Solving Triangles and Integer sided triangles

• Finding many numbers and properties of any triangle given 2 or 3 angles, 2 sides and an angle or all 3 sides or all three vertex coordinates
• the Pythagorean triangles (for which see much more on a separate page: a separate page )
• the Heronian triangles with integer sides and have a whole-number area and also
• Lattice Triangles which are those that can be drawn by joining three points with integer x-y coordinates.

Contents of this page

The icon means there is a Things to do section of questions to start your own investigations.
The calculator icon indicates that there is a live interactive calculator in that section.

Triangles and their properties

Names and Lines on the Triangle

• degrees: 360° in a full turn because this is traditional (it goes back to the Babylonians who used a base of 60 whereas we use base 10 numbers today)
• radians: measures the distance round the circumference of a circle of unit radius that the angle makes. 2 π radians are a full turn. This make most trig formulas easier otherwise a constant as a conversion factor is needed
• turns: a fraction of a whole turn.

More on the meanings of Acute, Right-angled, Obtuse triangles, Scalene, Isoceles, Equilateral triangles;
Perimeter, Area, Altitudes, Orthocentre, Medians, Centroid, Angle Bisector Theorem

Circles of the Triangle

Solving a triangle

We can often deduce the other sides and angles when we are only given some of them. This is called "solving the triangle".
There are three rules or theorems that are useful for deducing the missing sides and angles:

The angles in a triangle add up to to 180°
A + B + C = half a turn = 180° = Pi radians
This finds the third angle when we know the other two angles.

Informal Proof:
Place your pen along one side of the triangle. Note which way it is pointing.
There are two angles of the triangle on that side.
Turn the pen through one of the angles of the triangle (not one of the angles outside it) so that it lies along another side.
Repeat with another of the angles and then the third so that the pen again lies on the original side of the triangle.
Note which way the pen points now. It has turned a complete half-turn.

The Sine Rule
This is useful if we know both an angle and two sides one of which is the side opposite the angle. It can find the angle opposite the other known side.

a	=	b	=	c
sin(A)		sin(B)		sin(C)

Proof: In the diagram, draw the line h as the height of vertex B above the base b.
It divides the triangle into two triangles.
In the left hand triangle sin(C) = h/a so that h = a sin(C)
In the right hand triangle, sin(A) = h/c so that h = c sin(A)
Putting these two together and eliminating h gives:

a	=	c
sin(A)		sin(C)

By doing the same with another side as the base, we get the complete Sine Rule.

The Cosine Rule
This is useful if we know two sides and the angle between them

a² = b² + c² − 2 b c cos(A)
If we know sides b and c and angle A - the angle between them - then we can deduce the length of the remaining side a.
By rearranging the equation above we can find angle A if we know all three sides a, b and c:

cos(A) =	b2 + c2 − a2
	2 b c

Proof:
Using the same diagram and Pythagoras' Theorem in the left hand triangle we have
a² = h² + (b−x)²
Pythagoras in the right hand triangle gives
c² = h² + x²
If we substitute h from the second into the first, we can eliminate it:
a² = (c² − x²) + (b−x)²
Now we expand (b−x)² and find that the x² terms cancel:
a² = c² + b² − 2 b x
Finally, to remove x, we note that x/c = cos(A) in the right hand triangle so x = c cos (A):
substituting this in gives the Cosine Rule:
a² = b² + c² − 2 b c cos(A)

We need at least 3 of the 6 measurements (3 sides and 3 angles)

We can code the different sets of given measurements using A for angle and S for side and proceed, for example as shown here:

• Given 3 angles (AAA), the triangle can be any size so we need the length of one side too.
• Given 3 sides (SSS), use the Cosine rule to work out the three angles
• Given 2 sides and the angle between them (SAS), use the Cosine rule to find the third side and proceed as above
• Given 2 sides and a non-included angle (SSA), use the Sine rule to find the missing side. Sometimes there can be two triangles possible - see below.
• Given 2 angles and the side joining them (ASA), find the third angle (the angles sum to 180°) then use the Sine rule
• Given 2 angles and a side not joining them (AAS), find the third angle (the angles sum to 180°) then use the Sine rule

The ambiguous case of SSA

There is one case when we might have two different solutions: If we are given two sides and an angle which is not the included angle, say b and c and angle C, then there could be two possible places for vertex B as shown in the diagram here.
This is the case of two sides and the non-included angle meaning that the angle is not the one between the two given sides.
We can find the two solutions by noting that when we apply the Sine Rule, there may be another solution because:
sin( α ) = sin( 180°−α )

Less than 3 of the six measurements is not enough.

