词条 | List of triangle inequalities |
释义 |
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. Main parameters and notationThe parameters most commonly appearing in triangle inequalities are:
Side lengthsThe basic triangle inequality is or equivalently In addition, where the value of the right side is the lowest possible bound,[1]{{rp|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]{{rp|p.250,#82}} [1]{{rp|p. 260}} [1]{{rp|p. 261}} [1]{{rp|p. 261}} [1]{{rp|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]{{rp|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]{{rp|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]{{rp|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. Angles[1]{{rp|p. 286}} [2]{{rp|p.21,#836}} for semi-perimeter s, with equality only in the equilateral case.[2]{{rp|p.13,#608}} [4]{{rp|Thm.1}} [1]{{rp|p.286}} [1]{{rp|p. 286}} [5]{{rp|p. 203}} [2]{{rp|p.149,#3297}} where the golden ratio. [1]{{rp|p. 286}} [1]{{rp|p. 286}} [6] [2]{{rp|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°;[22]{{rp|Cor. 3}} and with equality if and only if the triangle is isosceles with apex angle less than or equal to 60°.[22]{{rp|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°.[22]{{rp|Prop. 5}} Further, any two angle measures A and B opposite sides a and b respectively are related according to[1]{{rp|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]{{rp|p. 261}} If a point D is in the interior of triangle ABC, then [1]{{rp|p. 263}} For an acute triangle we have[2]{{rp|p.26,#954}} with the reverse inequality holding for an obtuse triangle. Furthermore, for non-obtuse triangles we have[7]{{rp|Corollary 3}} with equality if and only if it is a right triangle with hypotenuse AC. AreaWeitzenböck's inequality is, in terms of area T,[1]{{rp|p. 290}} with equality only in the equilateral case. This is a corollary of the Hadwiger–Finsler inequality, which is Also, [31]{{rp|p. 138}} and[2]{{rp|p.192,#340.3}}[5]{{rp|p. 204}} From the rightmost upper bound on T, using the arithmetic-geometric mean inequality, is obtained the isoperimetric inequality for triangles: [5]{{rp|p. 203}} for semiperimeter s. This is sometimes stated in terms of perimeter p as with equality for the equilateral triangle.[8] This is strengthened by Bonnesen's inequality also strengthens the isoperimetric inequality: We also have [1]{{rp|p. 290}}[9]{{rp|p. 138}} with equality only in the equilateral case; [2]{{rp|p.111,#2807}} for semiperimeter s; and [2]{{rp|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.[10] 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]{{rp|p. 138}} Let the interior angle bisectors of A, B, and C meet the opposite sides at D, E, and F. Then[2]{{rp|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.[11] Medians and centroidLet 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 [12] 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]{{rp|p. 271}} Moreover,[2]{{rp|p.12,#589}} with equality only in the equilateral case, and for inradius r,[2]{{rp|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]{{rp|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]{{rp|p.17#723}} and in addition,[2]{{rp|p.156,#S56}} For an acute triangle we have[2]{{rp|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]{{rp|p.192,#339.3}} The three medians of any triangle can form the sides of another triangle:[13]{{rp|p. 592}} Furthermore,[14]{{rp|Coro. 6}} AltitudesThe altitudes ha , etc. each connect a vertex to the opposite side and are perpendicular to that side. They satisfy both[1]{{rp|p. 274}} and In addition, if then[2]{{rp|222,#67}} We also have[2]{{rp|p.140,#3150}} For internal angle bisectors ta, tb, tc from vertices A, B, C and circumcenter R and incenter r, we have[2]{{rp|p.125,#3005}} The reciprocals of the altitudes of any triangle can themselves form a triangle:[15] Internal angle bisectors and incenterThe 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]{{rp|pp. 271–3}} Further,[2]{{rp|p.224,#132}} in terms of the medians, and[2]{{rp|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]{{rp|p.11,#535}} with equality only in the equilateral case, and[2]{{rp|p.14,#628}} for circumradius R and inradius r, again with equality only in the equilateral case. In addition,.[2]{{rp|p.20,#795}} For incenter I (the intersection of the internal angle bisectors),[2]{{rp|p.127,#3033}} For midpoints L, M, N of the sides,[2]{{rp|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[16]{{rp|p.232}} and and we have the angle inequality[16]{{rp|p.233}} In addition,[16]{{rp|p.233,Lemma 3}} where v is the longest median. Three triangles with vertex at the incenter, OIH, GIH, and OGI, are obtuse:[16]{{rp|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:[17][18] The larger of two angles of a triangle has the shorter internal angle bisector:[19]{{rp|p.72,#114}} Perpendicular bisectors of sidesThese 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[20] and Segments from an arbitrary pointInterior pointConsider 