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A circular arc is a connected portion of a circle. For radius $R>0$ and subtended central angle $0<\theta <2\pi$ measured in radians, its arc length is $R\theta$. Two distinct endpoints determine two circular arcs: when they are not opposite, the minor arc has $\theta <\pi$ and the major arc has $\theta >\pi$. Opposite endpoints determine two semicircles. The straight line segment joining the endpoints is a chord, rather than the curved circular arc itself.

For a single-valued curve below the chord joining $(-a,0)$ to $(a,0)$, with length $2a<L\le \pi a$, the maximum enclosed area is attained by the lower minor circular arc. Equivalently it minimizes ${\int }_{-a}^{a}ydx$ at fixed length. Its equation and length areTo prove global minimality for $R>a$, augment the objective by $R\int \sqrt{1+y^{\prime 2}}dx$. The augmented integrand is linear in $y$ and strictly convex in $y^{\prime }$. The displayed arc satisfies its Euler-Lagrange equation, so the supporting-line inequality for the convex slope dependence, followed by integration by parts, proves it minimizes the augmented functional among curves with the prescribed endpoints. On the fixed-length class the multiplier term is constant. The semicircle endpoint follows by a limit; its endpoint tangents are vertical but its length is finite.

A circular segment is the part of a disk bounded by a chord and one of the two circular arcs joining its endpoints. For radius $R$ and the chosen arc's central angle $0\le \varphi \le 2\pi$ in radians, its area is $R^2(\varphi -\sin \varphi )/2$. Subtract the signed central triangle of area $R^2\sin \varphi /2$ from the circular sector of area $R^2\varphi /2$. Angles below $\pi$ give the smaller segment and angles above $\pi$ the larger one.

Choose a circular fillet that removes the sharp convex tip at $B$. The unit directions from $B$ toward $A$ and $C$ are $(-\sqrt{3}/2,-1/2)$ and $(-1,0)$. Therefore the prescribed tangent points areThe circle centre $K$ lies on the normal line to $BC$ through $N$, so $K=(3/2,k)$. Its radius to $M$ must be perpendicular to $AB$, whose direction vector is $(\sqrt{3},1)$. ThusConsequentlyIndeed $|M-K|=|N-K|=r$. The interior angle $ABC$ is $30^{\circ }$, and the same radius follows from $r=\frac{1}{2}\tan 15^{\circ }$. The selected circular arc runs from $M$ to $N$ through angles $-60^{\circ }\le \theta \le 90^{\circ }$ around $K$:The $150^{\circ }$ circular arc facing $B$ is the rounding; the complementary arc is not part of the boundary.

Define the disk and a half-plane through the chord $MN$ byBoth $M,N$ satisfy $g=0$, and $g(B)=1/2>0$. The clipping half-space $H_g$ retains the part away from $B$, while the disk restores the rounded cap. The required local constructive solid geometry tree isOnly the portion of $P$ on the $B$ side of $MN$ can be removed. To see the locality, $g$ is negative at every original vertex of a polygon except $B$, and vanishes on the two adjacent sides exactly at $M,N$. In the resulting triangle $BMN$, the disk keeps the curved cap next to $MN$ and discards the tip at $B$. A global set intersection with $D_r$ alone would destroy most of the seat. The outward normal vector on the new boundary is $\nabla f_c=2((x,y)-K)$; at $M,N$ it has the same direction as the outward normal vectors of $AB,BC$.

The tangent constraints alone do not insist on a circle; other conic sections can give different bulges. For example, a positive-weight rational quadratic Bézier curve with handles $M,B,N$ has the required tangent directions:Varying $w$ changes the bulge while fixing the tangent points; the circular choice has $w=\cos 75^{\circ }$. Once a circular fillet and these two tangent points are chosen, its centre and radius have no remaining freedom. The half-plane sign and arc selection are also essential: reversing them chooses a different retained region. This construction gives matching tangent directions, or first-order geometric continuity, at $M,N$; the curvature jumps from zero on each straight side to $1/r$ on the circle.

Insert the union subtree from part (f) as another child of the original intersection:Thus the constructive solid geometry tree has one intersection root, four original half-space leaves and two union children. The first union handles the inward corner at $C$; the second embeds the circular fillet at $B$. The line segments $AB$ and $BC$ now end at $M$ and start at $N$, with the chosen circular arc between them. All other boundary line segments are retained.

