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For the basic inequality a < b + c, see Triangle inequality.
For inequalities of acute or obtuse triangles, see Acute and obtuse triangles.

In geometry, triangle inequalities are inequalities involving the parameters of triangles, that hold for every triangle, or for every triangle meeting certain conditions. The inequalities give an ordering of two different values: they are of the form "less than", "less than or equal to", "greater than", or "greater than or equal to". The parameters in a triangle inequality can be the side lengths, the semiperimeter, the angle measures, the values of trigonometric functions of those angles, the area of the triangle, the medians of the sides, the altitudes, the lengths of the internal angle bisectors from each angle to the opposite side, the perpendicular bisectors of the sides, the distance from an arbitrary point to another point, the inradius, the exradii, the circumradius, and/or other quantities.

Unless otherwise specified, this article deals with triangles in the Euclidean plane.

Contents

Main parameters and notationEdit

The parameters most commonly appearing in triangle inequalities are:

  • the side lengths a, b, and c;
  • the semiperimeter s = (a + b + c) / 2 (half the perimeter p);
  • the angle measures A, B, and C of the angles of the vertices opposite the respective sides a, b, and c (with the vertices denoted with the same symbols as their angle measures);
  • the values of trigonometric functions of the angles;
  • the area T of the triangle;
  • the medians ma, mb, and mc of the sides (each being the length of the line segment from the midpoint of the side to the opposite vertex);
  • the altitudes ha, hb, and hc (each being the length of a segment perpendicular to one side and reaching from that side (or possibly the extension of that side) to the opposite vertex);
  • the lengths of the internal angle bisectors ta, tb, and tc (each being a segment from a vertex to the opposite side and bisecting the vertex's angle);
  • the perpendicular bisectors pa, pb, and pc of the sides (each being the length of a segment perpendicular to one side at its midpoint and reaching to one of the other sides);
  • the lengths of line segments with an endpoint at an arbitrary point P in the plane (for example, the length of the segment from P to vertex A is denoted PA or AP);
  • the inradius r (radius of the circle inscribed in the triangle, tangent to all three sides), the exradii ra, rb, and rc (each being the radius of an excircle tangent to side a, b, or c respectively and tangent to the extensions of the other two sides), and the circumradius R (radius of the circle circumscribed around the triangle and passing through all three vertices).

Side lengthsEdit

The basic triangle inequality is

 

or equivalently

 

In addition,

 

where the value of the right side is the lowest possible bound,[1]:p. 259 approached asymptotically as certain classes of triangles approach the degenerate case of zero area. The left inequality, which holds for all positive a, b, c, is Nesbitt's inequality.

We have

 [2]:p.250,#82
 [1]:p. 260
 [1]:p. 261
 [1]:p. 261
 [1]:p. 261

If angle C is obtuse (greater than 90°) then

 

if C is acute (less than 90°) then

 

The in-between case of equality when C is a right angle is the Pythagorean theorem.

In general,[2]:p.1,#74

 

with equality approached in the limit only as the apex angle of an isosceles triangle approaches 180°.

If the centroid of the triangle is inside the triangle's incircle, then[3]:p. 153

 

While all of the above inequalities are true because a, b, and c must follow the basic triangle inequality that the longest side is less than half the perimeter, the following relations hold for all positive a, b, and c:[1]:p.267

 

each holding with equality only when a = b = c. This says that in the non-equilateral case the harmonic mean of the sides is less than their geometric mean which in turn is less than their arithmetic mean.

AnglesEdit

  [1]:p. 286
 [2]:p.21,#836
 

for semi-perimeter s, with equality only in the equilateral case.[2]:p.13,#608

  [4]:Thm.1
  [1]:p.286
  [1]:p. 286
  [5]:p. 203
 [2]:p.149,#3297

where   the golden ratio.

