{
  "index": "1967-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-2. Let \\( 0 \\leqq p \\leqq 1 \\) and \\( 0 \\leqq r \\leqq 1 \\) and consider the identities\n(a) \\( (p x+(1-p) y)^{2}=A x^{2}+B x y+C y^{2} \\),\n(b) \\( (p x+(1-p) y)(r x+(1-r) y)=\\alpha x^{2}+\\beta x y+\\gamma y^{2} \\).\n\nShow that (with respect to \\( p \\) and \\( r \\) )\n(a) \\( \\max \\{A, B, C\\} \\geqq 4 / 9 \\),\n(b) \\( \\max \\{\\alpha, \\beta, \\gamma\\} \\geqq 4 / 9 \\).",
  "solution": "B-2 For part (a) one has immediately that \\( A=p^{2}, B=2 p(1-p) \\), and \\( C=(1-p)^{2} \\). The result follows by examination of the graphs for \\( A, B \\) and \\( C \\) on \\( 0 \\leqq p \\leqq 1 \\).\n\nFor part (b), \\( \\alpha=p r, \\beta=p(1-r)+r(1-p), \\gamma=(1-p)(1-r) \\). Consider the region \\( R \\) in the \\( p, r \\)-plane defined by \\( 0 \\leqq p \\leqq 1 \\) and \\( 0 \\leqq r \\leqq 1 \\). We will show that there is no point in \\( R \\) with \\( \\alpha<4 / 9, \\beta<4 / 9 \\) and \\( \\gamma<4 / 9 \\). If \\( \\alpha<4 / 9 \\) and \\( \\gamma<4 / 9 \\) then \\( (p, r) \\) is between the hyperbolas \\( p r=4 / 9 \\) and \\( (1-p)(1-r)=4 / 9 \\). These hyperbolas have vertices in \\( R \\) at \\( (2 / 3,2 / 3) \\) and \\( (1 / 3,1 / 3) \\), respectively. The symmetry about ( \\( \\frac{1}{2}, \\frac{1}{2} \\) ) suggests setting \\( p^{\\prime}=p-\\frac{1}{2} \\) and \\( r^{\\prime}=r-\\frac{1}{2} \\). Then \\( \\beta=\\frac{1}{2} \\) \\( -2 p^{\\prime} r^{\\prime} \\) and thus \\( \\beta<4 / 9 \\) if and only if \\( p^{\\prime} r^{\\prime}>1 / 36 \\). Note that the vertices for the hyperbola \\( p^{\\prime} r^{\\prime}=1 / 36 \\) are at \\( (p, r)=(1 / 3,1 / 3) \\) and \\( (2 / 3,2 / 3) \\). By looking at asymptotes, we see graphically that the region \\( \\beta<4 / 9 \\) does not overlap the region in \\( R \\) where \\( \\alpha<4 / 9 \\) and \\( \\gamma<4 / 9 \\).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "p",
    "r",
    "A",
    "B",
    "C",
    "R",
    "\\\\alpha",
    "\\\\beta",
    "\\\\gamma"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xvariable",
        "y": "yvariable",
        "p": "probpar",
        "r": "paramr",
        "A": "coeffa",
        "B": "coeffb",
        "C": "coeffc",
        "R": "regionr",
        "\\alpha": "alphaco",
        "\\beta": "betaco",
        "\\gamma": "gammaco"
      },
      "question": "B-2. Let \\( 0 \\leqq probpar \\leqq 1 \\) and \\( 0 \\leqq paramr \\leqq 1 \\) and consider the identities\n(a) \\( (probpar xvariable+(1-probpar) yvariable)^{2}=coeffa xvariable^{2}+coeffb xvariable yvariable+coeffc yvariable^{2} \\),\n(b) \\( (probpar xvariable+(1-probpar) yvariable)(paramr xvariable+(1-paramr) yvariable)=alphaco xvariable^{2}+betaco xvariable yvariable+gammaco yvariable^{2} \\).\n\nShow that (with respect to \\( probpar \\) and \\( paramr \\) )\n(a) \\( \\max \\{coeffa, coeffb, coeffc\\} \\geqq 4 / 9 \\),\n(b) \\( \\max \\{alphaco, betaco, gammaco\\} \\geqq 4 / 9 \\).",
