Table Of ContentLoci of Points Inspired by Viviani’s Theorem
7
1
0 EliasAbboud
2 Facultyofeducation, BeitBerlCollege,DoarBeitBerl44905,Israel
n [email protected]
a
J
January 26,2017
4
2
] Viviani (1622-1703),who was a student and assistant of Galileo, discoveredthat
O
equilateraltrianglessatisfythe followingproperty: the sumofthedistancesfromthe
H sidesofanypointinsideanequilateraltriangleisconstant.
. Viviani’stheoremcanbeeasilyprovedbyusingareas. JoiningapointP insidethe
h
t triangletoitsverticesdividesitintothreeparts.Thesumoftheirareaswillbeequalto
a
theareaoftheoriginalone. Therefore,thesumofthedistancesfromthesideswillbe
m
equaltothealtitude,andthetheoremfollows.
[
Givenatriangle(whichincludesbothboundaryandinnerpoints),∆,andapositive
1 constant,k,weshallconsiderthefollowingquestions:
v Question 1. What is the locus T (∆) of points P in ∆ such that the sum of the
k
9 distancesof P toeachofthesidesof ∆equalsk?
3
Question2. WhatisthelocusS (∆)ofpointsP intheplanesuchthatthesumof
3 k
thesquaresofthedistancesof P toeachofthesidesof ∆equalsk?
7
0 By Viviani’stheorem, theanswer to the firstquestionis straightforwardforequi-
. lateral triangles: If k equals the altitude of an equilateral triangle, then the locus of
1
0 pointsistheentiretriangle. On theotherhand,ifk isnotequaltothe altitudeofthe
7 equilateraltrianglethenthelocusofpointsisempty.
1 Sincethedistancefunctionislinearwithtwovariableandbecauseofsquaringthe
:
v distances,thelocusofpointsinthesecondquestion,ifnotempty,isaquadraticcurve.
i Inourcase,weshallseethatthisquadraticcurveisalwaysanellipse. Thisfactwillbe
X
thebasisofanewcharacterizationofellipses.
r
We start first with the locus of points which have constant sum of distances to
a
eachofthesidesofagiventriangleandconsiderseparatelythecasesofisoscelesand
scalene triangles. Then, we turn to the locus of points which have constant sum of
squareddistancestoeachofthesidesofagiventriangle. Weprovethemaintheorem
whichgivesacharacterizationofellipsesandconcludebydiscussingminimalsumof
squareddistances.
Constant sum of distances
Samelson[6,p. 225]gaveaproofofViviani’stheoremthatusesvectorsandChen&
Liang[3,p. 390-391]usedthisvectormethodtoproveaconverse: ifinsideatriangle
1
thereisacircularregioninwhichthesumofthedistancesfromthesidesisconstant,
thenthe triangleisequilateral. In[2], thisconverseisgeneralizedintheformoftwo
theoremsthatwewilluseinaddressingQuestion1. Tostatethesetheorems,weneed
thefollowingterminology:Let beapolygonconsistingofbothboundaryandinterior
points.DefinethedistancesumPfunction : R,whereforeachpointP the
V P → ∈P
value (P)isdefinedasthesumofthedistancesfromP tothesidesof .
V P
Theorem1 Any triangle can be divided into parallel line segments on which is
V
constant.Furthermore,thefollowingconditionsareequivalent:
isconstantonthetriangle∆.
• V
There are three non-collinearpoints, inside the triangle, at which takes the
• V
samevalue.
∆isequilateral.
•
Theorem2 (a)Anyconvexpolygon canbedividedintoparallellinesegments,on
P
which isconstant.
V
(b)If takesequalvaluesatthreenon-collinearpoints,insideaconvexpolygon,
V
then isconstanton .