• two sides are not enough since they can "hinge" where they join and so can have any angle between them
• two angles are not enough since we don't know how far apart they are
• one side and one angle at one end of the side will define a second side but we can then take any point on it as another vertex to join to the other end of the given side.
• one side and the angle opposite that side means the vertex with that angle can be anywhere on a circle segment.

About the Triangle Calculator

• From the coordinates of the vertices which you can give as numbers or exact expressions or as an expression to evaluate;
• From sides and angles if you give at least 3 of these the Solver will find the rest
• Find integer-sided triangles in a given range
• Find integer-coordinate (lattice) triangles: all non-congruent triangles are found (different sizes are shown) on a grid of given size.
  Pick's Theorem states that the Area is I−1+B/2 where I is the number of lattice points inside the triangle and B the number of lattice points on its bounding edges.

• Input: Show help...
  • Give the x-y coordinates of the 3 vertices. These need not be integers.
  • Fill in at least three of the sides or angles fields to Solve the Triangle.

  Coordinates, sides and angles may be input in any of these three formats:

  • a whole number or with a decimal point
  • an expression to evaluate
  • a simple symbolic expressions for the sides or for the sines or cosines of the angles:
    such as sqrt(18)/2 or (2+3sqrt(2))/2 which are then processed exactly by the Calculator.

  Note that writing 4.5 will give real number results only whereas using 9/2 will produce exact (symbolic) results, where possible.
  If numbers are too large for accurate symbolic results, change to real numbers. The easiest way to do this if the input is already symbolic is to include a decimal point in one number e.g: 35*sqrt(2000.0) instead of 35sqrt(2000).Note that * must be used for multiplication in expressions to evaluate as numbers but is NOT used in expressions to be used symbolically.

  • Integer sided triangles: use the selectors to make a sentence and then press .

    Heronian triangles

    are integer-sided triangles which have a whole number area.

    Primitive triangles

    have the smallest sides for that shape. Non-primitive triangles are a multiple of a primitive triangle, eg sides 6,8,10 are twice the sides of the primitive 3,4,5 triangle.

    Pythagorean triangles

    have integer sides and are right-angled.

    All integer sided triangles

    There are integer sided triangles which are neither right-angled (which are Heronian too) nor do they have an area that is a whole number such as 3,3,5 which has area (5√11)/4.

    Perimeter search

    The Calculator will search all integer-sided triangles with a given perimeter (sum of the 3 sides) or in a given range of perimeter values.

    Filters

    You can additionally select only those which contain a side of given size or other conditions. Leave the filter selection set to (no filter) if you have no additional conditions.

    When only counting the number of solutions, the Output choices will disappear. Uncheck the "Count only" checkbox to use them. Total counts will be given in Results even when "Count only" is not selected.
• Output: Show help...
  The Output section lets you select what properties of found triangles you want displayed in the RESULTS area. The sides are alway shown unless you have Count only checked.
  The output choices apply if you Solved a Triangle button and if you generated Integer Sided Triangles.
  You can just count the number with your given properties and filters (the other properties will be temporarily hidden until you uncheck the Count only checkbox).
  If the Count only checkbox is off, you can select any or all of the properties shown.
  If the sides are integers or sine or cosines were given for angles then exact symbolic expressions are given where possible.
  Where possible coordinates of the triangle can be given in whole numbers.
  Also a diagram of the triangle can be generated on the Drawing board.

  You can select if values are displayed with Show numbers and to how many decimal places.
  

A Triangle Calculator

Counts of Lattice triangles on an n×n grid

Congruent

means the triangles are identical, so they have the same angles and side lengths, such as those with vertices (0,0) (1,0) (0,3) and (0,0) (0,1) (3,0).
These two are therefore in the same congruence class and count as the same solution.