      "solution": "B-2 For part (a) one has immediately that \\( coeffa=probpar^{2}, coeffb=2 probpar(1-probpar) \\), and \\( coeffc=(1-probpar)^{2} \\). The result follows by examination of the graphs for \\( coeffa, coeffb \\) and \\( coeffc \\) on \\( 0 \\leqq probpar \\leqq 1 \\).\n\nFor part (b), \\( alphaco=probpar paramr, betaco=probpar(1-paramr)+paramr(1-probpar), gammaco=(1-probpar)(1-paramr) \\). Consider the region \\( regionr \\) in the \\( probpar, paramr \\)-plane defined by \\( 0 \\leqq probpar \\leqq 1 \\) and \\( 0 \\leqq paramr \\leqq 1 \\). We will show that there is no point in \\( regionr \\) with \\( alphaco<4 / 9, betaco<4 / 9 \\) and \\( gammaco<4 / 9 \\). If \\( alphaco<4 / 9 \\) and \\( gammaco<4 / 9 \\) then \\( (probpar, paramr) \\) is between the hyperbolas \\( probpar paramr=4 / 9 \\) and \\( (1-probpar)(1-paramr)=4 / 9 \\). These hyperbolas have vertices in \\( regionr \\) at \\( (2 / 3,2 / 3) \\) and \\( (1 / 3,1 / 3) \\), respectively. The symmetry about ( \\( \\frac{1}{2}, \\frac{1}{2} \\) ) suggests setting \\( probpar^{\\prime}=probpar-\\frac{1}{2} \\) and \\( paramr^{\\prime}=paramr-\\frac{1}{2} \\). Then \\( betaco=\\frac{1}{2} \\) \\( -2 probpar^{\\prime} paramr^{\\prime} \\) and thus \\( betaco<4 / 9 \\) if and only if \\( probpar^{\\prime} paramr^{\\prime}>1 / 36 \\). Note that the vertices for the hyperbola \\( probpar^{\\prime} paramr^{\\prime}=1 / 36 \\) are at \\( (probpar, paramr)=(1 / 3,1 / 3) \\) and \\( (2 / 3,2 / 3) \\). By looking at asymptotes, we see graphically that the region \\( betaco<4 / 9 \\) does not overlap the region in \\( regionr \\) where \\( alphaco<4 / 9 \\) and \\( gammaco<4 / 9 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lighthouse",
        "y": "pinecone",
        "p": "marshland",
        "r": "bookshelf",
        "A": "windstorm",
        "B": "dreamship",
        "C": "stargazer",
        "R": "moonlight",
        "\\alpha": "snowflake",
        "\\beta": "riverbank",
        "\\gamma": "buttercup"
      },
      "question": "B-2. Let \\( 0 \\leqq marshland \\leqq 1 \\) and \\( 0 \\leqq bookshelf \\leqq 1 \\) and consider the identities\n(a) \\( (marshland lighthouse+(1-marshland) pinecone)^{2}=windstorm lighthouse^{2}+dreamship lighthouse pinecone+stargazer pinecone^{2} \\),\n(b) \\( (marshland lighthouse+(1-marshland) pinecone)(bookshelf lighthouse+(1-bookshelf) pinecone)=snowflake lighthouse^{2}+riverbank lighthouse pinecone+buttercup pinecone^{2} \\).\n\nShow that (with respect to \\( marshland \\) and \\( bookshelf \\) )\n(a) \\( \\max \\{windstorm, dreamship, stargazer\\} \\geqq 4 / 9 \\),\n(b) \\( \\max \\{snowflake, riverbank, buttercup\\} \\geqq 4 / 9 \\).",
