AMC12 2015 A

Answer (D): There are 20 possible values for each of $a$ and $b$, namely those in the set $$ S=\left\{0,1,\frac12,\frac32,\frac13,\frac23,\frac43,\frac53,\frac14,\frac34,\frac54,\frac74,\frac15,\frac25,\frac35,\frac45,\frac65,\frac75,\frac85,\frac95\right\}. $$ If $x$ and $y$ are real numbers, then $(x+iy)^2=x^2-y^2+i(2xy)$ is real if and only if $xy=0$, that is, $x=0$ or $y=0$. Therefore $(x+iy)^4$ is real if and only if $x^2-y^2=0$ or $xy=0$, that is, $x=0$, $y=0$, or $x=\pm y$. Thus $(\cos(a\pi)+i\sin(b\pi))^4$ is a real number if and only if $\cos(a\pi)=0$, $\sin(b\pi)=0$, or $\cos(a\pi)=\pm\sin(b\pi)$. If $\cos(a\pi)=0$ and $a\in S$, then $a=\frac12$ or $a=\frac32$ and $b$ has no restrictions, so there are 40 pairs $(a,b)$ that satisfy the condition. If $\sin(b\pi)=0$ and $b\in S$, then $b=0$ or $b=1$ and $a$ has no restrictions, so there are 40 pairs $(a,b)$ that satisfy the condition, but there are 4 pairs that have been counted already, namely $\left(\frac12,0\right)$, $\left(\frac12,1\right)$, $\left(\frac32,0\right)$, and $\left(\frac32,1\right)$. Thus the total so far is $40+40-4=76$. Note that $\cos(a\pi)=\sin(b\pi)$ implies that $\cos(a\pi)=\cos\!\left(\pi\left(\frac12-b\right)\right)$ and thus $a\equiv \frac12-b\pmod 2$ or $a\equiv -\frac12+b\pmod 2$. If the denominator of $b\in S$ is 3 or 5, then the denominator of $a$ in simplified form would be 6 or 10, and so $a\notin S$. If $b=\frac12$ or $b=\frac32$, then there is a unique solution to either of the two congruences, namely $a=0$ and $a=1$, respectively. For every $b\in\left\{\frac14,\frac34,\frac54,\frac74\right\}$, there is exactly one solution $a\in S$ to each of the previous congruences. None of the solutions are equal to each other because if $\frac12-b\equiv -\frac12+b\pmod 2$, then $2b\equiv 1\pmod 2$; that is, $b=\frac12$ or $b=\frac32$. Similarly, $\cos(a\pi)=-\sin(b\pi)=\sin(-b\pi)$ implies that $\cos(a\pi)=\cos\!\left(\pi\left(\frac12+b\right)\right)$ and thus $a\equiv \frac12+b\pmod 2$ or $a\equiv -\frac12-b\pmod 2$. If the denominator of $b\in S$ is 3 or 5, then the denominator of $a$ would be 6 or 10, and so $a\notin S$. If $b=\frac12$ or $b=\frac32$, then there is a unique solution to either of the two congruences, namely $a=1$ and $a=0$, respectively. For every $b\in\left\{\frac14,\frac34,\frac54,\frac74\right\}$, there is exactly one solution $a\in S$ to each of the previous congruences, and, as before, none of these solutions are equal to each other. Thus there are a total of $2+8+2+8=20$ pairs $(a,b)\in S^2$ such that $\cos(a\pi)=\pm\sin(b\pi)$. The requested probability is $\frac{76-20}{400}=\frac{96}{400}=\frac{6}{25}$. Note: By de Moivre’s Theorem the fourth power of the complex number $x+iy$ is real if and only if it lies on one of the four lines $x=0$, $y=0$, $x=y$, or $x=-y$. Then the counting of $(a,b)$ pairs proceeds as above.