To tackle this problem, we need to delve into the properties of regular polygons and how they fit together at a common vertex. The key concept here is that the angles of the polygons must sum up to 360 degrees at that vertex. Let's break this down step by step.
Understanding the Angles of Regular Polygons
For a regular polygon with \( n \) sides, the measure of each interior angle can be calculated using the formula:
Interior Angle = (n - 2) × 180° / n
Now, if we have three regular polygons with \( a \), \( b \), and \( c \) sides, their respective interior angles will be:
- For polygon with \( a \) sides: \( \frac{(a - 2) \times 180°}{a} \)
- For polygon with \( b \) sides: \( \frac{(b - 2) \times 180°}{b} \)
- For polygon with \( c \) sides: \( \frac{(c - 2) \times 180°}{c} \)
Setting Up the Equation
Since these polygons meet at a common vertex, the sum of their interior angles must equal 360 degrees:
\(\frac{(a - 2) \times 180°}{a} + \frac{(b - 2) \times 180°}{b} + \frac{(c - 2) \times 180°}{c} = 360°\)
We can simplify this equation by dividing everything by 180°:
\(\frac{(a - 2)}{a} + \frac{(b - 2)}{b} + \frac{(c - 2)}{c} = 2\)
Now, rewriting each term gives us:
\(1 - \frac{2}{a} + 1 - \frac{2}{b} + 1 - \frac{2}{c} = 2\)
Combining like terms results in:
\(3 - \left(\frac{2}{a} + \frac{2}{b} + \frac{2}{c}\right) = 2\)
Rearranging this leads us to:
\(\frac{2}{a} + \frac{2}{b} + \frac{2}{c} = 1\)
Dividing through by 2 gives us the desired relationship:
\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}\)
Finding Possible Values for (a, b, c)
Next, we need to find all possible sets of integers \( (a, b, c) \) such that \( a \leq b \leq c \) and satisfy the equation \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2} \).
To do this, we can rearrange the equation:
\( \frac{bc + ac + ab}{abc} = \frac{1}{2}\)
Multiplying through by \( 2abc \) gives:
\( 2(bc + ac + ab) = abc\)
This is a Diophantine equation in three variables. We can start testing small integer values for \( a \), \( b \), and \( c \) to find valid combinations.
Testing Values
Let’s start with \( a = 3 \) (the smallest polygon, a triangle):
- For \( a = 3 \), we have \( \frac{1}{3} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2} \) leading to \( \frac{1}{b} + \frac{1}{c} = \frac{1}{6} \). This gives possible pairs like \( (6, 6) \).
- Next, try \( a = 4 \) (a square): \( \frac{1}{4} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2} \) leads to \( \frac{1}{b} + \frac{1}{c} = \frac{1}{4} \). Possible pairs include \( (4, 4) \) and \( (8, 8) \).
- Continuing this process, we can find combinations for \( a = 5, 6, \) and so on.
Summary of Valid Combinations
After testing various combinations, the valid sets of \( (a, b, c) \) that satisfy the equation and the condition \( a \leq b \leq c \) are:
- \( (3, 3, 6) \)
- \( (3, 4, 12) \)
- \( (4, 4, 4) \)
- \( (4, 6, 12) \)
- \( (5, 5, 10) \)
- \( (6, 6, 6) \)
In conclusion, the relationship \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2} \) holds true for these combinations of regular polygons meeting at a vertex. This exploration not only illustrates the beauty of geometry but also the elegance of mathematical relationships.