  [1]:p. 286
  [1]:p. 286
  [6]
 [2]:p.187,#309.2

For circumradius R and inradius r we have

 

with equality if and only if the triangle is isosceles with apex angle greater than or equal to 60°;[7]:Cor. 3 and

 

with equality if and only if the triangle is isosceles with apex angle less than or equal to 60°.[7]:Cor. 3

We also have

 

and likewise for angles B, C, with equality in the first part if the triangle is isosceles and the apex angle is at least 60° and equality in the second part if and only if the triangle is isosceles with apex angle no greater than 60°.[7]:Prop. 5

Further, any two angle measures A and B opposite sides a and b respectively are related according to[1]:p. 264

 

which is related to the isosceles triangle theorem and its converse, which state that A = B if and only if a = b.

By Euclid's exterior angle theorem, any exterior angle of a triangle is greater than either of the interior angles at the opposite vertices:[1]:p. 261

 

If a point D is in the interior of triangle ABC, then

 [1]:p. 263

For an acute triangle we have[2]:p.26,#954

 

with the reverse inequality holding for an obtuse triangle.

Furthermore, for non-obtuse triangles we have[8]:Corollary 3

 

with equality if and only if it is a right triangle with hypotenuse AC.

AreaEdit

Weitzenböck's inequality is, in terms of area T,[1]:p. 290

 

with equality only in the equilateral case. This is a corollary of the Hadwiger–Finsler inequality, which is

 

Also,

 [9]:p. 138

and[2]:p.192,#340.3[5]:p. 204

 

From the rightmost upper bound on T, using the arithmetic-geometric mean inequality, is obtained the isoperimetric inequality for triangles:

  [5]:p. 203

for semiperimeter s. This is sometimes stated in terms of perimeter p as

 

with equality for the equilateral triangle.[10] This is strengthened by

 

Bonnesen's inequality also strengthens the isoperimetric inequality:

 

We also have

  [1]:p. 290[9]:p. 138

with equality only in the equilateral case;

 [2]:p.111,#2807

for semiperimeter s; and

 [2]:p.88,#2188

Ono's inequality for acute triangles (those with all angles less than 90°) is

 

The area of the triangle can be compared to the area of the incircle:

 

with equality only for the equilateral triangle.[11]

If an inner triangle is inscribed in a reference triangle so that the inner triangle's vertices partition the perimeter of the reference triangle into equal length segments, the ratio of their areas is bounded by[9]:p. 138

 

Let the interior angle bisectors of A, B, and C meet the opposite sides at D, E, and F. Then[2]:p.18,#762

 

A line through a triangle’s median splits the area such that the ratio of the smaller sub-area to the original triangle’s area is at least 4/9.[12]

Medians and centroidEdit

Let ABC be a triangle, let G be its centroid, and let D, E, and F be the midpoints of BC, CA, and AB, respectively. For any point P in the plane of ABC then

 [13]

The three medians   of a triangle each connect a vertex with the midpoint of the opposite side, and the sum of their lengths satisfies[1]:p. 271

 

Moreover,[2]:p.12,#589

 

with equality only in the equilateral case, and for inradius r,[2]:p.22,#846

 

If we further denote the lengths of the medians extended to their intersections with the circumcircle as Ma , Mb , and Mc , then[2]:p.16,#689

 

The centroid G is the intersection of the medians. Let AG, BG, and CG meet the circumcircle at U, V, and W respectively. Then both[2]:p.17#723

 

and

 

in addition,[2]:p.156,#S56

 

For an acute triangle we have[2]:p.26,#954

 

in terms of the circumradius R, while the opposite inequality holds for an obtuse triangle.

