{
  "index": "2006-B-1",
  "type": "ALG",
  "tag": [
    "ALG",
    "GEO",
    "NT"
  ],
  "difficulty": "",
  "question": "Show that the curve $x^3 + 3xy + y^3 = 1$ contains only one set of three\ndistinct points, $A$, $B$, and $C$, which are vertices of an equilateral\ntriangle, and find its area.",
  "solution": "The ``curve'' $x^3 + 3xy + y^3 - 1 = 0$ is actually reducible, because the\nleft side factors as\n\\[\n(x+y-1)(x^2 -xy + y^2 + x + y + 1).\n\\]\nMoreover, the second factor is\n\\[\n\\frac{1}{2} ((x+1)^2 + (y+1)^2 + (x-y)^2),\n\\]\nso it only vanishes at $(-1,-1)$. Thus the curve in question consists of the\nsingle point $(-1,-1)$ together with the line $x+y=1$. To form a triangle\nwith three points on this curve, one of its vertices must be $(-1,-1)$.\nThe other two vertices lie on the line $x+y=1$, so the length of the altitude\nfrom $(-1,-1)$ is the distance from $(-1,-1)$ to $(1/2,1/2)$, or\n$3 \\sqrt{2}/2$. The area of an equilateral triangle of height $h$ is\n$h^2 \\sqrt{3}/3$, so the desired area is\n$3 \\sqrt{3}/2$.\n\n\\textbf{Remark:} The factorization used above is a special case of the fact\nthat\n\\begin{align*}\n&x^3 + y^3 + z^3 - 3xyz \\\\\n&= (x+y+z)(x+\\omega y + \\omega^2 z)(x + \\omega^2 y\n+ \\omega z),\n\\end{align*}\nwhere $\\omega$ denotes a primitive cube root of unity. That fact in turn follows\nfrom the evaluation of the determinant of the \\emph{circulant matrix}\n\\[\n\\begin{pmatrix} x & y & z \\\\ z & x & y \\\\ y & z & x\n\\end{pmatrix}\n\\]\nby reading off the eigenvalues of the eigenvectors\n$(1, \\omega^i, \\omega^{2i})$ for $i=0,1,2$.",
  "vars": [
    "x",
    "y",
    "z",
    "A",
    "B",
    "C",
    "h",
    "i"
  ],
  "params": [
    "\\\\omega"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "z": "applicate",
        "A": "vertexalpha",
        "B": "vertexbeta",
        "C": "vertexgamma",
        "h": "heighttri",
        "\\omega": "unityroot"
      },
      "question": "Show that the curve $abscissa^3 + 3\\,abscissa\\,ordinate + ordinate^3 = 1$ contains only one set of three distinct points, $vertexalpha$, $vertexbeta$, and $vertexgamma$, which are vertices of an equilateral triangle, and find its area.",
      "solution": "The ``curve'' $abscissa^3 + 3\\,abscissa\\,ordinate + ordinate^3 - 1 = 0$ is actually reducible, because the left side factors as\n\\[\n(abscissa+ordinate-1)(abscissa^2 -abscissa\\,ordinate + ordinate^2 + abscissa + ordinate + 1).\n\\]\nMoreover, the second factor is\n\\[\n\\frac{1}{2} ((abscissa+1)^2 + (ordinate+1)^2 + (abscissa-ordinate)^2),\n\\]\nso it only vanishes at $(-1,-1)$. Thus the curve in question consists of the single point $(-1,-1)$ together with the line $abscissa+ordinate=1$. To form a triangle with three points on this curve, one of its vertices must be $(-1,-1)$. The other two vertices lie on the line $abscissa+ordinate=1$, so the length of the altitude from $(-1,-1)$ is the distance from $(-1,-1)$ to $(1/2,1/2)$, or $3 \\sqrt{2}/2$. The area of an equilateral triangle of height $heighttri$ is $heighttri^2 \\sqrt{3}/3$, so the desired area is $3 \\sqrt{3}/2$.