Just hangin' around ......

A quick, cryptic solution ....

The rope will have the same slope on either side of the mass. This means that if the rope were to be extended downward from the top of the $100$ foot pole along the same slope it will have descended $\sqrt{100^{2} - 80^{2}} = 60$ feet, i.e., $30$ feet below the top of the $70$ foot pole. By symmetry, the mass must then be a vertical distance of $\frac{30}{2} = 15$ feet below the top of the shorter pole, (and $45$ feet below the top of the taller pole). Thus the mass will hang a vertical height of $Y = 55$ feet above the ground.

By way of similar triangles, the ratio of the horizontal distances of the mass from the taller and shorter poles will be $3 : 1$ , implying that the mass will be a horizontal distance of $X = 60$ feet from the $100$ foot pole.

We therefore have that $X + Y = \boxed{115}$ .

When the weight is at rest, the parts of the rope on the left and right of the weight has the same tension $T$ and form a same angle $\theta$ with the vertical, so that it has zero resultant horizontal force $T\sin {\theta} - T \sin {\theta} = 0$ .

We note that the length of the rope is given by:

$\quad \dfrac {X}{\sin{\theta}} + \dfrac {80- X}{\sin{\theta}} = 100\quad \Rightarrow \dfrac {80}{\sin{\theta}} = 100$

$\Rightarrow \sin{\theta} = \dfrac {80}{100} = \frac {4}{5} \quad \Rightarrow \cos {\theta} = \frac {3}{5} \quad \Rightarrow \tan {\theta} = \frac {4}{3}$

We also note that:

$\quad \begin {cases} \tan {\theta} = \dfrac {4}{3} = \dfrac {X}{100-Y} & \Rightarrow -4Y = 3X - 400 \\ \tan {\theta} = \dfrac {4}{3} = \dfrac {80-X}{70-Y} & \Rightarrow -4Y = 240-3X - 280 = -3X-40 \end {cases}$

Subtracting the two equations, we have:

$\quad 6X-360 =0 \quad \Rightarrow X = 60$

$\quad \Rightarrow -4Y = 3(60)-400\quad \Rightarrow Y = \dfrac {220} {4} = 55$

Therefore, $\quad X+Y = 60+55 =\boxed{115}$

My solution is a little bit brutal compared with Brian's elegant one.

Assuming that the tension in the rope is constant, the rope must make equal angles with the horizontal on either side of the weight. The two obvious right angled triangles are therefore similar, and we can say

$\frac{x}{100-y}=\frac{80-x}{70-y}$

And using Pythagoras theorem twice we can write the total length of the rope as

$\sqrt{x^2+(100-y)^2}+\sqrt{(80-x)^2+(70-y)^2}=100$

Asking Mathematica to solve these two equations simultaneously then gives $(x,y) = (60,55)$

and so $x+y=\boxed{115}$

$\text{Solve}\left[\frac{100-y}{x}=\frac{70-y}{80-x}\land \text{EuclideanDistance}[\{0,100\},\{x,y\}]+\text{EuclideanDistance}[\{x,y\},\{80,70\}]=100\land 0\leq x\leq 80\land 0\leq y\leq 70\right] \Rightarrow \\ x\to 60,y\to 55 \Rightarrow 115$ .

The first subexpression contains the information that the angles are equal, the second rope length and the third and fourth to preclude unreal answers.