On the Fermat Point of a Triangle Jakob Krarup, Kees Roos

280 NAW 5/18 nr. 4 december 2017 On the Fermat point of a triangle Jakob Krarup, Kees Roos

Jakob Krarup Kees Roos Department of Computer Science Faculty of EEMCS University of Copenhagen, Denmark Technical University Delft [email protected] [email protected]

Research On the Fermat point of a triangle

For a given triangle 9ABC, Pierre de Fermat posed around 1640 the problem of finding a lem has been solved but it is not known by 2 point P minimizing the sum sP of the Euclidean distances from P to the vertices A, B, C. whom. As will be made clear in the sec- Based on geometrical arguments this problem was first solved by Torricelli shortly after, tion ‘Duality of (F) and (D)’, Torricelli’s ap- by Simpson in 1750, and by several others. Steeped in modern optimization techniques, proach already reveals that the problems notably duality, however, Jakob Krarup and Kees Roos show that the problem admits a (F) and (D) are dual to each other. straightforward solution. Using Simpson’s construction they furthermore derive a formula Our first aim is to show that application expressing sP in terms of the given triangle. This formula appears to reveal a simple re- of duality in conic optimization yields the lationship between the area of 9ABC and the areas of the two equilateral triangles that solution of (F), straightforwardly. Devel- occur in the so-called Napoleon’s Theorem. oped in the 1990s, the field of conic optimization (CO) is a generalization of the more We deal with two problems in planar geo Problem (D) was posed in 1755 by well-known field of linear optimization (LO). metry. Both problems presuppose that T. Moss in the Ladies Diary [13], as shown in Its main source is [1]. No prior knowledge of three noncollinear points A, B and C are Figure 1. As this figure indicates, the prob- LO or CO is needed for reading the paper. given:

(F) Find a point P minimizing the sum sP of the Euclidean distances from P to the vertices of 9ABC. (D) Find an equilateral triangle T of max- imum height hT, with the points A, B and C on different sides of T. Fermat’s name is associated with various problems and theorems. Fermat posed problem (F) around 1640.1 In other literature (F) is often referred to as the 3-point Fermat Problem. For its long history we re- fer to, e.g., [2, 3, 11]. Using a result of his student Viviani, Tor- ricelli presented a geometrical construction of the Fermat point P, repeated below in the section ‘Solution of Fermat’s Problem’ and accompanied by his correctness proof in the section ‘Torricelli’s triangle’. Viviani’s result will be discussed in between those sections. Pierre de Fermat (1601?–1665) Evangelista Torricelli (1608–1647) Jakob Krarup, Kees Roos On the Fermat point of a triangle NAW 5/18 nr. 4 december 2017 281

of a convex optimization problem, namely a so-called conic optimization problem 0r a CO problem for short. Every convex optimization problem has a dual problem. This is a maximization problem with the property that the objective value of any dual feasible solution is less than or equal to the objective value of any ‘primal’ feasible solution. This is the property called weak duality. Finding a problem with this property can be a tedious task. For the class of CO problems there exists a simple recipe to get a dual problem. For the current purpose it is Figure 1 Original formulation of problem (D) in the Ladies Diary. not necessary to go into further detail. It suffices to mention that the theory of CO It turns out that two cases need be dis- diately yield the correctness of Torricelli’s yields the following dual problem of (2): tinguished depending on the largest angle construction of the Fermat point. 3 3 T in the triangle ABC: max //ayi i yi = 01,,yi # y * Analytic approach to Fermat’s problem i i ==1 i 1 (3) 9 Case F1: the largest angle in ABC is less Fermat’s problem as stated in (F) is a geo- i ! 12,,3 . than 120°; metric problem. To begin with we put it ",3 Case F2: the largest angle in 9ABC is at in analytic form. To each of the relevant To show that (2) and (3) considered togeth- least 120°. points we associate a vector as follows: er exhibit the desired property of weak du-