V P
The discussion in [2, p. 207-210]laysouta connectionbetweenlinear program-
ming and Theorems 1 and 2. Theorem 1 is proved explicitly by defining a suitable
linear programmingproblem and Theorem 2 is provedby means of analytic geome-
try,wherethedistancesumfunction iscomputeddirectlyandshowntobelinearin
V
twovariables. WeshallapplythesetheoremstofindthelociT (∆)forisoscelesand
k
scalenetriangles∆.
Isoscelestriangle
For isosceles triangles we exploit their reflection symmetry to find directly the line
segmentsonwhich isconstant,andforscalenetrianglesweapplyanyoftheabove
V
theorems to show that is constant on certain parallel line segments. To find the
V
directionoftheselinesegmentswecomputeexplicitlytheequationof . Theideais
V
illustratedthroughanexample.
Since an isosceles trianglehas a reflectionsymmetryacrossthe altitude, we con-
cludethat; ifthe sumofdistancesofpointP fromthesidesofthetriangleis k, then
thereflectionpointP acrossthealtitudesatisfiesthesameproperty.Forsuchtriangles
′
wehavethefollowingproposition.
Proposition1 If ∆ is a non-equilateralisosceles triangle and k ranges between the
lengths of the smallest and the largest altitudes of the triangle, then T (∆) is a line
k
segment,whoseendpointsareontheboundaryofthetriangle,paralleltothebase.
Proof. In Fig. 1, if the line segment DE is parallel to the base BC then, when P
movesalongDE,thelengtharemainsconstantandb+c=DPsinα+PEsinα=
2
DEsinα,whichisalsoconstant.Hence,a+b+c=kisconstantonthesegmentDE.
ByTheorem1, therearenootherpointsin∆withdistancesumk unlessthetriangle
isequilateral.
IfaapproacheszerothenthelengthofDE approachesthelengthofthebaseBC,
andk = a+DEsinαapproachesthelengthofthealtitudesfromB (orC). Onthe
otherhand, if the lengthof DE approacheszero thenk approachesthe length of the
altitudefromA. Hence, T (∆) is definedwheneverk rangesbetweenthe lengthsof
k
thesmallestandthelargestaltitudesofthetriangle.
Figure1: Thelocusofpointsisalinesegmentparalleltothebase.
Scalene triangle
Similarly,by Theorem1, the locusof pointsT (∆) fora scalene triangle∆ isa line
k
segment,providedkrangesbetweenthelengthsofthesmallestandthelargestaltitudes
of the triangle. Otherwise, T (∆) will be empty. Indeed, the connectionwith linear
k
programmingallows us to deduce the result. The convex polygonis taken to be the
feasibleregion,andthedistancesumfunction correspondstotheobjectivefunction.
V
The parallel line segments, on which is constant, correspondto the isoprofitlines.
V
The mathematical theory behind linear programming states that an optimal solution
to anyproblemwill lie at a cornerpointof the feasible region. If the feasible region
is bounded, then both the maximum and the minimum are attained at corner points.
Now,thedistancesumfunction isalinearcontinuousfunctionintwovariables.The
V
valuesof attheverticesofthefeasibleregion,whichisthetriangle∆,areexactlythe
V
lengthsofthealtitudes. Moreover,thisfunctionattainsitsminimumanditsmaximum
at the vertices of the triangle, and ranges continuously between its extremal values.
Hence,ittakesoneveryvaluebetweenitsminimumanditsmaximum. Therefore,
3
wehavethefollowingproposition.
Proposition2 If∆isascalenetriangleandkrangesbetweenthelengthsofthesmall-
estandthe largestaltitudesofthe triangle, thenT (∆) isa line segment,whose end
k
pointsareontheboundaryofthetriangle.
The question is: How can we determinethis segmentfor a generaltriangle? We
shallillustratethemethodbythefollowingexample. Giventheverticesofatriangle,
we firstcomputethe equationsofthe sidesthenwe find thedistancesfroma general
point(x,y) inside the triangle to each of the sides. In this way we obtain the corre-
sponding distance sum function . Taking = c, we get a family of parallel lines
V V
which, for certain values of the constantc, intersect the given triangle in the desired
linesegments.
