Altitude Reciprocals

While researching some properties of altitudes of triangles, I came across the theorem, “The reciprocals of the altitudes of any triangle can themselves form a triangle” (from Wikipedia).  Since the source of this theorem was from Wikipedia, and the theorem itself was not immediately obvious, I decided that I should prove its verity before passing it on to anyone else.  I not only found out that this theorem was true, but discovered an even stronger theorem, which is, “The reciprocals of the altitudes of any triangle can themselves form a triangle that is similar to the original triangle.”

To prove this theorem, first define an altitude as a function of the triangle’s side.  Given △ABC with sides a, b, and c, the law of cosines states that cos C = ^{a^2 + b^2 – c^2}/_2ab. 

Then the altitude h_a from ∠A to a, would be h_a = b sin C.  Substituting the law of cosine value of ∠C gives h_a = b sin(cos^-1(^{a^2 + b^2 – c^2}/_2ab)).  Using a right angle triangle and Pythagorean Theorem, sin(cos^-1(^{a^2 + b^2 – c^2}/_2ab)) = ^{√(4a^2b^2 – (a^2 + b^2 – c^2)^2)}/_2ab.

The radicand 4a²b² – (a² + b² – c²)² simplifies to 4a²b² – (a⁴ + b⁴ + c⁴ + 2a²b² – 2a²c² – 2b²c²) = 2a²b² + 2a²c² + 2b²c² – a⁴ – b⁴ – c⁴, and so h_a = b ^{√(2a^2b^2 + 2a^2c^2 + 2b^2c^2 – a^4 – b^4 – c^4)}/_2ab = ^{√(2a^2b^2 + 2a^2c^2 + 2b^2c^2 – a^4 – b^4 – c^4)} / _2a.  Letting k = ^{√(2a^2b^2 + 2a^2c^2 + 2b^2c^2 – a^4 – b^4 – c^4)}/₂, the altitude h_a = ^k/_a.

Using a similar argument obtains h_b = ^k/_b and h_c = ^k/_c, and so the reciprocals of the three altitudes are ¹/h_a = ^a/_k, ¹/h_b = ^b/_k, and ¹/h_c = ^c/_k.  These three reciprocals are proportional to the original three sides of the triangle a, b, and c by a factor of ¹/_k, and so the reciprocals of the altitudes form a triangle that is similar to its original triangle by SSS similarity.

Lattice Multiplication Method

In the Lattice Multiplication Method, a student is able to complete a multiplication problem of two large numbers by arranging numbers in a lattice grid.  For example, to multiply 48 x 12, the student would start out by creating a 2 x 2 lattice grid with the digits of 48 along the top as column headers, and the digits of 12 along the right side as row headers (Step 1).   The student would then multiply the digits in each column and row header and place each product in its corresponding position (using leading zeroes for single digit products), so in this case 4 x 1 = 04 (Step 2), 8 x 1 = 08 (Step 3), 4 x 2 = 08 (Step 4), and 8 x 2 = 16 (Step 5). 

Lattice Multiplication of 48 x 12

Finally, the student would add the digits along each diagonal, from the rightmost diagonal to the leftmost diagonal, and place each sum at the bottom left of each diagonal (carrying when necessary), so in this case 6 = 6 (Step 6), 8 + 1 + 8 = 17 which breaks down to 7 and a carried 1 (Step 7), 1 + 0 + 4 + 0 = 5 (Step 8), and 0 = 0 (Step 9).  The final answer can be determined by the right side column digits and the bottom row digits, in this case a final and correct answer of 576.  The Lattice Multiplication Method can also be extended to handle more digits by creating a larger lattice.

Lattice Multiplication Tutorial Video

The Lattice Multiplication Method is also known as Gelusia Multiplication, Sieve Multiplication, Shabakh, Venetian Squares, and the Chinese Lattice.  It is not known whether the method originated in Europe, the Middle East, or China, but it has been known since at least the 13^th century.

Validation

When it comes to multiplying two numbers of multiple digits, most people use the traditional Vertical Multiplication Method that is taught in most elementary schools.  For example, to multiply 48 x 12, the numbers are placed vertically, and the last digit of the top number is multiplied by the last digit of the second number to obtain 8 x 2 = 16, which is breaks down as 6 and a carried 1 (Step 1). 

