The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a

Q: Wait, why is the angle $\frac{\pi}{n}$ and not $\frac{2\pi}{n}$? I thought the central angle for one side was $\frac{2\pi}{n}$.

You are absolutely correct that the central angle subtended by one side is $\frac{2\pi}{n}$. However, when we drop a perpendicular from the center to the midpoint of the side, we are bisecting that central angle. Therefore, the angle in our right-angled triangle is exactly half of $\frac{2\pi}{n}$, which is $\frac{\pi}{n}$. Always remember to check if your triangle is the full isosceles triangle or the half-triangle!

Q: Can I use the Law of Sines instead of right-triangle trigonometry?

Yes, you absolutely can! The Law of Sines is a powerful tool. In the triangle formed by the center and two vertices, the sides are $R, R, a$ and the angles are $\frac{\pi}{n}, \frac{\pi}{n}, \pi - \frac{2\pi}{n}$. Using the Law of Sines, you would get $\frac{a}{\sin(2\pi/n)} = \frac{R}{\sin(\pi/n - \pi/n)}$. While this works, the right-triangle approach is often faster and less prone to algebraic errors in competitive exams like JEE.

Q: How do I know when to use half-angle identities?

Whenever you see an expression involving $\cot\theta + \csc\theta$ or similar combinations of trigonometric functions, it is a massive hint that simplification is possible using half-angle identities. In JEE problems, these identities are frequently the key to collapsing a complex-looking expression into a simple, elegant form. If you are stuck with a sum of trig functions, try converting to sine and cosine and looking for these identities.

Question

The sum of the radii of inscribed and circumscribed circles for an $n$ sided regular polygon of side $a$, is

Options:

(A) $\frac{a}{4} \cot\left(\frac{\pi}{2n}\right)$

(B) $a \cot\left(\frac{\pi}{n}\right)$

(C) $\frac{a}{2} \cot\left(\frac{\pi}{2n}\right)$

(D) $a \cot\left(\frac{2\pi}{n}\right)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a beautiful exploration of geometry. Today, we are not just solving a problem; we are peeling back the layers of a regular polygon to reveal the elegant relationship between its inscribed and circumscribed circles.

Imagine you are standing at the center of a regular $n$-sided polygon. You look out at one of its sides, which we will call $AB$, with a length of $a$.

By connecting the center $O$ to the vertices $A$ and $B$, we create an isosceles triangle $	riangle OAB$. This triangle is the heartbeat of our problem, containing all the information we need to find the radii of the circles that touch the sides and the vertices.

### The Apothem: Our Geometric Anchor

To make sense of this triangle, we need a reference line. Let us drop a perpendicular from the center $O$ to the side $AB$, meeting it at the midpoint $M$.

This line segment $OM$ is known as the apothem. Because $M$ is the midpoint, the length $AM$ is exactly $\frac{a}{2}$.

This simple act of dropping a perpendicular has transformed our isosceles triangle $	riangle OAB$ into two congruent right-angled triangles, $	riangle OAM$ and $	riangle OBM$. Now, let us focus on $	riangle OAM$.

The angle at the center, $\angle AOB$, is $\frac{2\pi}{n}$. Since $OM$ bisects this angle, our angle $\angle AOM$ is exactly $\frac{\pi}{n}$. This is the angle that will drive our trigonometric journey.

### Defining the Radii

Now, let us define our two protagonists: the inradius $r$ and the circumradius $R$. The inradius $r$ is the radius of the inscribed circle, which touches all sides.

Geometrically, this is the distance from the center $O$ to the side $AB$, which is the length of our apothem $OM$. In our right triangle $	riangle OAM$, we can use the definition of the cotangent function:

$$\cot\left(\frac{\pi}{n}ight) = \frac{	ext{Adjacent}}{	ext{Opposite}} = \frac{OM}{AM} = \frac{r}{a/2}$$

Rearranging this, we find that:

$$r = \frac{a}{2} \cot\left(\frac{\pi}{n}ight)$$

Next, the circumradius $R$ is the distance from the center $O$ to any vertex, which is the length $OA$. Using the cosecant function in the same triangle, we have:

$$\csc\left(\frac{\pi}{n}ight) = \frac{	ext{Hypotenuse}}{	ext{Opposite}} = \frac{OA}{AM} = \frac{R}{a/2}$$

Thus, we obtain:

$$R = \frac{a}{2} \csc\left(\frac{\pi}{n}ight)$$

### The Algebraic Dance

We are now ready to find the sum $r + R$. Substituting our expressions, we get:

$$r + R = \frac{a}{2} \cot\left(\frac{\pi}{n}ight) + \frac{a}{2} \csc\left(\frac{\pi}{n}ight)$$

Factoring out the common $\frac{a}{2}$, we have:

$$r + R = \frac{a}{2} \left[ \cot\left(\frac{\pi}{n}ight) + \csc\left(\frac{\pi}{n}ight) ight]$$

To simplify this, we convert to sine and cosine:

$$r + R = \frac{a}{2} \left[ \frac{\cos(\pi/n) + 1}{\sin(\pi/n)} ight]$$

This is where the magic happens. We invoke the half-angle identities: $1 + \cos	heta = 2\cos^2\left(\frac{	heta}{2}ight)$ and $\sin	heta = 2\sin\left(\frac{	heta}{2}ight)\cos\left(\frac{	heta}{2}ight)$. Substituting $	heta = \frac{\pi}{n}$, our expression becomes:

$$r + R = \frac{a}{2} \left[ \frac{2\cos^2(\pi/2n)}{2\sin(\pi/2n)\cos(\pi/2n)} ight]$$

### The Elegant Conclusion

Finally, we watch the terms cancel out. The factor of $2$ vanishes, and one factor of $\cos(\pi/2n)$ cancels from the numerator and denominator.

We are left with:

$$r + R = \frac{a}{2} \left[ \frac{\cos(\pi/2n)}{\sin(\pi/2n)} ight]$$

This simplifies beautifully to our final result:

**$$r + R = \frac{a}{2} \cot\left(\frac{\pi}{2n}ight)$$**

This result is not just an answer; it is a testament to the harmony of geometry and trigonometry. You have successfully navigated the complexity and arrived at the elegant truth.

Animated Solution for Mathematics - Trigonometry: The sum of the radii of inscribed and circumscribed circles for an $n$ sided regular polygon of side $a$ , is

Select Answer:

Visualized Solution

Visualizing the Regular Polygon Geometry

Identifying the Right Triangle $O A M$

Finding the Central Angle $∠ A O M$

Defining Inradius $r$ and Circumradius $R$

Expressing Inradius $r$

Expressing Circumradius $R$

Setting up the Sum $r + R$

Converting to Sine and Cosine

Applying Half-Angle Identities

Final Simplification

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Apothem

Defining the Radii

The Algebraic Dance

The Elegant Conclusion

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is