What are equiangular triangles

What is an equilateral triangle?

......As the name suggests, the equilateral triangle is a triangle with sides of equal length.
The statement that the interior angles are 60 ° is therefore equivalent.
When the triangle is mentioned on this page, the equilateral triangle is meant.

Formulas for the triangle Top
The sizes of the triangle are the side a, the height h, the radius of the circumference R, the radius of the inscribed circle r, the circumference U and the area A.

If side a is given, the other quantities can be calculated from it.
......According to the Pythagorean theorem (a / 2) ² + h² = a². This implies h = (1/2) sqrt (3) a.
The area A = (1/2) ah results in A = (1/4) sqrt (3) a². - The scope is 3a.

......In the triangle, the heights, the bisector (r), the bisector and the perpendicular (R) coincide. They intersect at the center M of the triangle.
The center lines (connecting line between two sides) are parallel to one side and are half their size.

......The side bisectors divide in a ratio of 1: 2.
Proof: ED is center line and parallel to AB. According to the second theorem of rays: MA: MD = AB: ED. From this, with AB = a and ED = a / 2, the relationship MA follows:MD = 2:1, qed.

......The radius of the perimeter and that of the inscribed circle are sections of the height.
The following applies: R = (2/3) h = (1/3) sqrt (3) a and r = (1/3) h = (1/6) sqrt (3) a.

One point in the triangle Top
Invariance of the sum of the distances

......Viviani's theorem reads: If P is any point inside the equilateral triangle, then the sum of the distances of this point from the sides is constant.
This theorem also applies to the center of the triangle, i.e. for P = M. In this case the distances are equal to the radii r of the inscribed circle.
So Viviani's theorem is called in formula language: s + t + u = 3r. It also applies 3r = h.

......The proof arises if one connects the point P with the corner points of the triangle and draws an area balance.

Same areas
There is one more sentence for the figure.
......If one draws the perpendiculars on the sides and the connecting lines to the corner points from a point in the triangle, then six triangles are created.
The sum of the areas of three separated triangles is the same.

......To prove this, draw the three parallels to the sides of the triangle through the point P. Then parallelograms are created at the corners and isosceles triangles in the middle. Both are halved, which results in the equality of area.

3-4-5 point
......A point P is given, which lies in an equilateral triangle ABC and which has the distances 3, 4 and 5 from the corner points. How big is the side of the triangle?
Solution: a = sqrt [25 + 12sqrt (3)]

Derivation at The Math Forum (Url below)

Square and triangle Top
Square in triangle 1

......In an equilateral triangle you can put a square on its tip so that it touches the sides.

......Let a be the side of the triangle and b the side of the square.
Then: b = (1/4) [3sqrt (2) -sqrt (6)] a (approximately 0.448a).
This formula is derived with the help of the ray theorem (blue) and the relationships
h = (1/2) sqrt (3) a and b = sqrt (2) x.
Square in triangle 2
......This is another square that fits into the triangle. It's on the bottom. It has the side length x = [2sqrt (3) -3] a or rounded x = 0.464a.
It's a little bigger than the square above at 0.448a.

Derivation of the formula
According to the second theorem of rays, h: (a / 2) = (h-x) :( x / 2) or h: a = (h-x): x. The product equation is hx = a (h-x). Then hx = ah-ax or (h + a) x = ah or x = (ah) / (h + a).
If one sets h = (1/2) sqrt (3) a, one obtains x = [(1/2) sqrt (3) a] / [(1/2) sqrt (3) +1] or x = [ sqrt (3) a] / [sqrt (3) +2)].
The denominator becomes rational when the fraction is expanded with 2-sqrt (3). This leads to x = [2sqrt (3) -3] a.

Triangle in square 1
.......You can put an equilateral triangle inside a square so that it shares one corner with the square and touches two sides.

......Let a be the side of the square and b the side of the triangle. Then: b = [sqrt (6) -sqrt (2)] a (approximately 1.035a).
This formula is derived from the Pythagorean theorem (blue and green). One arrives at a quadratic equation whose positive solution one has to take.

Square in triangle 2
......This is another triangle that fits into the square. Its base is the square side. It has sides a, so it's a little smaller than the triangle above at 1.035a.

