Calculating the Diameter of an Inscribed Circle in a Regular Dodecagon

The problem asks for the diameter of a circle that is inscribed within a regular polygon having 12 sides, where each side measures 1 cm in length. An inscribed circle in a regular polygon is tangent to all sides of the polygon. The radius of this inscribed circle is equal to the apothem of the regular polygon.

Understanding the Apothem

The apothem of a regular polygon is the distance from the center of the polygon to the midpoint of any side. This distance is perpendicular to the side. For an inscribed circle, the radius is exactly this apothem.

The relationship between the apothem (\(r\)), the side length (\(s\)) of a regular polygon, and the number of sides (\(n\)) is given by the formula:

\[ r = \frac{s}{2 \tan\left(\frac{180^\circ}{n}\right)} \]

Alternatively, using radians:

\[ r = \frac{s}{2 \tan\left(\frac{\pi}{n}\right)} \]

Applying the Formula to the 12-Sided Polygon

In this question, we have a regular polygon with:

• Number of sides, \(n = 12\)
• Side length, \(s = 1\) cm

We need to find the angle \( \frac{180^\circ}{n} \) or \( \frac{\pi}{n} \):

\[ \frac{180^\circ}{12} = 15^\circ \]

Or in radians:

\[ \frac{\pi}{12} \text{ radians} \]

Now, we calculate the apothem (which is the radius of the inscribed circle):

\[ r = \frac{1}{2 \tan(15^\circ)} \]

Calculating the Tangent of 15 degrees

To find the value of \( \tan(15^\circ) \), we can use the tangent subtraction formula: \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \). We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \).

\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ)\tan(30^\circ)} \]

We know that \( \tan(45^\circ) = 1 \) and \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).

\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \]

To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{3}-1 \).

\[ \tan(15^\circ) = \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{(\sqrt{3})^2 - 2\sqrt{3}(1) + (1)^2}{(\sqrt{3})^2 - (1)^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

So, \( \tan(15^\circ) = 2 - \sqrt{3} \).

Calculating the Apothem (Inscribed Circle Radius)

Now substitute the value of \( \tan(15^\circ) \) back into the apothem formula:

\[ r = \frac{1}{2(2 - \sqrt{3})} \]

To rationalize the denominator again, multiply by \( \frac{2 + \sqrt{3}}{2 + \sqrt{3}} \):

\[ r = \frac{1}{2(2 - \sqrt{3})} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{2((2)^2 - (\sqrt{3})^2)} = \frac{2 + \sqrt{3}}{2(4 - 3)} = \frac{2 + \sqrt{3}}{2(1)} = \frac{2 + \sqrt{3}}{2} \]

The radius of the inscribed circle is \( r = \frac{2 + \sqrt{3}}{2} \) cm.

Calculating the Diameter

The diameter of the inscribed circle is twice its radius.

\[ \text{Diameter} = 2r = 2 \times \frac{2 + \sqrt{3}}{2} = 2 + \sqrt{3} \text{ cm} \]

Thus, the diameter of the circle inscribed in a regular polygon of 12 sides, each of length 1 cm, is \( 2 + \sqrt{3} \) cm.

Summary of Calculation Steps

1. Identify the polygon type (regular dodecagon) and given properties (n=12, s=1 cm).
2. Recognize that the radius of the inscribed circle is the apothem of the polygon.
3. Use the apothem formula involving side length and number of sides: \( r = \frac{s}{2 \tan(\frac{180^\circ}{n})} \).
4. Calculate the angle \( \frac{180^\circ}{12} = 15^\circ \).
5. Calculate \( \tan(15^\circ) = 2 - \sqrt{3} \).
6. Substitute the value of \( \tan(15^\circ) \) into the apothem formula to find \( r = \frac{2 + \sqrt{3}}{2} \).
7. Calculate the diameter as \( 2r = 2 + \sqrt{3} \).

Revision Table: Regular Polygons and Inscribed Circles

Additional Information: Trigonometric Values and Dodecagons

The calculation involved finding \( \tan(15^\circ) \). Knowing common trigonometric values for angles like \( 30^\circ, 45^\circ, 60^\circ \) is essential. Values for \( 15^\circ \) (and \( 75^\circ \)) are often derived using sum/difference formulas.

• \( \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} \)
• \( \cos(15^\circ) = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} \)
• \( \tan(15^\circ) = \frac{\sin(15^\circ)}{\cos(15^\circ)} = \frac{(\sqrt{6} - \sqrt{2})/4}{(\sqrt{6} + \sqrt{2})/4} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} = \frac{\sqrt{2}(\sqrt{3}-1)}{\sqrt{2}(\sqrt{3}+1)} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2 - \sqrt{3} \)

A regular polygon with 12 sides is called a dodecagon. Regular dodecagons have unique geometric properties often explored in advanced geometry problems.