# Fat object (geometry)

In geometry, a fat object is an object in two or more dimensions, whose lengths in the different dimensions are similar. For example, a square is fat because its length and width are identical. A 2-by-1 rectangle is thinner than a square, but it is fat relative to a 10-by-1 rectangle. Similarly, a circle is fatter than a 1-by-10 ellipse and an equilateral triangle is fatter than a very obtuse triangle.
Fat objects are especially important in computational geometry. Many algorithms in computational geometry can perform much better if their input consists of only fat objects; see the applications section below.

## Global fatness

Given a constant R ≥ 1, an object o is called R-fat if its "slimness factor" is at most R. The "slimness factor" has different definitions in different papers. A common definition is:
    ${\frac {{\text{side of smallest cube enclosing}}\ o}{{\text{side of largest cube enclosed in}}\ o}}$
where o and the cubes are d-dimensional. A 2-dimensional cube is a square, so the slimness factor of a square is 1 (since its smallest enclosing square is the same as its largest enclosed disk). The slimness factor of a 10-by-1 rectangle is 10. The slimness factor of a circle is √2. Hence, by this definition, a square is 1-fat but a disk and a 10×1 rectangle are not 1-fat. A square is also 2-fat (since its slimness factor is less than 2), 3-fat, etc. A disk is also 2-fat (and also 3-fat etc.), but a 10×1 rectangle is not 2-fat. Every shape is ∞-fat, since by definition the slimness factor is always at most ∞.
The above definition can be termed two-cubes fatness since it is based on the ratio between the side-lengths of two cubes. Similarly, it is possible to define two-balls fatness, in which a d-dimensional ball is used instead. A 2-dimensional ball is a disk. According to this alternative definition, a disk is 1-fat but a square is not 1-fat, since its two-balls-slimness is √2.
An alternative definition, that can be termed enclosing-ball fatness (also called "thickness") is based on the following slimness factor:
    $\left({\frac {{\text{volume of smallest ball enclosing}}\ o}{{\text{volume of}}\ o}}\right)^{1/d}$
The exponent ⁠1/d⁠ makes this definition a ratio of two lengths, so that it is comparable to the two-balls-fatness.
Here, too, a cube can be used instead of a ball.
Similarly it is possible to define the enclosed-ball fatness based on the following slimness factor:
    $\left({\frac {{\text{volume of}}\ o}{{\text{volume of largest ball enclosed in}}\ o}}\right)^{1/d}$

### Enclosing-fatness vs. enclosed-fatness

The enclosing-ball/cube-slimness might be very different from the enclosed-ball/cube-slimness.
For example, consider a lollipop with a candy in the shape of a 1 × 1 square and a stick in the shape of a b × ⁠1/b⁠ rectangle (with b > 1 > ⁠1/b⁠). As b increases, the area of the enclosing cube (≈ b$^2$) increases, but the area of the enclosed cube remains constant (=1) and the total area of the shape also remains constant (=2). Thus the enclosing-cube-slimness can grow arbitrarily while the enclosed-cube-slimness remains constant (=√2). See this GeoGebra page for a demonstration.
On the other hand, consider a rectilinear 'snake' with width ⁠1/b⁠ and length b, that is entirely folded within a square of side length 1. As b increases, the area of the enclosed cube (≈ ⁠1/b$^2$⁠) decreases, but the total areas of the snake and of the enclosing cube remain constant (=1). Thus the enclosed-cube-slimness can grow arbitrarily while the enclosing-cube-slimness remains constant (=1).
With both the lollipop and the snake, the two-cubes-slimness grows arbitrarily, since in general:
$\left({{\text{enclosing-ball-}} \atop {\text{slimness}}}\right)\times \left({{\text{enclosed-ball-}} \atop {\text{slimness}}}\right)=\left({{\text{two-balls-}} \atop {\text{slimness}}}\right)$ $\left({{\text{enclosing-cube-}} \atop {\text{slimness}}}\right)\times \left({{\text{enclosed-cube-}} \atop {\text{slimness}}}\right)=\left({{\text{two-cubes-}} \atop {\text{slimness}}}\right)$
Since all slimness factor are at least 1, it follows that if an object o is R-fat according to the two-balls/cubes definition, it is also R-fat according to the enclosing-ball/cube and enclosed-ball/cube definitions (but the opposite is not true, as exemplified above).

