Animated Solution for Mathematics - Trigonometry: If $I_n$ is the area of $n$ sided regular polygon inscribed in a circle of unit radius and $O_n$ be the area of the polygon circumscribing the given circle, prove that $I_n=\frac{O_n}{2}\left(1+\sqrt{1-{\left(\frac{2I_n}{n}\right)}^2}\right)$.

Bridging Inscribed and Circumscribed Worlds

Imagine you are standing at the center of a unit circle, a perfect, pristine playground of radius $r=1$. You are tasked with a geometric challenge: to capture the essence of this circle using two different types of polygons, both with $n$ sides.

One is trapped inside, hugging the interior, and the other stands guard outside, perfectly tangent to the circle's edge. This is not just a problem of shapes; it is a beautiful dance of trigonometry.

Capturing the Interior

Let us first look at the inscribed polygon, $I_n$. Imagine drawing lines from the center of the circle to each vertex of the polygon. You have just sliced the polygon into $n$ identical, congruent isosceles triangles.

Each triangle has two sides equal to the radius of the circle, which is $1$. The central angle of each triangle is simply the full circle, $2\pi$, divided by the number of sides, $n$.

Using the classic area formula for a triangle, $\text{Area}=\frac{1}{2}ab\sin (\theta )$, where $a=1$, $b=1$, and $\theta =\frac{2\pi }{n}$, the area of one triangle becomes $\frac{1}{2}\sin \left(\frac{2\pi }{n}\right)$.

Since there are $n$ such triangles, the total area of our inscribed polygon is:

The Tangent Guard

Now, shift your perspective to the circumscribed polygon, $O_n$. This polygon sits outside the circle, where each side is a tangent line.

If you draw a radius to the point of tangency, it acts as the altitude of a triangle formed by the center and the side of the polygon. This radius, with length $1$, bisects the central angle, creating a right-angled triangle with an angle of $\frac{\pi }{n}$ at the center.

The base of this half-triangle is $1\cdot \tan \left(\frac{\pi }{n}\right)$. Therefore, the full base of the triangle is $2\tan \left(\frac{\pi }{n}\right)$.

The area of one such triangle is $\frac{1}{2}\cdot \text{base}\cdot \text{height}=\frac{1}{2}\left(2\tan \left(\frac{\pi }{n}\right)\right)(1)=\tan \left(\frac{\pi }{n}\right)$. With $n$ triangles, the total area is:

Finding the Ratio

We have our two areas. Now, let us find the relationship between them by taking their ratio:

The $n$ terms cancel out immediately, leaving us with $\frac{\frac{1}{2}\sin \left(\frac{2\pi }{n}\right)}{\tan \left(\frac{\pi }{n}\right)}$. This is where the magic happens.

We use the double-angle identity $\sin (2\theta )=2\sin (\theta )\cos (\theta )$ to expand the numerator: $\sin \left(\frac{2\pi }{n}\right)=2\sin \left(\frac{\pi }{n}\right)\cos \left(\frac{\pi }{n}\right)$. We also write $\tan \left(\frac{\pi }{n}\right)$ as $\frac{\sin (\pi /n)}{\cos (\pi /n)}$.

When we substitute these back, the $\sin (\pi /n)$ terms cancel out, and the $\cos (\pi /n)$ in the denominator flips to the numerator, giving us:

Unlocking the Identity

We are almost there. We need to reach the target expression involving a square root. We use the identity ${\cos }^2(\theta )=\frac{1+\cos (2\theta )}{2}$.

Applying this to our ratio, we get $\frac{I_n}{O_n}=\frac{1+\cos (2\pi /n)}{2}$, which rearranges to:

Now, look at the square root term in the question: $\sqrt{1-{\left(\frac{2I_n}{n}\right)}^2}$. From our initial formula for $I_n$, we know $\frac{2I_n}{n}=\sin \left(\frac{2\pi }{n}\right)$.

Substituting this into the square root gives $\sqrt{1-{\sin }^2\left(\frac{2\pi }{n}\right)}$, which is simply $\sqrt{{\cos }^2\left(\frac{2\pi }{n}\right)}=\cos \left(\frac{2\pi }{n}\right)$.

Substituting this back into our equation, we arrive at the final, elegant proof:

You have successfully bridged the gap between the interior and exterior worlds of the circle!