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Question Number 37804 by ajfour last updated on 17/Jun/18

Commented by ajfour last updated on 17/Jun/18

Find maximum area of quadrilateral MNOP in terms of radius R of circle. (O is the center of circle).

$${Find}\:{maximum}\:{area}\:{of} \\ $$$${quadrilateral}\:{MNOP}\:{in}\:{terms} \\ $$$${of}\:{radius}\:\boldsymbol{{R}}\:{of}\:{circle}. \\ $$$$\left({O}\:{is}\:{the}\:{center}\:{of}\:{circle}\right). \\ $$

Answered by MrW3 last updated on 18/Jun/18

Commented by MrW3 last updated on 18/Jun/18

OQ=R sin ϕ, QD=R cos ϕ OR=R sin θ, RA=R cos θ ((RM)/(RA))=((MQ)/(QD)) ((RM)/(RA))=((RQ−RM)/(QD)) ((RM)/(RA))=((OR+OQ−RM)/(QD)) ((RM)/(R cos θ))=((R(sin θ+sin ϕ)−RM)/(R cos ϕ)) (cos θ+cos ϕ) RM=cos θ (sin θ+sin ϕ)R ⇒RM=((cos θ (sin θ+sin ϕ)R)/((cos θ+cos ϕ))) OM=OR−RM=R sin θ−((cos θ (sin θ+sin ϕ)R)/((cos θ+cos ϕ))) OM=R ((sin θ cos θ+ sin θ cos ϕ−cos θ sin θ−cos θ sin ϕ)/((cos θ+cos ϕ))) ⇒OM= (( sin (θ−ϕ) R)/((cos θ+cos ϕ))) ...

$${OQ}={R}\:\mathrm{sin}\:\varphi,\:{QD}={R}\:\mathrm{cos}\:\varphi \\ $$$${OR}={R}\:\mathrm{sin}\:\theta,\:{RA}={R}\:\mathrm{cos}\:\theta \\ $$$$\frac{{RM}}{{RA}}=\frac{{MQ}}{{QD}} \\ $$$$\frac{{RM}}{{RA}}=\frac{{RQ}−{RM}}{{QD}} \\ $$$$\frac{{RM}}{{RA}}=\frac{{OR}+{OQ}−{RM}}{{QD}} \\ $$$$\frac{{RM}}{{R}\:\mathrm{cos}\:\theta}=\frac{{R}\left(\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi\right)−{RM}}{{R}\:\mathrm{cos}\:\varphi} \\ $$$$\left(\mathrm{cos}\:\theta+\mathrm{cos}\:\varphi\right)\:{RM}=\mathrm{cos}\:\theta\:\left(\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi\right){R} \\ $$$$\Rightarrow{RM}=\frac{\mathrm{cos}\:\theta\:\left(\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi\right){R}}{\left(\mathrm{cos}\:\theta+\mathrm{cos}\:\varphi\right)} \\ $$$${OM}={OR}−{RM}={R}\:\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta\:\left(\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi\right){R}}{\left(\mathrm{cos}\:\theta+\mathrm{cos}\:\varphi\right)} \\ $$$${OM}={R}\:\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta+\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\varphi−\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\:\mathrm{sin}\:\varphi}{\left(\mathrm{cos}\:\theta+\mathrm{cos}\:\varphi\right)} \\ $$$$\Rightarrow{OM}=\:\frac{\:\mathrm{sin}\:\left(\theta−\varphi\right)\:{R}}{\left(\mathrm{cos}\:\theta+\mathrm{cos}\:\varphi\right)} \\ $$$$ \\ $$$$... \\ $$

Commented by ajfour last updated on 18/Jun/18

Sir, will it be simply A_(max) = R^( 2) ?

$${Sir},\:{will}\:{it}\:{be}\:{simply}\:{A}_{{max}} =\:{R}^{\:\mathrm{2}} \:\:? \\ $$

Answered by MJS last updated on 19/Jun/18

A and B must be above C and D A= (((r(√(1−p^2 )))),((rp)) ) B= (((−r(√(1−p^2 )))),((rp)) ) C= (((−r(√(1−q^2 )))),((rq)) ) D= (((r(√(1−q^2 )))),((rq)) ) 0≤p<rp −p<q≤p (think it over, it′s easy to see why) ⇒ if C=B=N and D=A=P we get a triangle with NMP as base ⇒ area(ABO) must be maximal. so the problem is reduced to: find triangle of greatest area in a half circle ⇒ solution: p=q=((√2)/2), maximal area=(r^2 /2)

$${A}\:\mathrm{and}\:{B}\:\mathrm{must}\:\mathrm{be}\:\mathrm{above}\:{C}\:\mathrm{and}\:{D} \\ $$$${A}=\begin{pmatrix}{{r}\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }}\\{{rp}}\end{pmatrix}\:{B}=\begin{pmatrix}{−{r}\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }}\\{{rp}}\end{pmatrix} \\ $$$${C}=\begin{pmatrix}{−{r}\sqrt{\mathrm{1}−{q}^{\mathrm{2}} }}\\{{rq}}\end{pmatrix}\:{D}=\begin{pmatrix}{{r}\sqrt{\mathrm{1}−{q}^{\mathrm{2}} }}\\{{rq}}\end{pmatrix} \\ $$$$\mathrm{0}\leqslant{p}<{rp} \\ $$$$−{p}<{q}\leqslant{p} \\ $$$$\left(\mathrm{think}\:\mathrm{it}\:\mathrm{over},\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{why}\right) \\ $$$$\Rightarrow\:\mathrm{if}\:{C}={B}={N}\:\mathrm{and}\:{D}={A}={P}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:{NMP}\:\:\mathrm{as}\:\mathrm{base}\:\Rightarrow\:\mathrm{area}\left({ABO}\right)\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{maximal}.\:\mathrm{so}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{reduced}\:\mathrm{to}: \\ $$$$\mathrm{find}\:\mathrm{triangle}\:\mathrm{of}\:\mathrm{greatest}\:\mathrm{area}\:\mathrm{in}\:\mathrm{a}\:\mathrm{half}\:\mathrm{circle} \\ $$$$\Rightarrow\:\mathrm{solution}:\:{p}={q}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\:\mathrm{maximal}\:\mathrm{area}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}} \\ $$

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