\n\n\\textbf{Remark:} The factorization used above is a special case of the fact\n\\begin{align*}\n&abscissa^3 + ordinate^3 + applicate^3 - 3\\,abscissa\\,ordinate\\,applicate \\\\\n&= (abscissa+ordinate+applicate)(abscissa+unityroot\\,ordinate + unityroot^{2}\\,applicate)(abscissa + unityroot^{2}\\,ordinate + unityroot\\,applicate),\n\\end{align*}\nwhere $unityroot$ denotes a primitive cube root of unity. That fact in turn follows from the evaluation of the determinant of the \\emph{circulant matrix}\n\\[\n\\begin{pmatrix} abscissa & ordinate & applicate \\\\ applicate & abscissa & ordinate \\\\ ordinate & applicate & abscissa \\end{pmatrix}\n\\]\nby reading off the eigenvalues of the eigenvectors $(1, unityroot^{i}, unityroot^{2i})$ for $i=0,1,2$."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marshland",
        "y": "driftwood",
        "z": "hummingbird",
        "A": "shoreline",
        "B": "treeline",
        "C": "meadowlands",
        "h": "riverbank",
        "i": "lighthouse",
        "\\omega": "cornerstone"
      },
      "question": "Show that the curve $marshland^3 + 3 marshland driftwood + driftwood^3 = 1$ contains only one set of three\ndistinct points, $shoreline$, $treeline$, and $meadowlands$, which are vertices of an equilateral\ntriangle, and find its area.",
      "solution": "The ``curve'' $marshland^3 + 3 marshland driftwood + driftwood^3 - 1 = 0$ is actually reducible, because the\nleft side factors as\n\\[\n(marshland+driftwood-1)(marshland^2 - marshland driftwood + driftwood^2 + marshland + driftwood + 1).\n\\]\nMoreover, the second factor is\n\\[\n\\frac{1}{2} ((marshland+1)^2 + (driftwood+1)^2 + (marshland-driftwood)^2),\n\\]\nso it only vanishes at $(-1,-1)$. Thus the curve in question consists of the\nsingle point $(-1,-1)$ together with the line $marshland+driftwood=1$. To form a triangle\nwith three points on this curve, one of its vertices must be $(-1,-1)$.\nThe other two vertices lie on the line $marshland+driftwood=1$, so the length of the altitude\nfrom $(-1,-1)$ is the distance from $(-1,-1)$ to $(1/2,1/2)$, or\n$3 \\sqrt{2}/2$. The area of an equilateral triangle of height $riverbank$ is\n$riverbank^2 \\sqrt{3}/3$, so the desired area is\n$3 \\sqrt{3}/2$.\n\n\\textbf{Remark:} The factorization used above is a special case of the fact\nthat\n\\begin{align*}\n&marshland^3 + driftwood^3 + hummingbird^3 - 3 marshland driftwood hummingbird \\\\\n&= (marshland+driftwood+hummingbird)(marshland+cornerstone driftwood + cornerstone^2 hummingbird)(marshland + cornerstone^2 driftwood\n+ cornerstone hummingbird),\n\\end{align*}\nwhere $cornerstone$ denotes a primitive cube root of unity. That fact in turn follows\nfrom the evaluation of the determinant of the \\emph{circulant matrix}\n\\[\n\\begin{pmatrix} marshland & driftwood & hummingbird \\\\ hummingbird & marshland & driftwood \\\\ driftwood & hummingbird & marshland\n\\end{pmatrix}\n\\]\nby reading off the eigenvalues of the eigenvectors\n$(1, cornerstone^{lighthouse}, cornerstone^{2 lighthouse})$ for $lighthouse=0,1,2$. }"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "z": "planarcoordinate",
        "A": "nonvertexalpha",
        "B": "nonvertexbeta",
        "C": "nonvertexgamma",
        "h": "basewidth",
        "i": "realindex",
        "\\omega": "alphastart"
      },
      "question": "Show that the curve $verticalaxis^3 + 3verticalaxishorizontalaxis + horizontalaxis^3 = 1$ contains only one set of three\ndistinct points, $nonvertexalpha$, $nonvertexbeta$, and $nonvertexgamma$, which are vertices of an equilateral\ntriangle, and find its area.",
