Perspective, Silhouettes, and Taxi-Cab Geometry

In a drawing with one-point perspective, all lines perpendicular to the picture plane converge to a single point on the horizon called the vanishing point.  In this picture of some railroad tracks, for example, it would appear that the railroad tracks meet at some point far off in the distance, even though they are actually parallel.

one-point perspective of railroad tracks

Now imagine that you are looking down at some cubes on a level surface, and that you draw them using one-point perspective.  The top of each cube will not only appear larger than the bottom of each cube, but also be drawn further away from the vanishing point.

one-point perspective of some level cubes

  

Next, you create a silhouette of each cube by coloring inside each of the drawn outlines.  The resulting shapes are either irregular hexagons, pentagons, or quadrilaterals.

the silhouettes of some level cubes drawn with one-point perspective

Amazingly, the area of each silhouette can be determined by just four variables – the side length a of the top square of the drawn cube, the side length b of the bottom square of the drawn cube, the horizontal shift x of the cube from a cube directly over the vanishing point, and the vertical shift y of the cube from a cube directly over the vanishing point.

The center of the top square of a cube in the first quadrant would be drawn at (ax, ay) and its vertices at (a(x ± ½), a(y ± ½)).

Likewise, the center of the bottom square of a cube in the first quadrant would be drawn at (bx, by) and its vertices at (b(x ± ½), b(y ± ½)). 

Each silhouette is a hexagon (or a degenerate hexagon) that is the difference between the area of a rectangle with sides c = a(x + ½) – b(x – ½) and d = a(y + ½) – b(y – ½) and the areas of the two opposite triangles, one with sides c – a and d – b, and the other with sides c – b and d – a. 

In other words, the area of the silhouette is A = cd – ½(c – a)(d – b) – ½(c – b)(d – a), and letting p = ½(a² – b²) and q = ½(a² + b²), this simplifies to A = p(x + y) + q.  A similar argument can be made for the other quadrants to give the general formula of A = pd + q, where d = |x| + |y|. 

Surprisingly, d = |x| + |y| is also the equation for distances in taxicab geometry, a seemingly unrelated mathematical topic.  In taxicab geometry, segments are limited to being either vertical or horizontal, just like roads in a city block.  Since a taxicab cannot drive through city blocks along a hypotenuse, the distance it must travel to get to an intersection that is 4 blocks east and 3 blocks north is at least |3| + |4| = 7 blocks (whereas a flying bird not limited to the roads can do it in √(3² + 4²) = 5). 

taxicab geometry paths from (0, 0) to (4, 3)

In general, the distance a taxicab must travel to get to an intersection that is x blocks east or west and y blocks north or south is at least d = |x| + |y|.

Therefore, there is a surprising connection between one-point perspective drawings and taxicab geometry.  The area of the silhouette of a cube drawn in one-point perspective is A = pd + q, where d = |x| + |y|, the same distance a taxicab must at least travel to get to an intersection that is x blocks east or west and y blocks north or south.  This means that the area of the silhouette is a linear function of its own taxicab distance.

Equable Shapes with Integer Dimensions

An equable shape is a shape that has the same numerical perimeter and area.  For example, a square with side length of 4 is equable because its numerical perimeter and area are both 16 (P = 4 + 4 + 4 + 4 = 16 and A = 4² = 16).  There are an infinite number of equable shapes, but only a handful that have integer dimensions.

Circles

All circles have a perimeter (or circumference) of C = 2πr, and an area of A = πr².  An equable circle has the same numerical perimeter and area, so C = A.  Therefore,

01	C = A	definition of equable
02	2πr = πr2	substitution
03	2r = r2	divide by π
04	0 = r2 – 2r	subtract 2r
05	0 = r(r – 2)	factor r
06	r = 0 or r – 2 = 0	zero rule of multiplication
07	r = 0 or r = 2	add 2

Of course, a circle cannot have a radius of r = 0, so there is only one equable circle – a circle with a radius of r = 2.  Such a circle has both a perimeter and area of 4π (C = 2πr = 2π2 = 4π and A = πr² = π2² = 4π).