The right-handed material frame of a rod has ${\mathbf{d}}_2\times {\mathbf{d}}_1=-{\mathbf{d}}_3$ and ${\mathbf{d}}_2\times {\mathbf{d}}_3={\mathbf{d}}_1$. Hence the director kinematics of a rod reduce toSolving these constant-coefficient ordinary differential equations with the given initial frame givesThe inextensible rod has ${\mathbf{r}}^{\prime }={\mathbf{d}}_3$. An integration using $\mathbf{r}(0)=0$ yields, for $\hat{u}\ne 0$,Indeed,so the centreline is a circular arc of radius $1/|\hat{u}|$ in the plane perpendicular to ${\mathbf{e}}_2$. The continuous $\hat{u}\to 0$ limit is ${\mathbf{d}}_1={\mathbf{e}}_1$, ${\mathbf{d}}_3={\mathbf{e}}_3$, and $\mathbf{r}=S{\mathbf{e}}_3$.

The endpoint constraints are $y(-a)=y(a)=0$. The constants in the three costs contribute only $2aK$, $2a{K}^{\prime }$ and $2a{K}^{\prime \prime }$ respectively, so they do not affect the minimizing shapes. For companies B and C, the fixed arc length isThe original PDF has $(y^{\prime }{)}^2$ as company A's first variable term. The local TeX transcription has lost that derivative dot and incorrectly repeats $y^2$; the calculation below follows the PDF.

Company A. The relevant functional is $J_A=\int (y^{\prime 2}+y^2-2xy)dx$. The Euler-Lagrange equation givesIts general solution is $y=x+c_1\cosh x+c_2\sinh x$. Adding and subtracting the endpoint equations gives $c_1=0$ and $c_2=-a/\sinh a$. ThusThis curve is odd, negative on the left half and positive on the right half; company A has no lower-half-plane restriction. It is the unique global minimum. Indeed, for any endpoint-zero variation $\eta$, integration by parts and the differential equation cancel all linear terms, leavingunless $\eta =0$.

Company B. Minimize $\int ydx$ under the length constraint. Introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $y+\lambda \sqrt{1+y^{\prime 2}}$:The curvature is constant, so the required lower arc is a circular arc. Writing its radius as $R\ge a$ and imposing the endpoints givesSince $\sqrt{1+y_B^{\prime 2}}=R/\sqrt{R^2-x^2}$, its length determines the radius:Equivalently, introduce $0<\theta \le \pi /2$ with $a=R\sin \theta$. Then $L/(2a)=\theta /\sin \theta$, which increases strictly from $1$ to $\pi /2$. The stipulated range of $L$ therefore selects one minor arc. At $L=\pi a$ it becomes the lower semicircle, with vertical endpoint tangents; it is still a single-valued finite-length graph.

Global minimality follows from the convexity of the augmented slope cost with $\lambda =R$. For a competitor $y=y_B+\eta$,Consequently the augmented cost difference is at least $\int (\eta +x{\eta }^{\prime })dx=[x\eta {]}_{-a}^a=0$. Competitors have the same length, so the multiplier contribution cancels and the original cost is minimized. This proves that a circular arc maximizes area below a fixed chord: minimizing negative signed height maximizes the area between the fence and the chord.

Company C. The variable cost is $J_C=\int yds=\int y\sqrt{1+y^{\prime 2}}dx$. The augmented integrand is $(y+\lambda )\sqrt{1+y^{\prime 2}}$. Since it has no explicit $x$ dependence, the Beltrami identity givesThe hanging branch with $b>0$ integrates to $y+\lambda =b\cosh [(x-x_0)/b]$. Equal endpoint heights force $x_0=0$, and the endpoint value fixes $\lambda =b\cosh (a/b)$. ThusIts length isWriting $q=a/b$, the equation $L/(2a)=\sinh q/q$ has one positive solution for every $L>2a$: its derivative has numerator $q\cosh q-\sinh q>0$. The curve is a catenary, is even, and lies strictly below the chord except at the endpoints.

A direct argument proves this is a global minimum, rather than merely an extremum of an augmented functional. Parametrize any competitor by arc length $s\in [-L/2,L/2]$, from its left endpoint to its right endpoint. Then $x^{\prime 2}+y^{\prime 2}=1$, $\int x^{\prime }ds=2a$, and integration by parts givesThe Cauchy-Schwarz inequality gives the last step. Equality requires $x^{\prime }=b/\sqrt{b^2+s^2}$ and $y^{\prime }=s/\sqrt{b^2+s^2}$, whose integrals are precisely the displayed catenary with the stated length. Thus it attains a lower bound valid for every competitor. This proves that a catenary minimizes the integral of height along a fixed-length curve.

The sketch uses $a=1$ and $L=2.7$ for the two fixed-length cases. The vertical scale in the company A panel is enlarged to show its small odd deflection. No comparison between the three companies' absolute prices is required or possible from the shapes alone, since their additive cost constants are independent.

