{
  "index": "1955-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "5. If a parabola is given in the plane, find a geometric construction (ruler and compass) for the focus.",
  "solution": "Solution. Analysis: Consider the chords of the parabola determined by a family \\( \\mathcal{F} \\) of parallel lines. It is well known that the midpoints of these chords fall on a line parallel to the axis of the parabola. Such a line is called a diameter of the parabola. A diameter meets the parabola just once. If the diameter determined by \\( \\mathfrak{F} \\) meets the parabola at \\( P \\), then the member of \\( \\mathcal{F} \\) through \\( P \\) is tangent to the parabola. By the focussing property of the parabola, if the diameter through \\( P \\) is reflected in the tangent at \\( P \\), the resulting line passes through the focus. Hence we are led to the following construction.\n\nConstruction: Draw any chord of the parabola and construct another parallel to it. Find the midpoints \\( A \\) and \\( B \\) of these chords and let \\( \\overrightarrow{A B} \\) meet the parabola at \\( P \\). Draw \\( \\widehat{P Q} \\) parallel to the original chord and find \\( C \\) so that \\( \\angle A P Q=\\angle Q P C \\), with \\( C \\) on the side of \\( \\overrightarrow{P Q} \\) opposite to \\( A \\). The focus is on \\( \\overparen{P C} \\).\n\nRepeat the construction starting from a different chord to obtain another line through the focus. Assuming the two lines thus constructed are different their intersection is the focus.\n\nTo avoid the awkward possibility that the two lines through the focus coincide, be sure that the two starting chords are neither parallel nor perpendicular. To see that this suffices, note that if the starting chords for the two constructions make an angle \\( \\alpha \\), the constructed lines make an angle \\( 2 \\alpha \\); hence, if \\( \\alpha \\neq 0, \\pi / 2 \\), the constructed lines will have different directions.\n\nRemark. For the actual construction a number of shortcuts can be found. For example, once the first diameter has been found, the axis can quickly be located as the perpendicular bisector of any chord perpendicular to the diameter.",
  "vars": [
    "A",
    "B",
    "C",
    "P",
    "Q",
    "\\\\alpha"
  ],
  "params": [
    "F"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "pointa",
        "B": "pointb",
        "C": "pointc",
        "P": "pointp",
        "Q": "pointq",
        "\\alpha": "anglealpha",
        "F": "fixedfocus"
      },
      "question": "5. If a parabola is given in the plane, find a geometric construction (ruler and compass) for the focus.",
      "solution": "Solution. Analysis: Consider the chords of the parabola determined by a family \\( \\mathcal{fixedfocus} \\) of parallel lines. It is well known that the midpoints of these chords fall on a line parallel to the axis of the parabola. Such a line is called a diameter of the parabola. A diameter meets the parabola just once. If the diameter determined by \\( \\mathfrak{fixedfocus} \\) meets the parabola at \\( pointp \\), then the member of \\( \\mathcal{fixedfocus} \\) through \\( pointp \\) is tangent to the parabola. By the focussing property of the parabola, if the diameter through \\( pointp \\) is reflected in the tangent at \\( pointp \\), the resulting line passes through the focus. Hence we are led to the following construction.\n\nConstruction: Draw any chord of the parabola and construct another parallel to it. Find the midpoints \\( pointa \\) and \\( pointb \\) of these chords and let \\( \\overrightarrow{pointa\\, pointb} \\) meet the parabola at \\( pointp \\). Draw \\( \\widehat{pointp\\, pointq} \\) parallel to the original chord and find \\( pointc \\) so that \\( \\angle pointa\\, pointp\\, pointq = \\angle pointq\\, pointp\\, pointc \\), with \\( pointc \\) on the side of \\( \\overrightarrow{pointp\\, pointq} \\) opposite to \\( pointa \\). The focus is on \\( \\overparen{pointp\\, pointc} \\).\n\nRepeat the construction starting from a different chord to obtain another line through the focus. Assuming the two lines thus constructed are different their intersection is the focus.