Equable Circle

Squares

As mentioned above, a square with side lengths of s = 4 is equable, because its numerical perimeter and area are both 16 (P = 4 + 4 + 4 + 4 = 16 and A = 4² = 16).  It turns out that this is the only square that is equable.  Using P = 4s and A = s²,

01	P = A	definition of equable
02	4s = s2	substitution
03	0 = s2 – 4s	subtract 4s
04	0 = s(s – 4)	factor s
05	s = 0 or s – 4 = 0	zero rule of multiplication
06	s = 0 or s = 4	add 4

Since a square cannot have a side length of s = 0, the only equable square is one with a side length of s = 4.

Equable Square

Regular Polygons

Amazingly, both the equable circle and the equable square can be deduced from a general formula of regular polygons.   A regular polygon is a polygon with all equal sides and all equal angles, and a regular polygon with n sides and s side lengths has a perimeter of P = ns.  Since all regular polygons can be subdivided into n triangular pie slices, in which the vertex of each triangle is situated at the center of the regular polygon, the height of each triangle is the radius r of a circle inscribed in the regular polygon (or the apothem), and the base of each triangle is the side length s of the regular polygon, its area is A = n(½rs), since the area of a triangle is A = ½bh.  Therefore,

01	P = A	definition of equable
02	ns = n(½rs)	substitution
03	0 = n(½rs) – ns	subtract ns
04	0 = ns(½r – 1)	factor ns
05	n = 0 or s = 0 or ½r – 1 = 0	zero rule of multiplication
06	n = 0 or s = 0 or ½r = 1	add 1
07	n = 0 or s = 0 or r = 2	multiply by 2

Since a regular polygon cannot have n = 0 number of sides, or s = 0 side lengths, all equable regular polygons must have an inscribed circle radius of r = 2.  This was obviously seen above with the equable circle of r = 2, but also with the equable square of s = 4, since a square with an inscribed circle radius of r = 2 also has a side length of s = 4.  All other equable regular polygons (the equilateral triangle, the regular pentagon, the regular hexagon, and so on) all have an inscribed circle radius of r = 2.  (However, the square is the only equable regular polygon with an integer side length, which can be proved by using the trigonometric ratios and s = 4 tan ^180°/_n.)

Equable Regular Polygons

Regular Polyhedra

The same reasoning used to find equable regular polygons can be applied to the third dimension to find equable regular polyhedra, in which the numerical surface area is equal to the numerical volume.  A regular polyhedra is a solid with all regular faces, and a regular polyhedra with n faces and A face areas has a surface area of S = nA.  Since all regular polyhedra can be subdivided into n pyramid “pie slices”, in which the vertex of each pyramid is situated at the center of the regular polyhedra, the height of each pyramid is the radius r of a sphere inscribed in the regular polyhedra, and the base of each pyramid is the area A of each face of each regular polyhedra, its volume is V = n(¹/₃Ar), since the volume of a pyramid is ¹/₃Bh.  Therefore,

01	S = V	definition of equable
02	nA = n(1/3Ar)	substitution
03	0 = n(1/3Ar) – nA	subtract nA
04	0 = nA(1/3r – 1)	factor ns
05	n = 0, A = 0, or 1/3r – 1 = 0	zero rule of multiplication
06	n = 0, A = 0, or 1/3r = 1	add 1
07	n = 0, A = 0, or r = 3	multiply by 3

Since a regular polyhedra cannot have n = 0 number of sides, or A = 0 face areas, all equable regular polyhedra (the tetrahedron, the cube, the octahedron, the dodecagon, and the icosahedron) must have an inscribed sphere radius of r = 3.  For example, if a cube has an inscribed sphere radius of r = 3, then its side length is s = 6, its surface area is S = 6s² = 6·6² = 216, and its volume is also V = s³ = 6³ = 216.

The same logic can be applied to higher dimensions as well.  In general, all equable regular shapes in the n^th-dimension must have an inscribed n^th-dimensional circle radius of r = n. 