Step 1	Step 2	Step 3	Step 4	Step 5
148         x 12               6	148         x 12            96	48         x 12            96            80	48         x 12            96          480	48         x 12            96     + 1480          576

Vertical Multiplication of 48 x 12

Then the first digit of the top number 4 is multiplied by the last digit of the second number 2 to obtain 4 x 2 = 8, plus the carried 1 makes 8 + 1 = 9, and so the first product is 96 (Step 2).  Then the last digit of the first number 8 is multiplied by the first digit of the second number 1 to obtain 8 x 1 = 8 and placed beside a place holder of 0 (Step 3).  Then the first digit of the first number 4 is multiplied by the first digit of the second number 1 to obtain 4 x 1 = 4, so the second product is 480 (Step 4).  The two products 96 and 480 are then added together to obtain the final answer of 96 + 480 = 576 (Step 5).

The Lattice Multiplication Method is simply a rearrangement of the same numbers in the Vertical Multiplication Method.  In the Lattice Multiplication Method, the digits of 48 and 12 are found in the top row and right column, and the digits in 96 and 480 are found inside the lattice (recall that the 9 was actually 8 and 1), and the digits of 576 are found in the left column and bottom row.

The Lattice Multiplication Method and the Vertical Multiplication Method both work for the same reason – the distributive property.  In both methods, the student multiplies 40 x 10, 40 x 2, 8 x 10, and 8 x 2 in some order and adds all the results together.  This is the algebraic equivalent of (40 + 8)(10 + 2) = 40·10 + 40·2 + 8·10 + 8·2 = 400 + 80 + 80 + 16 = 576, which is the same order as the popular FOIL method (first, outer, inner, last) taught in most algebra classes to show the distributive property.  In other words, the integrity of the multiplication is maintained because each digit of the first number is multiplied by each digit of the second number (with appropriate positioning to preserve place value) and then added all together.

Conclusion

The Lattice Multiplication Method is an algorithm for multiplying two large numbers by arranging numbers in a lattice grid.  Each digit of the two numbers are separated and placed as column and row headers, then the product of each column and row header is found and positioned inside the grid, and then the sum of each diagonal is found placed at the bottom left of each diagonal, and finally these sums can be read to obtain the solution.  Because each digit of the first number is multiplied by each digit of the second number (with appropriate positioning to preserve place value) and then added all together, it is a valid algorithm for multiplying two numbers.  The Lattice Multiplication Method is both organized and visually appealing, making it an ideal algorithm for multiplying two large numbers.

Japanese Multiplication Method

In the Japanese Multiplication Method, a student is able to complete a multiplication problem of two large numbers by merely drawing a few lines and counting the points of intersections.  For example, to multiply 21 x 23, the student first represents 21 by 2 diagonals lines that are drawn up and to the right followed by 1 diagonal line that is drawn in the same direction just underneath this, and then 23 by 2 diagonal lines that are drawn down and to the right followed by 3 diagonals lines that are drawn in the same direction just above this, so that the four groups of lines form a diamond shape. 

Japanese Multiplication of 21 x 23

The lines intersect in a total of 4 points on the left side of the diamond, a total of 8 points in the top and bottom sides of the diamond, and a total of 3 times on the right side of the diamond for a final and correct answer of 483.

The Japanese Multiplication Method can be extended to handle more digits by creating a larger diamond, and handle a digit of zero by drawing a different colored line.  In some cases, carrying is required in the final addition steps.

Japanese Multiplication Method Tutorial Video

Despite its name, the origin of the Japanese Multiplication Method is unknown.  The method is also known as Indian Multiplication and Chinese Stick Multiplication, but it is not known if it actually did originate from Japan, India, China, or elsewhere.

Validation

When it comes to multiplying two numbers of multiple digits, most people use the traditional vertical method that is taught in most elementary schools.  For example, to multiply 21 x 23, the numbers are placed vertically, and the last digit of the top number is multiplied by the last digit of the second number to obtain 1 x 3 = 3 (Step 1). 