ality, assume that x and the triple (,yy12,)y3 xO==Pa,,12OA aO==Ba,.3 OC Our second aim is to show that in case F1 are feasible for (2) and (3), respectively. the problems (F) and (D) are each others Here we assume that the plane is equipped We may then write dual problem; the meaning of this sen- with a two-dimensional coordinate system, 3 tence will be made clear below. The corre- with origin O. Moreover, OP denotes the / axi - sponding result can be stated as follows. vector that goes from O to P. The distance i = 1 from P to A is the length of the vector PA. 3 $ axi - yi (4) Since OP +=PA OA we have PA =-ax1 . / Theorem 1. Let X be any point of a triangle i = 1 ABC of type F1 and T an equilateral trian- Hence, the distance from P to A is just 3 3 T T gle such that A, B and C lie on different the Euclidean norm of the vector ax1 - . $ //axi - yai = i yi . i ==1^hi 1 sides of T. Then one has As a consequence, the sum of the distances from P to A, B and C is equal to The first inequality in (4) is due to the fact 3 hsTX# . (1) ax- y /i = 1 i . This function will be called that i # 1 for each i and the second in- Equality holds if and only if X solves prob- our objective function. We need to mini- equality follows from the Cauchy–Schwarz lem (F) and T solves (D). mize this function when x runs through all inequality; the equality in (4) follows since 3 3 vectors in R2. So (F) can be reformulated , whence also T . /i = 1 yi = 0 xy/i = 1 i = 0 The inequality (1) expresses so-called as the following optimization problem: This shows that we have weak duality. weak duality for problems (F) and (D), 3 This is all we need to reveal a property 2 whereas the last statement in Theorem 1 min / axi - ;!x R . (2) of the Fermat point that enabled Torricelli says that also strong duality holds. The x *4i = 1 to geometrically construct this point in a proof of Theorem 1 will be given later. Each term in the above sum is nonnega- very simple way. As will become clear in As far as we know there does not exist tive, and since A, B and C are noncollinear, the next section the property in question a simple formula expressing sP in terms of at most one of them can be zero. Hence, follows by elementary means from (4). the given points A, B and C. Such a formu- the objective function is positive for every Since (3) maximizes a linear function la will be derived. It uses a simple geomet- x ! R2. Furthermore, each term is strictly over a bounded and closed convex do- ric construction of Simpson that yields a convex in x, as follows from the triangle in- main, an optimal solution certainly exists. line segment with length sP. We show that equality, and so will be their sum. We con- This, by the way, is predicted by the duality the sides of the so-called Napoleon’s out- clude that the objective function is strict- theory for CO, which also guarantees that ward triangle of 9ABC have length sP / 3. ly convex and positive. This implies that the optimal values of (2) and (3) are equal In the next section we reformulate (F) problem (2) has a unique solution, which and attained. as a conic optimization problem and in- necessarily is the Fermat point of 9ABC. troduce its dual problem. As will be shown Since the objective function is a sum of Solution of Fermat’s problem in the section thereafter, the optimality Euclidean norms of linear functions of x, In this section we show that we have also conditions for this pair of problems imme- and the domain is R2, (2) is a special case strong duality, i.e., the optimal values of 282 NAW 5/18 nr. 4 december 2017 On the Fermat point of a triangle Jakob Krarup, Kees Roos

