With $t$ as in i, the foot of a perpendicular lies strictly between the endpoints precisely when $0<t<1$. Since the denominator is positive, this givesEquivalently, both dot productsare positive: the angles of the triangle at $A$ and $B$ are both less than $\pi /2$. If “between” includes the endpoints, replace both strict inequalities by non-strict ones; equality gives $P=A$ or $P=B$.

Write $\mathbf{u}=\mathbf{b}-\mathbf{a}$. Since the triangle is nondegenerate, $\mathbf{u}\ne 0$. Every point on the line $AB$ has a unique position vector $\mathbf{p}=\mathbf{a}+t\mathbf{u}$. The condition that $CP$ is orthogonal to $AB$ is the dot product equationIts coefficient of $t$ is $-\Vert \mathbf{u}{\Vert }^2\ne 0$, so it has exactly one solution. The foot of a perpendicular is thereforeThis is the orthogonal projection of $C$ onto the line $AB$, expressed using the Euclidean norm.

Back-substitute in the Euclidean algorithm to obtain an integer linear combination:Multiplying by $3$ gives the particular solution $(x_0,y_0)=(-45,21)$ of the linear Diophantine equation. Subtracting it from any other solution and dividing by $9$ givesThe integers $34$ and $73$ are coprime, so $73$ divides $x+45$. Writing $x+45=73t$ forces $y-21=-34t$. Conversely these expressions satisfy the original equation for every integer $t$. Thus all solutions of a linear Diophantine equation here are

The successive division with remainder steps of the Euclidean algorithm areEach step preserves the greatest common divisor by part (a). The final nonzero remainder is $9$, so

The enclosed area of the ellipse is

The eccentricity of the ellipse is

A possible center $\mathbf{c}$ must be a fixed point, so $(I-R_{\theta })\mathbf{c}=\mathbf{b}$. Let $c=\cos \theta$ and $s=\sin \theta$. The determinant iswhen $\theta$ is not an integer multiple of $2\pi$. The unique center is thereforeThe fixed point equation implies $\mathbf{b}=\mathbf{c}-R_{\theta }\mathbf{c}$, whenceRelative to $\mathbf{c}$, this is precisely a planar rotation through $\theta$. This also explains the excluded case: when $\theta \in 2\pi \mathbb{Z}$, $T$ is a translation, with no fixed point unless $\mathbf{b}=0$.

Write $T_{\theta ,\mathbf{b}}(\mathbf{v})=R_{\theta }\mathbf{v}+\mathbf{b}$. The sine addition formula and cosine addition formula give $R_{\theta }{R}_{\psi }=R_{\theta +\psi }$. Therefore function composition satisfiesThe resulting angle is real and the new translation vector lies in ${\mathbb{R}}^2$, so $G$ is closed under function composition. Associativity holds because, for any functions $S,T,U$ and any $\mathbf{v}$, both $(S\circ T)\circ U$ and $S\circ (T\circ U)$ take $\mathbf{v}$ to $S(T(U(\mathbf{v})))$.

The identity element is $T_{0,0}$, since $R_0=I$. Every element has the two-sided inverse elementIndeed $R_{\theta }{R}_{-\theta }=R_{-\theta }{R}_{\theta }=I$, and substituting into the composition formula makes the translation vector zero in either order. These verify all the group axioms. Thus $G$ is a group under function composition, the special Euclidean group in two dimensions.

Use the convention $[\mathbf{a},\mathbf{b},\mathbf{c}]=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ for the scalar triple product. With $\mathbf{q},\mathbf{w}$ as above, $\mathbf{q}=\mathbf{n}\times \mathbf{w}$. The vector triple product identity givesThe Lagrange identity for the cross product, $|\mathbf{a}\times \mathbf{b}{|}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2-(\mathbf{a}\cdot \mathbf{b}{)}^2$, also givesTaking the dot product with the unit vector $\mathbf{n}$ proves

Use the vector triple product identitySince $\mathbf{n}$ is a unit vector, putThus $\mathbf{w}$ is the component of $\mathbf{v}$ in the orthogonal complement of $\mathbf{n}$, and $\mathbf{n}\cdot \mathbf{w}=0$. A cross product is orthogonal to both its factors, so $\mathbf{n}\cdot \mathbf{q}=0$ and $\mathbf{v}\cdot \mathbf{q}=0$. ConsequentlyAll three pairs are orthogonal. This remains true when $\mathbf{v}$ is parallel to $\mathbf{n}$: then $\mathbf{q}$ and $\mathbf{w}$ are zero vectors. Pairwise orthogonality does not assert linear independence in this case.

For $a<b$, choose the uniform partition of an interval into $n$ pieces and its midpoint tagsTheir arithmetic mean is exactly $(a+b)/2$. Apply the given Jensen inequality with equal weights ${\lambda }_i=1/n$:Multiplying by $b-a$ gives a midpoint Riemann sum on the left. Its mesh of a partition is $(b-a)/n\to 0$, so the Riemann sum convergence theorem givesFor $a=b$ the equality is immediate. Together with part i, this is the Hermite-Hadamard inequality with the directions appropriate to a concave function.