Square next to triangle
You place a suitable square in a triangle in such a way that one side lies on the base line and one corner point lies on the side of the triangle. Fitting means that the square has side length x = (1/2) [(3-sqrt (3)] a, or about 0.63a.

Description of a solution
The equations of the straight lines g and h are determined with the help of two marked points.
The straight line equations are g: y = x and h: y = -sqrt (3) x + sqrt (3).
The intersection point S has the x-value x = (1/2) [(3-sqrt (3)] a, wzbw.

Triangles in the triangle Top
Rotated and compressed triangle

......If you look for points that divide the sides in a ratio of 1: 2 and connect them,
one obtains a smaller (equilateral) triangle with side length x = (1/3) sqrt (3) a.

......The triangle side x in the yellow triangle is calculated using the cosine law. In the drawing, k = 1/3.
The side of the triangle is generally obtained as x = sqrt (3k²-3k + 1) a.

The inner triangle
......If you connect the points that divide the sides in a ratio of 1: 2 with the opposite corner points, an (equilateral) triangle with side length x = sqrt (7) / 7a is created in the middle. (3, page 67ff.)

If the number 1/3 is replaced by k, the side length is x = (1-2k) / sqrt (1-k + k²) a.

Proof of the general case, sent by Erhard Schuller
Let AB = a. Then BD = ka.
According to the law of cosines, CD² = BC² + BD²-2BC * BD * cos (60 °) or
CD² = a² + (ka) ²-2a (ka) (1/2) = a² + k²a²-ka² or CD = sqrt (1-k + k²) a.
The triangles BCD and BED are similar in that they match at corresponding angles.
There are two proportions, namely
BC: CD = BE: BD or a: [sqrt (1-k + k²) a] = BE: ka or BE = (ka) / sqrt (1-k + k²) and
BC: BD = BE: DE or a: ka = [(ka) / sqrt (1-k + k²)]: DE or DE = (k²a) / sqrt (1-k + k²).
x = CD-DE-FC = AE-BE-TE = sqrt (1-k + k²) a- (ka) / sqrt (1-k + k²) - (k²a) / sqrt (1-k + k²) = (1-2k) / sqrt (1-k + k²) * a, wzbw.

Napoleon's triangle
The following interesting figure is attributed to Napoleon.
......Draw any triangle (black).
Draw equilateral triangles over the three sides.
Find their centers.
Connect the centers of these triangles.
Result: An equilateral triangle was created.
The triangles can also be placed inside.
The circles of the equilateral triangles intersect at a common point (3, page 67ff.).

Morley's triangle
...Draw any triangle
Divide the three interior angles into three equal parts.
Connect matching intersections of the dividing lines.
Result: An equilateral triangle was created.

You get beautiful patterns,
when you get the triangle
divided into equal triangles.

......If you draw all heights in the triangle, you get six 30-60-90 triangles. They can be used as tangram stones. (2)

If you omit the shorter section of a height, you get three 30-120-30 triangles.
The figure on the left can be interpreted as a plan view of a tetrahedron.
If you add two 30-120-30 triangles to the triangle, you get a triangle made up of five similar triangles.

Circle and triangle Top
Circular figures
There are a number of circular figures that are related to an equilateral triangle.

You can find more on my circle parts page.

Malfatti's problem
The point is to find three touching circles in a triangle, which together have as large an area as possible. A special case is: The circles are the same size.

......We have h = 2r + sqrt (3) r + r. With h = (1/2) sqrt (3) a we get r = (1/4) [sqrt (3) -1] a.

Malfatti believed that this was the solution. That is not right.
For more details see e.g. at de.wikipedia under Malfatti circle (URL below)

The triangle on three parallels Top

......If there are three parallels with the distances a, b and a + b, it is easy to find a triangle whose corner points lie on the parallels. It is not that easy to find an equilateral triangle, or rather the equilateral triangle.

x is the length of the side of the triangle we are looking for.
......You can apply the sine formula twice and get two equations in x and alpha.
If you solve them, you get x = (2/3) sqrt (3) sqrt (a² + ab + b²) ..

Two fractals Top
Koch curve (snowflake curve)
The starting figure for the Koch curve is an equilateral triangle on side a (1st figure). If you divide the sides of the triangle into three equal parts and place an equilateral triangle with the side a / 3 on the middle segment, you get figure 2. You repeat this rule: You divide each of the 12 segments with the length a / 3 into three equal parts and place another triangle in the middle, this time with the length a / 9, this creates Figure 3. Figure 4 is created by the next step. You have to imagine that the rule is repeated as often as you like.