o (2) and (3) are equal. Using this we derive Note that mi > 0 for each i. The sum of generality we suppose that +C = c > 120 . the main result of this section, namely that these coefficients being1 , this means that Now let Q be an arbitrary point in R2 differ- in case F1 the Fermat point P satisfies xP is a convex combination of the vectors ent from C. In Figure 2 Q lies inside 9ABC, a . Since each m is positive, it follows that but the argument below is also valid when ++APBB==PC +CPA = 120o . (5) i i P lies in the interior of 9ABC. Q lies outside the triangle or on its bound- As was established in the previous section, Since the sum of the angles in the 9APC ary. With bc=-180o we turn the triangle (2) has a (unique) optimal solution, denot- is 180°, we have +++APCA++CP PAC = CQA clockwise around the vertex C over ed by x . We have strong duality if and P 180o . Since +APC = 120o and +PAC > 0 we the angle b. This yields a triangle CQ’'A only if there exists a dual feasible y such get +ACP < 60o. In the same way we get such that the vectors A’ C and CB point in that we have equalities throughout in (4), +BCP < 60o. Hence, ++CA=+CP +BCP the same direction and with xx= . This happens if and only if, P < 120o. The same holds for +A and +B. for each i, CQ ==CQ'',,AQ AQ''CA = CA . We conclude that if xaPi! for each i, we + o ax- = ax- y are in case F1. Since QC’ Q = b < 60 , the equal base iP iP i (6) T It remains to deal with the case where angles in the isosceles triangle Q’ CQ ex- = axiP- yi . ^h xaPi= for some i. Without loss of general- ceed b, which implies CQ > QQ' . By us- Let us first consider the case where xaPi! , ity we assume that this happens for i = 3. ing this and AQ = AQ'' we get for each i. Then we have axiP- ! 0 for In that case the optimal value is simply sQ =+QA QB + QC each i. The first equality in (6) then implies given by aa31- + aa32- . We next QA'' QB QQ' yi = 1, and then the second equality im- show that this happens only in case F2. > ++ . plies Since the points A, B and C are not col- The last expression is the length of the

axiP- linear, we have xaPi! for i ! 12, . Hence, polygonal path A’ Q’ QB, which is certainly yi = . (7) ", axiP- y1 and y2 are given by (7) and yy31=- - y2. larger than the length of the straight path Dual feasibility requires y # 1, which A’ CB. Since AC' = AC we obtain Thus, if x is not one of the vertices of 3 P y 2 T y 2 9 holds if and only if 1 ++2yy1 2 2 ABC we can conclude that (6) deter- sQC> AC +=CB s . # 1. Since yy12==1, this holds if and mines each of the vectors yi uniquely. We only if yyT #-=1 cos 120o. Therefore, Since this holds for any point Q in R2 dif- need still to verify whether these vectors 1 2 2 +APB $ 120o. Since xa= we have PC= ferent from C, it follows that C is the Fer- are dual feasible. This holds if and only if P 3 3 and therefore also +ACB $ 120o, proving mat point. / y = 0. Since yy+=- y it follows i = 1 i 12 3 that we are in case F2. From now on we restrict ourselves to that yy12+ ==y3 1. Since At this stage we may conclude that we case F1. In that case neither xP nor sP is 2 1 = yy12+ have completely solved (F) in case F2: yet known; we only have proved (5). But, 3 2 T 2 then the Fermat point is the vertex with the as observed by Torricelli , this proper- =+y1 2yy1 2 + y2 largest angle, and equals the sum of the ty already enables us to construct . His T sP P =+22yy1 2 , distances of that vertex to the other two surprisingly simple construction is demon- T 1 it follows that yy1 2 =-2 . Hence, y1 and y2 vertices. As is pointed out in [2] this case strated in Figure 3. The circumcircles of must make an angle of 120°. According to has either been overlooked or weakly treat- the outward equilateral triangles on the (7) the angle between the vectors y1 and y2 ed in the early literature. It was explicitly sides AC and BC intersect at the Fer- is the same as the angle between the vec- stated for the first time by Bonaventura mat point, because the three angles be- tors ax1 - and ax2 - , which is the angle Cavalieri (1598–1647) [4, pp. 504–510]. A tween the dashed lines PA, PB and PC all between PA and PB. Thus, +APB = 120o. simple geometric proof of its solution had equal 120°. The same argument applies to any two of to wait until 1976 [17]. Torricelli also found a geometric proof the vectors yi. So we may conclude that It is worthwhile to show Sokolowsky’s that (5) must hold. We present this later. the Fermat point P satisfies (5). geometric proof for case F2. Without loss of But first Viviani’s Theorem is presented. The Fermat point must lie in the interior of 9ABC. This can now be shown by us- 3 A ing once more. This relation is ′ /i = 1 yi = 0 equivalent to

ax1 - P ax2 - P ax3 - P + + = 0, C ax1 - P ax2 - P ax3 - P β which can be rewritten as β xaP =+mm11 22aa+ m33, (8) where Q′ Q

