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Question Number 64545 by ajfour last updated on 19/Jul/19
Commented by ajfour last updated on 19/Jul/19
Commented by aliesam last updated on 19/Jul/19
Answered by ajfour last updated on 19/Jul/19
let vertex angle be 2θ. ⇒ nφ=n((π/2)−θ)=2π ⇒ θ=(π/2)−((2π)/n) R=(1/(sin θ)) = (1/(cos (((2π)/n)))) This is clearly wrong, but i dont understand, how it came wrong; mrW Sir, please help. (I deleted my same previous answer and your commented answer got deleted alongwith).
$${let}\:{vertex}\:{angle}\:{be}\:\mathrm{2}\theta. \\ $$$$\Rightarrow\:{n}\phi={n}\left(\frac{\pi}{\mathrm{2}}−\theta\right)=\mathrm{2}\pi \\ $$$$\Rightarrow\:\:\theta=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}\pi}{{n}} \\ $$$${R}=\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{1}}{\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{{n}}\right)} \\ $$$${This}\:{is}\:{clearly}\:{wrong},\:{but}\:{i} \\ $$$${dont}\:{understand},\:{how}\:{it}\:{came} \\ $$$${wrong};\:{mrW}\:{Sir},\:{please}\:{help}. \\ $$$$\left({I}\:{deleted}\:{my}\:{same}\:{previous}\right. \\ $$$${answer}\:{and}\:{your}\:{commented} \\ $$$$\left.{answer}\:{got}\:{deleted}\:{alongwith}\right). \\ $$
Commented by aliesam last updated on 19/Jul/19
Commented by aliesam last updated on 19/Jul/19
this relationship is true for any n
$${this}\:{relationship}\:{is}\:{true}\:{for}\:{any}\:{n} \\ $$
Commented by mr W last updated on 19/Jul/19
na=2π is always true, but a=(π/2)−θ is not true! in your diagram you assume that both green lines are ⊥ to the sides of the star, because you take a=(π/2)−θ, but this is only the case with n=5. why don′t you check it when you draw a figur with n=7?
$${na}=\mathrm{2}\pi\:{is}\:{always}\:{true},\:{but} \\ $$$${a}=\frac{\pi}{\mathrm{2}}−\theta\:{is}\:{not}\:{true}! \\ $$$${in}\:{your}\:{diagram}\:{you}\:{assume}\:{that} \\ $$$${both}\:{green}\:{lines}\:{are}\:\bot\:{to}\:{the}\:{sides} \\ $$$${of}\:{the}\:{star},\:{because}\:{you}\:{take}\:{a}=\frac{\pi}{\mathrm{2}}−\theta, \\ $$$${but}\:{this}\:{is}\:{only}\:{the}\:{case}\:{with}\:{n}=\mathrm{5}. \\ $$$${why}\:{don}'{t}\:{you}\:{check}\:{it}\:{when}\:{you} \\ $$$${draw}\:{a}\:{figur}\:{with}\:{n}=\mathrm{7}? \\ $$
Commented by mr W last updated on 19/Jul/19
⇒ n((π/2)−θ)=2π is not generally true for a n−corner star. it′s only true for the displayed case with n=5. you′ll see the error when you draw a 7−corner star.
$$\Rightarrow\:\:\:{n}\left(\frac{\pi}{\mathrm{2}}−\theta\right)=\mathrm{2}\pi \\ $$$${is}\:{not}\:{generally}\:{true}\:{for}\:{a}\:{n}−{corner}\:{star}. \\ $$$${it}'{s}\:{only}\:{true}\:{for}\:{the}\:{displayed}\:{case} \\ $$$${with}\:{n}=\mathrm{5}. \\ $$$${you}'{ll}\:{see}\:{the}\:{error}\:{when}\:{you}\:{draw}\:{a} \\ $$$$\mathrm{7}−{corner}\:{star}. \\ $$
Commented by aliesam last updated on 20/Jul/19
are sides of the star tangent to the circle? if yes , then thr two sides in the star are perpenduclar to the green lines
$${are}\:{sides}\:{of}\:{the}\:{star}\:{tangent}\:{to}\:{the}\:{circle}? \\ $$$${if}\:{yes}\:,\:{then}\:{thr}\:{two}\:{sides}\:{in}\:{the}\:{star}\:{are} \\ $$$${perpenduclar}\:{to}\:{the}\:{green}\:{lines} \\ $$
Answered by mr W last updated on 19/Jul/19
Commented by mr W last updated on 19/Jul/19
CD^(⌢) is 1/n of the circumcircle ⇒4ϕ=((2π)/n) ⇒ϕ=(π/(2n)) sin ϕ=(1/R) ⇒R=(1/(sin ϕ))=(1/(sin (π/(2n))))
$$\overset{\frown} {{CD}}\:{is}\:\mathrm{1}/{n}\:{of}\:{the}\:{circumcircle} \\ $$$$\Rightarrow\mathrm{4}\varphi=\frac{\mathrm{2}\pi}{{n}} \\ $$$$\Rightarrow\varphi=\frac{\pi}{\mathrm{2}{n}} \\ $$$$\mathrm{sin}\:\varphi=\frac{\mathrm{1}}{{R}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{1}}{\mathrm{sin}\:\varphi}=\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}} \\ $$
Commented by ajfour last updated on 19/Jul/19
Thanks Sir, but i am still in doubt in the way i proceeded, i shall explain how..
$${Thanks}\:{Sir},\:{but}\:{i}\:{am}\:{still}\:{in} \\ $$$${doubt}\:{in}\:{the}\:{way}\:{i}\:{proceeded},\:{i} \\ $$$${shall}\:{explain}\:{how}.. \\ $$
Commented by ajfour last updated on 19/Jul/19
consider n=4 sir?
$${consider}\:{n}=\mathrm{4}\:\:\:\:{sir}? \\ $$
Commented by mr W last updated on 19/Jul/19
maybe we don′t mean the same with the ′star′ figur. you can not construct a 4−corner star. i assume you mean in the question a five corner star, not a pentagon. this is what i mean with n=7.
$${maybe}\:{we}\:{don}'{t}\:{mean}\:{the}\:{same}\:{with} \\ $$$${the}\:'{star}'\:{figur}.\:{you}\:{can}\:{not}\:{construct} \\ $$$${a}\:\mathrm{4}−{corner}\:{star}. \\ $$$${i}\:{assume}\:{you}\:{mean}\:{in}\:{the}\:{question}\:{a} \\ $$$${five}\:{corner}\:{star},\:{not}\:{a}\:{pentagon}. \\ $$$${this}\:{is}\:{what}\:{i}\:{mean}\:{with}\:{n}=\mathrm{7}. \\ $$
Commented by mr W last updated on 19/Jul/19

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