\n\nTo avoid the awkward possibility that the two lines through the focus coincide, be sure that the two starting chords are neither parallel nor perpendicular. To see that this suffices, note that if the starting chords for the two constructions make an angle \\( anglealpha \\), the constructed lines make an angle \\( 2\\, anglealpha \\); hence, if \\( anglealpha \\neq 0, \\pi / 2 \\), the constructed lines will have different directions.\n\nRemark. For the actual construction a number of shortcuts can be found. For example, once the first diameter has been found, the axis can quickly be located as the perpendicular bisector of any chord perpendicular to the diameter."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "sunflower",
        "B": "toothpick",
        "C": "raincloud",
        "P": "fireplace",
        "Q": "moonscape",
        "\\\\alpha": "drumstick",
        "F": "butterfly"
      },
      "question": "5. If a parabola is given in the plane, find a geometric construction (ruler and compass) for the focus.",
      "solution": "Solution. Analysis: Consider the chords of the parabola determined by a family \\( \\mathcal{butterfly} \\) of parallel lines. It is well known that the midpoints of these chords fall on a line parallel to the axis of the parabola. Such a line is called a diameter of the parabola. A diameter meets the parabola just once. If the diameter determined by \\( \\mathfrak{butterfly} \\) meets the parabola at \\( fireplace \\), then the member of \\( \\mathcal{butterfly} \\) through \\( fireplace \\) is tangent to the parabola. By the focussing property of the parabola, if the diameter through \\( fireplace \\) is reflected in the tangent at \\( fireplace \\), the resulting line passes through the focus. Hence we are led to the following construction.\n\nConstruction: Draw any chord of the parabola and construct another parallel to it. Find the midpoints \\( sunflower \\) and \\( toothpick \\) of these chords and let \\( \\overrightarrow{sunflower\\ toothpick} \\) meet the parabola at \\( fireplace \\). Draw \\( \\widehat{fireplace\\ moonscape} \\) parallel to the original chord and find \\( raincloud \\) so that \\( \\angle sunflower\\ fireplace\\ moonscape = \\angle moonscape\\ fireplace\\ raincloud \\), with \\( raincloud \\) on the side of \\( \\overrightarrow{fireplace\\ moonscape} \\) opposite to \\( sunflower \\). The focus is on \\( \\overparen{fireplace\\ raincloud} \\).\n\nRepeat the construction starting from a different chord to obtain another line through the focus. Assuming the two lines thus constructed are different their intersection is the focus.\n\nTo avoid the awkward possibility that the two lines through the focus coincide, be sure that the two starting chords are neither parallel nor perpendicular. To see that this suffices, note that if the starting chords for the two constructions make an angle \\( drumstick \\), the constructed lines make an angle \\( 2\\ drumstick \\); hence, if \\( drumstick \\neq 0, \\pi / 2 \\), the constructed lines will have different directions.\n\nRemark. For the actual construction a number of shortcuts can be found. For example, once the first diameter has been found, the axis can quickly be located as the perpendicular bisector of any chord perpendicular to the diameter."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "leftendpoint",
        "B": "rightendpoint",
        "C": "destroyedpoint",
        "P": "farawaypoint",
        "Q": "stillpoint",
        "\\alpha": "segmentlen",
        "F": "defocuspt"
      },
      "question": "5. If a parabola is given in the plane, find a geometric construction (ruler and compass) for the focus.",
