{"history":[],"article":{"id":"2f4ed09e-6b62-46d5-ac21-04b6c8b41d4b","title":"Fermat point","slug":"fermat-point","content":":::summary\n:::\n\n# \nIn [Euclidean geometry](https://bliptext.com/articles/euclidean-geometry), the **Fermat point** of a [triangle](https://bliptext.com/articles/triangle), also called the **Torricelli point** or Fermat–**Torricelli point**, is a point such that the sum of the three distances from each of the three vertices of the [triangle](https://bliptext.com/articles/triangle) to the point is the smallest possible or, equivalently, the [geometric median](https://bliptext.com/articles/geometric-median) of the three vertices.\nIt is so named because this problem was first raised by [Fermat](https://bliptext.com/articles/pierre-de-fermat) in a private letter to [Evangelista Torricelli](https://bliptext.com/articles/evangelista-torricelli), who solved it.\nThe Fermat point gives a solution to the [geometric median](https://bliptext.com/articles/geometric-median) and [Steiner tree problems](https://bliptext.com/articles/steiner-tree-problem) for three points.\n# [Fig 1. Construction of the first isogonic center, X(13). When no angle of the triangle exceeds 120°, this point is the Fermat point. | upload.wikimedia.org/wikipedia/commons/1/15/Fermat///Point.svg]\n\n# Construction\nThe Fermat point of a triangle with largest angle at most 120° is simply its **first isogonic center** or **X(13)**, which is constructed as follows:\nAn alternative method is the following:\nWhen a triangle has an angle greater than 120°, the Fermat point is sited at the obtuse-angled vertex.\nIn what follows \"Case 1\" means the triangle has an angle exceeding 120°.\n\"Case 2\" means no angle of the triangle exceeds 120°.\n\n# Location of X(13)\nFig. 2 shows the equilateral triangles\n△ARB, △AQC, △CPB\nattached to the sides of the arbitrary triangle\n△ABC\n. Here is a proof using properties of [concyclic points](https://bliptext.com/articles/concyclic-points) to show that the three lines RC, BQ, AP in Fig 2 all intersect at the point F and cut one another at angles of 60°.\nThe triangles\n△RAC, △BAQ\nare [congruent](https://bliptext.com/articles/congruence-geometry) because the second is a 60° rotation of the first about A. Hence\n∠ARF = ∠ABF\nand\n∠AQF = ∠ACF\n. By the converse of the [inscribed angle theorem](https://bliptext.com/articles/inscribed-angle) applied to the segment AF, the points ARBF are [concyclic](https://bliptext.com/articles/concyclic-points) (they lie on a circle).\nSimilarly, the points AFCQ are concyclic.\n∠ARB = 60°\n, so\n∠AFB = 120°\n, using the [inscribed angle theorem](https://bliptext.com/articles/inscribed-angle).\nSimilarly,\n∠AFC = 120°\n.\nSo\n∠BFC = 120°\n. Therefore,\n∠BFC + ∠BPC = 180°\n. Using the [inscribed angle theorem](https://bliptext.com/articles/inscribed-angle), this implies that the points BPCF are concyclic.\nSo, using the [inscribed angle theorem](https://bliptext.com/articles/inscribed-angle) applied to the segment BP,\n∠BFP = ∠BCP = 60°\n. Because\n∠BFP + ∠BFA = 180°\n, the point F lies on the line segment AP.\nSo, the lines RC, BQ, AP are [concurrent](https://bliptext.com/articles/concurrent-lines) (they intersect at a single point).\n[Q.E.D.](https://bliptext.com/articles/q-e-d)\nThis proof applies only in Case 2, since if\n∠BAC > 120°\n, point A lies inside the circumcircle of\n△BPC\nwhich switches the relative positions of A and F. However it is easily modified to cover Case 1.\nThen\n∠AFB = ∠AFC = 60°\nhence\n∠BFC = ∠AFB + ∠AFC = 120°\nwhich means BPCF is concyclic so\n∠BFP = ∠BCP = 60° = ∠BFA\n. Therefore, A lies on FP.