Fall 2021 AMC 10 B

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 $a+b+c=108+36+3=147$.

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Alternative solution.

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 $△BCK$ and then subtracting the areas of 4 triangles congruent to $△DJK$. The area of the square is $36$. The area of $△BCK$ is $$\frac{a^2\sqrt{3}}{2}=\frac{(3-\sqrt{3}{)}^2\sqrt{3}}{2}=6\sqrt{3}-9.$$ To find the area of $△DJK$, 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 $△DJK$ 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.