### Balls vs. cubes

The volume of a d-dimensional ball of radius r is: $V_{d}\cdot r^{d},$ where V$_d$ is a dimension-dependent constant: $V_{d}={\frac {\pi ^{d/2}}{\Gamma \left({\frac {d}{2}}+1\right)}}$
A d-dimensional cube with side-length 2a has volume (2a)$^d$. It is enclosed in a d-dimensional ball with radius $a{\sqrt {d}}$ whose volume is $V_{d}\left(a{\sqrt {d}}\right)^{d}.$ Hence for every d-dimensional object:
$\left({{\text{enclosing-ball-}} \atop {\text{slimness}}}\right)\leq \left({{\text{enclosing-cube-}} \atop {\text{slimness}}}\right)\times {\sqrt[{d}]{V_{d}}}\times {\frac {\sqrt {d}}{2}}$
For even dimensions (d = 2k), the factor simplifies to: ${\sqrt {\tfrac {\pi k}{2}}}\left/{\sqrt[{2k}]{k!}}\right..$ In particular, for two-dimensional shapes V$_2$ = π and the factor is: ${\sqrt {\tfrac {\pi }{2}}}\approx 1.25,$ so:
$\left({{\text{enclosing-disk-}} \atop {\text{slimness}}}\right)\leq \left({{\text{enclosing-square-}} \atop {\text{slimness}}}\right)\times 1.25$
From similar considerations:
$\left({{\text{enclosed-cube-}} \atop {\text{slimness}}}\right)\leq \left({{\text{enclosed-ball-}} \atop {\text{slimness}}}\right)\times {\sqrt[{d}]{V_{d}}}\times {\frac {\sqrt {d}}{2}}$ $\left({{\text{enclosed-square-}} \atop {\text{slimness}}}\right)\leq \left({{\text{enclosed-disk-}} \atop {\text{slimness}}}\right)\times 1.25$
A d-dimensional ball with radius a is enclosed in a d-dimensional cube with side-length 2a. Hence for every d-dimensional object:
$\left({{\text{enclosing-cube-}} \atop {\text{slimness}}}\right)\leq \left({{\text{enclosing-ball-}} \atop {\text{slimness}}}\right)\times {\frac {2}{\sqrt[{d}]{V_{d}}}}$
For even dimensions (d = 2k), the factor simplifies to: 2/²ᵏ√((k!)) /√π.{\textstyle 2\left/{\sqrt[{2k}]{(k!)}}\ \right/{\sqrt {\pi }}.} In particular, for two-dimensional shapes the factor is: ${\tfrac {2}{\sqrt {\pi }}}\approx 1.13,$ so:
$\left({{\text{enclosing-square-}} \atop {\text{slimness}}}\right)\leq \left({{\text{enclosing-disk-}} \atop {\text{slimness}}}\right)\times 1.13$
From similar considerations:
$\left({{\text{enclosed-ball-}} \atop {\text{slimness}}}\right)\leq \left({{\text{enclosed-cube-}} \atop {\text{slimness}}}\right)\times {\frac {2}{\sqrt[{d}]{V_{d}}}}$ $\left({{\text{enclosed-disk-}} \atop {\text{slimness}}}\right)\leq \left({{\text{enclosed-square-}} \atop {\text{slimness}}}\right)\times 1.13$
Multiplying the above relations gives the following simple relations:
$\left({{\text{two-balls-}} \atop {\text{slimness}}}\right)\leq \left({{\text{two-cubes-}} \atop {\text{slimness}}}\right)\times {\sqrt {d}}$ $\left({{\text{two-cubes-}} \atop {\text{slimness}}}\right)\leq \left({{\text{two-balls-}} \atop {\text{slimness}}}\right)\times {\sqrt {d}}$
Thus, an R-fat object according to the either the two-balls or the two-cubes definition is at most ⁠$R{\sqrt {d}}$⁠-fat according to the alternative definition.