\nThe lines joining the centers of the circles in Fig. 2 are perpendicular to the line segments AP, BQ, CR.\nFor example, the line joining the center of the circle containing\n△ARB\nand the center of the circle containing\n△AQC\n, is perpendicular to the segment AP.\nSo, the lines joining the centers of the circles also intersect at 60° angles.\nTherefore, the centers of the circles form an equilateral triangle.\nThis is known as [Napoleon's Theorem](https://bliptext.com/articles/napoleon-s-theorem).\n# [Fig 2. Geometry of the first isogonic center. | upload.wikimedia.org/wikipedia/commons/1/17/Fermat///Point///Proof.svg]\n\n# Location of the Fermat point\n\n## Traditional geometry\nGiven any Euclidean triangle\n△ABC\nand an arbitrary point P let The aim of this section is to identify a point\nP0\nsuch that d(P_0) 120°\nbecause there is no point O where a, b, c meet at angles of 120°.\nNevertheless, it is easily fixed by redefining\nk = − (i + j)\nand placing O at C so that\nc = 0\n. Note that\nbecause the angle between the unit vectors\ni, j\nis\n∠C\nwhich exceeds 120°.\nSince\nthe third inequality still holds, the other two inequalities are unchanged.\nThe proof now continues as above (adding the three inequalities and using\ni + j + k = 0\n) to reach the same conclusion that O (or in this case C) must be the Fermat point of\n△ABC\n.\n\n## Lagrange multipliers\nAnother approach to finding the point within a triangle, from which the sum of the distances to the [vertices](https://bliptext.com/articles/vertex-geometry) of the triangle is minimal, is to use one of the [mathematical optimization](https://bliptext.com/articles/mathematical-optimization) methods; specifically, the method of [Lagrange multipliers](https://bliptext.com/articles/lagrange-multipliers) and the [law of cosines](https://bliptext.com/articles/law-of-cosines).\nWe draw lines from the point within the triangle to its vertices and call them\nX, Y, Z\n. Also, let the lengths of these lines be x, y, z respectively.\nLet the angle between\nX\nand\nY\nbe α,\nY\nand\nZ\nbe β.\nThen the angle between\nX\nand\nZ\nis\nπ − α − β\n. Using the method of Lagrange multipliers we have to find the minimum of the Lagrangian L, which is expressed as:\nwhere a, b, c are the lengths of the sides of the triangle.\nEquating each of the five partial derivatives to zero and eliminating\nλ1, λ2, λ3\neventually gives\nsin α = sin β\nand\nsin(α + β) = − sin β\nso\nα = β = 120°\n. However the elimination is a long and tedious business, and the end result covers only Case 2.\n\n# Properties\n(A < 120°), (B < 120°), (C < 120°)\n.\n# [The two isogonic centers are the intersections of three vesicae piscis whose paired vertices are the vertices of the triangle | upload.wikimedia.org/wikipedia/commons/d/d6/Isogonic///centres///and///vesicae///piscis.png]\n\n# Aliases\nThe **isogonic centers** X(13) and X(14) are also known as the **first Fermat point** and the **second Fermat point** respectively.\nAlternatives are the **positive Fermat point** and the **negative Fermat point**.\nHowever these different names can be confusing and are perhaps best avoided.\nThe problem is that much of the literature blurs the distinction between the **Fermat point** and the first **Fermat point** whereas it is only in Case 2 above that they are actually the same.\n\n# History\nThis question was proposed by Fermat, as a challenge to [Evangelista Torricelli](https://bliptext.com/articles/evangelista-torricelli).\nHe solved the problem in a similar way to Fermat's, albeit using the intersection of the circumcircles of the three regular triangles instead.