Example1 Let ∆ be the right angled triangle with vertices (0,0),(0,3) and (4,0),
respectively(seeFig.2).Iftheconstantsumofdistancesfromthesidesisk,2.4 k
≤ ≤
4,thenthelocusisalinesegmentinsidethetriangleparalleltotheline2x+y =0.
In Fig. 2, T2.4(∆) = A ,T2.8(∆) = DG,T3.2(∆) = EH,T3.6(∆) = FI and
{ }
T4(∆) = C .Notethatattheextremalvaluesofk,thesegmentsshrinktoacorner
{ }
pointofthetriangle∆.
Indeed,thesmallestaltitudeofthetriangleis2.4andthelargestaltitudeis4.Hence,
by the previous proposition, the locus of points is a line segment. To find this line
segment,wecomputethedistancesumfunction .Theequationofthehypotenuseis
V
3x+4y =12.Hence,
3x+4y 12 2 1 12
=x+y − = x+ y+ .
V − 5 5 5 5
Therefore,thelines =careparalleltotheline2x+y =0andtheresultfollows.
V
NotethattheproofsofTheorems1and2followthesameline.
Constant sum of squares of distances
Motivatedbythepreviousresults,wedealwiththesecondquestionoffindingthelocus
ofpointswhichhaveaconstantsumofsquaresofdistancesfromthesidesofagiven
triangle. ReferringtoExample1,DE2+DF2+DG2 = 5(thenumber5ischosen
arbitrary)impliesthefollowingequationofaquadraticcurve;
2
3x+4y 12
x2+y2+ − =5.
(cid:18) 5 (cid:19)
Simplifying,onegetstheequivalentequation
34x2+41y2+24xy 72x 96y+19=0.
− −
ThisisexactlytheequationoftheellipseshowninFig. 3.
4
Figure2: T (∆),2.4 k 4,arelinesegmentsinsidethetriangleparalleltotheline
k
≤ ≤
2x+y =0.
Figure3:EachpointDontheellipsesatisfiesDE2+DF2+DG2 =5
5
Anew characterization ofthe ellipse
Inviewofthepreviousexamplewemayprovethefollowingtheorem,whichstatesthat
amongallquadraticcurvestheellipsecanbecharacterizedasthelocusofpointsthat
haveaconstantsumofsquaresofdistancesfromthesidesofanappropriatetriangle.
Theorem3 The locus S (∆ABC) of points that have a constant sum of squares of
k
distancesfromthesidesofanappropriatetriangleABC isanellipse(andviceversa).
Proof.Weshallprovethefollowingtwoclaims:
(a) Given a triangle, the locus of pointswhich have a constantsum of squaresof
distancesfromthesidesisanellipse. Theseloci,fordifferentvaluesoftheconstant,
arehomotheticellipseswithrespecttotheircommoncenter(theircorrespondingaxes
areproportionalwiththesamefactor).
(b) Given an ellipse, there is a triangle for which the sum of the squares of the
distancesfromthesides,forallpointsontheellipse,isconstant.
Usinganalyticgeometry,wechooseforpart(a)thecoordinatesystemsuchthatthe
verticesofthe trianglelie ontheaxes. Supposethecoordinatesoftheverticesofthe
triangle are A(0,a),B( b,0),C(c,0), where a,b,c > 0 (see Fig. 4). Let (x,y) be
−
anypointintheplaneandletd1,d2,d3 bethedistancesof(x,y)fromthesidesofthe
triangleABC. Itfollowsthat
3 (ax+cy ac)2 (ax by+ab)2
d2 = − + − +y2.
i a2+c2 a2+b2
Xi=1
3
Hence, d2 =k(constant)ifandonlyifthepoint(x,y)liesonthequadraticcurve:
i
iP=1
(ax+cy ac)2 (ax by+ab)2
− + − +y2 =k. (1)
a2+c2 a2+b2
Ingeneral,aquadraticequationintwovariables,
x2+ xy+ y2+ x+ y+ =0, (2)
A B C D E F
representsanellipseprovidedthediscriminantδ = 2 4 <0isnegative.
