HMMT February

Using polar coordinates, we can transform the problem to finding the intersections between $r=\cos 2 \theta$ and $r=2 \cos 3 \theta$. Drawing this out gives us a four-leaf clover and a large 3-leaf clover, which intersect at 7 points (one point being the origin). Note that since this graph is symmetric about the $x$ axis, we are only interested in finding the $x$-coordinates, which is $r \cos \theta=\cos 2 \theta \cos \theta=2 \cos ^{3} \theta-\cos \theta$. Now note that all points of intersection satisfy $$\cos 2 \theta=2 \cos 3 \theta \Longleftrightarrow 8 \cos ^{3} \theta-2 \cos ^{2} \theta-6 \cos \theta+1=0$$ Now, we want to compute the sum of $2 \cos ^{3} \theta-\cos \theta$ over all values of $\cos \theta$ that satisfy the above cubic. In other words, if the solutions for $\cos \theta$ to the above cubic are $a, b$, and $c$, we want $2 \sum_{\text {cyc }} 2 a^{3}-a$, since each value for $\cos \theta$ generates two solutions (symmetric about the $x$-axis). This is $$\sum_{\mathrm{cyc}} 4 a^{3}-2 a=\sum_{\mathrm{cyc}} a^{2}+a-\frac{1}{2}$$ where we have used the fact that $a^{3}=a^{2}+3 a-\frac{1}{2}$. By Vieta's formulas, $a+b+c=\frac{1}{4}$, while $$a^{2}+b^{2}+c^{2}=\left(\frac{1}{4}\right)^{2}+2 \cdot \frac{3}{4}=\frac{25}{16}$$ Thus the final answer is $\frac{5}{16}$.