Question

If r and R are respectively the radii of the inscribed and circumscribed circles of a regular polygon of n sides such that \[\dfrac{R}{r}=\sqrt{5}-1,\] then n is equal to:
(a) 5
(b) 6
(c) 10
(d) 18

Answer 1

Hint: To solve the question given above, first we will assume that the regular polygon has side ‘a’. Then we will find the values of r and R in terms of angle ‘a’ and minimum of sides n. Then, we will take their ratio and we will eliminate ‘a’ from it. From the ratio, we will determine the value of n.

Complete step-by-step answer:
It is given in the question that a polygon is inscribed in a circle of radius R. Let the side of the polygon be ‘a’. Thus, we have the following figure.

In the figure drawn, we can see that the polygon is divided into n isosceles triangles having equal dimensions. Now, the total length of AB = a. Now, we have drawn perpendicular to AB then OP will bisect AB (according to the perpendicular bisector theorem). So, we can say that,
\[AP=\dfrac{a}{2}\]
Now, we will consider the triangle APO. In triangle APO, we have,

\[\sin \alpha =\dfrac{AP}{OA}\]
\[\Rightarrow \sin \alpha =\dfrac{\left( \dfrac{a}{2} \right)}{R}\]
\[\Rightarrow R=\dfrac{a}{2\sin \alpha }.....\left( i \right)\]
Now, the total angle inside the circle is \[2\pi .\] Now, it is divided into n triangles and each triangle subtends an angle \[2\alpha .\] So,
\[n\left( 2\alpha \right)=2\pi \]
\[\Rightarrow \alpha =\dfrac{\pi }{n}....\left( ii \right)\]
In another situation, we are given a polygon that circumscribes a circle of radius r. The side of the polygon is ‘a’. Thus, we have,

Here, we can see that the angle subtended by each triangle is \[2\alpha \] and there are n triangles. So, we have,
\[n\left( 2{{\alpha }^{'}} \right)=2\pi \]
\[\Rightarrow {{\alpha }^{'}}=\dfrac{\pi }{n}....\left( iii \right)\]
Also, OP will be perpendicular to AB because AB is tangent to the given circle. So,
\[\angle OPA={{90}^{o}}\]
Thus, we have,

So, we can say that,
\[\tan {{\alpha }^{'}}=\dfrac{\left( \dfrac{a}{2} \right)}{r}\]
\[\Rightarrow r=\dfrac{a}{2\tan {{\alpha }^{'}}}....\left( iv \right)\]
Now, we will take the ratio of R and r. Thus, we have,
\[\dfrac{R}{r}=\dfrac{\dfrac{a}{2\sin \alpha }}{\dfrac{a}{2\tan {{\alpha }^{'}}}}\]
\[\Rightarrow \dfrac{R}{r}=\dfrac{\tan {{\alpha }^{'}}}{\sin \alpha }\]
Now, we will put the value of \[\alpha \text{ and }{{\alpha }^{'}}\] from (i) and (iii) to the above equation.
\[\Rightarrow \dfrac{R}{r}=\dfrac{\tan \left( \dfrac{\pi }{n} \right)}{\sin \left( \dfrac{\pi }{n} \right)}\]
\[\Rightarrow \dfrac{R}{r}=\dfrac{\dfrac{\sin \left( \dfrac{\pi }{n} \right)}{\cos \left( \dfrac{\pi }{n} \right)}}{\sin \left( \dfrac{\pi }{n} \right)}\]
\[\Rightarrow \dfrac{R}{r}=\dfrac{1}{\cos \left( \dfrac{\pi }{n} \right)}....\left( v \right)\]
It is given that,
\[\dfrac{R}{r}=\sqrt{5}-1.....\left( vi \right)\]
From (v) and (vi), we have,
\[\dfrac{1}{\cos \left( n \right)}=\sqrt{5}-1\]
\[\dfrac{1}{\cos \left( \dfrac{\pi }{n} \right)}=\dfrac{\sqrt{5}-1}{1}\times \dfrac{\sqrt{5}+1}{\sqrt{5}+1}\]
\[\dfrac{1}{\cos \left( \dfrac{\pi }{n} \right)}=\dfrac{5-1}{\sqrt{5}+1}\]
\[\Rightarrow \cos n=\dfrac{\sqrt{5}+1}{4}\]
\[\Rightarrow \cos \dfrac{\pi }{n}=\cos 36\]
\[\Rightarrow \dfrac{\pi }{n}=36\times \dfrac{\pi }{180}\]
\[\Rightarrow \dfrac{1}{n}=\dfrac{36}{180}\]
\[\Rightarrow n=5\]
Hence, option (a) is the right answer.

Note: In case, we do not know how we wrote \[\dfrac{\sqrt{5}+1}{4}\text{ as }\cos {{36}^{o}},\] we can calculate it as shown below,
\[\theta ={{36}^{o}}\]
\[\Rightarrow 5\theta =5\times {{36}^{o}}\]
\[\Rightarrow 5\theta ={{180}^{o}}\]
\[\Rightarrow \cos \left( 5\theta \right)=\cos {{180}^{o}}\]
\[\Rightarrow \cos 5\theta =0\]
\[\Rightarrow \cos \left( 2\theta +3\theta \right)=0\]
\[\Rightarrow \cos 2\theta \cos 3\theta -\sin 2\theta \sin 3\theta =0\]
\[\Rightarrow \left( 1-2{{\sin }^{2}}\theta \right)\left( 4{{\cos }^{3}}\theta -3\cos \theta \right)-\left( 2\sin \theta \right)\left( \cos \theta \right)\left( 3\sin \theta -4{{\sin }^{3}}\theta \right)=0\]
When we will solve this, we will get,
\[\cos \theta =\dfrac{\sqrt{5}+1}{4}\]