## Local fatness

The above definitions are all global in the sense that they don't care about small thin areas that are part of a large fat object.
For example, consider a lollipop with a candy in the shape of a 1 × 1 square and a stick in the shape of a 1 × ⁠1/b⁠ rectangle (with b > 1 > ⁠1/b⁠). As b increases, the area of the enclosing cube (=4) and the area of the enclosed cube (=1) remain constant, while the total area of the shape changes only slightly (= 1 + ⁠1/b⁠). Thus all three slimness factors are bounded:
${\begin{aligned}\left({{\text{enclosing-cube-}} \atop {\text{slimness}}}\right)&\leq 2,\\[4pt]\left({{\text{enclosed-cube-}} \atop {\text{slimness}}}\right)&\leq 2,\\[4pt]\left({{\text{two-cube-}} \atop {\text{slimness}}}\right)&=2.\end{aligned}}$
Thus by all definitions the lollipop is 2-fat. However, the stick-part of the lollipop obviously becomes thinner and thinner.
In some applications, such thin parts are unacceptable, so local fatness, based on a local slimness factor, may be more appropriate. For every global slimness factor, it is possible to define a local version. For example, for the enclosing-ball-slimness, it is possible to define the local-enclosing-ball slimness factor of an object o by considering the set B of all balls whose center is inside o and whose boundary intersects the boundary of o (i.e. not entirely containing o). The local-enclosing-ball-slimness factor is defined as:
${\frac {1}{2}}\cdot \sup _{b\in B}\left({\frac {{\text{volume}}(B)}{{\text{volume}}(B\cap o)}}\right)^{1/d}$
The 1/2 is a normalization factor that makes the local-enclosing-ball-slimness of a ball equal to 1. The local-enclosing-ball-slimness of the lollipop-shape described above is dominated by the 1 × ⁠1/b⁠ stick, and it goes to ∞ as b grows. Thus by the local definition the above lollipop is not 2-fat.