1 ax- m = iP , i 11++1 (9) A B ax33---PPPax ax3 i = 12,,3. Figure 2 Sokolowsky’s proof for case F2. Jakob Krarup, Kees Roos On the Fermat point of a triangle NAW 5/18 nr. 4 december 2017 283

have already seen how Torricelli constructed a point P in 9ABC that satisfies (5). B′ How did he know that the Fermat point has this property? 4 For that purpose he introduced another triangle whose sides contain the vertices A, C A ′ B and C and are perpendicular to AP, BP and CP, respectively, with P such that the three angles at P are 120°. In this way he obtained the dashed triangle A1 B1 C1 in Fig- ure 5, now called Torricelli’s triangle. Since o P +CPB is 120° and +PCAP11==BA 90 oo oo we have +A1 =-360 120 -=29# 060 . o For similar reasons also ++BC11==60 . 9 A B Hence, AB11C1 is equilateral. By the construction of Torricelli’s triangle Figure 3 Torricelli’s construction of the Fermat point in case F1. the sum of the distances of P to its sides is equal to ||PA ++||PB ||PC , which is sP. Viviani’s Theorem quite similar to Fermat’s problem but much By Viviani’s Theorem this sum is equal to In this section we deal with a lemma due to easier to solve, namely the height of Torricelli’s triangle, which we Torricelli’s student Vincenzo Viviani (1622– denote by h. Thus, shP = . 1703) [18] now known as Viviani’s Theorem. (V) Find the point V that minimizes (or Now let Q be any other point in 9ABC. maximizes) the sum xV of the distances Viviani’s Theorem tells us that the sum of Lemma 1 (Viviani’s Theorem). If T is an equi- of V to the sides of 9ABC. the distances from Q to the sides of Torri- lateral triangle then for each point X in T celli’s triangle also equals h. The grey tri- It will be convenient below to use the no- the sum of its distances to the sides of T angles in Figure 5 make clear that this sum tations equals the altitude hT of T. is less than or equal to ||QA ++||QB ||QC , aO||==Ab,,OB cO|= C. which is sQ. So we obtain shQP$ = s , Proof. The proof is easy and demonstrated proving that P minimizes the sum of the in Figure 4. The point X divides 9ABC into Lemma 2. The minimal (or maximal) value distances to A, B and C. It means that P three triangles: 9AXB, 9BXC and 9CXA. of xV is attained at a vertex of 9ABC, and solves (F), which implies that it is the Fer- Since the sum of their areas equals the it equals the height of 9ABC seen from mat point. area of 9ABC we get that vertex. B 1 1 1 1 1 2 AB h1 ++2 BC h2 2 CA h3 = 2 AB h. C Proof. Let nA be a unit vector that is or- A1 T Since AB ==BC CA , this implies h = thogonal to BC such that nbA ()- a > 0. P hh12++h3, proving the lemma. □ Then the distance to BC from any point in 9 is equal to T . Defining X ABC nbA ()- x Q It is worth noting that if X is on the nB and nC in a similar way, the sum of the A B boundary of T then one or two of the distances of X to the sides of 9ABC is three triangles in the above proof reduce given by to a single point. But the argument used fx()|=-nbT ()xn+-T ()cx+-naT ()x . in the above proof remains valid. This also A B C follows by considering a problem that is This function depends linearly on X. Hence, its minimal (or maximal) value is attained C1 C at a vertex of 9ABC and equals the dis- Figure 5 Torricelli’s proof of (5). tance of that vertex to the opposite side of 9ABC. □ Duality of (F) and (D) The height value of an equilateral tri- Torricelli’s triangle can also be used to angle seen from one of its vertices is in- prove the duality relation between the h dependent of that vertex. Hence, fx() is problems (F) and (D), as described in 2 h1 h constant on an equilateral triangle, which Theorem 1. In order to prove this theorem X yields a second proof of Viviani’s Theorem. we have also drawn another equilateral

h3 (dotted) triangle in Figure 6 such that the Torricelli’s triangle points A, B and C lie on different sides, as A B We now show how Torricelli used Viviani’s well as line segments from P orthogonal to