Similar

means they are the same shape (same angles) but need not be the same size.
The triangle with vertices (0,0) (2,0) (0,4) is similar to that with vertices at (0,0) (1,0) (0,2) but they are not congruent (the former has sides twice as long as those of the latter).
These two are therefore in the same similarity class but are in different congruency classes.

The number of congruency classes for a square grid of size 1, 2, 3, 4, ... is

1, 8, 29, 79, 172, 333, 587, 963, 1494, 2228, 3195, 4455, 6050, ... A028419

The number of similarity classes for a squre grid of size 1, 2, 3, 4, ... is

1, 6, 20, 55, 119, 229, 402, 667, 1019, 1536, 2216, 3049, 4168, ... A028492

Number of scalene triangles, distinct up to congruence, on a (n X n)-grid:

0, 3, 18, 57, 137, 280, ... A190313

Number of acute triangles, distinct up to congruence, on a (n X n)-grid

0, 2, 8, 23, 51, 101, 179, 295, ... A190021

Number of obtuse triangles, distinct up to congruence, on a (n X n)-grid

0, 2, 12, 39, 95, 193, 355, 597, ... A190022

Number of isosceles triangles, distinct up to congruence, on a (n X n)-grid

1, 5, 11, 22, 35, 53, 70, 100, ... A189978

Number of right-angled triangles, distinct up to congruence, on an n X n grid

0, 1, 4, 9, 17, 26, 39, 53, 71, ... A189979

Integer-Sided Triangles

Perimeter

• Round(r) is to the nearest integer to r and
• Floor(r) is the next integer equal to or below r
  which, since our values here are positive is the same as "forget anything after the decimal point"

• Mathematical Gems III Ross Honsberger, MAA (1985) Chapter 3, section 2 A Problem about Triangles pages 39-47
  which also has some more references to the first published formula (1979) for T(n).

Area

Non-integer area

The first two integer triangles with integer area are:

• the (right-angled) Pythagorean Triangle 3,4,5 with area 6
• 5,5,6 and area 12

Integer area

There are triangles with integer area that are not Heronian (their sides are not integers) eg. multiply the Pythagorean 3-4-5 triangle by √3: the triangle with sides 3√3, 4√3, 5√3 has an area of 18.

Heronian Properties

the sine, cosine and tangents of all three angles are rational.

The cosine rule a² = b² + c² − 2 b c cos(A) means that cos(A) = (b² + c² − a²)/(2 b c) and proves that all the cosines are rational because all the sides are whole numbers.
Since the area of a Heronian triangle is a whole number by definition and a formula for the area of a triangle is Area = b c sin(A)/2 then sin(A) = 2 Area/(b c) and this is rational. The same argument applies to all the angles.
If we allow 0 and infinity to be rational numbers (for cosine and tangent of 90°) then all the tangents will be rational also.

all the altitudes of a Heronian triangle are rational

Since the sides and area are whole numbers then using Area = ½ side×altitude we deduce all the altitudes are rational numbers.

one altitude is a whole number only if the Heronian triangle is primitive;

note this does not say that all Heronians triangles have an integer altitude - see below

if two or three altitudes are whole numbers the Heronian triangle is Pythagorean or is not primitive

The integer altitude of a non-Pythagorean Heronian triangle will divide it into two Pythagorean triangles.

Note that this does not say that all Heronian triangles are made from two Pythagorean triangles because there are primitive Heronian triangles with no whole number altitudes such as 5,29,30 with area 72 and altitudes 144/5, 144/29, 24/5.
The altitudes of such triangles are necessarily rational numbers so by multiplying the sides by a denominator of one of the altitudes we can always find a non-primitive Heronian triangle of the same shape that will be decomposable into two Pythagorean triangles.
The two Pythagorean triangles will have the altitude as the common leg but each Pythagorean triangle will have a different Heronian side as a hypotenuse. The Pythgorean triangles can overlap or be separate so that the sum or the difference of their areas is the Heronian area depending on whether the Heronian triangle is acute or obtuse.

No one knows if there is a Heronian triangle with 3 whole number medians.

52,102,146 is Heronian and has two integer medians 97, 35 but the third is 4√949; all its multiples will have 2 integer medians also.
Euler found the integer triangle 136,174,170 has medians 158, 127, 131 but its area is 60√2002.