We first show that a continuous injection $j:S^1\to S^1$ must be surjective. Otherwise its image misses a point $q$. Stereographic projection identifies $S^1\setminus \{q\}$ with $\mathbb{R}$, giving a continuous injection $u:S^1\to \mathbb{R}$. The extreme value theorem supplies distinct points $p_{-},p_{+}$ where $u$ attains its minimum $m$ and maximum $M$, with $m<M$ because $u$ is injective.

The two circular arcs from $p_{-}$ to $p_{+}$ each contain a point at which $u=(m+M)/2$, by the intermediate value theorem. Those points are distinct because they lie in the disjoint interiors of the two arcs. This contradicts injectivity, proving the claim about $j$.

Now suppose $F:S^2\to S^1$ were a continuous injection. Its restriction to the equatorial circle $E=\{(x,y,0):x^2+y^2=1\}$ would be a continuous injection from $S^1$ to itself, hence $F(E)=S^1$. The north pole $N=(0,0,1)$ is outside $E$, but $F(N)$ must equal $F(e)$ for some $e\in E$. This contradicts injectivity. Therefore

For the upper function graph, the vertical cross-section of the enclosed region has length $y(x)$, so its area is $J[y]={\int }_0^{2a}y(x)dx$. A maximizing nonnegative arch therefore uses $F=y$ and $G=\sqrt{1+y^{\prime 2}}$ in the normal Euler-Lagrange equation. It becomesIntegrating once gives $y^{\prime }/\sqrt{1+y^{\prime 2}}=(c-x)/\lambda$. The concave upper branch has $\lambda >0$ and integrates to $y=\sqrt{{\lambda }^2-(x-c{)}^2}+d$. The equal zero endpoint heights force $c=a$ and $d=-\sqrt{{\lambda }^2-a^2}$. HenceIt is a circular arc. To find its parameter, substitute into the length integral: $\sqrt{1+y^{\prime 2}}=\lambda /\sqrt{{\lambda }^2-(x-a{)}^2}$, givingThis scalar equation determines $\lambda$ numerically when needed; no explicit solution is required.

The arch is a global maximum in its smooth function graph regime. For $\lambda >0$, the integrand $y-\lambda \sqrt{1+y^{\prime 2}}$ is concave and strictly concave in its slope. Its supporting-tangent inequality, integrated against any competing function graph with the same endpoint values, has first-order term zero by the Euler-Lagrange equation. Thus $J[y]-\lambda I[y]\le J[y_{\ast }]-\lambda I[y_{\ast }]$. On the fixed-length class this is $J[y]\le J[y_{\ast }]$; strict slope concavity also gives uniqueness.

The source leaves the length range implicit. A twice continuously differentiable function graph of this form needs $\lambda >a$, giving $2a<l<\pi a$. At $l=2a$ the straight function graph is the degenerate solution. The semicircle limit $l=\pi a$ has vertical endpoint tangents and is outside that smooth function graph class. For $l>\pi a$, no arc in the claimed smooth function graph family exists, so the stated formula cannot describe a twice continuously differentiable maximizing function graph under those unrestricted data. Allowing overhanging curves changes the admissible class.

For the ray sketches in (i) and (ii), take a horizontally layered isotropic medium, a surface source and receiver, depth $z$ positive downwards, and wave speed $v(z)$. Snell's law conserves $p=\sin i/v$, where $i$ is measured from the vertical. A returning diving ray turns where $pv(z_t)=1$, withFor a uniform positive gradient $v=v_s+\beta z$, rays are circular arcs andThus ordinary travel time increases with distance but has a decreasing slope as deeper rays sample faster material. The following two leaves discuss the labelled changes to this reference structure.

For the OR alternative, the supplied fault-plane diagrams, assume the usual convention that shaded quadrants are compressional first motions on a lower-hemisphere projection. Each pair of nodal planes represents a fault plane and an auxiliary plane; first-motion polarity alone cannot choose which slipped. Geological mapping, aftershock locations or rupture observations can resolve that ambiguity.^[ref]

Diagram (i): normal faulting. The central unshaded band indicates the extensional dip-slip pattern. It is compatible with crustal extension, such as a continental rift or a spreading environment, with a hanging wall moving down relative to the footwall on the actual fault. Its two curved nodal traces are the possible fault and auxiliary planes.

Diagram (ii): strike-slip faulting. The two straight nodal diameters divide four alternating quadrants, as expected for two vertical candidate planes. It is compatible with a transform plate boundary or continental shear zone. One candidate is right lateral and the conjugate is left lateral; polarity alone does not select the actual plane. Without a north arrow, geographical strikes cannot be assigned.