Denoting as IA, IB, IC the distances of the incenter from the vertices, the following holds:[2]:p.192,#339.3

 

The three medians of any triangle can form the sides of another triangle:[14]:p. 592

 

Furthermore,[15]:Coro. 6

 

AltitudesEdit

The altitudes ha , etc. each connect a vertex to the opposite side and are perpendicular to that side. They satisfy both[1]:p. 274

 

and

 

In addition, if   then[2]:222,#67

 

We also have[2]:p.140,#3150

 

For internal angle bisectors ta, tb, tc from vertices A, B, C and circumcenter R and incenter r, we have[2]:p.125,#3005

 

The reciprocals of the altitudes of any triangle can themselves form a triangle:[16]

 

Internal angle bisectors and incenterEdit

The internal angle bisectors are segments in the interior of the triangle reaching from one vertex to the opposite side and bisecting the vertex angle into two equal angles. The angle bisectors ta etc. satisfy

 

in terms of the sides, and

 

in terms of the altitudes and medians, and likewise for tb and tc .[1]:pp. 271–3 Further,[2]:p.224,#132

 

in terms of the medians, and[2]:p.125,#3005

 

in terms of the altitudes, inradius r and circumradius R.

Let Ta , Tb , and Tc be the lengths of the angle bisectors extended to the circumcircle. Then[2]:p.11,#535

 

with equality only in the equilateral case, and[2]:p.14,#628

 

for circumradius R and inradius r, again with equality only in the equilateral case. In addition,.[2]:p.20,#795

 

For incenter I (the intersection of the internal angle bisectors),[2]:p.127,#3033

 

For midpoints L, M, N of the sides,[2]:p.152,#J53

 

For incenter I, centroid G, circumcenter O, nine-point center N, and orthocenter H, we have for non-equilateral triangles the distance inequalities[17]:p.232

 
 
 

and

 

and we have the angle inequality[17]:p.233

 

In addition,[17]:p.233,Lemma 3

 

where v is the longest median.

Three triangles with vertex at the incenter, OIH, GIH, and OGI, are obtuse:[17]:p.232

  >   > 90° ,   > 90°.

Since these triangles have the indicated obtuse angles, we have

 

and in fact the second of these is equivalent to a result stronger than the first, shown by Euler:[18][19]

 

The larger of two angles of a triangle has the shorter internal angle bisector:[20]:p.72,#114

 

Perpendicular bisectors of sidesEdit

These inequalities deal with the lengths pa etc. of the triangle-interior portions of the perpendicular bisectors of sides of the triangle. Denoting the sides so that   we have[21]

 

and

 

Segments from an arbitrary pointEdit

Interior pointEdit

Consider any point P in the interior of the triangle, with the triangle's vertices denoted A, B, and C and with the lengths of line segments denoted PA etc. We have[1]:pp. 275–7

 

and more strongly than the second of these inequalities is[1]:p. 278

 

We also have Ptolemy's inequality[2]:p.19,#770

 

for interior point P and likewise for cyclic permutations of the vertices.

If we draw perpendiculars from interior point P to the sides of the triangle, intersecting the sides at D, E, and F, we have[1]:p. 278

 

Further, the Erdős–Mordell inequality states that[22][23]

 

with equality in the equilateral case. More strongly, Barrow's inequality states that if the interior bisectors of the angles at interior point P (namely, of ∠APB, ∠BPC, and ∠CPA) intersect the triangle's sides at U, V, and W, then[24]

 

Also stronger than the Erdős–Mordell inequality is the following:[25] Let D, E, F be the orthogonal projections of P onto BC, CA, AB respectively, and H, K, L be the orthogonal projections of P onto the tangents to the triangle's circumcircle at A, B, C respectively. Then

 

With orthogonal projections H, K, L from P onto the tangents to the triangle's circumcircle at A, B, C respectively, we have[26]

 

where R is the circumradius.

Again with distances PD, PE, PF of the interior point P from the sides we have these three inequalities:[2]:p.29,#1045

 
 
 

For interior point P with distances PA, PB, PC from the vertices and with triangle area T,[2]:p.37,#1159

 

and[2]:p.26,#965

 

For an interior point P, centroid G, midpoints L, M, N of the sides, and semiperimeter s,[2]:p.140,#3164[2]:p.130,#3052

 

Moreover, for positive numbers k1, k2, k3, and t with t less than or equal to 1:[27]:Thm.1

 

while for t > 1 we have[27]:Thm.2

 

Interior or exterior pointEdit

There are various inequalities for an arbitrary interior or exterior point in the plane in terms of the radius r of the triangle's inscribed circle. For example,[28]:p. 109

 

Others include:[29]:pp. 180–1

 

for k = 0, 1, ..., 6;

 
 

and

 

for k = 0, 1, ..., 9.