Rectangles

Finding an equable rectangle is slightly more difficult than finding an equable square or circle because there are two variables instead of one – length and width.  The perimeter of a rectangle with width W and length L is P = 2W + 2L, and the area is A = LW.  Therefore,

01	A = P	definition of equable
02	LW = 2W + 2L	substitution
03	LW – 2L = 2W	subtract 2L
04	L(W – 2) = 2W	factor L
05	L = 2W/(W – 2)	divide W – 2

Therefore, all equable rectangles have a length and width relationship of L = ^2W/_{(W – 2)}, in which there are infinitely many possibilities.  However, there are only 2 integer solutions for an equable rectangle – the 4 x 4 rectangle (which is also an equable square and described above) and the 3 x 6 rectangle.  The 3 x 6 rectangle has a perimeter P = 2W + 2L = 2·3 + 2·6 = 18 and the same numerical area A = LW = 3·6 = 18.

Equable Rectangle

Rhombi

A rhombus, which is a quadrilateral with four equal sides, can also be defined by two variables, which in this case are the side length and the height.  The area of a rhombus with a side length s and a height h is A = sh, and the perimeter is P = 4s.  Therefore,

01	P = A	definition of equable
02	4s = sh	substitution
03	0 = sh – 4s	subtract 4s
04	0 = s(h – 4)	factor s
05	s = 0 or h – 4 = 0	zero rule of multiplication
06	s = 0 or h = 4	add 4

Since a rhombus cannot have a side length of s = 0, all equable rhombi have a height of h = 4.  (Note that it does not matter how long the sides are, although in order to have a height of h = 4, s ≥ 4.)  For example, a rhombus with side lengths of 5 and a height of 4 would have a perimeter of P = 4s = 4·5 = 20, and the same numerical area of A = sh = 5·4 = 20.  (Also note that when h = s = 4, the equable rhombus is the equable square mentioned above.)

Equable Rhombus

Right Triangles

A right angle triangle can also be defined by two variables – its base and height.  The area of a triangle with base b and a height h is A = ½bh.  Its perimeter is all of its sides added up, including the hypotenuse, which in this case according to Pythagorean’s Theorem is √(b² + h²), so P = b + h + √(b² + h²).  Therefore,

01	P = A	definition of equable
02	b + h + √(b2 + h2) = ½bh	substitution
03	h + √(b2 + h2) = ½bh – b	subtract b
04	√(b2 + h2) = ½bh – b – h	subtract h
05	b2 + h2 = (½bh – b – h)2	square both sides
06	b2 + h2 = b2 + h2 + ¼b2h2 + 2bh – b2h – bh2	multiply out square
07	0 = ¼b2h2+2bh–b2h–bh2	subtract b2 + h2
08	0 = bh(¼bh + 2 – b – h)	factor bh
09	b = 0, h = 0, or ¼bh + 2 – b – h = 0	zero rule of multiplication
10	¼bh + 2 – b – h = 0	b ≠ 0 and h ≠ 0
11	¼bh + 2 – h = b	add b
12	¼bh – h = b – 2	subtract 2
13	bh – 4h = 4(b – 2)	multiply by 4
14	h(b – 4) = 4(b – 2)	factor h
15	h = 4(b – 2)/(b – 4)	divide by b – 4

Therefore, all equable right triangles have a base and height relationship of h = ^{4(b – 2)}/_{(b – 4)}, in which there are infinitely many possibilities.  However, there are only 2 integer solutions for an equable right triangle – one with sides 5, 12, and 13 and another with sides 6, 8, and 10.  The (5, 12, 13) triangle has a perimeter P = a + b + c = 5 + 12 + 13 = 30 and the same numerical area A = ½bh = ½·5·12 = 30, and the (6, 8, 10) triangle has a perimeter P = a + b + c = 6 + 8 + 10 = 24 and the same numerical area A = ½bh = ½·6·8 = 24.