\nHis pupil, Viviani, published the solution in 1659.","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T22:20:11.621Z","updated_at":"2025-01-25T22:20:11.621Z","search_vector":"'fermat':1 'point':2","search_content":" \n \n\n# \nIn [Euclidean geometry](https://bliptext.com/articles/euclidean-geometry), the **Fermat point** of a [triangle](https://bliptext.com/articles/triangle), also called the **Torricelli point** or Fermat–**Torricelli point**, is a point such that the sum of the three distances from each of the three vertices of the [triangle](https://bliptext.com/articles/triangle) to the point is the smallest possible or, equivalently, the [geometric median](https://bliptext.com/articles/geometric-median) of the three vertices.\nIt is so named because this problem was first raised by [Fermat](https://bliptext.com/articles/pierre-de-fermat) in a private letter to [Evangelista Torricelli](https://bliptext.com/articles/evangelista-torricelli), who solved it.\nThe Fermat point gives a solution to the [geometric median](https://bliptext.com/articles/geometric-median) and [Steiner tree problems](https://bliptext.com/articles/steiner-tree-problem) for three points.\n# [Fig 1. Construction of the first isogonic center, X(13). When no angle of the triangle exceeds 120°, this point is the Fermat point. | upload.wikimedia.org/wikipedia/commons/1/15/Fermat///Point.svg]\n\n# Construction\nThe Fermat point of a triangle with largest angle at most 120° is simply its **first isogonic center** or **X(13)**, which is constructed as follows:\nAn alternative method is the following:\nWhen a triangle has an angle greater than 120°, the Fermat point is sited at the obtuse-angled vertex.\nIn what follows \"Case 1\" means the triangle has an angle exceeding 120°.\n\"Case 2\" means no angle of the triangle exceeds 120°.\n\n# Location of X(13)\nFig. 2 shows the equilateral triangles\n△ARB, △AQC, △CPB\nattached to the sides of the arbitrary triangle\n△ABC\n. Here is a proof using properties of [concyclic points](https://bliptext.com/articles/concyclic-points) to show that the three lines RC, BQ, AP in Fig 2 all intersect at the point F and cut one another at angles of 60°.\nThe triangles\n△RAC, △BAQ\nare [congruent](https://bliptext.com/articles/congruence-geometry) because the second is a 60° rotation of the first about A. Hence\n∠ARF = ∠ABF\nand\n∠AQF = ∠ACF\n. By the converse of the [inscribed angle theorem](https://bliptext.com/articles/inscribed-angle) applied to the segment AF, the points ARBF are [concyclic](https://bliptext.com/articles/concyclic-points) (they lie on a circle).\nSimilarly, the points AFCQ are concyclic.\n∠ARB = 60°\n, so\n∠AFB = 120°\n, using the [inscribed angle theorem](https://bliptext.com/articles/inscribed-angle).\nSimilarly,\n∠AFC = 120°\n.\nSo\n∠BFC = 120°\n. Therefore,\n∠BFC + ∠BPC = 180°\n. Using the [inscribed angle theorem](https://bliptext.com/articles/inscribed-angle), this implies that the points BPCF are concyclic.\nSo, using the [inscribed angle theorem](https://bliptext.com/articles/inscribed-angle) applied to the segment BP,\n∠BFP = ∠BCP = 60°\n. Because\n∠BFP + ∠BFA = 180°\n, the point F lies on the line segment AP.\nSo, the lines RC, BQ, AP are [concurrent](https://bliptext.com/articles/concurrent-lines) (they intersect at a single point).\n[Q.E.D.](https://bliptext.com/articles/q-e-d)\nThis proof applies only in Case 2, since if\n∠BAC > 120°\n, point A lies inside the circumcircle of\n△BPC\nwhich switches the relative positions of A and F. However it is easily modified to cover Case 1.\nThen\n∠AFB = ∠AFC = 60°\nhence\n∠BFC = ∠AFB + ∠AFC = 120°\nwhich means BPCF is concyclic so\n∠BFP = ∠BCP = 60° = ∠BFA\n. Therefore, A lies on FP.\nThe lines joining the centers of the circles in Fig. 2 are perpendicular to the line segments AP, BQ, CR.