PaperCut

Set up a coordinate system with origin at the bottom left corner, so that the square has vertices $(0,0)\,,\,(10,0)\,,\,(10,10)\,,\,(0,10)$ , and the ends of the cutaway edge have coordinates $(0,10-a)$ and $(b,10)$ , where $ab = 60$ .

The centroid $G_1$ of the removed triangle of paper is located at the algebraic mean of the triangle's vertices, and so has coordinates $(\tfrac13b,\tfrac13(30-a))$ . The centroid $G_2$ of the remaining piece of paper has coordinates $(6,y)$ . Thus $30 \left(\begin{array}{c} \tfrac13b \\ \tfrac13(30-a) \end{array} \right) + 70\left(\begin{array}{c} 6 \\ y \end{array}\right) \; = \; 100\left(\begin{array}{c} 5 \\ 5 \end{array} \right)$ Thus $10b + 420 = 500$ , so that $b=8$ , and hence $a = \tfrac{15}{2}$ . Since $10(30 - a) + 70y = 500$ , and hence $y = \tfrac{275}{70} = \tfrac{55}{14}$ , making the answer $55 + 14 = \boxed{69}$ .

Let's consider the picture rotated of $90°$ counterclockwise. It well-known that the centroid of a right triangle has coordinates $\displaystyle C_t=\bigg(x_c=\frac{b}{3}, y_c=\frac{h}{3}\bigg)$ , where $b$ is the base and $h$ the high. Let's prove it for $x_c$ :

$\displaystyle x_c=b-\frac{1}{A}\int_{A}^{} dA=b-\frac{2}{bh} \int_{0}^{b} \int_{0}^{\frac{h}{b}x} xdydx=b-\frac{2}{bh} \int_{0}^{b} \frac{h}{b}x^2 dx=b-\frac{2}{bh}\cdot \frac{h}{b} \cdot \frac{b^3}{3}=\frac{b}{3}$

Now, we know that the centroid of the square is $C_s=(5,5)$ and that the centroid of the blue are is $C_b=(\cdot,6)$ (in the initial configuration the $y$ -coordinate is 4). The area of the square is obviously $A_s=100$ , so, because the area of the triangle $A_t=30$ , the area of the blue area is $A_b=70$ . So, we have that

$\displaystyle \frac{1}{A_s} (A_t \cdot y_{c,t}+A_b \cdot y_{c,b})=5 \hspace{5pt} \Longrightarrow \hspace{5pt} \frac{1}{100} \bigg(\frac{h}{3} \cdot 30+ 6\cdot 70\bigg)=5 \hspace{5pt} \Longrightarrow \hspace{5pt} h=8$

So, since $\displaystyle \frac{bh}{2}=30$ , $\displaystyle b=\frac{15}{2}$ . Hence, $\displaystyle C_t=\bigg(x_c=\frac{5}{2}, y_c=\frac{8}{3}\bigg)$ . Eventually

$\displaystyle \frac{1}{A_s} (A_t \cdot x_{c,t}+A_b \cdot x_{c,b})=5 \hspace{5pt} \Longrightarrow \hspace{5pt} \frac{1}{100} \bigg(\frac{5}{3} \cdot 30+ x_{c,b}\cdot 70\bigg)=5 \hspace{5pt} \Longrightarrow \hspace{5pt} x_{c,b}=\frac{85}{14}$

In the initial configuration it corrisponds to

$\displaystyle 10-\frac{85}{14}=\frac{55}{14} \hspace{5pt} \Longrightarrow \hspace{5pt} 55+14=\boxed{69}$ ,