Step 1	Step 2	Step 3	Step 4	Step 5
21         x 23               3	21         x 23            63	21         x 23            63            20	21         x 23            63          420	21         x 23            63      + 420          483

Vertical Multiplication of 21 x 23

Then the first digit of the top number 2 is multiplied by the last digit of the second number 3 to obtain 2 x 3 = 6, and so the first product is 63 (Step 2).  Then the last digit of the first number 1 is multiplied by the first digit of the second number 2 to obtain 1 x 2 = 2 and placed beside a place holder of 0 (Step 3).  Then the first digit of the first number 2 is multiplied by the first digit of the second number 2 to obtain 2 x 2 = 4, so the second product is 420 (Step 4).  The two products 63 and 420 are then added together to obtain the final answer of 63 + 420 = 483 (Step 5).

In summary, in the traditional vertical method a student first multiplies 1 x 3, then 20 x 3, then 1 x 20, then 20 x 20, and then adds all the results together.  In other words, when multiplying numbers with multiple digits, each digit of the first number is multiplied by each digit of the second number (with appropriate positioning or zeroes to preserve place value) and then added all together.  Algebraically, this can be expressed as (20 + 1)(20 + 3) = 3·1 + 20·3 + 1·20 + 20·20 = 3 + 60 + 20 + 400 = 483.  Tweaking the order a little bit gives the algebraic equivalent of (20 + 1)(20 + 3) = 20·20 + 20·3 + 1·20 + 1·3 = 400 + 60 + 20 + 3 = 483, which is the same order as the popular FOIL method (first, outer, inner, last) taught in most algebra classes. 

The same multiplication problem can be visualized with a table using Base Ten Blocks.  The 21 can be represented as 2 rod blocks and 1 unit block as row headers, and the 23 can be represented as 2 rod blocks and 3 unit blocks as column headers. 

Base Ten Block Multiplication of 21 x 23

The table would then be filled in by 4 square blocks (hundreds), a group of 6 rod blocks and another group of 2 rod blocks for a total of 8 rod blocks (tens), and 3 unit blocks (ones) for a final answer of 483.  Once again, each digit of the first number is multiplied by each digit of the second number (this time with appropriate shapes to preserve place value) and then added all together.

The Japanese Multiplication Method is simply a transformation of the Base Ten Block Table.  Instead of 4 square blocks there are 4 points of intersection on the left side of the diamond of lines, instead of a group of 6 rod blocks and a group of 2 rod blocks there are 6 points of intersection on the top side of the diamond of lines and 2 points of intersection on the bottom side of the diamond of lines, and instead of 3 unit blocks there are 3 points of intersection on the right side of the diamond of lines.  The integrity of the multiplication is maintained because each digit of the first number is multiplied by each digit of the second number (with appropriate positioning to preserve place value) and then added all together.

Because of its similarities with the FOIL method, it should be noted that the Japanese Multiplication Method can also be used to multiply polynomials.  The above Japanese Multiplication diagram that shows 21 x 23 = 483 can also be used to multiply (2x + 1)(2x + 3) and obtain the result 4x² + 8x + 3.  (The equation 21 x 23 = 483 is a specific example of (2x + 1)(2x + 3) = 4x² + 8x + 3 when x = 10.)

Evaluation

After watching the video and examining the above example, it is tempting to conclude that the Japanese Multiplication Method is the superior algorithm for multiplying two numbers.  After all, it just requires drawing a few lines and counting its intersections.  Unfortunately, the above example is a bit misleading because all of the numbers used have small digits (3 and under).  Here is an example of multiplying some numbers with larger digits, 69 x 78, using the Japanese Multiplication Method:

Japanese Multiplication of 69 x 78

In this example, a lot more lines have to be drawn, and a lot more points of intersection are formed.  Counting the points of intersection becomes time-consuming and carrying is required.  In this example, the Japanese Multiplication Method takes longer than the traditional vertical method of multiplication.

Still, the Japanese Multiplication Method is a great way to visualize the multiplication process, especially for numbers with smaller digits.

Conclusion

The Japanese Multiplication Method is an algorithm for multiplying two large numbers by representing both numbers by a group of lines that form a diamond pattern.  The number of points of intersection near each vertex of the diamond are then counted in a certain order to obtain the solution.  Because each digit of the first number is multiplied by each digit of the second number (with appropriate positioning to preserve place value) and then added all together, it is a valid algorithm for multiplying two numbers.  Unfortunately, the Japanese Multiplication Method is too time-consuming for multiplying numbers with larger digits, but remains a great visual aid for the multiplication process.

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.