      "solution": "B-2 For part (a) one has immediately that \\( windstorm=marshland^{2}, dreamship=2 marshland(1-marshland) \\), and \\( stargazer=(1-marshland)^{2} \\). The result follows by examination of the graphs for \\( windstorm, dreamship \\) and \\( stargazer \\) on \\( 0 \\leqq marshland \\leqq 1 \\).\n\nFor part (b), \\( snowflake=marshland bookshelf, riverbank=marshland(1-bookshelf)+bookshelf(1-marshland), buttercup=(1-marshland)(1-bookshelf) \\). Consider the region \\( moonlight \\) in the \\( marshland, bookshelf \\)-plane defined by \\( 0 \\leqq marshland \\leqq 1 \\) and \\( 0 \\leqq bookshelf \\leqq 1 \\). We will show that there is no point in \\( moonlight \\) with \\( snowflake<4 / 9, riverbank<4 / 9 \\) and \\( buttercup<4 / 9 \\). If \\( snowflake<4 / 9 \\) and \\( buttercup<4 / 9 \\) then \\( (marshland, bookshelf) \\) is between the hyperbolas \\( marshland bookshelf=4 / 9 \\) and \\( (1-marshland)(1-bookshelf)=4 / 9 \\). These hyperbolas have vertices in \\( moonlight \\) at \\( (2 / 3,2 / 3) \\) and \\( (1 / 3,1 / 3) \\), respectively. The symmetry about ( \\( \\frac{1}{2}, \\frac{1}{2} \\) ) suggests setting \\( marshland^{\\prime}=marshland-\\frac{1}{2} \\) and \\( bookshelf^{\\prime}=bookshelf-\\frac{1}{2} \\). Then \\( riverbank=\\frac{1}{2}  -2 marshland^{\\prime} bookshelf^{\\prime} \\) and thus \\( riverbank<4 / 9 \\) if and only if \\( marshland^{\\prime} bookshelf^{\\prime}>1 / 36 \\). Note that the vertices for the hyperbola \\( marshland^{\\prime} bookshelf^{\\prime}=1 / 36 \\) are at \\( (marshland, bookshelf)=(1 / 3,1 / 3) \\) and \\( (2 / 3,2 / 3) \\). By looking at asymptotes, we see graphically that the region \\( riverbank<4 / 9 \\) does not overlap the region in \\( moonlight \\) where \\( snowflake<4 / 9 \\) and \\( buttercup<4 / 9 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedquantity",
        "y": "unvaryingterm",
        "p": "certainty",
        "r": "wholeness",
        "A": "voidness",
        "B": "emptiness",
        "C": "nullstate",
        "R": "pointmass",
        "\\alpha": "endingvalue",
        "\\beta": "centralvalue",
        "\\gamma": "startingvalue"
      },
      "question": "B-2. Let \\( 0 \\leqq certainty \\leqq 1 \\) and \\( 0 \\leqq wholeness \\leqq 1 \\) and consider the identities\n(a) \\( (certainty fixedquantity+(1-certainty) unvaryingterm)^{2}=voidness fixedquantity^{2}+emptiness fixedquantity unvaryingterm+nullstate unvaryingterm^{2} \\),\n(b) \\( (certainty fixedquantity+(1-certainty) unvaryingterm)(wholeness fixedquantity+(1-wholeness) unvaryingterm)=endingvalue fixedquantity^{2}+centralvalue fixedquantity unvaryingterm+startingvalue unvaryingterm^{2} \\).\n\nShow that (with respect to \\( certainty \\) and \\( wholeness \\) )\n(a) \\( \\max \\{voidness, emptiness, nullstate\\} \\geqq 4 / 9 \\),\n(b) \\( \\max \\{endingvalue, centralvalue, startingvalue\\} \\geqq 4 / 9 \\).",
