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If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).

I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?

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If you know a bit about complex numbers and roots of unity, here is another approach:

The $n^\text{th}$ roots of unity are vertices of a regular $n$-gon. Let $z=1$ denote $A_1$ and $w = e^{\frac{2\pi\iota}{n}}$ denote $A_2$ in the complex plane. Then $w^2$ denotes $A_3$ and $w^3$ denotes $A_4$. Now, the given equation can be written as

$$\frac{1}{|1-w|} = \frac{1}{|1-w^2|} + \frac{1}{|1-w^3|}$$ Multiplying with $|1-w|$ throughout, we get $$1 = \frac{1}{|1+w|} + \frac{1}{|1+w+w^2|}$$

Now since $|w|=1$, pulling out $|w|$ from the denominator of the second term in the R.H.S, $$1 = \frac{1}{|1+w|} + \frac{1}{|1+w+\bar{w}|}$$ Notice that $w+\bar{w}$ is purely real and can be written $w+\frac{1}{w}$, hence $$1 - \frac{w}{1+w+w^2} = \frac{1+w^2}{1+w+w^2} = \frac{1}{|1+w|}$$ Squaring both sides and using $|z|^2 = z\bar{z}$, $$\Bigg(\frac{1+w^2}{1+w+w^2}\Bigg)^2 = \frac{1}{(1+w)} \cdot \frac{1}{(1+\frac{1}{w})}= \frac{w}{(1+w)^2}$$ Cross multiply, $$(1+w+w^2+w^3)^2 = w (1+w+w^2)^2$$ and simplify to get $$1+w+w^2+w^3+w^4+w^5+w^6 = 0$$ Therefore, $$1-w^7 =0$$ and $w \neq 1$, so $w$ must be a $7$th root of unity. Argue that $n =7$.

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    $\begingroup$ Very interesting approach. Upvoting your answer so that more people look at this approach. Nice. $\endgroup$
    – Shailesh
    Aug 9 '15 at 9:27
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This problem can be trivialized by Ptolemy's theorem.

First we take the l.c.m and simplify both sides of the given equation. Let $A_1A_2=a$, $A_1A_3=b$, $A_1A_4=c$. Then we have

$$\frac{1}{a}=\frac{1}{b}+\frac{1}{c} \qquad\to\qquad b c = a b + a c \tag{1}$$

Now, note that regular polygons can always be inscribed in a circle. Take the quadrilateral $A_1A_3A_4A_5$, which is a cyclic quadrilateral. Applying Ptolemy's theorem we get, $$A_1A_3\cdot A_4A_5 + A_3A_4\cdot A_1A_5 = A_1A_4\cdot A_3A_5 \tag{2}$$ Now see that $A_1A_3=b$, $A_4A_5=a$, $A_3A_4=a$, $A_1A_4=c$, $A_3A_5=b$, so in $(2)$ we have, $$b a + a\cdot A_1A_5 = c b \tag{3}$$

Comparing equation $(1)$ with $(3)$, we see that $A_1A_5 = c = A_1A_4$. These are two diagonals of the regular polygon sharing a common vertex, so this diagonal must be a diagonal that bisects the area of the polygon. So on one side of the diagonal $A_1A_4$ there is $A_2$, $A_3$; and on the other side of the diagonal $A_1A_5$ there must be $A_6$, $A_7$. So there are 7 vertices of the polygon.

Hence $n=7$, leading to a heptagon.

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$Outline:$

I'm sorry I cannot draw a diagram and explain, though that would be very neat. Let $r$ be the radius of the circumcircle, and let $2\theta$ be the angle subtended at the center by the adjacent vertices. If $a$ is the length of any side of the regular polygon, then $a = 2r\sin\theta$. Now $\theta = \frac{\pi}{n}$, then the given equation becomes $\frac{1}{\sin\theta} = \frac{1}{\sin2\theta} + \frac{1}{\sin3\theta}$. Simplify this, you will get $\sin4\theta = \sin3\theta$, gives $\theta = \frac{\pi}{7}$. So $n = 7$ is the final answer. You may fill in the gaps in the solution.

