Animated Solution for Mathematics - Trigonometry: For a regular polygon, let $r$ and $R$ be the radii of the inscribed and the circumscribed circles. A false statement among the following is

Unlocking the Regular Polygon

Welcome, fellow traveler of the geometric realm! Today, we are going to peel back the layers of a beautiful problem involving regular polygons.

At first glance, it might seem like a simple exercise in trigonometry, but it is actually a profound exploration of how discrete constraints—the fact that a polygon must have an integer number of sides—dictate the physical reality of shapes.

Imagine you are standing at the center of a regular polygon with $n$ sides. You look out toward the vertices and the sides. The circle that perfectly kisses all the vertices is the circumcircle with radius $R$, and the circle that nestles inside, touching the midpoint of every side, is the incircle with radius $r$.

Our goal is to find the relationship between these two radii and the number of sides $n$.

Where Geometry Meets Algebra

To solve this, we must isolate the core geometric unit. Focus on the center of the polygon.

If we draw a line from the center to a vertex, we have the circumradius $R$. If we draw a line from the center to the midpoint of a side, we have the inradius $r$. These two lines, along with the segment connecting the midpoint to the vertex, form a right-angled triangle.

The total angle around the center is $2\pi$. Since there are $n$ sides, each side subtends an angle of $\frac{2\pi }{n}$ at the center.

Our inradius $r$, being the perpendicular bisector of the side, also bisects this central angle. Thus, the angle in our right-angled triangle is $\frac{\pi }{n}$.

Now, the trigonometry becomes elegant:

This is our master equation, the bridge between the physical shape and the numerical ratio.

Testing the Options

Now, we act as detectives. We have our master formula: $\cos \left(\frac{\pi }{n}\right)=\frac{r}{R}$. We need to check which of the given ratios is impossible.

Let's test them one by one:

For option (a), $\frac{r}{R}=\frac{1}{\sqrt{2}}$. We know that $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$, so $\frac{\pi }{n}=\frac{\pi }{4}$, which gives $n=4$. A square is a perfectly valid regular polygon.

For option (c), $\frac{r}{R}=\frac{\sqrt{3}}{2}$. We know that $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$, so $n=6$. A regular hexagon exists!

For option (d), $\frac{r}{R}=\frac{1}{2}$. We know that $\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$, so $n=3$. An equilateral triangle is the simplest regular polygon.

Finally, we arrive at option (b), where $\frac{r}{R}=\frac{2}{3}\approx 0.666$. We need to find an integer $n$ such that $\cos \left(\frac{\pi }{n}\right)=0.666$.

Let's look at the values we know: $\cos \left(\frac{\pi }{3}\right)=0.5$ and $\cos \left(\frac{\pi }{4}\right)\approx 0.707$. Our value of $0.666$ lies strictly between $0.5$ and $0.707$.

This implies that $n$ must lie between $3$ and $4$. But $n$ must be an integer! There is no polygon with $3.5$ sides.

This is the beauty of the problem: the math tells us that such a polygon cannot exist. Therefore, the statement in option (b) is the false one.

Keep this logic in your toolkit—whenever you see a problem involving discrete structures, look for the constraints that force the variables into the integer domain. You have mastered the geometry of the polygon!