Furthermore, for circumradius R,

 [30]:p. 227
 [30]:p. 233
 [30]:p. 233
 [30]:p. 233

Inradius, exradii, and circumradiusEdit

Inradius and circumradiusEdit

The Euler inequality for the circumradius R and the inradius r states that

 

with equality only in the equilateral case.[31]:p. 198

A stronger version[5]:p. 198 is

 

By comparison,[2]:p.183,#276.2

 

where the right side could be positive or negative.

Two other refinements of Euler's inequality are[2]:p.134,#3087

 

and

 

Another symmetric inequality is[2]:p.125,#3004

 

Moreover,

 [1]:288
 

in terms of the semiperimeter s;[2]:p.20,#816

 

in terms of the area T;[5]:p. 201

  [5]:p. 201

and

  [2]:p.17#708

in terms of the semiperimeter s; and

 
 

also in terms of the semiperimeter.[5]:p. 206[7]:p. 99 Here the expression   where d is the distance between the incenter and the circumcenter. In the latter double inequality, the first part holds with equality if and only if the triangle is isosceles with an apex angle of at least 60°, and the last part holds with equality if and only if the triangle is isosceles with an apex angle of at most 60°. Thus both are equalities if and only if the triangle is equilateral.[7]:Thm. 1

We also have for any side a[32]

 

where   if the circumcenter is on or outside of the incircle and   if the circumcenter is inside the incircle. The circumcenter is inside the incircle if and only if[32]

 

Further,

 [1]:p. 291

Blundon's inequality states that[5]:p. 206;[33][34]

 

We also have, for all acute triangles,[35]

 

For incircle center I, let AI, BI, and CI extend beyond I to intersect the circumcircle at D, E, and F respectively. Then[2]:p.14,#644

 

In terms of the vertex angles we have [2]:p.193,#342.6

 

Denote as   the radii of the tangent circles at the vertices to the triangle's circumcircle and to the opposite sides. Then[36]:Thm. 4

 

with equality only in the equilateral case, and[36]:Thm. 6

 

with equality only in the equilateral case.

Circumradius and other lengthsEdit

For the circumradius R we have[2]:p.101,#2625

 

and[2] :p.35,#1130

 

We also have[1]:pp. 287–90

 
 
 

in terms of the altitudes,

 

in terms of the medians, and[2]:p.26,#957

 

in terms of the area.

Moreover, for circumcenter O, let lines AO, BO, and CO intersect the opposite sides BC, CA, and AB at U, V, and W respectively. Then[2]:p.17,#718

 

For an acute triangle the distance between the circumcenter O and the orthocenter H satisfies[2]:p.26,#954

 

with the opposite inequality holding for an obtuse triangle.

The circumradius is at least twice the distance between the first and second Brocard points B1 and B2:[37]

 

Inradius, exradii, and other lengthsEdit

For the inradius r we have[1]:pp. 289–90

 
 

in terms of the altitudes, and

 

in terms of the radii of the excircles. We additionally have

 [2]:p.66,#1678

and

 [2]:p.183,#281.2

The exradii and medians are related by[2]:p.66,#1680

 

In addition, for an acute triangle the distance between the incircle center I and orthocenter H satisfies[2]:p.26,#954

 

with the reverse inequality for an obtuse triangle.

Also, an acute triangle satisfies[2]:p.26,#954

 

in terms of the circumradius R, again with the reverse inequality holding for an obtuse triangle.