Figure 4 Viviani’s Theorem: hh=+12hh+ 3 . Theorem to solve Fermat’s problem. We the sides of this triangle. 284 NAW 5/18 nr. 4 december 2017 On the Fermat point of a triangle Jakob Krarup, Kees Roos

ing F1 and its Fermat point P. So the three Formula for sP angles at P are all equal to 120°. Using the property of the Simpson lines We turn 9BPC over 60 degrees clock- just found, an expression of sP in terms of C wise around B, yielding 9BP''A , as shown 9ABC is immediately at hand, simply by in Figure 7. Then 9BCA' is isosceles. computing the length of the line segment Since +CBA' is 60 degrees it is even AA’ in Figure 7. To compute the point A’ P equilateral. The same argument yields we use that it connects the mid point of that also 9BPP' is equilateral. Hence, all BC by a line perpendicular to BC. Defining A B corners in both triangles are 60°. Since aO''|= A , we then have +APB = 120o it follows that +APP' = 180o. 1 This proves that APP’ is a straight line ab' =+2 ()cd+ , segment. The same argument yields where d is a vector perpendicular to BC that also PP’A’ is a straight line seg- whose length equals the distance of A’ to ment. It follows that APP’A’ is a straight BC. Since 9CBA' is equilateral, this dis- line segment. The length of this segment 1 tance equals 2 3 bc- . Moreover, the equals vector [;bc22--cb11] is orthogonal to BC Figure 6 Torricelli’s triangle maximizes hT. and has length bc- . Hence, ||AP ++||PP''||PA' =+||AP ||BP +||CP , Considering the grey triangles in the 1 bc22- d =! 2 3 . (10) figure it becomes clear that the sum of which is exactly sP, the sum of the distanc- >Hcb11- the distances of P to the sides of the es of the Fermat point P to the points A, One easily understands that d is an out- dotted triangle is less than the sum of B and C. ward direction at 1 ()bc+ if dbT ()- a > 0 the distances of P to the sides of Torri- The line AA’ is named a Simpson line. 2 and an inward direction if dbT ()- a < 0. It celli’s triangle. By Viviani’s Theorem these It can be easily constructed by ruler and is worth to have a closer look at the quan- sums are equal to the heights of the re- compass by first drawing the outward tity dbT ()- a . We may write spective triangles. The dotted triangle’s equilateral triangle BCA’. In the same way height is therefore less than the height of one may construct Simpson lines BB’ and bc- T ba- the Torricelli triangle. Since the height of CC’. The Simpson lines intersect at P. This 22 11 >cb11- H >ba22- H Torricelli’s triangle equals sP, this proves construction of the Fermat point is due 5 =-()bc()ba-+()cb--()ba Theorem 1. to Simpson (1710–1761) [16]. Strangely 2211 1122 Note that the above proof also makes enough Simpson did not observe that the =-ba21-+cb21 ca21+-cb12 ca12+ ba12 clear that Torricelli’s triangle for 9ABC lengths of the three Simpson lines are the =-detdcb et ca+ det ba 6 @ 66@@ same and equal to s . This was first shown 111 solves problem (D). So, when Torricelli P = det , was solving Fermat’s problem, i.e., prob- by Heinen [8]. =cbaG lem (F), he also solved problem (D), al- As a byproduct of the fact that 9BPP’ where the last equality sign follows by though most likely being unaware of this. is equilateral we get +BPP' = 60o, which evaluating the determinant det 111 to This is typical for problems that are each means that the Simpson line PA’ is the bi- =Gcba its first row. The absolute value of this others dual. At first sight they seem to sector of +BPC as was observed in [11]. determinant equals two times the area of be unrelated — though both are based This in turn implies that P is also the Fer- 9ABC. In the sequel we assume that the on the same input data, but when solv- mat point of 9AB''C', because of (5). Note determinant is positive; if this is not so, ing one of the two problems usually one that ||AA''++||BB ||CC' = 3s implies P interchanging the names of two vertices gets enough information to also solve the will make it positive. As a consequence d other problem. Harold W. Kuhn [12] states: ||AP ++||PA''||BP +||PB will be an outward direction at 1 ()bc+ . “Until further evidence is discovered this 2 ++||CP ||PC' = 3sP . So we must use the plus sign in (10). This must stand as the first instance of duali- gives ty in nonlinear programming” as was es- Hence, ||PA''++||PB ||PC' = 2sP, which is tablished in 1811–1812 by M. Vecten, pro- the sum of the distances from the Fermat 1 1 bc22- fessor of mathématiques spéciale at the point of 9AB''C' to its vertices. ab' =+2 ()c + 2 3 . >Hcb11- (11) Lycée de Nismes. In the next section we discuss another As before, we use |AB| to denote the length method to construct the Fermat point. This A′ of line segment AB. So ||AB = ab- . C method will enable us to find a expression P ′ We also use |ABC| to denote the area for sP in terms of the given 9ABC. This will of 9ABC. o be done in the section thereafter. 60 P o 60 Lemma 3. In case F1 one has Simpson lines sA=+1 (| BB||22CC||++AA|)2 23||BC . To introduce Simpson’s method we use Fig- A B P 2 ure 7, which shows a triangle ABC satisfy- Figure 7 Simpson line AA’. (12) Jakob Krarup, Kees Roos On the Fermat point of a triangle NAW 5/18 nr. 4 december 2017 285