      "solution": "The ``curve'' $verticalaxis^3 + 3verticalaxishorizontalaxis + horizontalaxis^3 - 1 = 0$ is actually reducible, because the\nleft side factors as\n\\[\n(verticalaxis+horizontalaxis-1)(verticalaxis^2 -verticalaxishorizontalaxis + horizontalaxis^2 + verticalaxis + horizontalaxis + 1).\n\\]\nMoreover, the second factor is\n\\[\n\\frac{1}{2} ((verticalaxis+1)^2 + (horizontalaxis+1)^2 + (verticalaxis-horizontalaxis)^2),\n\\]\nso it only vanishes at $(-1,-1)$. Thus the curve in question consists of the\nsingle point $(-1,-1)$ together with the line $verticalaxis+horizontalaxis=1$. To form a triangle\nwith three points on this curve, one of its vertices must be $(-1,-1)$.\nThe other two vertices lie on the line $verticalaxis+horizontalaxis=1$, so the length of the altitude\nfrom $(-1,-1)$ is the distance from $(-1,-1)$ to $(1/2,1/2)$, or\n$3 \\sqrt{2}/2$. The area of an equilateral triangle of height $basewidth$ is\n$basewidth^2 \\sqrt{3}/3$, so the desired area is\n$3 \\sqrt{3}/2$.\n\n\\textbf{Remark:} The factorization used above is a special case of the fact\nthat\n\\begin{align*}\n&verticalaxis^3 + horizontalaxis^3 + planarcoordinate^3 - 3verticalaxishorizontalaxisplanarcoordinate \\\\\n&= (verticalaxis+horizontalaxis+planarcoordinate)(verticalaxis+alphastart horizontalaxis + alphastart^2 planarcoordinate)(verticalaxis + alphastart^2 horizontalaxis\n+ alphastart planarcoordinate),\n\\end{align*}\nwhere $alphastart$ denotes a primitive cube root of unity. That fact in turn follows\nfrom the evaluation of the determinant of the \\emph{circulant matrix}\n\\[\n\\begin{pmatrix} verticalaxis & horizontalaxis & planarcoordinate \\\\ planarcoordinate & verticalaxis & horizontalaxis \\\\ horizontalaxis & planarcoordinate & verticalaxis\n\\end{pmatrix}\n\\]\nby reading off the eigenvalues of the eigenvectors\n$(1, alphastart^{realindex}, alphastart^{2realindex})$ for $realindex=0,1,2$. "
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hxvplkda",
        "z": "vrbjcqsm",
        "A": "nmbtkyse",
        "B": "xbqdhram",
        "C": "pslgnuyk",
        "h": "dncfrqle",
        "i": "ghtlsnzm",
        "\\omega": "\\wqvmdnla"
      },
      "question": "Show that the curve $qzxwvtnp^3 + 3 qzxwvtnp hxvplkda + hxvplkda^3 = 1$ contains only one set of three\ndistinct points, $nmbtkyse$, $xbqdhram$, and $pslgnuyk$, which are vertices of an equilateral\ntriangle, and find its area.",
      "solution": "The ``curve'' $qzxwvtnp^3 + 3 qzxwvtnp hxvplkda + hxvplkda^3 - 1 = 0$ is actually reducible, because the\nleft side factors as\n\\[\n(qzxwvtnp+hxvplkda-1)(qzxwvtnp^2 -qzxwvtnp hxvplkda + hxvplkda^2 + qzxwvtnp + hxvplkda + 1).\n\\]\nMoreover, the second factor is\n\\[\n\\frac{1}{2} ((qzxwvtnp+1)^2 + (hxvplkda+1)^2 + (qzxwvtnp-hxvplkda)^2),\n\\]\nso it only vanishes at $(-1,-1)$. Thus the curve in question consists of the\nsingle point $(-1,-1)$ together with the line $qzxwvtnp+hxvplkda=1$. To form a triangle\nwith three points on this curve, one of its vertices must be $(-1,-1)$.\nThe other two vertices lie on the line $qzxwvtnp+hxvplkda=1$, so the length of the altitude\nfrom $(-1,-1)$ is the distance from $(-1,-1)$ to $(1/2,1/2)$, or\n$3 \\sqrt{2}/2$. The area of an equilateral triangle of height $dncfrqle$ is\n$dncfrqle^2 \\sqrt{3}/3$, so the desired area is\n$3 \\sqrt{3}/2$.