### Global vs. local definitions

Local-fatness implies global-fatness. Here is a proof sketch for fatness based on enclosing balls. By definition, the volume of the smallest enclosing ball is ≤ the volume of any other enclosing ball. In particular, it is ≤ the volume of any enclosing ball whose center is inside o and whose boundary touches the boundary of o. But every such enclosing ball b is in the set B considered by the definition of local-enclosing-ball slimness. Hence:
${\begin{aligned}\left({{\text{enclosing-ball-}} \atop {\text{slimness}}}\right)^{d}&={\frac {\text{volume(smallest enclosing ball)}}{{\text{volume}}(o)}}\\[4pt]&\leq {\frac {{\text{volume}}(b\in B)}{{\text{volume}}(o)}}\\[4pt]&={\frac {{\text{volume}}(b\in B)}{{\text{volume}}(b\cap o)}}\\[4pt]&\leq \left[2\times \left({{\text{local-enclosing-}} \atop {\text{ball-slimness}}}\right)\right]^{d}\end{aligned}}$
Hence:
$\left({{\text{enclosing-ball-}} \atop {\text{slimness}}}\right)\leq \left({{\text{local-enclosing-}} \atop {\text{ball-slimness}}}\right)\times 2$
For a convex body, the opposite is also true: local-fatness implies global-fatness. The proof is based on the following lemma. Let o be a convex object. Let P be a point in o. Let b and B be two balls centered at P such that b is smaller than B. Then o intersects a larger portion of b than of B, i.e.:
${\frac {{\text{volume}}\ (b\cap o)}{{\text{volume}}\ (b)}}\geq {\frac {{\text{volume}}\ (B\cap o)}{{\text{volume}}\ (B)}}$
Proof sketch: standing at the point P, we can look at different angles θ and measure the distance to the boundary of o. Because o is convex, this distance is a function, say r(θ). We can calculate the left-hand side of the inequality by integrating the following function (multiplied by some determinant function) over all angles: $f(\theta )=\min {\left({\frac {r(\theta )}{{\text{radius}}\ (b)}},1\right)}$
Similarly we can calculate the right-hand side of the inequality by integrating the following function: $F(\theta )=\min {\left({\frac {r(\theta )}{{\text{radius}}\ (B)}},1\right)}$
By checking all 3 possible cases, it is possible to show that always f(θ) ≥ F(θ). Thus the integral of f is at least the integral of F, and the lemma follows.
The definition of local-enclosing-ball slimness considers all balls that are centered in a point in o and intersect the boundary of o. However, when o is convex, the above lemma allows us to consider, for each point in o, only balls that are maximal in size, i.e., only balls that entirely contain o (and whose boundary intersects the boundary of o). For every such ball b:
${\text{volume}}\ (b)\leq C_{d}\cdot {\text{diameter}}\ (o)^{d}$
where C$_d$ is some dimension-dependent constant.
The diameter of o is at most the diameter of the smallest ball enclosing o, and the volume of that ball is: $C_{d}\cdot \left({\frac {{\text{diameter(smallest ball enclosing}}\ o)}{2}}\right)^{d}$
Combining all inequalities gives that for every convex object:
$\left({{\text{local-enclosing-}} \atop {\text{ball-slimness}}}\right)\leq \left({{\text{enclosing-ball-}} \atop {\text{slimness}}}\right)$
For non-convex objects, this inequality of course doesn't hold, as exemplified by the lollipop above.

## Examples

The following table shows the slimness factor of various shapes based on the different definitions. The two columns of the local definitions are filled with "*" when the shape is convex (in this case, the value of the local slimness equals the value of the corresponding global slimness):
| Shape | two-balls | two-cubes | enclosing-ball | enclosing-cube | enclosed-ball | enclosed-cube | local-enclosing-ball | local-enclosing-cube |
| square | ${\sqrt {2}}$ | $1$ | ${\sqrt {\frac {\pi }{2}}}\approx 1.25$ | $1$ | ${\sqrt {\frac {4}{\pi }}}\approx 1.13$ | $1$ | * | * |
| b × a rectangle with b > a | ${\sqrt {1+{\frac {b^{2}}{a^{2}}}}}$ | ${\frac {b}{a}}$ | ${\frac {\sqrt {\pi }}{2}}\left({\frac {a}{b}}+{\frac {b}{a}}\right)$ | ${\sqrt {\frac {b}{a}}}$ | $2{\sqrt {\frac {b}{a\pi }}}$ | ${\sqrt {\frac {b}{a}}}$ | * | * |
| disk | $1$ | ${\sqrt {2}}$ | $1$ | ${\sqrt {\frac {4}{\pi }}}\approx 1.13$ | $1$ | ${\sqrt {\frac {\pi }{2}}}\approx 1.25$ | * | * |
| ellipse with radii b > a | ${\frac {b}{a}}$ | $>{\frac {b}{a}}$ | ${\sqrt {\frac {b}{a}}}$ | $>{\sqrt {\frac {b}{2\pi a}}}$ | ${\sqrt {\frac {b}{a}}}$ | $>{\sqrt {\frac {\pi b}{a}}}$ | * | * |
| semi-ellipse with radii b > a, halved in parallel to b | ${\frac {2b}{a}}$ | $>{\frac {2b}{a}}$ | ${\sqrt {\frac {2b}{a}}}$ | $>{\sqrt {\frac {4b}{\pi a}}}$ | ${\sqrt {\frac {2b}{a}}}$ | $>{\sqrt {\frac {2\pi b}{a}}}$ | * | * |
| semidisk | $2$ | ${\sqrt {5}}$ | ${\sqrt {2}}$ | ${\sqrt {\frac {8}{\pi }}}\approx 1.6$ | ${\sqrt {2}}$ | ${\sqrt {\frac {5\pi }{8}}}\approx 1.4$ | * | * |
| equilateral triangle | | $1+{\frac {2}{\sqrt {3}}}\approx 2.15$ | ${\sqrt {\frac {\pi }{\sqrt {3}}}}\approx 1.35$ | ${\sqrt {\frac {4}{\sqrt {3}}}}\approx 1.52$ | | ${\frac {\sqrt[{4}]{3}}{2}}+{\frac {1}{\sqrt[{4}]{3}}}\approx 1.42$ | * | * |
| isosceles right-angled triangle | ${\frac {1}{{\sqrt {2}}-1}}\approx 2.4$ | $2$ | | ${\sqrt {2}}$ | | ${\sqrt {2}}$ | * | * |
| 'lollipop' made of unit square and b × a stick, b > 1 > a | | $b+1$ | | ${\sqrt {\frac {(b+1)^{2}}{ab+1}}}$ | | ${\sqrt {ab+1}}$ | | ${\sqrt {\frac {b}{a}}}$ |