\nFor example, the line joining the center of the circle containing\n△ARB\nand the center of the circle containing\n△AQC\n, is perpendicular to the segment AP.\nSo, the lines joining the centers of the circles also intersect at 60° angles.\nTherefore, the centers of the circles form an equilateral triangle.\nThis is known as [Napoleon's Theorem](https://bliptext.com/articles/napoleon-s-theorem).\n# [Fig 2. Geometry of the first isogonic center. | upload.wikimedia.org/wikipedia/commons/1/17/Fermat///Point///Proof.svg]\n\n# Location of the Fermat point\n\n## Traditional geometry\nGiven any Euclidean triangle\n△ABC\nand an arbitrary point P let The aim of this section is to identify a point\nP0\nsuch that d(P_0) 120°\nbecause there is no point O where a, b, c meet at angles of 120°.\nNevertheless, it is easily fixed by redefining\nk = − (i + j)\nand placing O at C so that\nc = 0\n. Note that\nbecause the angle between the unit vectors\ni, j\nis\n∠C\nwhich exceeds 120°.\nSince\nthe third inequality still holds, the other two inequalities are unchanged.\nThe proof now continues as above (adding the three inequalities and using\ni + j + k = 0\n) to reach the same conclusion that O (or in this case C) must be the Fermat point of\n△ABC\n.\n\n## Lagrange multipliers\nAnother approach to finding the point within a triangle, from which the sum of the distances to the [vertices](https://bliptext.com/articles/vertex-geometry) of the triangle is minimal, is to use one of the [mathematical optimization](https://bliptext.com/articles/mathematical-optimization) methods; specifically, the method of [Lagrange multipliers](https://bliptext.com/articles/lagrange-multipliers) and the [law of cosines](https://bliptext.com/articles/law-of-cosines).\nWe draw lines from the point within the triangle to its vertices and call them\nX, Y, Z\n. Also, let the lengths of these lines be x, y, z respectively.\nLet the angle between\nX\nand\nY\nbe α,\nY\nand\nZ\nbe β.\nThen the angle between\nX\nand\nZ\nis\nπ − α − β\n. Using the method of Lagrange multipliers we have to find the minimum of the Lagrangian L, which is expressed as:\nwhere a, b, c are the lengths of the sides of the triangle.\nEquating each of the five partial derivatives to zero and eliminating\nλ1, λ2, λ3\neventually gives\nsin α = sin β\nand\nsin(α + β) = − sin β\nso\nα = β = 120°\n. However the elimination is a long and tedious business, and the end result covers only Case 2.\n\n# Properties\n(A < 120°), (B < 120°), (C < 120°)\n.\n# [The two isogonic centers are the intersections of three vesicae piscis whose paired vertices are the vertices of the triangle | upload.wikimedia.org/wikipedia/commons/d/d6/Isogonic///centres///and///vesicae///piscis.png]\n\n# Aliases\nThe **isogonic centers** X(13) and X(14) are also known as the **first Fermat point** and the **second Fermat point** respectively.\nAlternatives are the **positive Fermat point** and the **negative Fermat point**.\nHowever these different names can be confusing and are perhaps best avoided.\nThe problem is that much of the literature blurs the distinction between the **Fermat point** and the first **Fermat point** whereas it is only in Case 2 above that they are actually the same.\n\n# History\nThis question was proposed by Fermat, as a challenge to [Evangelista Torricelli](https://bliptext.com/articles/evangelista-torricelli).\nHe solved the problem in a similar way to Fermat's, albeit using the intersection of the circumcircles of the three regular triangles instead.\nHis pupil, Viviani, published the solution in 1659.","revisionCount":0},"hasMore":false,"nextCursor":null}