Equable Right Triangles

Triangles

Finding all equable triangles with integer sides is much more difficult, because it is defined by at least three variables.  Using the three sides as variables a, b, and c, the perimeter is simply P = a + b + c, but the area, using Heron’s area formula, is A = √(s(s – a)(s – b)(s – c)) where s = ½(a + b + c).  Because of the difficulty of solving such an equation, a trial and error method using a computer program is much more efficient to find all equable triangles with integer sides, such as the following program written in Python:

01	# Equable Triangles
02	# Python 2.7.3
03	# Finds all integer side solutions of an equable triangle (where the
04	#   numerical area is equal to its numerical perimeter) in a given range.
05
06	# maximum integer side to test
07	intMax = 100
08
09	# loop through all the different integer side possibilities
10	#   where a <= b <= c
11	for a in range(1, intMax + 1):
12	for b in range(a, intMax + 1):
13	for c in range (b, intMax + 1):
14	# find the perimeter
15	P = a + b + c
16
17	# find the area
18	#   using Heron's formula: sqrt(s(s - a)(s - b)(s - c))
19	#   where s = (a + b + c) / 2
20	s = (a + b + c) / 2
21
22	# only calculate the area if there are no negatives
23	#   in the square root (so s > a, s > b, and s > c)
24	if (s > a and s > b and s > c):
25	A = (s * (s - a) * (s - b) * (s - c)) ** (0.5)
26	else:
27	A = -1
28
29	# if the perimeter and area are equal, display the sides
30	if (P == A):
31	print (a, b, c)

which outputs the following five equable triangles:

>>

(5, 12, 13)

(6, 8, 10)

(6, 25, 29)

(7, 15, 20)

(9, 10, 17)

The first two equable triangles, the (5, 12, 13) and the (6, 8, 10) triangles, are also right triangles, and were mentioned above.  The last three equable triangles are obtuse triangles.  The (6, 25, 29) equable triangle has a numerical perimeter and area of 60, the (7, 15, 20) equable triangle has a numerical perimeter and area of 42, and the (9, 10, 17) equable triangle has a numerical perimeter and area of 36.  According to mathematicians Whitworth and Biddle, these 5 triangles are the only equable triangles with integer sides.

Equable Triangles

Parallelograms

Parallelograms are also defined by at least three variables, but choosing the two sides a and b and the height h as the three variables makes it fairly easy to find equable parallelograms.  The perimeter is P = 2a + 2b and the area is A = bh.  Therefore,

01	A = P	definition of equable
02	bh = 2a + 2b	substitution
03	h = 2a/b + 2	divide by b

This means there are an infinite number of equable parallelograms with integer sides, as long as b is a factor of 2a, and h ≤ a.  (This is because a parallelogram of a fixed perimeter can have differing heights and therefore differing areas.)  For example, a parallelogram with sides a = 5 and b = 10 and a height of h = ^2a/_b + 2 = ^2·5/₁₀ + 2 = 1 + 2 = 3 is equable, because its perimeter is P = 2a + 2b = 2·5 + 2·10 = 10 + 20 = 30 and its area is A = bh = 10·3 = 30.  Note that for a rhombus, which is a special parallelogram in which a = b, the height is fixed at h = ^2a/_b + 2 = ^2a/_a + 2 = 2 + 2 = 4, which was described above.  Also note that for a rectangle, which is a special parallelogram in which h = a, h = ^2h/_b + 2 can be rearranged to h = ^2b/_{(b – 2)}, which was also described above.

Equable Parallelogram

Conclusion

There are many equable shapes that exist with integer sides, including a circle with a radius of r = 2, a square with a side of s = 4, regular polygons with an inscribed circle radius of r = 2, regular polyhedra with an inscribed sphere radius of r = 3, the  3 x 6 rectangle, and rhombi with a height of h = 4.  In addition, there are five equable triangles with integer sides, including the (5, 12, 13) and the (6, 8, 10) right triangles, and the (6, 25, 29), (7, 15, 20), and (9, 10, 17) triangles.  There are plenty of other equable shapes with integer sides, including trapezoids, irregular quadrilaterals, irregular pentagons, and so on, but all of these shapes are all defined by four or more variables and are much more difficult to find.