Since the answer got a few upvotes, I looked up a little bit more and found that this problem was asked in IIT JEE 1994 (entrance exam for Indian Institute of Technology) and again a variant of it in 2011. Prof K. D. Joshi, in his book Educative JEE, has outlined 3 solutions, one of them is like mine (only better), then the Ptolemy one given by user260674 as well as the one using complex numbers given by Seven. You can refer to page 18 and 19 of the original script here

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I was casting about for a method that didn't require a lot of work with multiple-angles and trigonometric identities. Up to now, I had a different -- though ultimately related -- argument (without using a circumscribed circle) which led me to the same equation, $ \ \frac{1}{\sin\theta} \ = \ \frac{1}{\sin2\theta} \ + \ \frac{1}{\sin3\theta} $ , which Shailesh already produced. Here's something using more basic triangle geometry than what I'd had.

[diagram to be inserted -- need to use another browser for that]

For convenience, we will call the lengths of the sides of the polygon $ \ s \ $ , making $ \ A_1A_2 \ = \ s \ $ and call the other lengths of interest $ \ A_1A_3 \ = \ t \ $ and $ \ A_1A_4 \ = \ u \ $ .

"Drop perpendiculars" to $ \ \overline{A_1A_4} \ $ from $ \ A_2 \ $ to produce point $ \ P \ $ and from $ \ A_3 \ $ to produce $ \ Q \ $ . Since we are working with a regular polygon, it is straightforward to show that $ \ A_1A_2A_3A_4 \ $ is a trapezoid and that $ \ PQ \ = \ s \ $ . Extend, say , $ \ A_4A_3 \ $ to a point $ \ R \ $ : since $ \ \angle A_2A_3R \ $ is an exterior angle of the polygon, $ \ m(\angle A_2A_3R) \ = \ \frac{2 \pi}{n} \ $ . We have $ \ \overline{A_2A_3} \ \ \Vert \ \ \overline{A_1A_4} \ $ , so corresponding angle $ \ \angle QA_4A_3 \ $ also has measure $ \ \frac{2 \pi}{n} \ $ .

Consequently, $ \ A_4Q \ = \ A_1P \ = \ s \ \cos \left(\frac{2 \pi}{n} \right) \ $ [it is clear that the two segments are congruent] and so

$$ u \ = \ s \ \left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right) \ \ . $$

Although $ \ \Delta A_1A_2A_3 \ $ is isosceles, and $ \ m(\angle A_1A_2A_3) \ = \ \pi \ - \ \frac{2 \pi}{n} \ $ by a familiar theorem (or because it is supplementary to an exterior angle), we will actually not exploit the Law of Cosines here to find $ \ t \ $ . Instead, we obtain that $ \ m(\angle A_1A_3A_2 ) \ = \ \frac{1}{2} \ ( \ \pi \ - \ [ \pi \ - \ \frac{2 \pi}{n} ] \ ) \ = \ \frac{\pi}{n} \ $ . From this, we find $ \ m(\angle A_1A_3A_4 ) \ = \ \left(\pi \ - \ \frac{2 \pi}{n} \right) \ - \frac{\pi}{n} $ $ \ = \ \left(\pi \ - \ \frac{3 \pi}{n} \right) \ $ .

By the Law of Sines,

$$ \frac{t}{\sin \left( \frac{2 \pi}{n} \right)} \ = \ \frac{u}{\sin \left( \ \pi \ - \ \frac{3 \pi}{n} \ \right) \ } \ \ \Rightarrow \ \ \frac{t}{\sin \left( \frac{2 \pi}{n} \right)} \ = \ \frac{s \ \left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)}{\sin \left( \frac{3 \pi}{n} \right) } $$

[using the identity "sine of an angle equals the sine of its supplement"].

At last coming to the equation under discussion, we obtain

$$ \frac{1}{s} \ = \ \frac{1}{t} \ + \ \frac{1}{u} \ = \ \frac{1}{s} \ \left[ \ \frac{1 }{\left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)} \ \left( \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right) \ \right] \ \left[ \ 1 \ + \ \left( \ \frac{\sin \left( \frac{2 \pi}{n} \right)}{\sin \left( \ \frac{3 \pi}{n} \ \right)} \ \right) \ \right] $$

$$ \Rightarrow \ \ 1 \ = \ \frac{1 }{\left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)} \ \ \left[ \ \left( \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right) \ + \ 1 \ \right] $$

$$ \Rightarrow \ \ 1 \ + \ 2 \ \cos \left( \frac{2 \pi}{n} \right) \ = \ 1 \ + \ \left[ \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right] $$ $$\Rightarrow \ \ 2 \ \sin \left( \frac{2 \pi}{n} \right) \ \cos \left( \frac{2 \pi}{n} \right) \ = \ \sin \left( \frac{3 \pi}{n} \right) \ \ \Rightarrow \ \ \sin \left( \frac{4 \pi}{n} \right) \ = \ \sin \left( \frac{3 \pi}{n} \right) \ \ , $$

applying the "double-angle formula" for sine at the end.