      "solution": "Solution. Analysis: Consider the chords of the parabola determined by a family \\( \\mathcal{F} \\) of parallel lines. It is well known that the midpoints of these chords fall on a line parallel to the axis of the parabola. Such a line is called a diameter of the parabola. A diameter meets the parabola just once. If the diameter determined by \\( \\mathfrak{F} \\) meets the parabola at \\( farawaypoint \\), then the member of \\( \\mathcal{F} \\) through \\( farawaypoint \\) is tangent to the parabola. By the focussing property of the parabola, if the diameter through \\( farawaypoint \\) is reflected in the tangent at \\( farawaypoint \\), the resulting line passes through the focus. Hence we are led to the following construction.\n\nConstruction: Draw any chord of the parabola and construct another parallel to it. Find the midpoints \\( leftendpoint \\) and \\( rightendpoint \\) of these chords and let \\( \\overrightarrow{leftendpoint\\, rightendpoint} \\) meet the parabola at \\( farawaypoint \\). Draw \\( \\widehat{farawaypoint\\, stillpoint} \\) parallel to the original chord and find \\( destroyedpoint \\) so that \\( \\angle leftendpoint\\, farawaypoint\\, stillpoint = \\angle stillpoint\\, farawaypoint\\, destroyedpoint \\), with \\( destroyedpoint \\) on the side of \\( \\overrightarrow{farawaypoint\\, stillpoint} \\) opposite to \\( leftendpoint \\). The focus is on \\( \\overparen{farawaypoint\\, destroyedpoint} \\).\n\nRepeat the construction starting from a different chord to obtain another line through the focus. Assuming the two lines thus constructed are different their intersection is the focus.\n\nTo avoid the awkward possibility that the two lines through the focus coincide, be sure that the two starting chords are neither parallel nor perpendicular. To see that this suffices, note that if the starting chords for the two constructions make an angle \\( segmentlen \\), the constructed lines make an angle \\( 2\\, segmentlen \\); hence, if \\( segmentlen \\neq 0, \\pi / 2 \\), the constructed lines will have different directions.\n\nRemark. For the actual construction a number of shortcuts can be found. For example, once the first diameter has been found, the axis can quickly be located as the perpendicular bisector of any chord perpendicular to the diameter."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "mxntrqwe",
        "P": "zlkjhfdp",
        "Q": "vbnmcxqp",
        "\\alpha": "sldkfjwe",
        "F": "plmoknij"
      },
      "question": "5. If a parabola is given in the plane, find a geometric construction (ruler and compass) for the focus.",
      "solution": "Solution. Analysis: Consider the chords of the parabola determined by a family \\( \\mathcal{F} \\) of parallel lines. It is well known that the midpoints of these chords fall on a line parallel to the axis of the parabola. Such a line is called a diameter of the parabola. A diameter meets the parabola just once. If the diameter determined by \\( \\mathfrak{F} \\) meets the parabola at \\( zlkjhfdp \\), then the member of \\( \\mathcal{F} \\) through \\( zlkjhfdp \\) is tangent to the parabola. By the focussing property of the parabola, if the diameter through \\( zlkjhfdp \\) is reflected in the tangent at \\( zlkjhfdp \\), the resulting line passes through the focus. Hence we are led to the following construction.\n\nConstruction: Draw any chord of the parabola and construct another parallel to it. Find the midpoints \\( qzxwvtnp \\) and \\( hjgrksla \\) of these chords and let \\( \\overrightarrow{qzxwvtnp hjgrksla} \\) meet the parabola at \\( zlkjhfdp \\). Draw \\( \\widehat{zlkjhfdp vbnmcxqp} \\) parallel to the original chord and find \\( mxntrqwe \\) so that \\( \\angle qzxwvtnp zlkjhfdp vbnmcxqp=\\angle vbnmcxqp zlkjhfdp mxntrqwe \\), with \\( mxntrqwe \\) on the side of \\( \\overrightarrow{zlkjhfdp vbnmcxqp} \\) opposite to \\( qzxwvtnp \\). The focus is on \\( \\overparen{zlkjhfdp mxntrqwe} \\).\n\nRepeat the construction starting from a different chord to obtain another line through the focus. Assuming the two lines thus constructed are different their intersection is the focus.\n\nTo avoid the awkward possibility that the two lines through the focus coincide, be sure that the two starting chords are neither parallel nor perpendicular. To see that this suffices, note that if the starting chords for the two constructions make an angle \\( sldkfjwe \\), the constructed lines make an angle \\( 2 sldkfjwe \\); hence, if \\( sldkfjwe \\neq 0, \\pi / 2 \\), the constructed lines will have different directions.\n\nRemark. For the actual construction a number of shortcuts can be found. For example, once the first diameter has been found, the axis can quickly be located as the perpendicular bisector of any chord perpendicular to the diameter."