      "solution": "B-2 For part (a) one has immediately that \\( voidness=certainty^{2}, emptiness=2 certainty(1-certainty) \\), and \\( nullstate=(1-certainty)^{2} \\). The result follows by examination of the graphs for \\( voidness, emptiness \\) and \\( nullstate \\) on \\( 0 \\leqq certainty \\leqq 1 \\).\n\nFor part (b), \\( endingvalue=certainty\\,wholeness,\\; centralvalue=certainty(1-wholeness)+wholeness(1-certainty),\\; startingvalue=(1-certainty)(1-wholeness) \\). Consider the region \\( pointmass \\) in the \\( certainty, wholeness \\)-plane defined by \\( 0 \\leqq certainty \\leqq 1 \\) and \\( 0 \\leqq wholeness \\leqq 1 \\). We will show that there is no point in \\( pointmass \\) with \\( endingvalue<4 / 9,\\; centralvalue<4 / 9 \\) and \\( startingvalue<4 / 9 \\). If \\( endingvalue<4 / 9 \\) and \\( startingvalue<4 / 9 \\) then \\( (certainty, wholeness) \\) is between the hyperbolas \\( certainty\\,wholeness=4 / 9 \\) and \\( (1-certainty)(1-wholeness)=4 / 9 \\). These hyperbolas have vertices in \\( pointmass \\) at \\( (2 / 3,2 / 3) \\) and \\( (1 / 3,1 / 3) \\), respectively. The symmetry about \\( \\left( \\frac{1}{2}, \\frac{1}{2} \\right) \\) suggests setting \\( certainty^{\\prime}=certainty-\\frac{1}{2} \\) and \\( wholeness^{\\prime}=wholeness-\\frac{1}{2} \\). Then \\( centralvalue=\\frac{1}{2}-2 certainty^{\\prime} wholeness^{\\prime} \\) and thus \\( centralvalue<4 / 9 \\) if and only if \\( certainty^{\\prime} wholeness^{\\prime}>1 / 36 \\). Note that the vertices for the hyperbola \\( certainty^{\\prime} wholeness^{\\prime}=1 / 36 \\) are at \\( (certainty, wholeness)=(1 / 3,1 / 3) \\) and \\( (2 / 3,2 / 3) \\). By looking at asymptotes, we see graphically that the region \\( centralvalue<4 / 9 \\) does not overlap the region in \\( pointmass \\) where \\( endingvalue<4 / 9 \\) and \\( startingvalue<4 / 9 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "lkjhgfds",
        "y": "poiuytre",
        "p": "qazwsxed",
        "r": "plmoknij",
        "A": "nbvcxzas",
        "B": "mksajdhe",
        "C": "hgfdsrty",
        "R": "zxcvbnml",
        "\\alpha": "qzxwvtnp",
        "\\beta": "hjgrksla",
        "\\gamma": "mnbvcxqr"
      },
      "question": "B-2. Let \\( 0 \\leqq qazwsxed \\leqq 1 \\) and \\( 0 \\leqq plmoknij \\leqq 1 \\) and consider the identities\n(a) \\( (qazwsxed lkjhgfds+(1-qazwsxed) poiuytre)^{2}=nbvcxzas lkjhgfds^{2}+mksajdhe lkjhgfds poiuytre+hgfdsrty poiuytre^{2} \\),\n(b) \\( (qazwsxed lkjhgfds+(1-qazwsxed) poiuytre)(plmoknij lkjhgfds+(1-plmoknij) poiuytre)=qzxwvtnp lkjhgfds^{2}+hjgrksla lkjhgfds poiuytre+mnbvcxqr poiuytre^{2} \\).\n\nShow that (with respect to \\( qazwsxed \\) and \\( plmoknij \\) )\n(a) \\( \\max \\{nbvcxzas, mksajdhe, hgfdsrty\\} \\geqq 4 / 9 \\),\n(b) \\( \\max \\{qzxwvtnp, hjgrksla, mnbvcxqr\\} \\geqq 4 / 9 \\).",