## Fatness of a triangle

Slimness is invariant to scale, so the slimness factor of a triangle (as of any other polygon) can be presented as a function of its angles only. The three ball-based slimness factors can be calculated using well-known trigonometric identities.

### Enclosed-ball slimness

The largest circle contained in a triangle is called its incircle. It is known that:
$\Delta =r^{2}\cdot \left(\cot {\frac {\angle A}{2}}+\cot {\frac {\angle B}{2}}+\cot {\frac {\angle C}{2}}\right)$
where Δ is the area of a triangle and r is the radius of the incircle. Hence, the enclosed-ball slimness of a triangle is:
${\sqrt {\frac {\cot {\frac {\angle A}{2}}+\cot {\frac {\angle B}{2}}+\cot {\frac {\angle C}{2}}}{\pi }}}$

### Enclosing-ball slimness

The smallest containing circle for an acute triangle is its circumcircle, while for an obtuse triangle it is the circle having the triangle's longest side as a diameter.
It is known that:
$\Delta =R^{2}\cdot 2\sin A\sin B\sin C$
where again Δ is the area of a triangle and R is the radius of the circumcircle. Hence, for an acute triangle, the enclosing-ball slimness factor is:
${\sqrt {\frac {\pi }{2\sin A\sin B\sin C}}}$
It is also known that:
$\Delta ={\frac {c^{2}}{2(\cot \angle {A}+\cot \angle {B})}}={\frac {c^{2}(\sin \angle {A})(\sin \angle {B})}{2\sin(\angle {A}+\angle {B})}}$
where c is any side of the triangle and A, B are the adjacent angles. Hence, for an obtuse triangle with acute angles A and B (and longest side c), the enclosing-ball slimness factor is:
${\sqrt {\frac {\pi \cdot (\cot \angle {A}+\cot \angle {B})}{2}}}={\sqrt {\frac {\pi \cdot \sin(\angle {A}+\angle {B})}{2(\sin \angle {A})(\sin \angle {B})}}}$
Note that in a right triangle, $\sin {\angle {C}}=\sin(\angle {A}+\angle {B})=1,$ so the two expressions coincide.

