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In the math puzzle book by Gardner, the maximum area ${\cal A}$ of the rectangle inscribed in the sector of unit-radius circle (see the Fig.) with angle $0 < \theta \le \pi/2$ has been asked to show as $${\cal A }=\frac{1-\cos \theta}{2\sin \theta}~~~~~~(1).$$ By considering point B critically on the arc along the angle bisector OB, we can prove the maximal result (1) as:

If O is the origin then $OD=AD \cot \theta$, $AD=BC=\sin(\theta/2)$, $OC=\cos(\theta/2)$. Max area ${\cal A}$ of the rectangle $${\cal A}= BC.DC=BC.[OC-OD]= \sin(\theta/2)[\cos(\theta/2)-\sin(\theta/2).\cot \theta]$$ $$=[\sin(\theta/2)\cos(\theta/2)-\sin^2(\theta/2)\cot(\theta)]=\frac{1}{2}[ \sin \theta-(1-\cos\theta) \cot \theta]$$ $$=\frac{1}{2}\left(\frac{\sin^2\theta-\cos\theta +\cos^2 \theta}{\sin\theta}\right)=\frac{1-\cos \theta}{2 \sin \theta},~~ 0<\theta \le \pi/2.$$ The question is as to what are other proofs of (1) given a fixed acute angle $\theta$.

Z Ahmed
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6 Answers6

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EDIT1:

As to your supplementary question about global, it was felt that a separate question may be better.

However it is clear that the area fraction Rectangle/ Sector does not change when we consider the rectangles symmetrical about previous sector's base. In other words, bisecting the sector into two parts and then again bisecting each half into two, making four equal pies furnishes the combination solution below needing no new proof.

$$ \dfrac{A_{rect 1}}{A_{sector 1}}+\dfrac{A_{rect 2}}{A_{sector 2}}=\dfrac{A_{rect}}{A_{sector}}$$

As the areas are symmetrically disposed, maximum fraction of area is ensured.

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Narasimham
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  • @Narasimhan You may see https://math.stackexchange.com/questions/3725856/global-max-of-the-area-of-a-rectangle-put-any-which-way-inside-a-unit-radius-acu – Z Ahmed Jun 19 '20 at 05:07
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Let $\angle BOC=u$

Area = $$\sin u ( \cos u- \sin u \cot \theta)$$

Differentiate with respect to $u$ and simplify

$$ \dfrac {\sin u }{ ( \cos u- \sin u \cot \theta) } =- \dfrac {\cos u}{\sin u+ \cos u \cot \theta} $$ $$ \cot \theta = \cot 2 u $$

$$ u=\dfrac{\theta}{2}$$ In other words $OB$ is the bisector of sector angle for maximum rectangle area. Since $\theta$ is assumed constant, it is bisector of any chosen sector angle.

Next for second query plug this into Area and simplify $$ \dfrac12 \tan (\theta/2)$$

which equals (1), answer is yes globally, meaning any acute angle chosen for sector.

Narasimham
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  • @Narsimhan Will the area a rectangle put inside the sector in a way other than discussed here also be less than or equal to${\cal A}$,for instance when B and C are on the arc and AD is not perpendicular tp OC? – Z Ahmed Jun 18 '20 at 13:42
  • It is a new question, sounds like a different goalpost. A condition that the sides are either vertical/ horizontal has been given by implication as understood by me and the others who responded so far. Imho it is better to ask this different question (when sides are free to tilt etc.) after closing this one. Hope this would be in order. – Narasimham Jun 18 '20 at 14:48
  • @Narsimhan OK, but by global i had meant this.So now I am going to remove this part in this question. – Z Ahmed Jun 18 '20 at 15:23
  • ok, way to go.. – Narasimham Jun 18 '20 at 17:40
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Let $\angle BOC=\phi$, $|AB|=|CD|=a(\phi),\ |BC|=|AD|=b(\phi)$.

