国外数学问题

Useful Identities and Inequalities in Geometry ThreadRegular Notation for a , with sides:Let the area be .Letbe any point in. Let the cevians throughandintersectat.Let the semiperimeter , inradius, and circumradius be respectively.Let the heights protruding from be respectively, which meet at the orthocenter . Let the feet ofthe perpendiculars from toberespectively.Let the medians protruding from be respectively, which meet at the centroid. Let themidpoints ofberespectively.Let the angle bisectors protruding from be respectively, which meet at the incenter . Let the feet ofthe perpendiculars from toberespectively.Let the centers of the excircles tangent toberespectively, and the excircles be tangent toat. Let the radii of the excircles tangent toberespectively.Result 1. Common methods of obtaining the area of a triangle(Heron's Theorem)Result 2. Highly common lengths associated with a triangle in terms of(Apollonius' Theorem)Result 3. Highly common radii associated with a triangle in terms ofResult 4. Symmetric sums ofin terms ofResult 5. Symmetric sums of in terms ofResult 6. Symmetric sums of in terms of.This leads to the inequalities and .Angle Bisector Theorem:where is the angle bisector ofIf is the orthic triangle of thenwhere are the excentresStewart's Theorem:Ptolemy's Theorem: If is a quadrilateral then .Equality occurs if and only if is cyclic.Result 43 :Result 45 :Result 46 :Result 47 :Result 48 :Result 49 :Result 51 :is acute-angledis obtuseis right-angledResult 52 :Law of tangents :Result 53.Result 54.Let be the sides of a triangle ABC. Let be any reals and let be any point. Prove that.1. If , then2. If and is the circumcenter, then it's Weitzenbock.3. If and is the circumcenter then .Result 55.Result 56., which amazingly coincides with Result 6a.Result 57.Result 58. Brahmagupta's FormulaIn cyclic quadrilateral with sides ,(produces Heron's Formula as )Result 59. Bretschneider's FormulaIn any quadrilateral with sides and opposite angles ,(produces Brahmagupta's Formula when is cyclic )Result 60. Euler's InequalityResult 61. Gerretsen's Inequality (often teamed up with Euler)Result 62. Blundon's Inequality (not usually necessary, but useful when Gerretsen is not strong enough)Result 63. Law of CosinesResult 64. Law of SinesResult 66. More methods of finding the area of a triangleResult 67.Result 68.Result 69.Result 70.Result 71.Result 72.Result 72.Result 73.Result 74.(credits to luisgeometra)Result 75. Triangle Inequality, and cyclic versions hold as usual, with equality holding if and only if , or in other words the triangle is degenerate as the segment . As such, the symbol is more popular.Result 76. Polygon InequalityLet there be an -sided polygon with side lengths represented by the sequence . Let. Then it holds that for each .Equality holds if and only if the polygon is degenerate as the segment . Once again, the symbol is more popular.This is the Triangle Inequality when .Result 77 : (Pedoe's Inequality)Let , , be the side-lengths of a triangle with area , and , , be the side-lengths of a triangle with area ,then :Result 78 : (Euler's Formula)Result 79 : (Remarkable Distances)Where is Nagel's point and is Gergonne's point.Result 80 : (Useful Identities)Result 81 : (Geometric Identities for every point)For any point in the plane :Result 82 : (Trigonometric Identity)Result 83 : (Useful Inequality, stronger than Gerretsen's one)Result 84 : (Other Inequalities with , , )(Walker's Inequality : for an acute-angled triangle)Result 85. Reciprocal IdentitiesResult 86. Pythagorean IdentitiesResult 87. Quotient IdentitiesResult 88. Co-function Transformation IdentitiesResult 89. Reflection IdentitiesResult 90. Period Identities (best constants obviously),Result 91. Even-odd IdentitiesResult 92. Sum IdentitiesResult 93. Difference IdentitiesResult 94. Double Angle IdentitiesResult 95. Half-angle Identities(credits to luisgeometra) Result 96. Product-to-sum IdentitiesResult 97. Sum-to-Product Identities_____________________________________________________________________________A final few symmetric sums (the first uses WakeUp's post)Result 98. ExradiiResult 99._____________________________________________________________________________Let such that:So it is easy to convert all symmetric sums of into symmetric sums of . Result 100.In a cyclic quadrilateral , .Result 101.For any -sided polygon, the sum of the interior angles is .For any regular -sided polygon, each interior angle is .For any convex polygon, the sum of the exterior angles is .Result 102.For any regular -sided polygon with side length , the inradius is .For any regular -sided polygon with side length , the circumradius is .Result 103. Ceva's TheoremFollowing regular notation, ._____________________________________________________________________________Constantin PM-ed me some interesting identities, the best of which are:Result 104.Result 105.Result 106.Result 107. How sides are split by common concurrent cevians,,,Let the symmedians intersect sides respectively at respectively.,The angle bisector case is not included because it is ugly and unused.Result 108. Ravi Transformation.In words, the sides of an arbitrary triangle can be converted into where .Conversely, an arbitrary triple forms the triple of a triangle. This is very important.Let .Result 109..In very basic cases we use , in more complex cases we use , and in some equally or more complex cases we use (usually in problems with Gergonne or Nagel).Let the Gergonne cevians be , and the Nagel cevians be .Result 110. More cevians lengths.Result 111. Side lengths of common cevian triangles(Orthic Triangle)(Medial Triangle)(Intouch Triangle)(Extouch Triangle)I did not include the Incentral Triangle because it is very ugly.Note these are all easily derived using the Cosine Law and then simplified using manipulation and additional identities. Result 112. Klamkin's InequalityLet there be arbitrary . For an arbitrary :Extension of result 111.Let , and be the classical elements of - the orthic triangle of an acute-angled triangle .Then :Also, note that is the