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HandWiki. Fat Object. Encyclopedia. Available online: https://encyclopedia.pub/entry/36879 (accessed on 25 April 2024).
HandWiki. Fat Object. Encyclopedia. Available at: https://encyclopedia.pub/entry/36879. Accessed April 25, 2024.
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HandWiki. (2022, November 28). Fat Object. In Encyclopedia. https://encyclopedia.pub/entry/36879
HandWiki. "Fat Object." Encyclopedia. Web. 28 November, 2022.
Fat Object

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.

computational geometry ellipse equilateral

## 1. Global Fatness

https://handwiki.org/wiki/index.php?curid=1651769

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[1] is:

$\displaystyle{ \frac{\text{side of smallest cube enclosing} \ o}{\text{side of largest cube enclosed in} \ o} }$

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.[2] 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"[3]) is based on the following slimness factor:

$\displaystyle{ \left(\frac{\text{volume of smallest ball enclosing} \ o}{\text{volume of} \ o}\right)^{1/d} }$

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:

$\displaystyle{ \left(\frac{\text{volume of} \ o}{\text{volume of largest ball enclosed in} \ o}\right)^{1/d} }$

### 1.1. 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 (≈b2) 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/b2) 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:

enclosing-ball-slimness ⋅ enclosed-ball-slimness = two-balls-slimness
enclosing-cube-slimness ⋅ enclosed-cube-slimness = two-cubes-slimness

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).

### 1.2. Balls vs. Cubes

The volume of a d-dimensional ball of radius r is: $\displaystyle{ V_d\cdot r^d }$, where Vd is a dimension-dependent constant:

$\displaystyle{ V_d = \frac{\pi^{d/2}}{\Gamma(\frac{d}{2} + 1)} }$

A d-dimensional cube with side-length 2a has volume (2a)d. It is enclosed in a d-dimensional ball with radius a√d whose volume is Vd(a√d)d. Hence for every d-dimensional object:

enclosing-ball-slimness ≤ enclosing-cube-slimness ⋅ $\displaystyle{ {V_d}^{1/d}\cdot {\sqrt d} / 2 }$.

For even dimensions (d=2k), the factor simplifies to: $\displaystyle{ \sqrt{0.5 \pi k} / {{(k!)}^{1/2k}} }$. In particular, for two-dimensional shapes V2=π and the factor is: √(0.5 π)≈1.25, so:

enclosing-disk-slimness ≤ enclosing-square-slimness ⋅ 1.25

From similar considerations:

enclosed-cube-slimness ≤ enclosed-ball-slimness ⋅ $\displaystyle{ {V_d}^{1/d}\cdot {\sqrt d} / 2 }$
enclosed-square-slimness ≤ enclosed-disk-slimness ⋅ 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:

enclosing-cube-slimness ≤ enclosing-ball-slimness ⋅ $\displaystyle{ 2 / {V_d}^{1/d} }$

For even dimensions (d=2k), the factor simplifies to: $\displaystyle{ 2 / {(k!)}^{1/2k} / {\sqrt{\pi}} }$. In particular, for two-dimensional shapes the factor is: 2/√π≈1.13, so:

enclosing-square-slimness ≤ enclosing-disk-slimness ⋅ 1.13

From similar considerations:

enclosed-ball-slimness ≤ enclosed-cube-slimness ⋅ $\displaystyle{ 2 / {V_d}^{1/d} }$
enclosed-disk-slimness ≤ enclosed-square-slimness ⋅ 1.13

Multiplying the above relations gives the following simple relations:

two-balls-slimness ≤ two-cubes-slimness ⋅ √d
two-cubes-slimness ≤ two-balls-slimness ⋅ √d

Thus, an R-fat object according to the either the two-balls or the two-cubes definition is at most Rd-fat according to the alternative definition.

## 2. 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: enclosing-cube-slimness≤2, enclosed-cube-slimness≤2, two-cube-slimness=2. 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:[3][4]

$\displaystyle{ \frac{1}{2}\cdot \sup_{b \in B}\left(\frac{\text{volume of} \ B}{\text{volume of} \ B\cap o}\right)^{1/d} }$

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.