This is Problem 14.2 in Unsolved Problems...see the reference below.

• Heronian Triangles Are Almost Everywhere K. Robin McLean, The Mathematical Gazette Vol. 72, No. 459 (1988), pages 49-51
• Construction of Indecomposable Heronian Triangles Paul Yiu, Rocky Mountain Journal of Mathematics vol 28 (1998) pages 1189-1202.
• Old and New Unsolved Problems in Plane Geometry and Number Theory V Klee, S Wagon, Mathematical Assoc of America, 2nd edition (1991)

Applications of Integer Triangles

Alcuin's Sequence

• 1 of each kind: 0 solutions;
• 2 of each kind has 1 solution;
• 3 of each kind also has 1 solution
• 4 of each has 2 solutions
• 5 barrels has 1 solution
• 6 barrels has 3 solutions
• and so on

• Alcuin's Sequence , Donald J. Bindner and Martin Erickson,The American Mathematical Monthly Vol. 119, No. 2 (February 2012), pages 115-121 pdf

Another application - partitions of a number

This correspondence can be related directly to the number of integer triangles with a given perimeter and is a nice exercise for the interested reader. The excellent reference below details the investigation of this problem by a school geometry class.
In order to find the connections yourself, you might find the integer triangles by perimeter table above a useful starting point. The reference below contains a fuller development and is not advanced.

• Counting Integer Triangles Nicholas Krier and Bennet Manvel in Mathematics Magazine Vol. 71, No. 4 (1998), pages 291-295

Lattice Triangles

A triangle that has lattice points for all of its vertices is called a lattice triangle.
Not all of these have integer sides. For example the 45°-45°-90° right-angled triangle with sides 1, 1, √2 can be drawn using lattice points as its vertices: (0,0), (1,0), (1,1).
However, a moment's thought shows that all Pythagorean triangles can be drawn with integer coordinates because their two legs are integers.
What is not obvious is that all Heronian triangles can be drawn on the grid lattice with integer coordinates too.
But there are integer-sided triangles that are not lattice triangles!
Let's explore this a little more.

A special case of a lattice triangle is the Pythagorean triangle a-b-h which has both a right-angle and whole number sides. We can therefore draw it with its vertices at the integer coordinates (0,0), (a,0) and (b,0). A property of Pythagorean triangles is that the two legs cannot both be odd and therefore the area of a Pythagorean triangle will always be a whole number (half the product of the two legs).

What is also true is that every lattice triangle has an area that is an integer or half an integer. More...

James Tanton has a nice webpage on this.
Any lattice triangle can be enclosed within a rectangle as shown here. If we subtract from that rectangle the three triangles, we have the area of our lattice triangle. The rectangle area is an integer; the areas of the three right-angled triangles are integers or half integers (their hypotenuses need not be whole numbers so now the two legs of the right-angles may be odd).
When we remove the triangles from the rectangle the area left must be an integer or half an integer.

A right-angled triangle with legs 1 and n will have an area of n/2 so we can always find an integer triangle for every integer or half-integer.

A Heronian triangle has integer sides and integer area. All of them can be drawn as lattice triangles. A special subset of these are the Pythagorean triangles. There are other Heronian triangles that are not Pythagorean, for example join two 3,4,5 triangles along their 4 sides to make a 5,5,6 triangle with area of 12.
All this can be summarised in the Venn diagram on the right above.

There are integer-sided triangles that are not lattice triangles, such as the equilateral triangle of any (integer) side length. Such triangles never have a whole-number area. E Lucas in 1878 proved that no matter how big an equilateral triangle is, if its sides are whole numbers then it is never a lattice triangle.

Area of a Lattice Triangle

If all the coordinates' components are integers then the area is half an integer.
A more immediate proof that any lattice triangle has such an area is given by the diagram on the right.

• Lattice triangles can be flipped or rotated so that one vertex is always in the lower right of a rectangle surrounding the triangle. There are two types, one where the three vertices are on three sides of th rectangle and another when only two of the vertices are.
• In each case, the rectangle has an integer area (all the vertices of the triangle are at lattice points).
• The silver triangles are removed from the rectangle to make the lattice tyiangles
• Each of the silver triangles has integer base and height so they have an area which is a half-integer.
• The second case also needs the darker rectangle removed from the bigger rectangle but it too has integer sides so its area is an integer.
• The area of the white lattice triangle in both cases is thus an integer (the area of the rectangle or rectangle minus blue rectangle) minus half-integers and integers and so will be a half-integer.

