Equilateral triangle

An equilateral triangle. It has equal sides (${\displaystyle a=b=c}$), equal angles (${\displaystyle \alpha =\beta =\gamma }$), and equal altitudes (${\displaystyle h_{a}=h_{b}=h_{c}}$).

Denoting the common length of the sides of the equilateral triangle as ${\displaystyle a}$, we can determine using the Pythagorean theorem that:

• The area is ${\displaystyle A={\frac {\sqrt {3}}{4}}a^{2}}$,
• The perimeter is ${\displaystyle p=3a\,\!}$
• The radius of the circumscribed circle is ${\displaystyle R={\frac {a}{\sqrt {3}}}}$
• The radius of the inscribed circle is ${\displaystyle r={\frac {\sqrt {3}}{6}}a}$ or ${\displaystyle r={\frac {R}{2}}}$
• The geometric center of the triangle is the center of the circumscribed and inscribed circles
• The altitude (height) from any side is ${\displaystyle h={\frac {\sqrt {3}}{2}}a}$

Denoting the radius of the circumscribed circle as R, we can determine using trigonometry that:

• The area of the triangle is ${\displaystyle \mathrm {A} ={\frac {3{\sqrt {3}}}{4}}R^{2}}$

Many of these quantities have simple relationships to the altitude ("h") of each vertex from the opposite side:

• The area is ${\displaystyle A={\frac {h^{2}}{\sqrt {3}}}}$
• The height of the center from each side, or apothem, is ${\displaystyle {\frac {h}{3}}}$
• The radius of the circle circumscribing the three vertices is ${\displaystyle R={\frac {2h}{3}}}$
• The radius of the inscribed circle is ${\displaystyle r={\frac {h}{3}}}$

In an equilateral triangle, the altitudes, the angle bisectors, the perpendicular bisectors, and the medians to each side coincide.

A triangle ABC that has the sides a, b, c, semiperimeter s, area T, exradii r_a, r_b, r_c (tangent to a, b, c respectively), and where R and r are the radii of the circumcircle and incircle respectively, is equilateral if and only if any one of the statements in the following nine categories is true. Thus these are properties that are unique to equilateral triangles, and knowing that any one of them is true directly implies that we have an equilateral triangle.

Visual proof of Viviani's theorem

1.	Nearest distances from point P to sides of equilateral triangle ABC are shown.
2.	Lines DE, FG, and HI parallel to AB, BC and CA, respectively, define smaller triangles PHE, PFI and PDG.
3.	As these triangles are equilateral, their altitudes can be rotated to be vertical.
4.	As PGCH is a parallelogram, triangle PHE can be slid up to show that the altitudes sum to that of triangle ABC.

Morley's trisector theorem states that, in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle.

Napoleon's theorem states that, if equilateral triangles are constructed on the sides of any triangle, either all outward, or all inward, the centers of those equilateral triangles themselves form an equilateral triangle.

A version of the isoperimetric inequality for triangles states that the triangle of greatest area among all those with a given perimeter is equilateral.[12]

Viviani's theorem states that, for any interior point P in an equilateral triangle with distances d, e, and f from the sides and altitude h,

${\displaystyle d+e+f=h,}$

independent of the location of P.[13]

Pompeiu's theorem states that, if P is an arbitrary point in the plane of an equilateral triangle ABC but not on its circumcircle, then there exists a triangle with sides of lengths PA, PB, and PC. That is, PA, PB, and PC satisfy the triangle inequality that the sum of any two of them is greater than the third. If P is on the circumcircle then the sum of the two smaller ones equals the longest and the triangle has degenerated into a line, this case is known as Van Schooten's theorem.

By Euler's inequality, the equilateral triangle has the smallest ratio R/r of the circumradius to the inradius of any triangle: specifically, R/r = 2.[14]^:p.198

The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral.[15]

The ratio of the area of the incircle to the area of an equilateral triangle, ${\displaystyle {\frac {\pi }{3{\sqrt {3}}}}}$, is larger than that of any non-equilateral triangle.[16]^{:Theorem 4.1}

The ratio of the area to the square of the perimeter of an equilateral triangle, ${\displaystyle {\frac {1}{12{\sqrt {3}}}},}$ is larger than that for any other triangle.[12]

If a segment splits an equilateral triangle into two regions with equal perimeters and with areas A₁ and A₂, then[11]^:p.151,#J26

${\displaystyle {\frac {7}{9}}\leq {\frac {A_{1}}{A_{2}}}\leq {\frac {9}{7}}.}$

If a triangle is placed in the complex plane with complex vertices z₁, z₂, and z₃, then for either non-real cube root ${\displaystyle \omega }$ of 1 the triangle is equilateral if and only if[17]^{:Lemma 2}

${\displaystyle z_{1}+\omega z_{2}+\omega ^{2}z_{3}=0.}$

Given a point P in the interior of an equilateral triangle, the ratio of the sum of its distances from the vertices to the sum of its distances from the sides is greater than or equal to 2, equality holding when P is the centroid. In no other triangle is there a point for which this ratio is as small as 2.[18] This is the Erdős–Mordell inequality; a stronger variant of it is Barrow's inequality, which replaces the perpendicular distances to the sides with the distances from P to the points where the angle bisectors of ∠APB, ∠BPC, and ∠CPA cross the sides (A, B, and C being the vertices).

