To find the largest possible value of \( c \) when a regular \( a \)-gon, \( b \)-gon, \( c \)-gon, and \( d \)-gon fit perfectly around a point, we start by considering the conditions of the polygons fitting around a point. The sum of the interior angles around that point must equal \( 360^\circ \).
The interior angle of a regular \( n \)-gon can be calculated using the formula:
\[ \text{Interior angle} = \frac{(n-2) \times 180^\circ}{n} = 180^\circ - \frac{360^\circ}{n} \]
Thus, the exterior angle of a regular \( n \)-gon is given by:
\[ \text{Exterior angle} = 180^\circ - \text{Interior angle} = \frac{360^\circ}{n} \]
For the polygons to fit perfectly around a point, their exterior angles must sum to \( 360^\circ \). Therefore, we have:
\[ \frac{360^\circ}{a} + \frac{360^\circ}{b} + \frac{360^\circ}{c} + \frac{360^\circ}{d} = 360^\circ \]
Dividing through by \( 360^\circ \) gives:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 1 \]
To maximize \( c \), we need to minimize the other variables \( a \), \( b \), and \( d \). The smallest values for \( a \), \( b \), and \( d \) while maintaining their status as polygons (where \( n \geq 3 \) for an \( n \)-gon) are:
- \( a = 3 \) (triangle)
- \( b = 3 \) (triangle)
- \( d = 3 \) (triangle)
Plugging in these values into the equation:
\[ \frac{1}{3} + \frac{1}{3} + \frac{1}{c} + \frac{1}{3} = 1 \]
This simplifies to:
\[ 1 + \frac{1}{c} = 1 \]
Subtracting 1 from both sides gives:
\[ \frac{1}{c} = 0 \]
However, since \( c \) must be a finite positive integer representing a polygon, we must choose different values for \( a \), \( b \), and \( d \). Next, we can set \( a = 3 \), \( b = 3 \), and \( d = 4 \) (a triangle and a square):
\[ \frac{1}{3} + \frac{1}{3} + \frac{1}{c} + \frac{1}{4} = 1 \]
Calculating the left-hand side:
\[ \frac{2}{3} + \frac{1}{4} + \frac{1}{c} = 1 \]
Finding a common denominator for \( \frac{2}{3} \) and \( \frac{1}{4} \):
\[ \frac{8}{12} + \frac{3}{12} = \frac{11}{12} \]
Thus, we have:
\[ \frac{11}{12} + \frac{1}{c} = 1 \]
Subtracting \( \frac{11}{12} \):
\[ \frac{1}{c} = \frac{1}{12} \implies c = 12 \]
To verify, if \( a = 3 \), \( b = 3 \), \( d = 4 \), and \( c = 12\):
\[ \frac{1}{3} + \frac{1}{3} + \frac{1}{12} + \frac{1}{4} = \frac{2}{3} + \frac{1}{12} + \frac{3}{12} = \frac{2}{3} + \frac{4}{12} = \frac{8}{12} + \frac{4}{12} = 1 \]
This holds true. Now, to see if we can go higher, we will test combinations with \( a = 3 \), \( b = 4 \), and \( d = 4 \) (two squares).
This gives:
\[ \frac{1}{3} + \frac{1}{4} + \frac{1}{4} + \frac{1}{c} = 1 \]
Calculating:
\[ \frac{1}{3} + \frac{1}{2} + \frac{1}{c} = 1 \]
Finding a common denominator:
\[ \frac{2}{6} + \frac{3}{6} + \frac{1}{c} = 1 \implies \frac{5}{6} + \frac{1}{c} = 1 \]
Subtracting \( \frac{5}{6} \):
\[ \frac{1}{c} = \frac{1}{6} \implies c = 6 \]
Thus confirming:
The greatest feasible value at this point remains confirmed at \( c \).
Concluding, the largest possible value of \( c \) such that the regular polygons can fit perfectly around a point is:
\[ \boxed{12} \]
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