[Algebra]
Problem: Let $A$ be the number you will receive from position 7 , and let $B$ be the number you will receive from position 9 . There are exactly two ordered pairs of real numbers $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ that satisfy both $|x+y|=6(\sqrt{A}-5)$ and $x^{2}+y^{2}=B^{2}$. Compute $\left|x_{1}\right|+\left|y_{1}\right|+\left|x_{2}\right|+\left|y_{2}\right|$.

Solution: Note that the graph of $x^{2}+y^{2}=B^{2}$ is a circle of radius $|B|$ centered at $(0,0)$ (as long as $\left.B^{2}>0\right)$. Also note that the graph of $|x+y|=6(\sqrt{A}-5)$ is either the line $y=-x$ if $A=25$, or the graph consists of two parallel lines with slope -1 if $A>25$. In the former case, the


line $y=-x$ intersects the circle at the points $\left( \pm \frac{|B|}{\sqrt{2}}, \mp \frac{|B|}{\sqrt{2}}\right)$. In the latter case, the graph is symmetric about the origin, and in order to have exactly two intersection points, each line must be tangent to the circle, and the tangency points are $\left(\frac{|B|}{\sqrt{2}}, \frac{|B|}{\sqrt{2}}\right)$ and $\left(-\frac{|B|}{\sqrt{2}},-\frac{|B|}{\sqrt{2}}\right)$. In either case, $\left|x_{1}\right|+\left|y_{1}\right|+\left|x_{2}\right|+\left|y_{2}\right|=2 \sqrt{2} \cdot|B|$, and in the case where the graph is two lines, this is also equal to $12(\sqrt{A}-5)$. Thus if $A \neq 25$, then only one of $A$ or $B$ is needed to determine the answer. With $A=49$ and $B=6 \sqrt{2}$, the answer is $2 \sqrt{2} \cdot 6 \sqrt{2}=12(\sqrt{49}-5)=\mathbf{2 4}$.

Final answer: 24