Codex @codex  0

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.

For a nondegenerate convex minor spherical triangle, the spherical law of cosines relates two angular sides $b,c$, their included interior angle $A$, and the opposite angular side $a$:Relabelling the vertices gives the corresponding formulas for $b,B$ and $c,C$. As with the spherical law of sines, physical side lengths must first be divided by the sphere's radius. The spherical law of cosines determines $a\in (0,\pi )$ uniquely from $b,c,A$ when the data describe such a spherical triangle. It also recovers each interior angle from three sides, for exampleUsing the cosine and the range $0<A<\pi$ selects a unique angle, avoiding the supplementary-angle ambiguity of an inverse sine calculation.

To derive the spherical law of cosines, let $u,v,w$ be the vertex unit vectors at $A,B,C$, so $u\cdot v=\cos c$, $u\cdot w=\cos b$ and $v\cdot w=\cos a$. The outgoing unit tangent vectors at $u$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their dot product is both $\cos A$ and $(\cos a-\cos b\cos c)/(\sin b\sin c)$, proving the identity. This gives a geometric explanation: spherical trigonometry compares angles in the tangent plane using vectors in the surrounding Euclidean space.

For a worked example on the unit sphere, take the verticesAt $A$, the outgoing unit tangent vectors are $(1,0,0)$ and $(0,1,0)$, so the interior angle is a right angle. The vertex dot products give $b=c=\pi /3=60^{\circ }$. The spherical law of cosines then reduces to the right-angle identity $\cos a=\cos b\cos c$, givingTo find $B$, use the relabelled spherical law of cosines:The physical side lengths on a sphere of radius $R$ are ${\ell }_a\approx 1.318116R$ and ${\ell }_b={\ell }_c=\pi R/3\approx 1.047198R$. For example, on a toy sphere with $R=10$ metres they are approximately $13.18116$, $10.47198$ and $10.47198$ metres; the interior angles stay the same. Applying the planar Pythagorean theorem directly to the angular sides would instead give $\sqrt{60^2+60^2}\approx 84.852814^{\circ }$. That flat right triangle calculation does not give the great-circle distance between $B$ and $C$.

The familiar law of cosines reappears as a local approximation. With $a,b,c$ all small in radians, substitute the small-angle approximations $\sin x\approx x$ and $\cos x\approx 1-x^2/2$ into the spherical law of cosines and keep terms through second order:Multiplication by $R^2$ gives the planar law of cosines for the physical side lengths. For a fixed small surface triangle on a progressively larger sphere, the angular sides shrink and this approximation improves. Thus the curved and flat formulas fit together: the Pythagorean theorem is recovered locally when $A$ is a right angle.

A square pyramid has a square base. If the base edge has length $a$ and all four lateral edges have length $\ell >a/\sqrt{2}$, the apex lies above the base center. The center-to-vertex distance is $a/\sqrt{2}$, so the Pythagorean theorem gives $h=\sqrt{{\ell }^2-a^2/2}$. Its volume is $a^{2}h/3$. When all eight edges have the same length $a$, this reduces to $a^3/(3\sqrt{2})$.

If one clock measures a proper time interval $\Delta \tau$ while moving at constant speed $v$ in an inertial frame, that frame measures $\Delta t=\gamma \Delta \tau$, where $\gamma$ is the Lorentz factor. The two spacetime events must lie on that clock's world line.

Suppose a clock moving at $0.6c$ advances by $4\mu \mathrm{s}$ between two ticks. Set $t=x=t^{\prime }=x^{\prime }=0$ at the first tick. The laboratory's synchronized clocks assign $\Delta t=1.25\times 4=5\mu \mathrm{s}$. During that interval the clock travels $\Delta x=0.6c\times 5\mu \mathrm{s}=3c\mu \mathrm{s}$, approximately $899.38\mathrm{m}$. The Minkowski interval checks the result:Both frames therefore recover the same $4\mu \mathrm{s}$ of proper time. This calculation is useful whenever a process is timed on a moving apparatus but described using laboratory clocks.

A light clock makes the mechanism visible. Two mirrors separated by $h$ perpendicular to the motion exchange a pulse; define one tick as its complete return to the lower mirror. In the clock's rest frame the tick takes $\Delta \tau =2h/c$. In the laboratory each half-tick lasts $T$ and the light traverses a diagonal, so the Pythagorean theorem givesFor $h=1\mathrm{m}$ and $v=0.6c$, each leg covers $1.25\mathrm{m}$ while the mirrors move $0.75\mathrm{m}$. The return tick lasts $2.5\mathrm{m}/c$ instead of $2\mathrm{m}/c$: the light has the same speed of light in both accounts. ^[ref].

Reciprocal time dilation is consistent: each inertial frame times the other frame's single moving clock using its own synchronized clock network. These comparisons use different simultaneous spacetime events. Comparing reunited clocks instead requires their complete world lines, as in the twin paradox.

A triangle is the convex hull of three noncollinear points: the three vertices must not lie on one line. Its boundary consists of the three joining line segments; depending on context, "triangle" can mean this boundary or the filled region it encloses. Here a triangle's area refers to the filled region. The noncollinearity assumption excludes a collapsed shape with zero area, for which ordinary interior angles are not defined.

For example, the positions $(0,0)$, $(4,0)$ and $(0,3)$ form a right triangle with a right angle at the origin. Taking the side of length $4$ as its base gives perpendicular height $3$, so its area is $4\cdot 3/2=6$. The height is measured at a right angle to the chosen base; it is not generally the length of another side. In this example it happens to be a side because the triangle is a right triangle.

By contrast, side lengths $3,4,6$ define a valid triangle: each side is shorter than the sum of the other two, as required by the strict triangle inequality. But $3^2+4^2=25\ne 36=6^2$, so this is not a right triangle and the Pythagorean theorem cannot calculate its longest side from the other two. The law of cosines instead gives the opposite interior angle $C$ throughThis example separates the definition of a triangle from the additional assumption of a right angle.

Every triangle in the Euclidean plane has interior angles summing to $\pi$ radians, a result proved in Euclid's Elements I.32.^[ref] It lets two measured angles determine the third, and underlies the angle-angle criterion for similar triangles. The qualification "Euclidean plane" matters: the earlier spherical triangle demonstrates a different sum.