\n\n\\textbf{Remark:} The factorization used above is a special case of the fact\nthat\n\\begin{align*}\n&qzxwvtnp^3 + hxvplkda^3 + vrbjcqsm^3 - 3 qzxwvtnp hxvplkda vrbjcqsm \\\\\n&= (qzxwvtnp+hxvplkda+vrbjcqsm)(qzxwvtnp+\\wqvmdnla hxvplkda + \\wqvmdnla^2 vrbjcqsm)(qzxwvtnp + \\wqvmdnla^2 hxvplkda\n+ \\wqvmdnla vrbjcqsm),\n\\end{align*}\nwhere $\\wqvmdnla$ denotes a primitive cube root of unity. That fact in turn follows\nfrom the evaluation of the determinant of the \\emph{circulant matrix}\n\\[\n\\begin{pmatrix} qzxwvtnp & hxvplkda & vrbjcqsm \\\\ vrbjcqsm & qzxwvtnp & hxvplkda \\\\ hxvplkda & vrbjcqsm & qzxwvtnp\n\\end{pmatrix}\n\\]\nby reading off the eigenvalues of the eigenvectors\n$(1, \\wqvmdnla^{ghtlsnzm}, \\wqvmdnla^{2 ghtlsnzm})$ for $ghtlsnzm=0,1,2$. }.",
      "error": ""
    },
    "kernel_variant": {
      "question": "Show that the plane curve\n\\[\n   x^{3}+6xy+y^{3}=8\n\\]\ncontains exactly one triple of distinct points \\(A,\\,B,\\,C\\) that are the vertices of an equilateral triangle, and determine the area of that triangle.",
      "solution": "Write the equation as\n  x³ + y³ + (–2)³ – 3xy(–2) = 0.\nBy the classical identity\n  x³ + y³ + z³ – 3xyz\n    = (x+y+z)(x+ωy+ω²z)(x+ω²y+ωz),   (ω³=1, ω≠1),\nwith z=−2 we obtain the factorization\n  x³+6xy+y³−8 = (x+y−2)(x²−xy+y²+2x+2y+4).\nThe quadratic factor is\n  ½[(x+2)²+(y+2)²+(x−y)²] ≥ 0,\nvanishing only at (x,y) = (−2,−2).  Hence the curve is the line L: x+y=2 together with the isolated point P = (−2,−2).\n\n1. Any triangle with three vertices on the curve must use P and two distinct points B,C on L.\n\n2. The foot M of the perpendicular from P to L is found by intersecting x=y with x+y=2, giving M=(1,1).  The distance (altitude) is\n  h = dist(P,L)\n    = |(−2)+(−2)−2|/√(1²+1²)\n    = 6/√2 = 3√2.\n\n3. In an equilateral triangle of side a the altitude is a√3/2, so\n  a = 2h/√3 = 2⋅3√2/√3 = 6√(2/3).\nAlong L the unit direction is (1,−1)/√2, so the two vertices at distance a/2 from M are\n  B = M + (a/2)(1,−1)/√2,\n  C = M − (a/2)(1,−1)/√2.\nThis triangle is unique.\n\n4. Its area is\n  A = h²√3/3 = (3√2)²√3/3 = 18√3/3 = 6√3.\n\nTherefore there is exactly one such equilateral triangle, and its area is 6√3.",
      "_meta": {
        "core_steps": [
          "Apply identity x^3+y^3+z^3-3xyz to factor x^3+3xy+y^3-1 into (x+y-1)(positive-definite quadratic)",
          "Observe the quadratic factor is ≥0 and vanishes only at one point, so the curve is that point plus the line x+y=1",
          "Any triangle on the curve must use the isolated point together with two points on the line",
          "Compute altitude as the distance from the isolated point to the line",
          "Convert altitude to area via the equilateral-triangle formula A = h^2√3/3"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Constant c in a more general equation x^3 + 3c·xy + y^3 = c^3 (here c=1); changing c just translates/scales the line x+y=c and the special point (−c,−c)",
            "original": 1
          },
          "slot2": {
            "description": "Coefficient 3c of the xy-term (equals 3 when c=1); it must vary consistently with slot1 but its exact value is otherwise irrelevant to the logical chain",
            "original": 3
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}