• The Area of a Triangle in terms of the coordinates of the vertices F W Flisher Math Gaz 52 (1968) pages 149-150 PDF has a simple proof of both formulae above.

Dividing any lattice Polygon into triangles

Pick's Theorem

Applications of Pick's Theorem

• For all polygons with lattice points as vertices the area is a multiple of 1/2 which is sometimes written as

  area ∈	ℕ
  	2

  
  A lattice polygon's area is half a whole number and, since some numbers are even, this includes whole numbers too
• It is impossible to draw an equilateral triangle using lattice points only.
  With sides of length n its area would be ⁿ/₂√3 and n√3 is never a whole number because √3 is irrational.
• Though Pick's Theorem was published in 1899 it seems it was only in the 1980's that it became popular. One popularisation was through a conference of mathematicians who invited a forester to show how mathematics was useful in his work.
  He showed that to estimate the area of any shape on a map, place a transparrent square grid of dots over it using what was essentially Pick's Theorem. You can read more about this in the reference below.
• A convex lattice polygon is one where there are no vertices with an internal polygon angle bigger than a half turn. If a convex lattice shapes with any number of sides has a boundary of B lattice points on its edges including vertices then B can be no bigger than 2I+7 where I is the number of lattice points inside the polygon.
  This leads to a nice Geoboard game called Stretch. A geoboard is a board of pins (nails) in a square grid on which elastic bands can be placed in various mathematical activities. The game is to start with a convex shape and two players take turns expanding the shape but without introducing any extra internal points (new points on the edge are ok) and without making it convex (having a dimple or cave at a vertex). The first player who cannot move is the loser.

• Pick's Formula D DeTemple Mathematical Notes from Washington State University Vol 32, (1989)
  Notes on a Northwest Mathematics conference in the 1980s where a forester was invited to show how mathematics is used in the forest industry. It included a method of calculating area by overlaying a lattice of grid points on a map and counting the number of points inside and on the boundary of a forest to estimate its area.
  This use is quoted in Pick's Theorem by B Grunbaum and G C Shephard in The American Mathematical Monthly Vol. 100 (1993), page 150.
• Stretch: A Geoboard Game D B Coleman Mathematics Magazine vol 51 (1978) pages 49-54

Pick's Theorem and Lattice Triangles

Range b: :0..50 :0-120 :0-250

i=0, b≥3 Show these

Points along the i-axis in the plot on the right. These are triangles with no internal lattice points. They are the 'fundamental triangles' since every other triangle (and polygons too) can be parititioned into such basic shapes. All such triangles have an area of ½.
There is a triangle with no internal points (i=0) and with b points on its boundary for every value of b≥3 with vertices at (0,0) (b-2,0) (b-2,1)

b=3, i≥0 Show these

There are triangles with only 3 boundary points but with i internal points for all values of i≥0. Examples include the triangles with vertices (0,0) (1,0) (2, 2i + 1)

b=4, i≥0 Show these

includes the triangles with vertices (0,0 (1,0) (2, 2i + 2)

b=5, i must be a multiple of 3 Show these

such as triangles with vertices (0,0) (1,0) (3, 2i + 3) where i=3n

Scott's Inequality says that there is only one triangle with b = 2i + 7 Show it

the triangle (0,0) (3,0) (0,3) or congruent lattice triangles where i=1 and b=9.

The lowest sloping line: b=2i+6 Show it

For all other triangles except the one above, we must have b ≤ 2i + 6. The triangles at this limit have b = 2i + 6 such as the triangles with vertices (0,0) (2,0) (2,2i + 2).