### Two-balls slimness

The inradius r and the circumradius R are connected via a couple of formulae which provide two alternative expressions for the two-balls slimness of an acute triangle:
${\frac {R}{r}}={\frac {1}{4\sin {\frac {\angle {A}}{2}}\sin {\frac {\angle {B}}{2}}\sin {\frac {\angle {C}}{2}}}}={\frac {1}{\cos \angle {A}+\cos \angle {B}+\cos \angle {C}-1}}$
For an obtuse triangle, c/2 should be used instead of R. By the law of sines:
${\frac {c}{2}}=R\sin {\angle {C}}$
Hence the slimness factor of an obtuse triangle with obtuse angle C is:
${\frac {c/2}{r}}={\frac {\sin {\angle {C}}}{4\sin {\frac {\angle {A}}{2}}\sin {\frac {\angle {B}}{2}}\sin {\frac {\angle {C}}{2}}}}={\frac {\sin {\angle {C}}}{\cos \angle {A}+\cos \angle {B}+\cos \angle {C}-1}}$
Note that in a right triangle, $\sin {\angle {C}}=1,$ so the two expressions coincide.
The two expressions can be combined in the following way to get a single expression for the two-balls slimness of any triangle with smaller angles A and B:
${\frac {\sin {\max(\angle {A},\angle {B},\angle {C},\pi /2)}}{4\sin {\frac {\angle {A}}{2}}\sin {\frac {\angle {B}}{2}}\sin {\frac {\pi -\angle {A}-\angle {B}}{2}}}}={\frac {\sin {\max(\angle {A},\angle {B},\angle {C},\pi /2)}}{\cos \angle {A}+\cos \angle {B}-\cos(\angle {A}+\angle {B})-1}}$
To get a feeling of the rate of change in fatness, consider what this formula gives for an isosceles triangle with head angle θ when θ is small: ${\frac {\sin {\max(\theta ,\pi /2)}}{4\sin ^{2}{\frac {\pi -\theta }{4}}\sin {\frac {\theta }{2}}}}\approx {\frac {1}{4{\sqrt {1/2}}^{2}\theta /2}}={\frac {1}{\theta }}$
The following graphs show the 2-balls slimness factor of a triangle:
* Slimness of a general triangle when one angle (a) is a constant parameter while the other angle (x) changes.
* Slimness of an isosceles triangle as a function of its head angle (x).

## Fatness of circles, ellipses and their parts

The ball-based slimness of a circle is of course 1 - the smallest possible value.
For a circular segment with central angle θ, the circumcircle diameter is the length of the chord and the incircle diameter is the height of the segment, so the two-balls slimness (and its approximation when θ is small) is: ${\frac {\text{length of chord}}{\text{height of segment}}}={\frac {2R\sin {\frac {\theta }{2}}}{R\left(1-\cos {\frac {\theta }{2}}\right)}}={\frac {2\sin {\frac {\theta }{2}}}{\left(1-\cos {\frac {\theta }{2}}\right)}}\approx {\frac {\theta }{\theta ^{2}/8}}={\frac {8}{\theta }}$
For a circular sector with central angle θ (when θ is small), the circumcircle diameter is the radius of the circle and the incircle diameter is the chord length, so the two-balls slimness is: ${\frac {\text{radius of circle}}{\text{length of chord}}}={\frac {R}{2R\sin {\frac {\theta }{2}}}}={\frac {1}{2\sin {\frac {\theta }{2}}}}\approx {\frac {1}{2\theta /2}}={\frac {1}{\theta }}$
For an ellipse, the slimness factors are different in different locations. For example, consider an ellipse with short axis a and long axis b. the length of a chord ranges between $2a\sin {\tfrac {\theta }{2}}$ at the narrow side of the ellipse and $2b\sin {\tfrac {\theta }{2}}$ at its wide side; similarly, the height of the segment ranges between $b{\bigl1-\cos {\tfrac {\theta }{2}}{\bigr )}$ at the narrow side and $a{\bigl1-\cos {\tfrac {\theta }{2}}{\bigr )}$ at its wide side. So the two-balls slimness ranges between: ${\frac {2a\sin {\frac {\theta }{2}}}{b\left(1-\cos {\frac {\theta }{2}}\right)}}\approx {\frac {8a}{b\theta }}$
and: ${\frac {2b\sin {\frac {\theta }{2}}}{a\left(1-\cos {\frac {\theta }{2}}\right)}}\approx {\frac {8b}{a\theta }}$
In general, when the secant starts at angle Θ the slimness factor can be approximated by:
${\frac {2\sin {\frac {\theta }{2}}}{\left(1-\cos {\frac {\theta }{2}}\right)}}$ $\left({\frac {b}{a}}\cos ^{2}(\Theta +{\frac {\theta }{2}})+{\frac {a}{b}}\sin ^{2}(\Theta +{\frac {\theta }{2}})\right)$

