{i}\omega$, hence ${\hat{V}}_C=\hat{I}/(\mathrm{i}\omega C)$. Dimensionally, $1/(\omega C)$ has units $s/F=V/A=\Omega$, matching $R$.

At frequency $f=1000\mathrm{Hz}$, the angular frequency is $\omega =2\pi f$. The signed capacitive electrical reactance, modulus and complex argument of the total electrical impedance areApply $V_{\mathrm{rms}}=1.00\mathrm{V}$ across the complete series circuit. Its root mean square electric current isWith voltage as the zero-phase reference, the current phasor has complex argument $+57.858092^{\circ }$: electric current leads. The phasor drops have moduli $V_{R,\mathrm{rms}}=0.532018\mathrm{V}$ and $V_{C,\mathrm{rms}}=0.846733\mathrm{V}$; these add as complex numbers, not as positive magnitudes. Their squared magnitudes sum to $(1.00\mathrm{V}{)}^2$.

The mean absorbed power uses the root mean square values:The ideal capacitor has zero cycle-averaged absorbed power; all the mean absorbed power becomes Joule heating in the resistor.^[ref] To see why the phase factor appears, average $p(t)=v(t)i(t)$ over a full cycle with $v(t)=\sqrt{2}{V}_{\mathrm{rms}}\cos \omega t$ and $i(t)=\sqrt{2}{I}_{\mathrm{rms}}\cos (\omega t-\varphi )$. The cosine addition formula reduces the average to $V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$; the term at $2\omega$ averages to zero. Independently, $p_C=d(C{v}_C^2/2)/dt$, and the capacitor energy returns to its starting value after each cycle.

At fixed applied $V_{\mathrm{rms}}$, the limit $\omega \to 0^{+}$ gives $|Z|\to \infty$, $\varphi \to -\pi /2$ and $I_{\mathrm{rms}}\to 0$: the ideal capacitor blocks steady electric current. The limit $\omega \to \infty$ gives $Z\to R$, $\varphi \to 0$ and $I_{\mathrm{rms}}\to V_{\mathrm{rms}}/R$. These are limits of the ideal lumped electrical circuit; unwanted lead inductance and component losses can invalidate that model at high frequency. At $\omega RC=1$, the equal legs of the impedance right triangle give $|Z|=\sqrt{2}R$ and $\varphi =-\pi /4$ without numerical evaluation. Measuring both the magnitude and the sign of the phase difference distinguishes this capacitor response from an equally large positive electrical reactance.

Two triangles are similar if a Euclidean similarity maps one onto the other. Their corresponding angles agree, and corresponding side lengths have a common positive ratio.

Specify the correspondence first: $A$ matches $A^{\prime }$, $B$ matches $B^{\prime }$ and $C$ matches $C^{\prime }$. For similar triangles, the common scale factor $k>0$ satisfiesThe ratios compare matching sides in the same order. A Euclidean similarity can also move, turn or reflect the triangle; its orientation on the page is irrelevant. In the Euclidean plane, matching just two interior angles is enough: their sums force the third to match, and equal corresponding angles force the side ratios to agree. This is the angle-angle (AA) criterion for similar triangles. Matching just one angle is insufficient.

Scale the right triangle with vertices $(0,0)$, $(4,0)$ and $(0,3)$ by $k=2$. Its new vertices are $(0,0)$, $(8,0)$ and $(0,6)$, and its side lengths are $6,8,10$. The matching side ratios are $6/3=8/4=10/5=2$. Both triangles have a right angle and the same two other interior angles, approximately $36.87^{\circ }$ and $53.13^{\circ }$. However, their areas are $3\cdot 4/2=6$ and $6\cdot 8/2=24$: doubling lengths multiplies area by four.

Generally, a Euclidean similarity multiplies both a base and its perpendicular height by $k$, so the area becomes $\frac{1}{2}(kb)(kh)=k^2\frac{1}{2}bh$. This explains why an enlarged drawing needs four times as much surface when each length doubles. It also enables indirect measurement: under assumptions of level ground, vertical objects and parallel light rays, an object of height $3$ casting a shadow of length $4$ provides a right triangle. Another object's shadow of length $8$ gives a similar triangle and height $3(8/4)=6$. The shared angles justify the inference; visual resemblance alone does not.

Euclid's Elements VI, Definition 1 describes similarity using corresponding angles and proportional sides. VI.4 establishes the implication from matching angles to matching side ratios, while VI.19 states the squared relation between side ratios and areas. These are documented ancient formulations of the principles used above, without establishing a first discoverer.^{[ref][ref][ref]}

For a real angle $\theta$, $\sin \theta$ is the vertical coordinate on the unit circle reached by a counterclockwise rotation through $\theta$ radians. For an acute angle in a right triangle, it is the opposite side divided by the hypotenuse. In particular,The first-quarter rise and second-quarter fall are visible in the circle projection; the lower semicircle gives negative values.

Reflecting the unit circle across the horizontal axis gives $\sin (-\theta )=-\sin \theta$, so sine is an odd function. A full turn gives $\sin (\theta +2\pi )=\sin \theta$, making it a periodic function. For real arguments, its range is $[-1,1]$ and its zeros of a function occur at $\theta =k\pi$ for integers $k$. The cosine describes the perpendicular coordinate of the same point, and the Pythagorean trigonometric identity keeps their two values tied together.