AMC12 2013 A

Answer (C): Assume that the vertices of $ABCD$ are labeled in counterclockwise order. Let $A'$, $B'$, $C'$, and $D'$ be the images of $A$, $B$, $C$, and $D$, respectively, under the rotation. Because $\triangle A'PA$ and $\triangle C'PC$ are isosceles right triangles, points $A'$ and $C'$ are on lines $AB$ and $CD$, respectively. Moreover, because $AP=\sqrt{2}$ and $PC=AC-AP=\sqrt{2}(\sqrt{3}+1)-\sqrt{2}=\sqrt{6}$, it follows that $AA'=\sqrt{2}AP=2$ and $CC'=\sqrt{2}CP=2\sqrt{3}$. By symmetry, points $B'$ and $D'$ are on lines $CD$ and $AB$, respectively. Let $X\ne B$ and $Y\ne D'$ be the intersections of $BC$ and $C'D'$, respectively, with the circle centered at $P$ with radius $PB$. Note that $PD'=PD=PB$, so this circle also contains $D'$. Therefore the required region consists of sectors $APA'$, $BPX$, $CPC'$, and $YPD'$, and triangles $BPA'$, $CPX$, $YPC'$, and $APD'$. Sector $APA'$ has area $\frac14\cdot(\sqrt{2})^2\pi=\frac{\pi}{2}$, and sector $CPC'$ has area $\frac14\cdot(\sqrt{6})^2\pi=\frac{3\pi}{2}$. Let $H$ and $I$ be the midpoints of $AA'$ and $BX$, respectively. Then $PH=AH=\frac{\sqrt{2}}{2}AP=1$, and $PI=HB=AB-AH=\sqrt{3}$. Thus $\triangle BPH$ is a 30-60-90° triangle, implying that $PB=2$ and $\triangle XPB$ is equilateral. Therefore congruent sectors $BPX$ and $YPD'$ each have area $\frac16\cdot2^2\pi=\frac{2\pi}{3}$. Congruent triangles $BPA'$ and $D'PA$ each have altitude $PH=1$ and base $A'B=AB-AH-HA'=\sqrt{3}-1$, so each has area $\frac12(\sqrt{3}-1)$. Congruent triangles $CPX$ and $C'PY$ each have altitude $PI=\sqrt{3}$ and base $XC=BC-BX=\sqrt{3}-1$, so each has area $\frac12(3-\sqrt{3})$. The area of the entire region is $$ \frac{\pi}{2}+\frac{3\pi}{2}+2\cdot\frac{2\pi}{3}+2\left(\frac{\sqrt{3}-1}{2}\right)+2\left(\frac{3-\sqrt{3}}{2}\right)=\frac{10\pi+6}{3}, $$ and $a+b+c=10+6+3=19$.