For any point P in the plane, with distances p, q, and t from the vertices A, B, and C respectively,[19]

${\displaystyle \displaystyle 3(p^{4}+q^{4}+t^{4}+a^{4})=(p^{2}+q^{2}+t^{2}+a^{2})^{2}.}$

For any point P in the plane, with distances p, q, and t from the vertices,[20]

${\displaystyle \displaystyle p^{2}+q^{2}+t^{2}=3(R^{2}+L^{2})}$

and

${\displaystyle \displaystyle p^{4}+q^{4}+t^{4}=3[(R^{2}+L^{2})^{2}+2R^{2}L^{2}],}$

where R is the circumscribed radius and L is the distance between point P and the centroid of the equilateral triangle.

For any point P on the inscribed circle of an equilateral triangle, with distances p, q, and t from the vertices,[21]

${\displaystyle \displaystyle 4(p^{2}+q^{2}+t^{2})=5a^{2}}$

and

${\displaystyle \displaystyle 16(p^{4}+q^{4}+t^{4})=11a^{4}.}$

For any point P on the minor arc BC of the circumcircle, with distances p, q, and t from A, B, and C respectively,[13]

${\displaystyle \displaystyle p=q+t}$

and

${\displaystyle \displaystyle q^{2}+qt+t^{2}=a^{2};}$

moreover, if point D on side BC divides PA into segments PD and DA with DA having length z and PD having length y, then [13]^:172

${\displaystyle z={\frac {t^{2}+tq+q^{2}}{t+q}},}$

which also equals ${\displaystyle {\tfrac {t^{3}-q^{3}}{t^{2}-q^{2}}}}$ if t ≠ q; and

${\displaystyle {\frac {1}{q}}+{\frac {1}{t}}={\frac {1}{y}},}$

which is the optic equation.

There are numerous triangle inequalities that hold with equality if and only if the triangle is equilateral.

An equilateral triangle is the most symmetrical triangle, having 3 lines of reflection and rotational symmetry of order 3 about its center. Its symmetry group is the dihedral group of order 6 D₃.

Equilateral triangles are the only triangles whose Steiner inellipse is a circle (specifically, it is the incircle).

The integer-sided equilateral triangle is the only triangle with integer sides and three rational angles as measured in degrees.[22]

The equilateral triangle is the only acute triangle that is similar to its orthic triangle (with vertices at the feet of the altitudes) (the heptagonal triangle being the only obtuse one).[23]^{:p. 19}

A regular tetrahedron is made of four equilateral triangles.

Equilateral triangles are found in many other geometric constructs. The intersection of circles whose centers are a radius width apart is a pair of equilateral arches, each of which can be inscribed with an equilateral triangle. They form faces of regular and uniform polyhedra. Three of the five Platonic solids are composed of equilateral triangles. In particular, the regular tetrahedron has four equilateral triangles for faces and can be considered the three-dimensional analogue of the shape. The plane can be tiled using equilateral triangles giving the triangular tiling.

Construction of equilateral triangle with compass and straightedge

An equilateral triangle is easily constructed using a straightedge and compass, because 3 is a Fermat prime. Draw a straight line, and place the point of the compass on one end of the line, and swing an arc from that point to the other point of the line segment. Repeat with the other side of the line. Finally, connect the point where the two arcs intersect with each end of the line segment

An alternative method is to draw a circle with radius r, place the point of the compass on the circle and draw another circle with the same radius. The two circles will intersect in two points. An equilateral triangle can be constructed by taking the two centers of the circles and either of the points of intersection.

In both methods a by-product is the formation of vesica piscis.

The proof that the resulting figure is an equilateral triangle is the first proposition in Book I of Euclid's Elements.

The area formula ${\displaystyle A={\frac {\sqrt {3}}{4}}a^{2}}$ in terms of side length a can be derived directly using the Pythagorean theorem or using trigonometry.

Using the Pythagorean theorem

The area of a triangle is half of one side a times the height h from that side:

${\displaystyle A={\frac {1}{2}}ah.}$

An equilateral triangle with a side of 2 has a height of √3, as the sine of 60° is √3/2.

The legs of either right triangle formed by an altitude of the equilateral triangle are half of the base a, and the hypotenuse is the side a of the equilateral triangle. The height of an equilateral triangle can be found using the Pythagorean theorem

${\displaystyle \left({\frac {a}{2}}\right)^{2}+h^{2}=a^{2}}$

so that

${\displaystyle h={\frac {\sqrt {3}}{2}}a.}$

Substituting h into the area formula (1/2)ah gives the area formula for the equilateral triangle:

${\displaystyle A={\frac {\sqrt {3}}{4}}a^{2}.}$

Equilateral triangles have frequently appeared in man made constructions:

• The shape occurs in modern architecture such as the cross-section of the Gateway Arch.[24]
• Its applications in flags and heraldry includes the flag of Nicaragua[25] and the flag of the Philippines.[26]
• It is a shape of a variety of road signs, including the yield sign.[27]

• Almost-equilateral Heronian triangle
• Isosceles triangle
• Ternary plot
• Trilinear coordinates

• Weisstein, Eric W. "Equilateral Triangle". MathWorld.