incenter of and a.s.o.Let , , be the classical elements of - the triangle formed by the centers of excircles ( represents the center of excircle w.r.t. vertex a.s.o.)Then :Also, note that is the orthocenter of and a.s.o.In addition, the following beautiful identity holds:where and represent the areas of triangles and respectively. This equality points out the followinginteresting implication: Weitzenbock's inequality Hadwiger-Finsler's inequality (the converse implication being obvious)These two triangles play a very important role in creating geometric inequalities and identities as well. Specifically, one can obtain inequalities that hold in any triangle by applying an inequality which is valid in any or acute triangle to- since the latter triangle is obviously acute-angled. Similarly, we can derive an inequality for an acute-angledtriangle from one holding in any triangle by applying the latter to the orthic triangle i.e. . Of course, one can use the converse transformations of the aforementioned ones when solving geometric/trigonometric inequalities. Thus, when dealing with an inequality in any triangle, we can transform it into one restricted to an acute-angled triangle byusing the substitutions : (in other words, the formerinequality - in any triangle - has been applied to , thus obtaining the one confined to an acute triangle). A similar reasoning applies to the transformation acute-angled triangle any triangle.Here is a simple way of obtaining a really nice identity. It is well known that in any triangle the following identity is true :. Now we want to derive a new identity from it, by applying the previous identity to .Thus, one gets: which is equivalent to the following one:(this identity allows us to obtain many other inequalities with andbecause if we have a lower/upper bound of then we easily obtain an upper/lower bound ofand vice versa.)Result 113. A triangle whose sides are , and exists. We shall use the following notations for itsclassical elements, namely : , , , , , , , (and the cyclic versions as well)Then :Result 114.By using the above identities and the obvious equality we can derive the following amazing identity which points out the fact that the sum can be expressed onlyin terms of , , , and :Result 115.Result 116.Result 117.Result 118..In other words, the equationhas the roots .Result 119.Equalizing the first two identities and doing the same with the last two one obtains:where: . Furthermore, by using Result 115 we can also derivethe following identities : .Result 120.Result 121.Result 122.Result 123.Yeah two of the double-angle sums are "ok", but really, forget the rest.Result 124.Result 125. (derived by Constantin)Result 126. Hadwiger-Finsler Inequality - implied by Pedoe's InequalityThe reverse side is .Result 127. Weitzenböck's inequality - implied by Hadwiger-Finsler InequalityResult 128.Inside the -sided convex polygon , there exists an arbitrary point , and sides . Thedistance from to side , extended if necessary, is . Then .A couple of unknown results to accompany Stewart's TheoremResult 129.Let three cevians concur at a point inside . Then(Gergonne's Theorem), CVH (Van Aubel's Theorem)Result 130.Let the internal bisector of meet the circumcircle of again at . Then. CVH.Result 131.Let the radius of Euler's circle (the 9-point circle) be . .Let the center of the 9-point circle be . Note that since Feuerbach's Theorem states that the 9-point circle is externallytangent to the excircles and internally tangent to the incircle, the remarkable distances between and each of can be easily calculated.Result 132. Generalization of Angle Bisector TheoremResult 133. Trigonometric Ceva's Theoremfollowing Regular Notation. The converse holds as well.Result 134. Erdös-Mordell InequalityIf is a point inside triangle and are the lengths of the perpendiculars from from to sides andrespectively, then.Equality holds if and only if is equilateral and is the center.Specifically,Alternate, Trigonometric InterpretationLet the foot of the perpendiculars and be denoted by and respectively. ThenWe also have thatSimilarly, andAdding these inequalities gives usEquality holds if and only if and which essentially means that triangle isequilateral and lies at the center.Result 135Let in triangle . Then, as BigSams said:some other nice ones:This can be useful for both solving and creating problems quickly. These can be used for a quick proof of Gerretsen's Inequality, for example.Edit: Some results the opposite way also:Result 136. Carnot's TheoremThe perpendiculars from three points to concur if and only ifResult 137.Let be arbitrary points. .Result 138.Given arbitrary points ,with equality if and only if is a parallelogram.Result 139. Euler's Theorem for Pedal Triangles.Let be the projections of a point onto , respectively, where is a triangle. Thenwhere denote as usual the circumcenter of circumradius of .Result 140. (Emmerich's Inequality)In any non-acute the following inequality holds: .Result 141. Advanced identities with multiple consequences (OWN)Comment. These five identities, although not looking aesthetically pleasing, are a valuable source of obtaining a large amount of geometrical inequalities. Specifically, if the angles of a triangle fulfill one of the following conditions:OR OR OR- where is an appropriately chosen angle - then each of the above products are either positive or negative, depending on the condition used (obviously, this follows from the monotonicity of the trigonometric functions).E1.A well known particular case is obtained for i.e. in any non-obtuse the following inequality holds:.E2.This inequality is a consequence of the identity 141.2E3.This inequality is a consequence of the identity 141.4Result 142. For any positive real numbers , , there exists atriangle with sides of lengths , , the area of which is equal to . Also, ifdenotes the measure of the angle opposite to side of length then: and.Result 143. Different ways of expressing in terms of other elements.21 / 21Result 144. Nice upper bound for .In anyit holds that:, with equality iff . Result 145.。