B − AC
Toproveourclaim,notethatfromequation(1)wehave:
a2 a2 2ac 2ab c2 b2
= + , = , = + +1 (3)
A p q B p − q C p q
where,p=a2+c2andq =a2+b2.Therefore,
a2
δ = 2 4 = 4 b2+2bc+c2+p+q ,
B − AC − pq
(cid:0) (cid:1)
andtheresultfollows.
Now,weprovepart(b).Applyingrotationortranslationoftheaxeswemayassume
that the ellipse has a canonicalform x2 + y2 = 1, α β > 0. If we find positive
α2 β2 ≥
6
real numbers a,b with α2 = a2 + 3b2,β2 = 2a2, then the isosceles triangle with
vertices A(0,a l),B ( b, l),C (b, l), gives the required property in part (b),
′ ′ ′
wherel = 2ab2−. − − −
a2+3b2
Indeed,becauseofthesymmetryoftheellipse,wechoosefirstanisoscelestriangle
withverticesA(0,a),B( b,0),C(b,0)andcomputethesumofthesquareddistances
−
from its sides. Substituting c = b in equation (1), gives the equation of the locus
S (∆ABC)as
k
(ax+by ab)2 (ax by+ab)2
− + − +y2 =k.
a2+b2 a2+b2
Equivalently,wegetatranslationofacanonicalellipse:
x2 + (y− a22+ab32b2)2 = (a2+b2)k−2a2b2 + 2b4 . (4)
a2+3b2 2a2 2a2(a2+3b2) (a2+3b2)2
Thus,itisenoughtotake
(a2+b2)k 2a2b2 2b4
− + =1,
2a2(a2+3b2) (a2+3b2)2
andtherefore,
2a2(a4+7a2b2+10b4)
k = . (5)
(a2+b2)(a2+3b2)
Translating downwardby l = 2ab2 , we get that the ellipse x2 + y2 = 1
a2+3b2 a2+3b2 2a2
is the locus of points, S (∆AB C ), which have a constant sum of squares of dis-
k ′ ′ ′
tancesfromthesidesofthetrianglewithverticesA(0,a l),B ( b, l),C (b, l).
′ ′ ′
− − − −
Moreover,thisconstantisgivenbyequation(5).
Figure4:Thelocusisanellipse
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Table1includestwoexamplesthatdemonstrateTheorem3.
a b α2 =a2+3b2 β2 =2a2 Ellipse k Figure
1 1 4 2 x2 + y2 =1 9 5
4 2 2
√3 1 6 6 x2+y2 =6 10 6
Table1:Examples
Noticethat,thoughagiventriangleandagivenconstantkyieldatmostoneellipse,
the same ellipse can be obtainedusing differenttriangles. This can be easily seen in
Fig. 5. By the above computations, the ellipse x2 + y2 = 1 is the locus of points
4 2
S4.5(∆), where ∆ is the triangle with vertices A′ = (0,0.5),B′ = ( 1, 0.5) and
− −
C =(1, 0.5).Ifwereflectthewholefigureacrossthex axis,thenthesameellipse
′
x42 +y22 =−1willbethelocusofpointsS4.5(∆),where∆−isthereflectionof∆across
thex axiswithvertices:(0, 0.5),( 1,0.5)and(1,0.5).
− − − e e
Noticealsothat,inthesecondexample,thelocusofpointsisacircle. Ingeneral,
thelocusofpointsisacircleexactlywhen,inequation(2),
= and =0.
