Let A1, A2, ..., An be the vertices of an n-sided regular polygon such that 1/A1A2 = 1/A1A3 + 1/A1A4

Q: Wait, why can we assume the polygon is inscribed in a circle? Is that always true?

Excellent question! A regular polygon is cyclic by definition—it possesses a circumcircle. By introducing the radius $R$, we aren't changing the problem; we are creating a geometric framework that makes the math elegant. The $R$ will eventually cancel out, proving that the relationship between the sides and diagonals is independent of the circle's size.

Q: Why can't we just use the Law of Cosines for these triangles?

You absolutely could! But imagine the algebraic nightmare of dealing with square roots and nested cosines for $A_1 A_4$. The chord formula $2R \sin(\frac{\theta}{2})$ is the 'JEE shortcut' that keeps the algebra clean. Always look for the path of least resistance in geometry problems.

Q: In the final step, why do we reject the factor $\cos(\frac{\pi}{2n}) - \cos(\frac{3\pi}{2n}) = 0$?

Great catch. If that term were zero, then $\frac{\pi}{2n} = \frac{3\pi}{2n}$, which implies $\pi = 3\pi$, an impossibility. More generally, for any valid polygon where $n \ge 3$, the angles $\frac{\pi}{2n}$ and $\frac{3\pi}{2n}$ are distinct and lie within the first quadrant where the cosine function is strictly decreasing. Thus, they cannot be equal.

Question

Let $A_1, A_2, \dots, A_n$ be the vertices of an $n$-sided regular polygon such that $\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1 A_4}$. Find the value of $n$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to unravel a beautiful geometric puzzle. We are looking at a regular $n$-sided polygon, and we are given a curious relationship between its side and two of its diagonals:

$$\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1 A_4}$$

At first glance, this looks like a simple algebraic equation, but it is actually a gateway into the elegant world of trigonometry. Let us begin by visualizing our polygon inscribed perfectly inside a circle of radius $R$.

By placing the polygon in a circle, we gain access to the powerful chord length formula. If a chord subtends an angle $	heta$ at the center, its length is $2R \sin(\frac{	heta}{2})$. For a regular $n$-gon, the angle subtended by one side at the center is $\frac{2\pi}{n}$.

Therefore, the distance between the first vertex $A_1$ and any other vertex $A_{k+1}$ is simply:

$$A_1 A_{k+1} = 2R \sin\left(\frac{k\pi}{n}ight)$$

### The Trigonometric Bridge

Now, let us translate the problem into the language of trigonometry. For the side $A_1 A_2$, we have $k=1$, so $A_1 A_2 = 2R \sin(\frac{\pi}{n})$. For the diagonal $A_1 A_3$, $k=2$, giving us $2R \sin(\frac{2\pi}{n})$.

For $A_1 A_4$, $k=3$, resulting in $2R \sin(\frac{3\pi}{n})$. Substituting these into our original equation, we get:

$$\frac{1}{2R \sin(\frac{\pi}{n})} = \frac{1}{2R \sin(\frac{2\pi}{n})} + \frac{1}{2R \sin(\frac{3\pi}{n})}$$

Notice the beauty here? The radius $R$ appears in every denominator. Since $R$ is a non-zero length, we can simply cancel it out to obtain:

$$\frac{1}{\sin(\frac{\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})} + \frac{1}{\sin(\frac{3\pi}{n})}$$

### The Algebraic Dance

Now, we must manipulate this equation. Combining the fractions on the right-hand side gives us:

$$\frac{1}{\sin(\frac{\pi}{n})} = \frac{\sin(\frac{3\pi}{n}) + \sin(\frac{2\pi}{n})}{\sin(\frac{2\pi}{n}) \sin(\frac{3\pi}{n})}$$

Cross-multiplying, we arrive at:

$$\sin\left(\frac{2\pi}{n}ight) \sin\left(\frac{3\pi}{n}ight) = \sin\left(\frac{\pi}{n}ight) \left[ \sin\left(\frac{3\pi}{n}ight) + \sin\left(\frac{2\pi}{n}ight) ight]$$

This looks intimidating, but we have a secret weapon: the product-to-sum identities. By multiplying the entire equation by $2$, we use the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$.

Applying this to every term transforms our products into a string of cosines:

$$\cos\left(\frac{\pi}{n}ight) - \cos\left(\frac{5\pi}{n}ight) = \cos\left(\frac{2\pi}{n}ight) - \cos\left(\frac{4\pi}{n}ight) + \cos\left(\frac{\pi}{n}ight) - \cos\left(\frac{3\pi}{n}ight)$$

### The Final Resolution

Look closely at the equation. The $\cos(\frac{\pi}{n})$ terms on both sides cancel out perfectly! Rearranging the remaining terms, we get:

$$\cos\left(\frac{3\pi}{n}ight) + \cos\left(\frac{4\pi}{n}ight) = \cos\left(\frac{2\pi}{n}ight) + \cos\left(\frac{5\pi}{n}ight)$$

Now, we use the sum-to-product identity: $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$. Applying this to both sides, we find:

$$2 \cos\left(\frac{7\pi}{2n}ight) \cos\left(\frac{\pi}{2n}ight) = 2 \cos\left(\frac{7\pi}{2n}ight) \cos\left(\frac{3\pi}{2n}ight)$$

Factoring out the common term, we are left with:

$$2 \cos\left(\frac{7\pi}{2n}ight) \left[ \cos\left(\frac{\pi}{2n}ight) - \cos\left(\frac{3\pi}{2n}ight) ight] = 0$$

As we discussed, the bracketed term cannot be zero for a valid polygon. Thus, we must have $\cos(\frac{7\pi}{2n}) = 0$. This implies $\frac{7\pi}{2n} = \frac{\pi}{2}$, which leads us directly to **$n = 7$**.

Animated Solution for Mathematics - Trigonometry: Let $A_{1}, A_{2}, \dots, A_{n}$ be the vertices of an $n$ -sided regular polygon such that $\frac{1}{A _{1} A _{2}} = \frac{1}{A _{1} A _{3}} + \frac{1}{A _{1} A _{4}}$ . Find the value of $n$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Polygon Chords

The Chord Length Formula

Expressing Lengths in Sine

Substituting into the Given Equation

Simplifying the Fractions

Cross Multiplication

Product-to-Sum Identities

Applying the Identity

Canceling Common Terms

Sum-to-Product Transformation

Applying Sum-to-Product

Factorizing the Equation

Solving for n

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Trigonometric Bridge

The Algebraic Dance

The Final Resolution

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