Problem 911

Let $O$ be the center of the polygon, and label 11 points as shown in the figure. Let $a = AB = BC$.

Triangle $BCK$ is a $30$-$60$-$90^\circ$ triangle, so $BK = 2a$ and $CK = KG = a\sqrt{3}$. Then $AG = 3a + a\sqrt{3} = 6$, so $a = 3 - \sqrt{3}$. The area of the 24-sided polygon can be computed as $12$ times the area of kite $OBCD$. The longer diagonal of this kite is $OC$, half of a diagonal of the square, so $OC = 3\sqrt{2}$. The shorter diagonal of the kite is $BD$, the hypotenuse of isosceles right triangle $BCD$ with leg $a = 3 - \sqrt{3}$. The area of a kite is half the product of the lengths of its diagonals, so the area of the 24-sided polygon is
$$12 \cdot \frac{1}{2} \cdot 3\sqrt{2} \cdot (3 - \sqrt{3}) \sqrt{2} = 108 - 36\sqrt{3}.$$
Therefore $a + b + c = 108 + 36 + 3 = 147$.

Label the points as in the first solution, where it was shown that $a = AB = BC = 3 - \sqrt{3}$. The area of the 24-sided polygon can be found by adding to the area of square $AGHI$ the areas of 8 triangles congruent to $\triangle BCK$ and then subtracting the areas of 4 triangles congruent to $\triangle DJK$. The area of the square is $36$. The area of $\triangle BCK$ is
$$\frac{a^2 \sqrt{3}}{2} = \frac{(3 - \sqrt{3})^2 \sqrt{3}}{2} = 6\sqrt{3} - 9.$$
To find the area of $\triangle DJK$, note that it is an isosceles triangle with vertex angle $120^\circ$ and with base $JK = 6 - 2a\sqrt{3} = 12 - 6\sqrt{3}$. The length of the altitude to the base is then
$$\frac{12 - 6\sqrt{3}}{2\sqrt{3}} = 2\sqrt{3} - 3.$$
Thus the area of $\triangle DJK$ is
$$\frac{1}{2} \cdot (12 - 6\sqrt{3}) \cdot (2\sqrt{3} - 3) = 21\sqrt{3} - 36.$$
Finally, the area of the polygon is
$$36 + 8(6\sqrt{3} - 9) - 4(21\sqrt{3} - 36) = 108 - 36\sqrt{3}.$$
Therefore $a + b + c = 147$, as above.