Since all of the angles discussed here have measure less than $ \ \pi \ $ , it is either the case that

$$ \frac{4 \pi}{n} \ = \frac{3 \pi}{n} \ \ , $$

which would require $ \ \frac{ \pi}{n} \ = \ 0 \ $ , or that

$$ \frac{4 \pi}{n} \ = \ \pi \ - \frac{3 \pi}{n} \ \ \Rightarrow \ \ \frac{7 \pi}{n} \ = \ \pi \ \ \Rightarrow \ \ n \ = \ 7 \ \ . $$

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Hint: Let, each side of the regular polygon $A_1A_2A_3\ldots A_n$ be $a$ then we have $$A_1A_2=a$$ $$A_1A_3=2a\sin\left(\frac{(n-2)\pi}{2n}\right)$$ $$A_1A_4=a-2a\cos\left(\frac{(n-2)\pi}{n}\right)$$

Now, we have $$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$$

$$\frac{1}{a}=\frac{1}{2a\sin\left(\frac{(n-2)\pi}{2n}\right)}+\frac{1}{a-2a\cos\left(\frac{(n-2)\pi}{n}\right)}$$ Let, $\frac{(n-2)\pi}{2n}=\alpha$ then we have

$$1-\frac{1}{1-2\cos2\alpha}=\frac{1}{2\sin\alpha}$$

$$\frac{1-2\cos2\alpha-1}{1-2\cos2\alpha}=\frac{1}{2\sin\alpha}$$ $$8\sin^3\alpha-4\sin^2\alpha-4\sin\alpha+1=0$$

I hope you can take it from here.

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    $\begingroup$ Perhaps the one weakness of this approach is that it does not provide an immediate way to pick out the value for $ \ n \ $ . While one of the roots corresponds to an obtuse angle and so can be rejected in the context of the geometrical situation, the other two roots produce $ \ \sin \alpha \ \approx \ 0.9010 \ $ and $ \ \sin \alpha \ \approx \ 0.2225 \ $ . The latter gives us $ \ \alpha \ = \ \frac{\pi}{14} \ $ , which looks promising, but leads to a non-integral result for $ \ n \ $ . It is actually the other value that turns out to work, yielding $ \ \alpha \ = \ \frac{5 \pi}{14} \ $ . $\endgroup$ Aug 10 '15 at 7:14
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    $\begingroup$ Yes, you are right $\endgroup$ Aug 10 '15 at 7:39
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Here's a solution using just the law of cosines. The identity does not hold unless $n\ge 5$. Label the edges and diagonals as in the diagram below, so that by hypothesis $$\frac1a = \frac1b + \frac1c.\tag1$$

The segments $A_2A_3$, $A_3A_4$, and $A_4A_5$ are concyclic and have the same length $a$, so they subtend the same angle $\theta$ at $A_1$. Apply the law of cosines to triangles $\triangle A_1A_2A_3$, $\triangle A_1A_3A_4$, $\triangle A_1A_4A_5$ respectively: $$ a^2=a^2+b^2-2ab\cos\theta\tag2 $$ $$ a^2=b^2+c^2-2bc\cos\theta\tag3 $$ $$ a^2=c^2+d^2-2cd\cos\theta\tag4 $$ Solve equation (2) for $\cos\theta$ and apply (1): $$ \cos\theta=\frac b{2a}\stackrel{(1)}=\frac{b+c}{2c}.\tag5 $$ Equate the right sides of (3) and (4): $$ b^2-2bc\cos\theta=d^2-2cd\cos\theta, $$ substitute (5) for $\cos\theta$, and simplify to obtain $(b-d)(d-c)=0$. We rule out $b=d$ [which merely makes equation (4) a duplicate of (3)], leaving $d=c$. This means $\triangle A_1A_4A_5$ is isosceles, so the polygon has five internal triangles meeting at $A_1$, and has seven vertices.

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