3f:null 40:["$","title",null,{"children":"Polarity with respect to a triangle | skydog23 | Observable"}] 46:T5f4,// Thanks to Steven De Keninck for making the ganja.js library polarityWRTTriangle = Algebra(2,0,1,()=>{ // Using Geometric/Clifford algebra with signature (2,0,1), based on the work of Charles Gunn (that's me) var pt = (x,y)=>1e12-x*1e02+y*1e01; // start with an equilateral triangle and its center point as X var A=pt(-0.5,0), B=pt(0, -.866), C=pt(0.5,0); var P=pt(0,-.288), X = P; return this.graph(()=>{ var a = B&C, b = C&A, c = A&B; // sides of triangle var ax = X&A, bx = X&B, cx = X&C; // joining lines to three corners var AX = ax^a, BX = bx^b, CX = cx^c; // intersection points with three sides var am = BX&CX, bm = CX&AX, cm = AX&BX; var AM = am^a, BM = bm^b, CM = cm^c; var p = AM&BM; // create the polar line return ["Polar line of a point with respect to a triangle. Drag P, A, B, C.", 0xff0000].concat([ 0x444444, A,'A', B,'B', C,'C']).concat([ 0x008800, a,'a', b,'b', c,'c']).concat([ 0x0000aa, ax,'a\'', bx,'b\'', cx,'c\'']).concat([ 0x884400, AX,'A\'', BX,'B\'', CX,'C\'']).concat([ 0x008888, am,'a\'', bm,'b\'', cm,'c\'']).concat([ 0x888800, AM,'A\'\'', BM,'B\'\'', CM,'C\'\'']).concat([ 0xff0000, p.Normalized,'p',P,'P']) }, { // more render properties for the default items. pointRadius:1.5, // point radius lineWidth:2, // line width fontSize:1.5, // font size grid:true, // grid width:window.innerWidth, animate:true}); })47:T556,texmd`The polarity with respect to a triangle might seem like a curiosity of limited value. Against this understandable position we present below a construction adapted from an example on the ganja.js coffeeshop webpage devoted to the theme of triangle centers. We have drawn in the polar lines for the three centers centroid, orthocenter, and circumcenter. It's not hard to show that the polar line of the centroid is the ideal line (since the points $D$, $E$, and $F$ in the above construction form a triangle whose sides are parallel to the original triangle). The polar line of the orthocenter provides more interesting behavior. First of all, it appears to always be perpendicular to the Euler line (watch the red point marking the intersection of the two lines). And secondly, it appears to be identical to the 3-way symmetric product in the geometric algebra: $abc+bca+cab+acb+bac+cba$ (drawn in red,overlaying the polar line of the orthocenter). Such phenomena may turn out to be trivial to prove -- but independent of the proof process they illustrate the richness of geometric formulas that PGA offers even (or especially?) when applied to simple, classical topics. The geometric significance of the polar line of the circumcenter is not immediately obvious.`48:T81a,Algebra(2,0,1,()=>{ // The Euler line connects several important points in any non equilateral triangle. // Not many identities needed.. var point = (x,y)=>1e12-x*1e02+y*1e01; // Three points. var A=point(-1.5,0.5).Normalized, B=point(0.5,-1.5).Normalized, C=point(1,0.5).Normalized, // The edges a=()=>B&C,b=()=>C&A,c=()=>A&B, // The orthocenter oa=()=>a<b<c<oa^ob, // The circumcenter ca=()=>c<<(A+B), cb=()=>a<<(B+C), cc=()=>b<<(C+A), circum=()=>ca^cb, // The centroid centroid = ()=>A+B+C, // The Euler line euler = ()=>ortho&circum, orthoL = polarLine(ortho), // symmetric product: appears equal to polar line of orthocenter // which is furthermore perpendicular to the euler line s = ()=>a*b*c+b*c*a+c*a*b+a*c*b+b*a*c+c*b*a; function polarLine(X){ var ax = ()=> X&A, bx =()=> X&B, cx =()=> X&C; // joining lines to three corners var AX = ()=> ax^a, BX =()=> bx^b, CX = ()=> cx^c; // intersection points with three sides var am = ()=> BX&CX, bm =()=> CX&AX, cm =()=> AX&BX; var AM = ()=> am^a, BM = ()=> bm^b, CM = ()=> cm^c; return ()=>AM&BM; } return this.graph(()=>{ return ["Polarity wrt a triangle and triangle centers", 0xFFFFEE,[A,B,C], // triangle 0xAA8888,oa,ob,oc,ortho,"O", polarLine(ortho), "o", // orthocenter 0x448844,ca,cb,cc,circum,"C",0x00ff00,polarLine(circum),"c", // circumcenter 0x8888AA,centroid,"M",polarLine(centroid),"m",euler,"Euler Line", // centroid and euler line 0x444444,A,B,C, // triangle vertices 0xff0000,()=>orthoL^euler,"X",s,"s"] }, { // more render properties for the default items. pointRadius:1.5, // point radius lineWidth:2, // line width fontSize:1.5, // font size grid:true, // grid width:window.innerWidth, animate:true}); })49:[] 44:{"id":"1a75f299abdf9391","slug":"polarity-with-respect-to-a-triangle","title":"Polarity with respect to a triangle","thumbnail":"95e6cd8210ec0b1a07c6319dede34fd36661b15512eb0e883066ff8a2e567fc0","access_level":"public","create_time":"$D2018-08-08T10:31:01.964Z","update_time":"$D2020-07-17T09:08:52.148Z","view_time":null,"first_public_time":"$D2018-08-08T19:04:22.234Z","has_unpublished_edits":null,"fork":false,"stars":5,"starred":false,"owner":{"type":"team","id":"c12ededdbf97bde4","login":"skydog23","name":"","avatar_url":"https://avatars-next.observableusercontent.com/57d55b329e0de79c4f1a8c1447c6cb3689b05343c2cde413e6c55225ab16ea26","tier":"starter_2024","bio":"","following":false},"version":342,"publish_version":342,"role":null,"authored":false,"authors":[{"type":"individual","id":"8ebee8d27ef26f03","login":"skydog23","name":"","avatar_url":"https://avatars-next.observableusercontent.com/57d55b329e0de79c4f1a8c1447c6cb3689b05343c2cde413e6c55225ab16ea26","tier":null,"bio":"","following":false}],"members":[{"type":"individual","id":"8ebee8d27ef26f03","login":"skydog23","name":"","avatar_url":"https://avatars-next.observableusercontent.com/57d55b329e0de79c4f1a8c1447c6cb3689b05343c2cde413e6c55225ab16ea26","tier":null,"role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"lapsed":null,"subscribed":null,"forks":3,"collection":null,"type":"notebook","body":{"titling":"auto","title":"Polarity