If the internal angle bisectors of angles A, B, C meet the opposite sides at U, V, W then[2]:p.215,32nd IMO,#1

 

If the internal angle bisectors through incenter I extend to meet the circumcircle at X, Y and Z then [2]:p.181,#264.4

 

for circumradius R, and[2]:p.181,#264.4[2]:p.45,#1282

 

If the incircle is tangent to the sides at D, E, F, then[2]:p.115,#2875

 

for semiperimeter s.

Inscribed figuresEdit

Inscribed hexagonEdit

If a tangential hexagon is formed by drawing three segments tangent to a triangle's incircle and parallel to a side, so that the hexagon is inscribed in the triangle with its other three sides coinciding with parts of the triangle's sides, then[2]:p.42,#1245

 

Inscribed triangleEdit

If three points D, E, F on the respective sides AB, BC, and CA of a reference triangle ABC are the vertices of an inscribed triangle, which thereby partitions the reference triangle into four triangles, then the area of the inscribed triangle is greater than the area of at least one of the other interior triangles, unless the vertices of the inscribed triangle are at the midpoints of the sides of the reference triangle (in which case the inscribed triangle is the medial triangle and all four interior triangles have equal areas):[9]:p.137

 

Inscribed squaresEdit

An acute triangle has three inscribed squares, each with one side coinciding with part of a side of the triangle and with the square's other two vertices on the remaining two sides of the triangle. (A right triangle has only two distinct inscribed squares.) If one of these squares has side length xa and another has side length xb with xa < xb, then[38]:p. 115

 

Moreover, for any square inscribed in any triangle we have[2]:p.18,#729[38]

 

Euler lineEdit

A triangle's Euler line goes through its orthocenter, its circumcenter, and its centroid, but does not go through its incenter unless the triangle is isosceles.[17]:p.231 For all non-isosceles triangles, the distance d from the incenter to the Euler line satisfies the following inequalities in terms of the triangle's longest median v, its longest side u, and its semiperimeter s:[17]:p. 234,Propos.5

 

For all of these ratios, the upper bound of 1/3 is the tightest possible.[17]:p.235,Thm.6

Right triangleEdit

In right triangles the legs a and b and the hypotenuse c obey the following, with equality only in the isosceles case:[1]:p. 280

 

In terms of the inradius, the hypotenuse obeys[1]:p. 281

 

and in terms of the altitude from the hypotenuse the legs obey[1]:p. 282

 

Isosceles triangleEdit

If the two equal sides of an isosceles triangle have length a and the other side has length c, then the internal angle bisector t from one of the two equal-angled vertices satisfies[2]:p.169,# 44

 

Equilateral triangleEdit

For any point P in the plane of an equilateral triangle ABC, the distances of P from the vertices, PA, PB, and PC, are such that, unless P is on the triangle's circumcircle, they obey the basic triangle inequality and thus can themselves form the sides of a triangle:[1]:p. 279

 

However, when P is on the circumcircle the sum of the distances from P to the nearest two vertices exactly equals the distance to the farthest vertex.

A triangle is equilateral if and only if, for every point P in the plane, with distances PD, PE, and PF to the triangle's sides and distances PA, PB, and PC to its vertices,[2]:p.178,#235.4

 

Two trianglesEdit

Pedoe's inequality for two triangles, one with sides a, b, and c and area T, and the other with sides d, e, and f and area S, states that

 

with equality if and only if the two triangles are similar.

The hinge theorem or open-mouth theorem states that if two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle. That is, in triangles ABC and DEF with sides a, b, c, and d, e, f respectively (with a opposite A etc.), if a = d and b = e and angle C > angle F, then

 

The converse also holds: if c > f, then C > F.

The angles in any two triangles ABC and DEF are related in terms of the cotangent function according to[6]

 

Non-Euclidean trianglesEdit

In a triangle on the surface of a sphere, as well as in elliptic geometry,

 

This inequality is reversed for hyperbolic triangles.

See alsoEdit

ReferencesEdit

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