The other two cases with b=2i+k Show them

k=4: Vertices for triangles with i internal points and b=2i+4 include (0,0) (1,0) (2i+1, 4i+2)
k=2: Vertices for triangles with i internal points are b=2i+2 include (0,0) (1,0) (2i,4i)

The lower boundary of the next "streak" of points with b=i+k...Show them

b=i+8: (0,0) (3,0) (3,3n+3) has b=3n+9 and i=3n+1

...also has 2 more prominent parallel lines in that same streak:

b=i+5: the triangles (0,0) (1,0) (3n+1,9n+3) have b=3n+5 and i=3n and alternate with (0,0) (1,0) (3n+3, 9n+6) having b=3n+6 and i=3n+1
b=i+2: the other line has triangles (0,0) (1,4) (i+1,i+4) with b=i+2 if b is not a multiple of 3 OR else (0,0) (1,3) (3i,0) where b=3i

They have a starting point Show them

If the lowest cone has parameter c=1 the apex of cone c is b=2c²+2c+2 and i=c³-c
an example triangle that fits this specification is (0,0) (2c²+2c,0) (1,c)

c	apex b,i	example tri vertices
1	6, 0	(0, 0), (4, 0), (1, 1)
2	14, 6	(0, 0), (12, 0), (1, 2)
3	26, 24	(0, 0), (24, 0), (1, 3)
4	42, 60	(0, 0), (40, 0), (1, 4)
5	62, 120	(0, 0), (60, 0), (1, 5)
6	86, 210	(0, 0), (84, 0), (1, 6)
7	114, 336	(0, 0), (112, 0), (1, 7)

• Ehrhart Polynomials of Lattice Triangles J Hofscheier, B Nill, D Öberg Electronic Journal of Combinatorics vol 25 Issue 1 (2018), #P1.3 pages 1-8 PDF

Pick's Theorem in 3D?

First the bad news:

There is no simple 3D generalization to find the volume of a flat sided, straight edged polyhedron (a solid with volume) whose vertices are on 3D lattice points if we want the volume solely in terms of the number of lattice points inside it and the number on its surface.
Reeve (1957) (see references below) gives a simple example to show this by considering a tetrahedron defined by the 4 vertices (0,0,0), (1,0,0), (0,1,0) and (1,1,k). Its volume is k/6 but it has no points inside it nor on its surface nor its edges apart from the vertices yet its volume depends on k.

and now the good news...

We can draw a perfect equilateral triangle using 3D lattice points!
(0,0,1), (0,1,0) and (1,0,0) define such a triangle and we can multiply these coordinates by any value we like to obtain an infinite set of equilateral triangles.
Imagine drawing three lines, one on the floor and two on the two walls of a room where the three points are each 1 metre from the same corner.

and more good news....

the hexagon is also a 3D lattice polygon.
More ...

:outline :midpoints

The answer lies in a nice puzzle. You can find a regular hexagon in a cube. How?
By holding the cube so that a diagonal through opposite corners of the cube is in line with your eye.
If we now join the mid points of the sides as we go round the visible hexagon, we will get a flat hexagon. If the cube has sides of length 2 and the origin is a corner of the cube, its coordinates are
(2, 1, 0), (2, 0, 1), (1, 0, 2), (0, 1, 2), (0, 2, 1), (1, 2, 0)
or, if the origin is the centre of the cube:
(1, 0, -1), (1, -1, 0), (0, -1, 1), (-1, 0, 1), (-1, 1, 0), (0, 1, -1)
These coordinates are just the 6 possible sequences of numbers 0, 1 and 2 or -1, 0 and 1.

but more bad news...

The equilateral triangle, square and hexagon are the only regular polygons that can be drawn in a lattice of any number of dimensions! By regular here, we mean all sides are the same length.
More on what a higher dimension lattice is...

A "graph-paper" 2D lattice is the set of points (x,y) where x and y are integers.
The distance between (x,y) and (p,q) is √(p−x)² + (q−y)².

When we extend to 3D we have (x,y,z) and again all three coordinates are integers and the distance between (x,y,z) and (p,q,r) is √(p−x)² + (q−y)² + (r−q)².

Mathematically there is no reason not to extend this to a list of 4 coordinates (x,y,z,w) or 5 or 6 ... and we measure distances between two "points" as the square-root of the squares of the differences in each dimension.

although there is even more good news....

If we mean all angles in the polygon are the same but not necessarily all the edges of the same size then we have equiangular polygons and these include all regular polygons too. Rectangles are equiangular polygons and they are lattice polygons if their side lengths are whole numbers.
But the good news is that there is another equiangular polygon that we can draw on our integer (2D) lattice that has neither 3, 4 nor 6 sides.
What is it?
More ...