### 2.1. 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 is in the set B considered by the definition of local-enclosing-ball slimness. Hence:

enclosing-ball-slimnessd =
= volume(smallest-enclosing-ball)/volume(o)
≤ volume(enclosing-ball-b-in-B)/volume(o)
= volume(enclosing-ball-b-in-B)/volume(bo)
≤ (2 local-enclosing-ball-slimness)d

Hence:

enclosing-ball-slimness ≤ 2⋅local-enclosing-ball-slimness

For a convex body, the opposite is also true: local-fatness implies global-fatness. The proof[3] 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.:

$\displaystyle{ \frac{\text{volume} \ (b \cap o)}{\text{volume} \ (b)} \geq \frac{\text{volume} \ (B \cap o)}{\text{volume} \ (B)} }$

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:

$\displaystyle{ f(\theta) = \min{(\frac{r(\theta)}{\text{radius} \ (b)},1)} }$

Similarly we can calculate the right-hand side of the inequality by integrating the following function:

$\displaystyle{ F(\theta) = \min{(\frac{r(\theta)}{\text{radius} \ (B)},1)} }$

By checking all 3 possible cases, it is possible to show that always $\displaystyle{ f(\theta)\geq F(\theta) }$. 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:

$\displaystyle{ \text{volume} \ (b)\leq C_d\cdot \text{diameter} \ (o)^d }$

where $\displaystyle{ C_d }$ 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: $\displaystyle{ C_d\cdot(\text{diameter(smallest ball enclosing} \ o)/2)^d }$. Combining all inequalities gives that for every convex object:

local-enclosing-ball-slimness ≤ enclosing-ball-slimness

For non-convex objects, this inequality of course doesn't hold, as exemplified by the lollipop above.

## 3. 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 √2 1 √(π/2)≈1.25 1 √(4/π) ≈ 1.13 1 * *
b×a rectangle with b>a √(1+b^2/a^2) b/a 0.5√π(a/b+b/a)[3] √(b/a) 2√(b/aπ) √(b/a) * *
disk 1 √2 1 √(4/π)≈1.13 1 √(π/2)≈1.25 * *
ellipse with radii b>a b/a >b/a √(b/a) >√(b/2πa) √(b/a) >√(πb/a) * *
semi-ellipse with radii b>a, halved in parallel to b 2b/a >2b/a √(2b/a) >√(4ba) √(2b/a) >√(2πb/a) * *
semidisk 2 √5 √2 √(8/π)≈1.6 √2 √(5π/8)≈1.4 * *
equilateral triangle   1+2/√3≈2.15 √(π/√3)≈1.35 √(4/√3)≈1.52   √√3/2+1/√√3≈1.42 * *
isosceles right-angled triangle 1/(√2-1)≈2.4 2   √2   √2 * *
'lollipop' made of unit square and b×a stick, b>1>a   b+1   √((b+1)^2/(ab+1))   √(ab+1)   √(b/a)

## 4. 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.

### 4.1. Enclosed-Ball Slimness

The largest circle contained in a triangle is called its incircle. It is known that:

$\displaystyle{ \Delta = r^2\cdot(\cot \frac{\angle A}{2} + \cot \frac{\angle B}{2} + \cot \frac{\angle C}{2}) }$

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:

$\displaystyle{ \sqrt{ \frac{\cot \frac{\angle A}{2} + \cot \frac{\angle B}{2} + \cot \frac{\angle C}{2}}{\pi} } }$

### 4.2. 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.[5]

It is known that:

$\displaystyle{ \Delta = R^2 \cdot 2 \sin A \sin B \sin C }$

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:

$\displaystyle{ \sqrt{\frac{\pi}{2 \sin A \sin B \sin C}} }$

It is also known that:

$\displaystyle{ \Delta = \frac{c^{2}}{2(\cot \angle{A} + \cot \angle{B})} = \frac{c^{2} (\sin \angle{A})(\sin \angle{B})}{2\sin(\angle{A} + \angle{B})} }$

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:

$\displaystyle{ \sqrt{\frac{\pi\cdot(\cot \angle{A} + \cot \angle{B})}{2}} = \sqrt{\frac{\pi\cdot\sin(\angle{A} + \angle{B})}{2 (\sin \angle{A}) (\sin \angle{B})}} }$

Note that in a right triangle, $\displaystyle{ \sin{\angle{C}}=\sin{\angle{A}+\angle{B}}=1 }$, so the two expressions coincide.