    },
    "kernel_variant": {
      "question": "Focus–construction for a parabola by means of two prescribed translations\n\nLet Γ be a fixed non-degenerate parabola drawn in the Euclidean plane.  A positively oriented straight line ℓ and a unit segment u are given on the sheet; the segment u may be copied with ruler and compasses whenever a “translation through the signed distance 1” is required (distance is measured along the indicated oriented line).\n\nA.  (the “90° pair”)\n   •  Choose a chord k of Γ whose supporting line is perpendicular to ℓ.\n   •  Translate that supporting line through the signed distance 1 measured along ℓ.  The image is a line k₀ parallel to k.  Sliding k slightly along Γ if necessary (this is always possible because Γ is unbounded in every direction that is not parallel to its axis) we may—and shall—assume that k₀ meets Γ again.  Put k′ := k₀ ∩ Γ.  Thus k and k′ are parallel chords whose supporting lines are exactly one unit apart.\n   •  Using only the two chords k and k′ construct a straight line r that is guaranteed to pass through the focus of Γ.\n\nB.  (the “60° pair”)\n   •  Choose a chord m of Γ which makes an angle of 60° with ℓ.\n   •  Translate the supporting line of m through the signed distance 1 in the direction perpendicular to m.  As before, a slight shift of the first chord ensures that the translate meets Γ; set m′ := m₀ ∩ Γ.  Hence m and m′ are parallel chords at mutual distance 1.\n   •  From the two chords m and m′ alone construct a straight line s that passes through the focus of Γ.\n\nProve that the two lines r and s are distinct and that their intersection is the focus of Γ.  Consequently the above scheme furnishes a classical ruler-and-compass construction of the focus of an arbitrary parabola.",
      "solution": "Notation and preliminaries\n———————————————\nΓ is a fixed non-degenerate parabola with axis A and focus F.  A “parallel family” is the set of all chords whose supporting lines have the same direction u.  For every family the mid-points of its chords are collinear; this common line is called the diameter of the family and it is parallel to the axis A.\n\nPart I.  From two parallel chords to a line through the focus\n————————————————————————————————————————\nLet q and q′ be two distinct, parallel chords of Γ with common direction u, and let M and N be their mid-points.\n\nStep 1 – The diameter.\nThe segment MN belongs to all the mid-points of the family, whence the line d := MN is the diameter associated with direction u.  Consequently d ∥ A.\n\nStep 2 – The unique point where the diameter meets the parabola.\nBecause a parabola meets every line parallel to its axis in at most one point, d intersects Γ in exactly one point; call it P.\n\nStep 3 – The chord through P having direction u and its rôle.\nThrough P draw the line t parallel to q (and q′).  Two situations occur.\n   (i)  If u is not perpendicular to the axis A, t is tangent to Γ at P.\n   (ii) If u is perpendicular to A (the vertical-chord family), then P is the vertex V of Γ.  The diameter of this family is the axis itself (d = A), whereas the tangent at V is the line t parallel to the chords; hence t ⟂ d and t ≠ d.  (A full analytic verification appears after Step 6.)\n\nStep 4 – Producing a line through the focus.\nLet ρ be the reflection of the whole line d in the line t.  We shall prove that ρ contains the focus F.\n\nStep 5 – A correct “optical” lemma.\nLemma 1.  Let Γ be the parabola y² = 2 p x (p > 0) with focus F = (p/2, 0).\nLet P = (x₀, y₀) ∈ Γ (so y₀ ≠ 0), let m be the slope of the tangent t at P (hence m = p/y₀), and let d be the diameter of the family of chords parallel to t.  Then for every point X = (x, y₀) on d the reflection X′ of X in t is collinear with P and F.\n\nProof.\nThe tangent at P has equation\n      y − y₀ = m(x − x₀)                              (1)\nwith standard form  m x − y + b = 0,  where b = y₀ − m x₀.\nWrite a = m, b₁ = −1, c = b.  For an arbitrary point X = (x, y₀) on d the quantity\n      S := a x + b₁ y + c = m(x − x₀).                (2)\nWith the classical reflection formula\n      x′ = x − 2 a S /(a² + b₁²),                      \n      y′ = y − 2 b₁ S /(a² + b₁²),                    (3)\nwe find, because a² + b₁² = m² + 1,\n      x′ = x − 2 m²(x − x₀)/(m² + 1),\n      y′ = y₀ + 2 m (x − x₀)/(m² + 1).               (4)\nHence\n      slope(PX′) = (y′ − y₀)/(x′ − x₀)\n                 = 2 m/(1 − m²).                     (5)\nOn the other hand, because x₀ = y₀²/(2 p) and p = m y₀, the slope of PF is\n      slope(PF) = (0 − y₀)/(p/2 − x₀)\n                 = −2 y₀ /(p − 2 x₀)\n                 = 2 m/(1 − m²),                     (6)\nwhich is identical with (5).  Therefore P, X′, F are collinear, proving the lemma. ∎\n\nStep 6 – Why ρ passes through the focus.\nBecause every point of d has its reflection on PF, the image of the whole line d is precisely the line PF, i.e. ρ = PF.  In particular ρ contains the focus F.\n\nRemark on case (ii).\nWhen u ⟂ A we have d = A and t ⟂ d.  