Problem (Geometry):
Let $T=T N Y W R$. The parabola $y=x^{2}+T x$ is tangent to the parabola $y=-(x-2 T)^{2}+b$. Compute $b$.

Solution:
In this case, the two parabolas are tangent exactly when the system of equations has a unique solution. (Query: Is this the case for every pair of equations representing parabolas?) So set the right sides equal to each other: $x^{2}+T x=-(x-2 T)^{2}+b$. Then $x^{2}+T x=$ $-x^{2}+4 T x-4 T^{2}+b$, or equivalently, $2 x^{2}-3 T x+4 T^{2}-b=0$. The equation has a double root when the discriminant is 0 , so set $(-3 T)^{2}-4\left(4 T^{2}-b\right)(2)=0$ and solve: $9 T^{2}-32 T^{2}+8 b=0$ implies $-23 T^{2}+8 b=0$, or $b=23 T^{2} / 8$. Using $T=8$ yields $b=\mathbf{1 8 4}$.

Final answer: 184