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]{{rp|pp. 275–7}} and more strongly than the second of these inequalities is[1]{{rp|p. 278}} We also have Ptolemy's inequality[2]{{rp|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]{{rp|p. 278}} Further, the Erdős–Mordell inequality states that[21] [22]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[23] Also stronger than the Erdős–Mordell inequality is the following:[24] 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[25] 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]{{rp|p.29,#1045}} For interior point P with distances PA, PB, PC from the vertices and with triangle area T,[2]{{rp|p.37,#1159}} and[2]{{rp|p.26,#965}} For an interior point P, centroid G, midpoints L, M, N of the sides, and semiperimeter s,[2]{{rp|p.140,#3164}}[2]{{rp|p.130,#3052}} Moreover, for positive numbers k1, k2, k3, and t with t less than or equal to 1:[26]{{rp|Thm.1}} while for t > 1 we have[26]{{rp|Thm.2}} Interior or exterior pointThere 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,[27]{{rp|p. 109}} Others include:[28]{{rp|pp. 180–1}} for k = 0, 1, ..., 6; and for k = 0, 1, ..., 9. Furthermore, for circumradius R, [29]{{rp|p. 227}} [29]{{rp|p. 233}} [29]{{rp|p. 233}} [29]{{rp|p. 233}} Inradius, exradii, and circumradiusInradius and circumradiusThe Euler inequality for the circumradius R and the inradius r states that with equality only in the equilateral case.[30]{{rp|p. 198}} A stronger version[5]{{rp|p. 198}} is By comparison,[2]{{rp|p.183,#276.2}} where the right side could be positive or negative. Two other refinements of Euler's inequality are[2]{{rp|p.134,#3087}} and Another symmetric inequality is[2]{{rp|p.125,#3004}} Moreover, [1]{{rp|288}} in terms of the semiperimeter s;[2]{{rp|p.20,#816}} in terms of the area T;[5]{{rp|p. 201}} [5]{{rp|p. 201}} and [2]{{rp|p.17#708}} in terms of the semiperimeter s; and also in terms of the semiperimeter.[5]{{rp|p. 206}}[31]{{rp|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.[31]{{rp|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]{{rp|p. 291}}Blundon's inequality states that[5]{{rp|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]{{rp|p.14,#644}} In terms of the vertex angles we have [2]{{rp|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]{{rp|Thm. 4}} with equality only in the equilateral case, and[36]{{rp|Thm. 6}} with equality only in the equilateral case. Circumradius and other lengthsFor the circumradius R we have[2]{{rp|p.101,#2625}} and[2] {{rp|p.35,#1130}} We also have[1]{{rp|pp. 287–90}} in terms of the altitudes, in terms of the medians, and[2]{{rp|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]{{rp|p.17,#718}} For an acute triangle the distance between the circumcenter O and the orthocenter H satisfies[2]{{rp|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 lengthsFor the inradius r we have[1]{{rp|pp. 289–90}} in terms of the altitudes, and in terms of the radii of the excircles. We additionally have [2]{{rp|p.66,#1678}} and [2]{{rp|p.183,#281.2}} The exradii and medians are related by[2]{{rp|p.66,#1680}} In addition, for an acute triangle the distance between the incircle center I and orthocenter H satisfies[2]{{rp|p.26,#954}} with the reverse inequality for an obtuse triangle. Also, an acute triangle satisfies[2]{{rp|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]{{rp|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]{{rp|p.181,#264.4}} for circumradius R, and[2]{{rp|p.181,#264.4}}[2]{{rp|p.45,#1282}} If the incircle is tangent to the sides at D, E, F, then[2]{{rp|p.115,#2875}} for semiperimeter s. Inscribed figuresInscribed hexagonIf 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]{{rp|p.42,#1245}} Inscribed triangleIf 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]{{rp|p.137}} Inscribed squaresAn 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]{{rp|p. 115}} Moreover, for any square inscribed in any triangle we have[2]{{rp|p.18,#729}}[38] Euler lineA 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.[16]{{rp|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:[16]{{rp|p. 234,Propos.5}} For all of these ratios, the upper bound of 1/3 is the tightest possible.[16]{{rp|p.235,Thm.6}} Right triangleIn right triangles the legs a and b and the hypotenuse c obey the following, with equality only in the isosceles case:[1]{{rp|p. 280}} In terms of the inradius, the hypotenuse obeys[1]{{rp|p. 281}} and in terms of the altitude from the hypotenuse the legs obey[1]{{rp|p. 282}} Isosceles triangleIf 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]{{rp|p.169,#44}} Equilateral triangleFor 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]{{rp|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]{{rp|p.178,#235.4}} Two trianglesPedoe'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 trianglesIn a triangle on the surface of a sphere, as well as in elliptic geometry, This inequality is reversed for hyperbolic triangles. See also
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