Reflecting d in t leaves d invariant, so ρ = d again, and therefore ρ contains the focus in this exceptional case as well.\n\nConclusion of Part I.\nFrom two parallel chords of Γ one can construct, with ruler and compasses, a line through the focus F – namely the line ρ obtained in Step 4.\n\nAnalytic verification of Steps 1–3.\nChoose coordinates as above so that Γ has equation y² = 2 p x.  If the common direction u has finite slope m, an arbitrary line of slope m is y = m x + b.  Intersecting with Γ gives\n      m² x² + 2 m b x + b² − 2 p x = 0.\nIts roots have sum 2(p − m b)/m²; hence the mid-points of the two intersection points lie on y = p/m – a horizontal line, which is therefore the diameter d of the family.  The unique point P = (p/(2 m²), p/m) of d ∩ Γ satisfies dy/dx|_P = p/y₀ = m, so the line through P parallel to the family is indeed tangent – this is case (i).\n\nIf u is perpendicular to A, the supporting lines are vertical x = c (formally m = ∞).  Solving x = c with y² = 2 p x gives intersection points (c, ±√(2 p c)); their mid-point is (c, 0), so the diameter is the x-axis (d = A).  The unique point of A ∩ Γ is (0,0), the vertex V, and the tangent there is the vertical line x = 0 – the direction of the family, i.e. case (ii).\n\nPart II.  Carrying out the two prescribed instances\n————————————————————————————————\n1.  “90° pair”.  Apply Part I to the chords k and k′.  Call the resulting line r.\n\n2.  “60° pair”.  Apply Part I to the chords m and m′.  Call the resulting line s.\n\nPart III.  The lines r and s are different\n—————————————————————————————\nLet u be the direction of k and k′ and v the direction of m and m′.  By construction u is perpendicular to ℓ, whereas v makes an angle of 60° with ℓ, so the angle between u and v is φ = 30°.\n\nEvery diameter is parallel to the axis; denote this common direction by a.  In the regular case (u not perpendicular to A) each of r and s is obtained by reflecting the direction a in two distinct tangents t_u and t_v whose directions are u and v.  Let α_u and α_v be the oriented angles between a and t_u, t_v.  Then α_v − α_u = φ, and reflection doubles the angle with the mirror, so the angle between r and s equals 2(α_v − α_u) = 2 φ = 60°.  Hence r ≠ s.\n\nExceptional case (u ⟂ A).  Then r = A by Part I, whereas v forms an angle of 30° with u, so v is not parallel to A; consequently s cannot coincide with A, and again r ≠ s.\n\nPart IV.  Their intersection is the focus\n————————————————————————\nEach of r and s contains F, and Part III shows that they are distinct lines.  Hence their unique intersection point is F.\n\nThis completes the required construction of the focus from the data stated in the problem.\n\nRemarks\n———\n1.  The slight sliding of the initial chord in Parts A and B is always possible because for every direction that is not parallel to the axis a parabola is unbounded in both directions perpendicular to that line.\n2.  Once the focus has been located, the directrix is obtained as the locus of points equidistant from F and Γ, giving the classical elements (vertex, axis, focus, directrix) of the parabola.",
      "_meta": {
        "core_steps": [
          "Midpoints of any family of parallel chords lie on a diameter parallel to the axis of the parabola.",
          "That diameter meets the parabola once; the chord of the family through this point is the tangent there.",
          "Reflecting the diameter across that tangent yields a line that necessarily passes through the focus (focal property).",
          "Construct such a ‘reflected’ line from one chosen pair of parallel chords by locating their mid-points, the diameter, the tangent, and its reflection.",
          "Repeat with a second, differently oriented chord family; the intersection of the two reflected lines is the focus."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Orientation of the first family of parallel chords (any direction that produces non-degenerate construction).",
            "original": "unspecified; simply “draw any chord … and another parallel to it.”"
          },
          "slot2": {
            "description": "Separation between the two parallel chords in each family.",
            "original": "unspecified distance between the first chord and the second, parallel chord."
          },
          "slot3": {
            "description": "Choice of the second chord family used for the repeat step (must differ in direction from the first but is otherwise arbitrary).",
            "original": "advice to pick the second starting chords so that they are neither parallel nor perpendicular to the first family."
          },
          "slot4": {
            "description": "Point-label nomenclature and the particular way the reflection is carried out (e.g., using equal angles, perpendicular bisectors, or another equivalent ruler-and-compass reflection).",
            "original": "labels A, B, P, Q, C and the equal-angle construction “∠APQ = ∠QPC”."
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}