Proof. From Section 7 we know that sP = aa- ' . Therefore, Theorem 2. The triangles XYZ and X’Y’Z’ are equilateral. The cen- 2 22T sP =+aa' -2aa'. Since tres of these triangles coincide with the centre of ABC. Moreover,

2 1 2 3 2 1 T bc22- sXP = 3 | Y | ||a' = 4 bc+ + 4 bc- ++2 3 ()bc >cb11- H and 22T 1 221 222 =+bc-+bc 3 ((bc11+-)(bc22)(++bc22)(cb11- )) ||XY +=||XY'' ||AB ++||BC ||CA , (13) 2 3 _ i 22T =+bc-+bc 3 ()bc21- bc12 224 =+bc22-+bcT 3 det cb. ||XY -=||XYll ||ABC . (14) 6 @ 3 and bc- 23aaTT' =+abc + 22 Proof. Let us defineaO = A, xO= X, et cetera. Then aO''= A is e >cb11- Ho 1 given by (11). Since xb=+3 ()ca+ ' it therefore follows that TT =+ab ac+ 3 ab12--ab21 ac12+ ac21 ^ h bc- bc- =+abTTac+ 3 detdab- et ac, 1 1 1 22 1 1 22 xb=+3 cb++2 ()c + 2 33=+2 ()bc+ 6 . ^ 66@@h eo>>cb11- HHcb11- it follows that The centre X’ of the inward equilateral triangle on side BC is ob- 2 2 22TTT sP = ab++-c ab--bc ac tained by mirroring X in the line BC, which gives + 3 detdcb-+et ab det ac ^h66@@6 @ 1 1 bc22- 1 2 2 2 111 xb' =+2 ()c - 6 3 . = ()ab- + bc- + ca- + 3 det . >Hcb11- 2 =cbaG This gives the lemma. □ In the same way we get the following expressions for y, y’, z and z’:

1 1 ca22- 1 1 ca22- Surprisingly enough the formula for sP just found also appears yc=+2 ()a + 6 33,(yc' =+2 a),- 6 >ac11- H >ac11- H when dealing with so-called Napoleon triangles. This will be the 1 1 ab22- 1 1 ab22- subject of the next section, where we find a line segment of length za=+2 ()b + 6 33,(za' =+2 b).- 6 >ba11- H >ba11- H sP / 3. From the above expressions one readily obtains xy++z = Napoleon’s triangles xy''++za' =+bc+ , which implies that the centres of XYZ, X’Y’Z’ Figure 8 shows a triangle ABC and the three outward equilateral tri- and ABC coincide. angles on its sides. The centres of the triangles BCA’, CAB’ and ABC’ One has ||XY = xy- and ||XY''= xy''- . We start by com- are denoted by X, Y and Z, respectively. Napoleon’s Theorem 6 states puting xy- . We may write that the triangle XYZ is equilateral. Moreover, the same holds for the bc- ca- triangle X’Y’Z’, where X’, Y’ and Z’ are the centres of the inward equi- xy-=1 ()ba-+1 22- 22 2 23 cb- ac- lateral triangles on the sides of ABC. The triangle XYZ is called the e>>11H 11Ho outward Napoleon triangle and X’Y’Z’ the inward Napoleon tri- 1 1 ab22+-2c2 =-2 ()ba+ . angle. 23>2ca11--b1H The first part of the next theorem is just Napoleon’s Theorem. As a consequence we have Part of the second statement is also known, namely that XYZ and 2 2 2 1 3 1 3 X’Y’Z’ have the same centre [9]. xy- = ()ba-+ ab22+-22c2 + ()ba-+ ca11--b1 a 2 11 6 ^ hk a 2 22 6 ^ hk =-1 2+-1 2++1 -+2 1 -- 2 B′ 4 ()ba11 4 ()ba22 12 ()ab2222cc2 12 ()11ab1 3 3 +-6 ()ba11()ab22+-22cb2 +-6 ()22ac()11--ab1 A C ′ =-1 ((33ba)(2+-ba)(2++ab-+22cc)(2 --ab))2 Y 12 11 22 22 2 111 X 3 +-((ba11)(ab22+-22cb22)(+-ac21)( --ab11)) Z′ 6

X′ 1 2 2 2 2 2 2 =+3 ()aabb1 2+++1 2 cc1 +-2 ac11--ac22 ab11--bc11 ab22-bc22 Y ′ 3 A B +-3 ((ab21cb12)(+-cb11ca21+-bc22+ ))

1 2 22TT T =+3 ()ab+-c cb--baac Z +-1 ()detdba et ca+ det cb ) 3 66@@6 @ 111 = 1 ab- 2+ bc- 2+ ca- 2 + 1 det . 6 _ i 3 =cbaG 2 Comparison of the last expression with the expression for sP in (12) C 2 2 ′ yields sP = 3 xy- , which gives the value for XY in the theo- Figure 8 Napoleon’s triangles. rem. Since the expression is invariant under cyclic permutations of 286 NAW 5/18 nr. 4 december 2017 On the Fermat point of a triangle Jakob Krarup, Kees Roos

a, b and c it follows that xy- = yz- The area of an equilateral triangle with matical model for (D). This is remarkable 3 2 = zx- , which implies that 9XYZ is side length s equals 4 s . Moreover, the and seems to deserve further investigation. area |ABC| of 9ABC is equal to 1 det 111 . Different variants of Fermat’s original equilateral. 2 =Gcba Now consider the triangle X’ Y’ Z’. We As a consequence of (14) we therefore get problem have been investigated by sever- have a surprisingly simple relation between the al authors. Among the more obvious ex- areas of Napoleon’s triangles and the area tensions are weighted versions where the 1 1 bc22- ca22- of 9ABC, namely (cf. [5] and [9]) ||ABC = distances are multiplied by positive or neg- xy''-=2 ()ba-- + 23eo>>cb11- H ac11- H ||XYZX-||''YZ' . ative weights associated with the vertices 1 1 ab22+-2c2 of 9ABC as in e.g. [7] and [9]. Other lines =-2 ()ba- . 23>2ca11--b1H Concluding remarks of research have focused upon the number As was made clear the key to the solution of given points and/or the dimension of The only difference with the expression for of Fermat’s problem is property (5), which the space. For an overview the interested xy- is the sign of the second term. It is we obtained straightforwardly by using the reader is referred to [3]. easy to check that a similar derivation as s conic dual of Fermat’s problem. It may be for xy- 2 leads to worth pointing out that at first (and sec- Acknowledgement 2 1 ond!) sight there is no obvious relation Thanks are due Professor Marijana Zeki -Su ac, xy''- = ab- 2 + bc- 2 + ca- 2 ć š 6 _ i between the ‘analytic’ dual (3) and the the chief organizer of the Croatian Operations 111 Research Society’s conference held in Osijek, - 1 det . ‘geometric’ dual (D) of Fermat’s problem. 3 =cbaG Croatia, September 2016. Without her invita- Whereas (2) easily is identified as a mathe- tion to both of us to attend this conference, the 2 From this and the expression for xy- matical model for Fermat’s problem (F), it present paper, let alone its potential successors, we readily obtain (13) and (14). □ seems harder to recognize (3) as a mathe- would never have materialized.