The stated weighted inequality is the Jensen inequality for a concave function. Assume $a<b$; if $a=b$, both sides of the required inequality are zero. Write $x=\lambda a+(1-\lambda )b$ with $\lambda =(b-x)/(b-a)$. Applying the weighted inequality at the two endpoints givesThus the affine function $\ell$ joining the endpoint values lies below $f$. By monotonicity of the Riemann integral,This is the lower bound in the Hermite-Hadamard inequality for a concave function.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is

SetThe points strictly increase from zero to $x_{n,n}=n/(n+1)$. With $f(t)=\sin t$, the given expression is exactly the midpoint Riemann sumfor the partition of an interval ${\mathcal{P}}_n=(x_{0,n},\dots ,x_{n,n})$ of $[0,n/(n+1)]$. The sine argument is precisely the midpoint tag, since it is half the sum of the consecutive endpoints.

For a fixed integration interval $[0,1]$, append $x_{n+1,n}=1$ and choose the final tag ${\xi }_{n,n}=1$. The resulting Riemann sum $T_n$ satisfiesThis makes explicit the small final interval that the original sum omits.

A maximum or minimum can occur at an endpoint. The rule that a derivative vanishes at an extremum requires an interior extremum and differentiability there. The hypotheses give no two-sided derivative at the endpoints, and even an existing one-sided derivative need not vanish there.

For example, $f(x)=x$ on $[0,1]$ has $m=0$ at $x_m=0$ and $M=1$ at $x_M=1$, but $f^{\prime }(x)=1$ at every interior point; its endpoint one-sided derivatives are also one. Its graph meets both horizontal bounding lines without being tangent to either. Thus the claimed conclusion about two zeros of the derivative fails.

There is a second logical issue: the argument never establishes two distinct extremizing points. For a constant function, the maximum and minimum coincide and may be represented by the same point. The constant function does have many stationary points, but that needs its own argument; it does not follow from counting the same selected point twice. In Rolle's theorem, equal endpoint values ensure an interior extremum when the function is nonconstant, which is the missing mechanism here.

The extreme value theorem gives finite real numbersSince these values are attained, the set of strict upper bounds in the definition of $M$ is precisely $(\beta ,\infty )$, and the set of strict lower bounds in the definition of $m$ is precisely $(-\infty ,\alpha )$. They are nonempty and bounded on the required sides. By the definitions of infimum and supremum,

We give the explicit point $\boxed{x=1/4}$. Repeatedly choose the left remaining third and then the right remaining third. After $2m$ choices, the resulting closed interval is $[L_m,R_m]$, whereThese formulas follow from a finite geometric series. They show $L_m<1/4<R_m$, and that both endpoints tend to $1/4$. The point $1/4$ is also strictly inside each intermediate left third: its distance from $L_m$ is one quarter of the current length. It therefore never enters any assigned closed middle third or its endpoints, so $1/4\in C$.

Endpoint values in the Cantor function construction obey the same left/right choices, with halving in place of thirding. ThusAll these endpoints are in the domain of the partial function: $L_m$ is the right endpoint of a middle third assigned at round $2m$, and $R_m$ is the left endpoint of a middle third assigned at round $2m-1$.

Given $\epsilon >0$, choose $m$ so that $(2/3)4^{-m}<\epsilon$, and put $\delta =9^{-m}/4$. If $t\notin C$ and $|t-1/4|<\delta$, then $L_m<t<R_m$. The assumed monotonicity givesConsequently $|f(t)-1/3|<\epsilon$. This proves directly, on the stated partial domain,There are defined points arbitrarily close to $1/4$, for example $L_m$ and $R_m$, so the limit of a function is not vacuous. No continuous extension has been assumed. The residual $C$ here omits the endpoints that the question assigns values to; it is the usual Cantor set with those assigned endpoints removed.

The construction is the partial middle-third construction of the Cantor function. After $n$ rounds of assigning middle thirds, every remaining open interval has endpoint valuesfor some $k=0,\dots ,2^n-1$, and every such consecutive pair occurs. This follows by mathematical induction. Initially the endpoint values are $0,1$. An interval with those consecutive endpoint values receives the valueon its closed middle third. Its two remaining open intervals therefore have consecutive endpoint values at denominator $2^{n+1}$. All previously assigned endpoint values persist. Hence all even numerators at the next denominator were already present, and all odd numerators have just been assigned. Including the initial values zero and one provesThese are precisely the dyadic rational values between zero and one.

An explicit piecewise linear function isThe adjacent formulas agree at $1/3$ and $2/3$, so the function is continuous, and $f(0)=f(1)=0$. Any pair of points at distance $2/3$ can be written $t,t+2/3$ with $0\le t\le 1/3$. For every such $t$,Thus their values are never equal. The horizontal chord of reciprocal-integer length conclusion does not extend to this distance.

For $n=1$, choose $x=0$, $y=1$. For $n\ge 2$, define the continuous functionIts values at the grid points have the telescoping sumIf one summand is zero, we already have a zero of $g$. Otherwise the summands cannot all have the same sign, so two of these values have opposite signs. The intermediate value theorem between their arguments gives $t\in [0,1-1/n]$ with $g(t)=0$. Set $x=t$, $y=t+1/n$. ThenThis is the horizontal chord of reciprocal-integer length property; the telescoping sum explains why reciprocal integers appear.