### 4.3. 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:[6]

$\displaystyle{ \frac{R}{r} = \frac{1}{4 \sin(\frac{\angle{A}}{2}) \sin(\frac{\angle{B}}{2}) \sin(\frac{\angle{C}} {2})} = \frac{1}{\cos \angle{A} + \cos \angle{B} + \cos \angle{C} - 1} }$

For an obtuse triangle, c/2 should be used instead of R. By the Law of sines:

$\displaystyle{ \frac{c}{2} = R \sin{\angle{C}} }$

Hence the slimness factor of an obtuse triangle with obtuse angle C is:

$\displaystyle{ \frac{c/2}{r} = \frac{\sin{\angle{C}}}{4 \sin(\frac{\angle{A}}{2}) \sin(\frac{\angle{B}}{2}) \sin(\frac{\angle{C}} {2})} = \frac{\sin{\angle{C}}}{\cos \angle{A} + \cos \angle{B} + \cos \angle{C} - 1} }$

Note that in a right triangle, $\displaystyle{ \sin{\angle{C}}=1 }$, 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:

$\displaystyle{ \frac{\sin{\max(\angle{A},\angle{B},\angle{C},\pi/2)}}{4 \sin(\frac{\angle{A}}{2}) \sin(\frac{\angle{B}}{2}) \sin(\frac{\pi-\angle{A}-\angle{B}} {2})} = \frac{\sin{\max(\angle{A},\angle{B},\angle{C},\pi/2)}}{\cos \angle{A} + \cos \angle{B} - \cos (\angle{A}+\angle{B}) - 1} }$

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:

$\displaystyle{ \frac{\sin{\max(\theta,\pi/2)}}{4 \sin^2(\frac{\pi-\theta}{4}) \sin(\frac{\theta}{2})} \approx \frac{1}{4 {\sqrt{1/2}}^2 \theta/2}=\frac{1}{\theta} }$

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).

## 5. Fatness of Circles, Ellipses and Their Parts

The ball-based slimness of a circle is of course 1 - the smallest possible value.

https://handwiki.org/wiki/index.php?curid=2000587

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:

$\displaystyle{ \frac{\text{length of chord}}{\text{height of segment}}=\frac{2R\sin\frac{\theta}{2}}{R\left(1-\cos\frac{\theta}{2}\right)}=\frac{2\sin\frac{\theta}{2}}{\left(1-\cos\frac{\theta}{2}\right)} \approx \frac{\theta}{\theta^2/8} = \frac{8}{\theta} }$
https://handwiki.org/wiki/index.php?curid=2020651

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:

$\displaystyle{ \frac{\text{radius of circle}}{\text{length of chord}}=\frac{R}{2R\sin\frac{\theta}{2}}=\frac{1}{2\sin\frac{\theta}{2}} \approx \frac{1}{2 \theta / 2} = \frac{1}{\theta} }$

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 $\displaystyle{ 2a\sin\frac{\theta}{2} }$ at the narrow side of the ellipse and $\displaystyle{ 2b\sin\frac{\theta}{2} }$ at its wide side; similarly, the height of the segment ranges between $\displaystyle{ b\left(1-\cos\frac{\theta}{2}\right) }$ at the narrow side and $\displaystyle{ a\left(1-\cos\frac{\theta}{2}\right) }$ at its wide side. So the two-balls slimness ranges between:

$\displaystyle{ \frac{2a\sin\frac{\theta}{2}}{b\left(1-\cos\frac{\theta}{2}\right)} \approx \frac{8a}{b\theta} }$

and:

$\displaystyle{ \frac{2b\sin\frac{\theta}{2}}{a\left(1-\cos\frac{\theta}{2}\right)} \approx \frac{8b}{a\theta} }$

