a dog tied to a pole by a rope A square hole of depth $h$ whose base is of length $a$ is given.
A dog is tied to the center of the square at the bottom of the hole by a rope of length 
$L>\sqrt{2a^2+h^2}$ ,and walks on the ground around the hole.
The edges of the hole are smooth, so that the rope can freely slide along it. Find the shape and area of the territory accessible to the dog (whose size is neglected).
 A: It can move any where inside the pit 
now for L > $ \sqrt{2*a^2 +h^2}$
so it can very well peep out of the pit (from being inside ) because $ \sqrt{2*a^2 +h^2}$ is the length of hypotenuse with one leg as diagonal of square and other being height .
so shape will be circle on top , but shape of area it is  a cone with lateral height greater than $ \sqrt{a^2/2 +h^2}$  
If the dog cannot jump , then it is circle on top . Just it is a circle area around square top .
But practical case , is that , as the distance from centre of square to its circumference is the maximum at corner , then it decreases as it moves from one corner to another , then again reaches maximum when it reaches other corner so this happens for all through the square . so at the corner the dog sweeps less area(as more rope gets used even to peep out) as it moves along side area swept increases and it is maximum at mid point of the side of square then again starts decreasing then it is back to minimum at the next corner . so the top are swept will in real sense cannot be a circle , but a irregular figure may be like an ellipse 

A: Consider the rope touches the edge of the hole at point $p$ with distance $x$ from the corner. Then the distance from the projection of the point $p$ on the edge to the point on the bottom $p^\prime$ is $a$.
The distance between $p^\prime$ to the center of the edge of bottom it lands on is $a/2$. Hence the distance from $o$ to $p^\prime$ is
$$\sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2} - x\right)^2}.$$
Therefore the distance from $o$ to $p$ is
$$\sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2} - x\right)^2 + a^2}.$$
Hence, on the ground, the dog can move around the half disk centered at point $p$ with radius $$L - \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2} - x\right)^2 + a^2}.$$
Here is the construction, but the solution is the union of the half disks, plus the bottom.
A way to think of the union, is consider the radius changes smoothly along each egdge.
A: I tried a simplified version of this problem, where the dog is facing a cliff of height $h$, which I represent as the line $y=\frac{a}{2}$. In other words, considering only one side of the square, and extending it to form a line.
After a bunch of hastily scribbled algebra, I arrived at the following curve describing the border.
$$
L-\sqrt{h^2+\left(\frac{\frac{a}{2}x}{y}\right)^2+\left(\frac{a}{2}\right)^2} = \sqrt{\left(x-\frac{\frac{a}{2}x}{y}\right)^2+(y-\frac{a}{2})^2}
$$
A: Let ABCD be the upper square of the hole. We will determine the locus along a side at one time. For this, let O be the centre of the bottom square. Let T be the variable point on the edge AB through which the string passes. And let P be the point where the goat is standing, such that the string remains taut.
The edges are slippery, so for the string to remain at the position, it has to be such that the points O,P,T lie on a straight line when viewed from the top. So, let the centre of the top square be O'.
When viewed from top, $O'\equiv O$ and therefore O'PT is a straight line.
Note that the length of the string inside the hole remains $\sqrt{\frac{a^2}{4}+h^2}=\frac 12\sqrt{a^2+4h^2}$ by Pythagoras' theorem. So the length of the string on the top plane is always constant, say $\ell$. Now we're going to find the locus of P such that $TP=\ell$ with coordinate geometry. 
Let $O'\equiv (0,0)$ and the axes are parallel to the sides of the square. If $P\equiv (h,k)$ then $T\equiv PO'\cap AB\equiv \{hy=kx\} \cap \{x=\frac a2\} \equiv \left(\frac a2,\frac{ka}{2h}\right)$. Then $PT^2=\ell^2=\left(h-\frac a2\right)^2+\left(k-\frac{ka}{2h}\right)^2$; so that the locus turns out to be (where a=2b)$
(x-b)^2+\frac{y^2(x-b)^2}{x^2}=\ell^2$
The attachment is a plot for $b=1, \ell =2$, thanks to WolframAlpha.
So all we need to do is, reflect the right branch of this plot about the axes to get the desired locus. $\Box$
I don't know what this shape is called, or how you can even find this shape without a plotting software...