with respect to a triangle","cells":[{"id":0,"value":"md`# Polarity with respect to a triangle`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":4,"value":"md `A polarity in the projective plane is an involutive correlation. Correlation means that it is a projectivity that maps points to lines and vice-versa; involutive means that performing it twice yields the identity map. The traditional polarity is defined with respect to a conic section and is discussed in this Observable notebook. There is also a simpler version of polarity defined with respect to a triangle that is the subject of this notebook. This simpler form is not strictly speaking a projectivity since it is not defined for elements incident with the triangle, that is, for the vertices and sides of the triangle and points (lines) incident with these sides (vertices). But on the other points and lines it is an involutive correlation.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":52,"value":"texmd `Begin with a triangle $ABC$ with sides $abc$. For a variable point $P$ we construct the polar line $p$. First connect $P$ with the vertices: $$ \\small a' := A \\vee P;~~~b' := B\\vee P;~~~c':=C\\vee P$$ Next, intersect these lines with the corresponding triangle side: $$ \\small A' := a \\wedge a';~~~B' := b\\wedge b';~~~C':=c\\wedge c'$$ Then form the triangle sides for the triange $A'B'C'$: $$ \\small a'':= B' \\vee C';~~~b'' := C'\\vee A';~~~c'':=A'\\vee B'$$ Finally form the intersections of these sides with the corresponding sides of the original triangle: $$ \\small A'' := a \\wedge a'';~~~B'' := b \\wedge b'';~~~C'' := c \\wedge c''$$ Then $A''$, $B''$, and $C''$ are collinear and their common line is the *polar* line $p$ of $P$. (To see that $A''$, $B''$, and $C''$ are collinear, notice that the triangles $ABC$ and $A''B''C''$ are perspective in the point $P$. Applying the Theorem of Desargues yields the desired result.)`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":91,"value":"md `In the figure below, we begin with a very symmetric position. The polar line of the red point is the ideal line. Moving any vertex of the triangle or the red point breaks the symmetry and moves the polar line into the euclidean plane.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":327,"value":"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":11,"value":"$46","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":279,"value":"md`## Applications of polarity with respect to a triangle`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":276,"value":"$47","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":326,"value":"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":240,"value":"$48","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":308,"value":"texmd`These results may seem surprising considering that the polarity with respect to a triangle is a purely projective construction, independent of metric, while all the triangle centers are defined in terms of the euclidean metric. It's not so surprising however when one recalls the projective definition of the euclidean metric. That is, the choice of the ideal line and an elliptic involution determines distance and angle measurement. The centroid for example is defined in terms of the midpoints of the sides. Projectively a point M is the midpoint of a segment AB when the harmonic fourth point with respect to $M$, $A$, and $B$ is ideal. The polarity construction is permeated with harmonic sets, so the fact that the polar line of the centroid is ideal is certainly plausible. Similar remarks probably apply to the fact that the polar line of the orthocenter cuts the Euler line at right angles, since the orthocenter and circumcenter are both based on right angle constructions.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":312,"value":"md`Open question: If this polarity were a true projectivity than incidence would be preserved. That would imply that the polar lines of three collinear points (the triangle centers) would pass through the same point. Since one of the lines is the ideal line, this would mean that the other two polar lines are parallel. But the construction makes clear that they are not. What's going on?`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":317,"value":"md`And, it would be very cool if the infrastructure were in place to check out these constructions in the elliptic and hyperbolic planes to see if the same or similar results hold there.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":324,"value":"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":9,"value":"Algebra = require('ganja.js')","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":325,"value":"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":53,"value":"import {texmd} from '@kelleyvanevert/katex-within-markdown'","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"}],"files":"$W49","stdlib":"1"}} 45:"$44:body"