Notes 1 There seems to be uncertainty about the 4 Some authors who ask this question leave amuse the reader to know that Napoleon’s birth year of Fermat, see e.g. [11]. it unanswered, suggesting that it was based Theorem is claimed to be one of the most 2 The Ladies’ Diary: or, Woman’s Almanack on Torricelli’s intuition [3, p. 88; 7, p. 444]. often rediscovered results in triangular ge- appeared annually in London from 1704 to 5 Thomas Simpson (1710–1761) was an En- ometry. Thus, no less than 32 sources relat- 1841 after which it was succeeded by The glish mathematician. He taught mathematics ed to it are mentioned in [19]. Lady’s and Gentleman’s Diary. It was a re- at the Royal Military Academy and became spectable place to pose mathematical prob- known for the Simpson rule for numerical lems (brain teasers!) and sustain debate. integration [6]. 3 Torricelli was a famous French number theo- 6 According to [11] and others, any connec- rist, student of Galileo and discoverer of the tion between Napoleon’s Theorem and the barometer. French emperor is highly doubtful. It may

References 1 A. Ben-Tal and A. Nemirovski, Lectures on 7 S. Gueron and R. Tessler, The Fermat–Steiner 13 T. Moss, Ladies Diary 620, Question CCCXCVI, modern convex optimization, MPS/SIAM Se- problem, Amer. Math. Monthly 109(5) (2002), 1755. ries on Optimization, SIAM, 2001. 443–451. 14 R. T. Rockafellar, Convex Analysis, Prentice- 2 V. Boltyanski, H. Martini and V. Soltan, Geo- 8 F. A. Heinen, Über systeme von Kräften, Hall, 1970. metric Methods and Optimization Problems, Gymnasium zu Cleve, 1834. 15 C. Roos, M. Balvert, B. L. Gorissen and D. Combinatorial Optimization, Vol. 4, Kluwer, 9 R. Honsberger, Mathematical Gems from den Hertog, A universal and structured way 1999. Elementary Combinatorics, Number Theory, to derive dual optimization problem formu- 3 G. Bruno, A. Genovese and G. Improta, A his- and Geometry, The Mathematical Associa- lations. Submitted. torical perspective on location problems, tion of America, 1973. 16 T. Simpson, The Doctrine and Application of Journal of the British Society for the History 10 G. Jalal and J. Krarup, Geometrical solution to Fluxions, 1750. of Mathematics 29(2) (2014), 83–97. the Fermat problem with arbitrary weights, 17 D. Sokolowsky, A note on the Fermat prob- 4 B. Cavalieri, Exercitationes geometricae, Ann. Oper. Res. 123 (2003), 67–104. lem, Amer. Math. Monthly 83(4) (1976), 276. 1657. 11 J. Krarup and S. Vajda, On Torricelli’s geo- 18 V. Viviani, De maximis et minimis divinatio 5 H. S. M. Coxeter and S. L. Greitzer, Geometry metrical solution to a problem of Fermat, in quintum Conicorum Apollonii Pergaei. Revisited, New Mathematical Library, Vol. 19, IMA J. Math. Appl. Bus. Indust. 8(3) (1997), Part II, Firenze, 1659. Random House, 1967. 215–224. 19 J. E. Wetzel, Converses of Napoleon’s the- 6 S. I. Gass and A. A. Assad, An annotated 12 H. W. Kuhn, Nonlinear programming: A his- orem, Amer. Math. Monthly 99(4) (1992), timeline of operations research: an informal torical note, in: J. K. Lenstra, A. H. G. Rinnooy 339–351. history, International Series in Operations Kan and A. Schrijver, eds., History of Math- Research & Management Science, Vol. 75, ematical Programming. A Collection of Per- Kluwer, 2005. sonal Reminiscences, CWI, 1991, pp. 82–96.