Then

\begin{align} b(\phi)&=\sin\phi ,\\ a(\phi)&=\cos\phi-\sin\phi\cot\theta ,\\ S(\phi)&=a(\phi)b(\phi) =(\cos\phi-\sin\phi\cot\theta)\,\sin\phi ,\\ S'(\phi)&= \cos^2\phi-\sin^2\phi-2\sin\phi\cos\phi\,\cot\theta =\cos2\phi-\sin2\phi\,\cot\theta . \end{align}

$S'(\phi)=0$ results in

\begin{align} \cos2\phi&=\sin2\phi\,\cot\theta ,\\ \cot2\phi&=\cot\theta ,\\ \phi&=\tfrac12\,\theta ,\\ \text{for }\theta\in(0,\tfrac\pi2)\quad \max_{\phi\in(0,\theta)} S(\phi) &= S(\tfrac12\,\theta) =\frac{1-\cos\theta}{2\,\sin\theta} . \end{align}

g.kov
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  • @ g. kov Will the area a rectangle put inside the sector in a way other than discussed here also be less than or equal to${\cal A}$,for instance when B and C are on the arc and AD is not perpendicular tp OC? – Z Ahmed Jun 18 '20 at 13:43
  • @Dr Zafar Ahmed DSc: That's a really good question, it deserves to be asked separately. – g.kov Jun 18 '20 at 15:21
  • You may see https://math.stackexchange.com/questions/3725856/global-max-of-the-area-of-a-rectangle-put-any-which-way-inside-a-unit-radius-acu – Z Ahmed Jun 19 '20 at 05:04
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Let the unit-radius sector have measure $2\alpha$, and let the vertex of the inscribed rectangle determine an angle $\alpha-\phi$ (where $\phi$ could be negative), as shown:

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Then the area of the rectangle is given by

$$\begin{align} &\;\sin(\alpha-\phi)\left(\cos(\alpha-\phi)-\sin(\alpha-\phi)\cot2\alpha \right) \\[4pt] =&\;\frac{\sin(\alpha-\phi)}{\sin2\alpha}\left(\sin2\alpha\cos(\alpha-\phi)-\cos2\alpha\sin(\alpha-\phi)\right) \\[4pt] =&\;\frac{\sin(\alpha-\phi)\sin(\alpha+\phi)}{\sin2\alpha} \\[4pt] =&\;\frac{\cos 2\phi - \cos 2\alpha}{2\sin2\alpha} \end{align}$$ which is clearly maximized when $\phi=0$. $\square$

Blue
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Let the co-ordinate of A be (x,y) and then that of B is $(\sqrt{1-y^2},y)$ So the area of ABCD is given by $$a(y)=y(\sqrt{1-y^2}-y \cot \theta)~~~~(1)$$ Setting $a'(y)=0$ for max, we get $$a'(y)=\sqrt{1-y^2}+y\frac{-2y}{2\sqrt{1-y^2}}-2\cot \theta=0 \implies 1-2y^2=2y\sqrt{1-y^2} \cot \theta $$ $$\implies 4 \csc^2 \theta~ y^4-4 \csc^2 \theta ~y^2+1=0$$ Discarding $+$ sign in the root as it gives $y(0)=\ne 0$, we get $y^2=\frac{ 1- \cos \theta}{2}$, putting this in (1), we get $${\cal A}= \frac{1-\cos \theta}{2 \sin \theta}.$$

Z Ahmed
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Let $\alpha = \angle BOC$. Then, the area is

\begin{align} Area & = AD \cdot AB = \sin\alpha\cdot (\cos\alpha - \sin\alpha \cot \theta) \\ & = \frac{\cos(2\alpha -\theta)}{2\sin\theta} - \frac12\cot\theta\\ & \le \frac{1}{2\sin\theta} - \frac12\cot\theta = \frac{1-\cos\theta}{2\sin\theta} \end{align}

where the maximum area occurs at $\cos(2\alpha - \theta)=1$, or $\alpha = \frac12\theta$.

Quanto
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