A regular a-gon, a regular b-gon, a regular c-gon, and a regular d-gon fit perfectly around a point. What is the largest possible value of c?

Answer 1

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 $n\ge 3$ 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 $\frac{11}{12}$:

$$\frac{1}{c}=\frac{1}{12}⟹c=12$$

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:

$$\frac{2}{6}+\frac{3}{6}+\frac{1}{c}=1⟹\frac{5}{6}+\frac{1}{c}=1$$

Subtracting $\frac{5}{6}$:

$$\frac{1}{c}=\frac{1}{6}⟹c=6$$

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:

$$\boxed{{\displaystyle 12}}$$