If one accompanies the pictures with calculations, one arrives at two strange statements: The circumference U (n) increases over all limits, the area A (n) approaches the number 8 / 5A.
The name Koch curve goes back to the Swedish mathematician Helge von Koch (1870-1924), who was the first to describe it in 1904.

Sierpinski triangle
The starting point for the Sierpinski triangle is an equilateral triangle (Figure 1). Cut out the center triangle (Figure 2). Three (red) triangles remain. From them you cut out the center triangles. The result is Figure 3. In Figure 4, the center triangles are removed again. You repeat this procedure as often as you like, so that the starting triangle is perforated more and more.
The calculation leads to strange statements: The circumference of the triangles grows beyond all boundaries, the area tends to zero.
Waclaw Franciszek Sierpinski (1882-1969) was a Polish mathematician who dealt with (set theoretical) topology.

Create a triangle Top
It is common to construct an equilateral triangle with a compass and ruler.

......You draw a line that will later become the side of the triangle, then a circular line around its endpoints with the radius of the line. An intersection of the circles is connected to the end points of the route.

You can also create a triangle by folding it.
1 Given is a strip of paper.
2 Halve the strip.
3 Place the upper left corner on the center line and at the same time make sure that the fold line goes through the lower left corner.
4 Continue to fold along the red line and undo the two folds.
5 An equilateral triangle has emerged. (4)
That leaves the proof that the triangle is equilateral. The blue auxiliary line provides it:
......If one specifies the angle alpha, then alpha1 = alpha because of the folding on the straight line AB and alpha2 = alpha because of the folding on the center line m. Together, the three angles form a straight angle of 180 °. So there remains 60 ° for each angle.
By the way, this method can be used to fold a strip of equilateral triangles. You need the strip for flexagons.

You can make a triangle.
> Three matches, toothpicks or kebab sticks, connected with glue, putty or kneading or Bostik balls, form a triangle.
> You put three drinking straws into each other with a "joint". The inner stalks are cut lengthways.
> You solder three wires together or you bend a wire to form a triangle. Then you only have to solder once.
> You connect three bar magnets of equal length with balls.

The rabbit window in Paderborn Top
The rabbit window is in the cloister of Paderborn Cathedral. It is the tracery of a 16th century Gothic window.

"Three rabbits and the spoon three - and yet everyone has two"
The ears form an equilateral triangle.

Equilateral triangles on my homepage
Level figures

Tripartite a triangle

Figures in a triangle

Homogeneous tiling


Triominos puzzles

Platonic solids

Archimedean solids

Convex deltahedron

Equilateral triangle on the internet Top


Alexander Bogomolny (Cut The Knot)
Napoleon's theorem, Morley's Miracle

Ask Dr. Math (The Math Forum)
Triangle Formulas

efg's Computer Lab
Sierpinski Triangle

Eric W. Weisstein (World of Mathematics)
Equilateral triangle, Napoleon triangle, Routh's theorem, Malfatti's problem

Kevin Brown (mathpages.com)
Napoleon's theorem

The Math Forum
Area of ​​Equilateral Triangle

Torsten Sillke
grid triangles

Equilateral triangle, Viviani's theorem, Napoleon's theorem, Morley's trisector theorem,
Koch snowflake, Sierpinski triangle, Malfatti circles


Jürgen Kummer
Equilateral triangle calculator

Equilateral triangle, Viviani's theorem, Napoleon triangle, Morley triangle, Koch curve, Sierpinski triangle, Malfatti circle

credentials Top
(1) Martin Gardner: Mathematisches Labyrinth, Braunschweig 1971 (ISBN 3-528-08402-2)
(2) Karl-Heinz Koch: ... lege Spiele, Cologne 1987 (ISBN 3-7701-2097-3)
(3) Martin Gardner: Mathematical Circus, Berlin 1988 (ISBN 3550076924)
(4) Kunihiko Kasahara, Origami - figurative and geometrical, Munich 2000 (ISBN 3-8043-0664-0)

I would like to thank Torsten Sillke for support.

Feedback: Email address on my main page

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© 2003 Jürgen Köller