In general, when the secant starts at angle Θ the slimness factor can be approximated by:[7]

$\displaystyle{ \frac{2\sin\frac{\theta}{2}}{\left(1-\cos\frac{\theta}{2}\right)} }$ $\displaystyle{ \left(\frac{b}{a}\cos^2(\Theta+\frac{\theta}{2}) + \frac{a}{b}\sin^2(\Theta+\frac{\theta}{2})\right) }$

## 6. 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 $\displaystyle{ O(1/r) }$.[8]:7-8

A convex polygon is called k,r-separated if:

1. It does not have parallel edges, except maybe two horizontal and two vertical.
2. Each non-axis-parallel edge makes an angle of at least r with any other edge, and with the x and y axes.
3. If there are two horizontal edges, then diameter/height is at most k.
4. 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 $\displaystyle{ O(\max(k,1/r)) }$.[9] improve the upper bound to $\displaystyle{ O(d) }$.

## 7. 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 Vdad. Hence, by definition of enclosing-ball-fatness, the volume of an R-fat object with diameter 2a must be at least: Vdad/Rd. 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:
$\displaystyle{ V_d\cdot (C a)^d / (V_d \cdot a^d / R^d) = (R C)^d }$

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 32=9. (Actually, it is at most 7).

If we consider local-fatness instead of global-fatness, we can get a stronger lemma:[3]

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:
$\displaystyle{ (4 R \cdot (C+1))^d }$

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 42=16 (this is not a tight bound since in this case it is easy to prove an upper bound of 5).

## 8. Generalizations

The following generalization of fatness were studied by [2] 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.[2]

## 9. 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.[3]
• 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.[10]
• More applications can be found in the references below.

### References

1. Katz, M. J. (1997). "3-D vertical ray shooting and 2-D point enclosure, range searching, and arc shooting amidst convex fat objects". Computational Geometry 8 (6): 299–316. doi:10.1016/s0925-7721(96)00027-2. , Agarwal, P. K.; Katz, M. J.; Sharir, M. (1995). "Computing depth orders for fat objects and related problems". Computational Geometry 5 (4): 187. doi:10.1016/0925-7721(95)00005-8.  https://dx.doi.org/10.1016%2Fs0925-7721%2896%2900027-2
2. Efrat, A.; Katz, M. J.; Nielsen, F.; Sharir, M. (2000). "Dynamic data structures for fat objects and their applications". Computational Geometry 15 (4): 215. doi:10.1016/s0925-7721(99)00059-0.  https://dx.doi.org/10.1016%2Fs0925-7721%2899%2900059-0
3. Van Der Stappen, A. F.; Halperin, D.; Overmars, M. H. (1993). "The complexity of the free space for a robot moving amidst fat obstacles". Computational Geometry 3 (6): 353. doi:10.1016/0925-7721(93)90007-s.  https://dx.doi.org/10.1016%2F0925-7721%2893%2990007-s
4. Berg, M.; Groot, M.; Overmars, M. (1994). "New results on binary space partitions in the plane (extended abstract)". Algorithm Theory — SWAT '94. Lecture Notes in Computer Science. 824. pp. 61. doi:10.1007/3-540-58218-5_6. ISBN 978-3-540-58218-2. , Van Der Stappen, A. F.; Overmars, M. H. (1994). "Motion planning amidst fat obstacles (extended abstract)". Proceedings of the tenth annual symposium on Computational geometry - SCG '94. pp. 31. doi:10.1145/177424.177453. ISBN 978-0897916486. , Overmars, M. H. (1992). "Point location in fat subdivisions". Information Processing Letters 44 (5): 261–265. doi:10.1016/0020-0190(92)90211-d. , Overmars, M. H.; Van Der Stappen, F. A. (1996). "Range Searching and Point Location among Fat Objects". Journal of Algorithms 21 (3): 629. doi:10.1006/jagm.1996.0063.  https://dx.doi.org/10.1007%2F3-540-58218-5_6
5. "How fat is a triangle?". https://math.stackexchange.com/q/915187. Retrieved 28 September 2014.