## Fatness of a convex polygon

A convex polygon is called r-separated if the angle between each pair of edges (not necessarily adjacent) is at least r.
Lemma: The enclosing-ball-slimness of an r-separated convex polygon is at most $O(1/r)$.$^{:7–8}$
A convex polygon is called k,r-separated if:
* It does not have parallel edges, except maybe two horizontal and two vertical.
* Each non-axis-parallel edge makes an angle of at least r with any other edge, and with the x and y axes.
* If there are two horizontal edges, then diameter/height is at most k.
* If there are two vertical edges, then diameter/width is at most k.
Lemma: The enclosing-ball-slimness of a k,r-separated convex polygon is at most $O(\max(k,1/r))$. improve the upper bound to $O(d)$.

## Counting fat objects

If an object o has diameter 2a, then every ball enclosing o must have radius at least a and volume at least $V_{d}a^{.}$ Hence, by definition of enclosing-ball-fatness, the volume of an R-fat object with diameter 2a must be at least ${\tfrac {V_{d}a^{d}}{R^{d}}}.$ Hence:
    Lemma 1: Let R ≥ 1 and C ≥ 0 be two constants. Consider a collection of non-overlapping d-dimensional objects that are all globally R-fat (i.e. with enclosing-ball-slimness ≤ R). The number of such objects of diameter at least 2a, contained in a ball of radius C⋅a, is at most:
$V_{d}\cdot {\frac {(Ca)^{d}}{V_{d}\cdot {\frac {a^{d}}{R^{d}}}}}=(RC)^{d}$
For example (taking d = 2, R = 1 and C = 3): The number of non-overlapping disks with radius at least 1 contained in a circle of radius 3 is at most 3$^2$ = 9. (Actually, it is at most 7).
If we consider local-fatness instead of global-fatness, we can get a stronger lemma:
    Lemma 2: Let R ≥ 1 and C ≥ 0 be two constants. Consider a collection of non-overlapping d-dimensional objects that are all locally R-fat (i.e. with local-enclosing-ball-slimness ≤ R). Let o be a single object in that collection with diameter 2a. Then the number of objects in the collection with diameter larger than 2a that lie within distance 2C⋅a from object o is at most:
$(4R\cdot (C+1))^{d}$
For example (taking d = 2, R = 1 and C = 0): the number of non-overlapping disks with radius larger than 1 that touch a given unit disk is at most 4$^2$ = 16 (this is not a tight bound since in this case it is easy to prove an upper bound of 5).

## Generalizations

The following generalization of fatness were studied by for 2-dimensional objects.
A triangle ∆ is a (β, δ)-triangle of a planar object o (0 < β ≤ π/3, 0 < δ < 1), if ∆ ⊆ o, each of the angles of ∆ is at least β, and the length of each of its edges is at least δ·diameter(o). An object o in the plane is (β, δ)-covered if for each point P ∈ o there exists a (β, δ)-triangle ∆ of o that contains P.
For convex objects, the two definitions are equivalent, in the sense that if o is α-fat, for some constant α, then it is also (β, δ)-covered, for appropriate constants β and δ, and vice versa. However, for non-convex objects the definition of being fat is more general than the definition of being (β, δ)-covered.

## Applications

Fat objects are used in various problems, for example:
* Motion planning - planning a path for a robot moving amidst obstacles becomes easier when the obstacles are fat objects.
* Fair cake-cutting - dividing a cake becomes more difficult when the pieces have to be fat objects. This requirement is common, for example, when the "cake" to be divided is a land-estate.
* More applications can be found in the references below.