Diagram (iii): oblique-slip faulting with a reverse component. The off-centre nodal intersection and curved traces distinguish it from pure vertical strike slip, while shading through the centre indicates the reverse component under the stated convention. It is compatible with transpression, such as oblique convergence at a plate margin or a compressional strike-slip bend. Both strike-slip and dip-slip components are required; it is not merely a rotated pure strike-slip pattern. Exact strike, dip, rake, and slip-plane selection cannot be recovered numerically from this uncalibrated printed sketch. Reversing the shading convention would reverse the normal/reverse interpretation, so the convention is essential.

For the specified functional,Since $y>0$, the Beltrami identity gives $y\sqrt{1+y^{\prime 2}}=R$ for some $R>0$. This suggests a circular arc, but it is worth deriving its equation without dividing by $y^{\prime }$: such division would lose the point where the arc is horizontal.

Substitute the displayed partial derivative and $F_y=-\sqrt{1+y^{\prime 2}}/y^2$ in the full Euler-Lagrange equation. Combining terms over the common denominator givesThus $(y{y}^{\prime }{)}^{\prime }=-1$, so $y{y}^{\prime }=h-x$ and a second integration givesThis establishes the semicircular extremals of the hyperbolic length functional and excludes constant $y$, which would satisfy the first-order Beltrami identity for a suitable $R$ but not the full Euler-Lagrange equation.

The two prescribed endpoints giveSubtracting yields $4h=4(d^2-c^2)$. ThereforeIn particular the centre is $\boxed{(d^2-c^2,0)}$. Admissibility requires $c,d>0$, as the prescribed heights must lie in $y>0$. Under that condition the radicand is positive throughout $[-1,1]$: it is a concave quadratic polynomial whose two endpoint values are $4c^2$ and $4d^2$. The positive root consequently gives one smooth admissible arc, uniquely determined by the endpoints.

When an endpoint's horizontal coordinate is fixed but its height is free, the natural boundary conditions for a free endpoint require $F_{y^{\prime }}=0$ there. In this problem,Every admissible extremal is an upper circular arc with centre on $y=0$, by part (b). Its tangent is horizontal precisely above its centre. Thus imposing the natural boundary conditions for a free endpoint at $x=1$ forces $h=1$. The fixed height at $x=-1$ gives $R^2=4+4c^2$, soIts free endpoint height is $y(1)=2\sqrt{1+c^2}$. Positivity on the whole interval follows because the radicand is at least $4c^2$.

There can be no solution satisfying the natural boundary conditions for a free endpoint at both ends: an upper circular arc has only one point with a horizontal tangent, whereas the two endpoints have distinct horizontal coordinates. Therefore

A weighted quadratic Bézier curve with positive middle weight $w$ retains endpoint interpolation and endpoint tangent directions while changing the bulge. For a circular arc of angle $\theta$ less than $180^{\circ }$, choose endpoints on the circle, their tangent intersection as the middle control point, and $w=\cos (\theta /2)$. Unlike an ordinary nonconstant polynomial Bézier curve, this rational representation can describe a circular arc exactly.

For the minor circular arc cut off by a chord of length $0\le L\le 2R$ on a circle of radius $R>0$, the sagitta is the maximum perpendicular distance from the chord to that arc:The perpendicular bisector of the chord passes through its midpoint and the center. The Pythagorean theorem gives the center-to-chord distance $\sqrt{R^2-(L/2{)}^2}$; subtracting it from $R$ proves the formula. The major circular arc instead gives $R+\sqrt{R^2-L^2/4}$, so choosing the arc matters.

For $L/R\ll 1$, a Taylor expansion gives $s=L^2/(8R)+O(L^4/R^3)$. In a uniform magnetic field perpendicular to a charged track's motion, $R=p/(|q|B)$, so the sagitta measures the momentum $p$.

For positive twice differentiable functions $y$, the Euler-Lagrange equation for this functional reduces to $y{y}^{\prime \prime }+1+y^{\prime 2}=0$. Thus $(y{y}^{\prime }{)}^{\prime }=-1$ and integration gives $(x-h{)}^2+y^2=R^2$. The upper circular arc is a graph wherever its radicand is positive. A fixed horizontal endpoint with free height imposes $F_{y^{\prime }}=y^{\prime }/(y\sqrt{1+y^{\prime 2}})=0$, so that endpoint must lie above the centre. Distinct horizontal endpoints cannot both satisfy this natural boundary conditions for a free endpoint. The Beltrami identity alone also admits constant functions, which are excluded by the full Euler-Lagrange equation.

A circular arch of chord length $2a$ described as a smooth single-valued graph has radius $R>a$ and length $2R\arcsin (a/R)$, ranging strictly between $2a$ and $\pi a$. The straight-line limit has length $2a$. At the semicircle limit the endpoint slopes are infinite. Longer circular arcs require overhangs and cannot be graphs in this class.