The octagon:

Unfortunately this is all there is, even in higher dimensions. The proof goes back to 1963. Here is a PDF on the Dynamics Of Polygons site that explains more.

Rational Triangles

However, there is at least one problem where the rational triangle/integer triangle correspondence is not applicable: tiling the plane with distinct rational triangles, that is, completely fill the infinite 2D plane with rational triangles, no two of which are the same shape (similar). There will clearly be an infinite number of tiles, each a rational triangle and so we cannot find the largest denominator of all of them to scale up any solution for integer triangles.
J H Conway posed this problem for rational triangles in 1965. The published solution of another solver involves taking one simple triangle that will tile the plane on its own, namely the 3-4-5 Pythagorean triangle and then showing how it can be split into two rational triangles in an infinite series of different ways. By applying this series of splits to each 3-4-5 triangle in the (infinite) plane tiling, we will have a complete tiling where no two rational triangle shape is ever repeated.

• Problem 5328 J H Conway American Math. Monthly vol 72 (1965) page 915
  "Show how the plane may be filled with rational-sided triangles in such a way as to use just one of each type. (The triangles should be taken as closed and "filled" means covered in such a way that the overlap has zero area)."
  
• 5328 solution D C Kay American Math. Monthly vol 73 (1966) page 903-904.

Links and References

• Triangles with Vertices on Lattice Points Michael J. Beeson The American Mathematical Monthly Vol. 99, No. 3 (Mar., 1992), pp. 243-252. pdf
  A complete investigation into which n-dimensional triangles are lattice triangles. There are triangles we can put on a 3D lattice but not 2D, and in 5D but not 4D but the 3D lattice triangles are exactly the same as the 4D ones. He also proves the only lattice polygons in any dimension are triangles, squares and hexagons.
• Heronian Triangles Are Lattice Triangles Paul Yiu The American Mathematical Monthly Vol. 108, No. 3 (2001), pages. 261-263
• Heronian Tetrahedra are Lattice Tetrahedra S H Marshall, A R Perlis American Math Monthly Vol 120 (2013) pages 140-149.
  Not only does this paper give a method using quaternions to show all integer-sided tetrahedra (3D solids with 4 trianglular faces) can be realised on a 3D lattice of integer points, it also shows how complex numbers can be used to prove all Heronian triangles fit on the 2D integer lattice.
• Counting Heron Triangles With Constraints P Stanica, S Sarkar, S S Gupta, S Maitra and N Kar Integers Vol 13 (2013), A3 pdf
• Geometrisches zur ZahlenlehreG Pick, Sitzungber. Lotos, Naturwissen Zeitschrift (Prague, 1899) vol 19, pages 311-319. This is the first appearance of what is now called Pick's Theorem.
• Mathematical Snapshots H Steinhaus, (2011, 3rd edition Dover paperback reprint).
  This is an excellent paperback of the 1983 edition; the original version was published in 1950. I first came across Pick's Theorem in this book (my 1969 Oxford Press hardback version) when I was a teenager but also so much more that does not appear in any other popular mathematics book either before or since. It covers so many interesting topics in a largely visual way at a level that an interested school student could understand and in much more depth than practically all other more modern maths-and-its-applications books. It is timeless and I highly recommended it.
• A Colorful Proof of Pick's Theorem J E Graver, Y A Monachino Math Horizons, Vol. 18, No. 2 (2010), pages 14-16
  is a very different approach to proving Pick's Theorem by flipping pieces of a polygon about the mid-point of the sides to build up filled squares and half-squares.
• The Scarcity of Regular Polygons on the Integer Lattice D J O'Loughlin Mathematics Magazine Vol. 75, No. 1 (2002), pages 47-51.
  Here is a fairly elementary-level trigonometric proof that the only regular lattice polygons in 2D are the squares.
• On the Volume of Lattice Polyhedra J E Reeve Proceedings of the London Mathematical Society, s3-7 (1957) pages 378 - 395.
  Here Reeve introduces the tetrahedron mentioned above.

version 2: created 27 February 2016, updated 30 July 2018
version 1: 21 Nov 2003