thedudemanguyperson

"Mock AIME" nah its the real aime cheater.

If you draw it, the diameter of the sphere is equal to the edge of the cube.

Cross multiply to get $a-3=-a^2\implies a^2+a-3=0$. Solve this with the quadratic formula.

Let $x$ be the distance from the lower-left vertex to the intersection of the diagonals. Let $y$ be the same distance just for the lower right. Similar triangles gives

$$\frac{x}{35-x}=\frac{y}{120-y}=\frac{75}{50}\implies x=21, y=72$$

Heron's formula tells us that the area of the lower similar triangle (larger one) is $756$. Hence the height of the triangle is $\frac{504}{25}$. Then the height of the other one is $\frac{336}{25}$. So the whole height is $\frac{168}{5}$. The average of the bases is $\frac{125}{2}$. Thus the area of the trapezoid is $2100$.

Let $x$ be the distance from the lower-left vertex to the intersection of the diagonals. Let $y$ be the same distance just for the lower right. Similar triangles gives

$$\frac{x}{35-x}=\frac{y}{120-y}=\frac{75}{50}\implies x=21, y=72$$

Heron's formula tells us that the area of the lower similar triangle (larger one) is $756$. Hence the height of the triangle is $\frac{504}{25}$. Then the height of the other one is $\frac{336}{25}$. So the whole height is $\frac{168}{5}$. The average of the bases is $\frac{125}{2}$. Thus the area of the trapezoid is $2100$.

Just case on the possibilities, there are only $15$ possibilities.

It's not true for all $x$, take $x=\frac{1}{2}$ and $a=b=c=1$. I'm assuming you want to find the minimum $x$ then. I claim this is $x=1$, to do this note that $(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$ so it suffices to show that $a^2+b^2+c^2\ge ab+bc+ac$. This is left as an exercise to the reader, it's easy with AM-GM. 

Since equality occurs at $a=b=c$, we're done.

Hint: Use the triangle inequality, $a+b>c$ for all sides $a,b,c$ of a triangle.