A C B
Equivalently,from(3)wehave;
a2 a2 c2 b2 2ac 2ab
+ = + +1and =0. (6)
p q p q p − q
Substitutingthevaluesofpandqandsimplifyingweget;
2ac 2ab
=0 (a2 bc)(c b)=0.
p − q ⇔ − −
Consequently,twocaseshavetobeconsidered:
First, a2 = bc,in whichcase the triangle∆ABC isrightangled. Thiscase does
notoccur,sincetheconditionsin(6)leadtoacontradictionasfollows: 2ac 2ab =0
p − q
implies c = b or equivalently bc = b2. Since, a2 = bc we get a2 = b2. Hence,
p q p q p q
thefirstconditionin(6)simplifiesinto a2 = c2 +1.Thisequationtogetherwiththe
q p
relationsa2 =bcand c = b,yieldacontradiction.
p q
Second, c = b, in which case the triangle ∆ABC is isosceles. In this case, we
havep = q. Substitutingin the firstconditionof (6) we get a2 +3b2 = 2a2, which
isequivalenttoa = √3b.Hence,theverticesofthetriangleareA(0,√3b),B( b,0)
−
andC(b,0). Thisimpliesthatthetriangleisequilateral.
Therefore,wehavethefollowingresult.
Conclusion1 ThelocusS (∆)ofpointsthathaveaconstantsumofsquaresofdis-
k
tancesfrom thesides ofa giventriangle∆ isa circle ifandonlyif the triangle∆ is
equilateral.
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Figure5:Thelocusisanellipsewithk =4.5
Figure6:Thelocusisacirclewithk=10
9
Minimalsum ofsquared distances
Anotherinterestingquestionistofindtheminimalsumofsquareddistances,fromthe
sides of a given triangle. For the isosceles triangle with vertices A(0,a),B( b,0)
−
andC(b,0),equation(4)indicatesthat,astherighthandsideapproaches0,theellipse
degeneratestoonepoint:(0, 2ab2 ).Thus,thelocusofpointsisdefinedexactlywhen
a2+3b2
(a2+b2)k 2a2b2 2b4
− + 0.
2a2(a2+3b2) (a2+3b2)2 ≥
Equivalently,
2a2b2
k .
≥ a2+3b2
Thereforethefollowingconsequenceholds.
Conclusion2 Fortheisoscelestrianglewith verticesA(0,a),B( b,0)andC(b,0),
the minimal sum of squared distances from the sides is 2a2b2 a−ttained at the point
a2+3b2
(0, 2ab2 ),insidethetriangle.
a2+3b2
Whena2 = 3b2 thenthetriangleisequilateralwithverticesA(0,√3b),B( b,0)
and C(b,0). In this case, the minimal sum of squared distances is b2 attained−at the
point(0,√3b)whichisexactlytheincenteroftheequilateraltriangle.InFig. 7,b=1
3
and the loci of points are circles which degenerate to the incenter of the equilateral
triangleaskapproaches1.
Concluding remarks
Other related results are the following: Kawasaki [4, p. 213], with a proof without
words, used only rotations to establish Viviani’s theorem. Polster [5], considered a
naturaltwisttoabeautifulone-glanceproofofViviani’stheoremanditsimplications
forgeneraltriangles.
The restrictionof the locus, in the first question, to subsets of the closed triangle
is intended to avoid the use of the signed distances. These signed distances, when
considered,allowsustosearchlociofpointsoutsidethetrianglefor”large”valuesof
theconstantk.Thereaderisencouragedtodosomeexamples.
Theorem2,allowsustoaddressQuestion1toanyconvexpolygonintheplane.
The questionaboutloci of points, with constantsum of squaresof distances, can
begeneralizedtoanypolygonoranysetoflinesintheplane. Inthiscase,oneshould
distinguishwhetherallthelinesareparallelornot. Thereaderisencouragedtowork
outsomeexamplesusingGeoGebra.
Conclusion 2 can be restated for generaltriangles, by a changeof the coordinate
system. This demands familiarity with the following theorem (which is beyond our
discussion): Everyrealquadraticformq = XTAX withsymmetricmatrixAcanbe
reducedbyanorthogonaltransformationtoacanonicalform.
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