      "solution": "B-2 For part (a) one has immediately that \\( nbvcxzas=qazwsxed^{2}, mksajdhe=2 qazwsxed(1-qazwsxed) \\), and \\( hgfdsrty=(1-qazwsxed)^{2} \\). The result follows by examination of the graphs for \\( nbvcxzas, mksajdhe \\) and \\( hgfdsrty \\) on \\( 0 \\leqq qazwsxed \\leqq 1 \\).\n\nFor part (b), \\( qzxwvtnp=qazwsxed plmoknij, hjgrksla=qazwsxed(1-plmoknij)+plmoknij(1-qazwsxed), mnbvcxqr=(1-qazwsxed)(1-plmoknij) \\). Consider the region \\( zxcvbnml \\) in the \\( qazwsxed, plmoknij \\)-plane defined by \\( 0 \\leqq qazwsxed \\leqq 1 \\) and \\( 0 \\leqq plmoknij \\leqq 1 \\). We will show that there is no point in \\( zxcvbnml \\) with \\( qzxwvtnp<4 / 9, hjgrksla<4 / 9 \\) and \\( mnbvcxqr<4 / 9 \\). If \\( qzxwvtnp<4 / 9 \\) and \\( mnbvcxqr<4 / 9 \\) then \\( (qazwsxed, plmoknij) \\) is between the hyperbolas \\( qazwsxed plmoknij=4 / 9 \\) and \\( (1-qazwsxed)(1-plmoknij)=4 / 9 \\). These hyperbolas have vertices in \\( zxcvbnml \\) at \\( (2 / 3,2 / 3) \\) and \\( (1 / 3,1 / 3) \\), respectively. The symmetry about ( \\( \\frac{1}{2}, \\frac{1}{2} \\) ) suggests setting \\( qazwsxed^{\\prime}=qazwsxed-\\frac{1}{2} \\) and \\( plmoknij^{\\prime}=plmoknij-\\frac{1}{2} \\). Then \\( hjgrksla=\\frac{1}{2} -2 qazwsxed^{\\prime} plmoknij^{\\prime} \\) and thus \\( hjgrksla<4 / 9 \\) if and only if \\( qazwsxed^{\\prime} plmoknij^{\\prime}>1 / 36 \\). Note that the vertices for the hyperbola \\( qazwsxed^{\\prime} plmoknij^{\\prime}=1 / 36 \\) are at \\( (qazwsxed, plmoknij)=(1 / 3,1 / 3) \\) and \\( (2 / 3,2 / 3) \\). By looking at asymptotes, we see graphically that the region \\( hjgrksla<4 / 9 \\) does not overlap the region in \\( zxcvbnml \\) where \\( qzxwvtnp<4 / 9 \\) and \\( mnbvcxqr<4 / 9 \\)."
    },
    "kernel_variant": {
      "question": "Let \\(\\tfrac14\\le p\\le \\tfrac34\\) and \\(\\tfrac14\\le r\\le \\tfrac34\\).  Define\n\\[(p x+(1-p) y)^2=A x^{2}+B x y+C y^{2},\\]\n\\[(p x+(1-p) y)(r x+(1-r) y)=\\alpha x^{2}+\\beta x y+\\gamma y^{2} .\\]\nProve that\n(a) \\(\\max\\{A,B,C\\}\\ge\\dfrac25,\\qquad\\)\n(b) \\(\\max\\{\\alpha,\\beta,\\gamma\\}\\ge\\dfrac25.\\)",
      "solution": "Solution.  We set\nA=p^2,\nB=2p(1−p),\nC=(1−p)^2\nand\nα=pr,\nβ=p(1−r)+r(1−p)=p+r−2pr,\nγ=(1−p)(1−r),\nwith p,r in [¼,¾].\n\n(a)  Let f(p)=max{A,B,C}.  We check f at the critical points p=1/4,1/3,1/2,2/3,3/4 where two of A,B,C coincide or at the endpoints:\n p=1/4: (A,B,C)=(1/16,3/8,9/16) so f=9/16=0.5625;\n p=1/3: (A,B,C)=(1/9,4/9,4/9) so f=4/9≈0.4444;\n p=1/2: (A,B,C)=(1/4,1/2,1/4) so f=1/2;\n p=2/3: (A,B,C)=(4/9,4/9,1/9) so f=4/9;\n p=3/4: (A,B,C)=(9/16,3/8,1/16) so f=9/16.\nBy continuity each local minimum of f on [¼,¾] occurs where two of A,B,C coincide, and those values are as above.  The smallest of these maxima is 4/9, which exceeds 2/5=0.4.  Hence ∀p∈[¼,¾], max{A,B,C}≥4/9>2/5.\n\n(b)  Suppose for contradiction that α<2/5, β<2/5, γ<2/5.  Then\n pr<2/5,\n (1−p)(1−r)<2/5.\nSet p=½+u, r=½+v with u,v∈[−1/4,1/4].  Then\n β=½−2uv,\n so β<2/5 ⇒ ½−2uv<2/5 ⇒ uv>1/20.\nThus u,v have the same sign and |u|,|v|>√(1/20)≈0.2236.  If u,v>0 then\n pr=(½+u)(½+v)=¼+(u+v)/2+uv\n    ≥¼+√{uv}+uv     (since u+v≥2√{uv})\n    >¼+1/√20+1/20\n    ≈0.5236>0.4=2/5,\ncontradicting pr<2/5.  If u,v<0 then set u′=−u,v′=−v>0 with u′v′>1/20 and similarly\n (1−p)(1−r)=(½−u)(½−v)=¼+(−u−v)/2+uv\n    =¼+(u′+v′)/2+u′v′\n    >¼+1/√20+1/20\n    ≈0.5236>2/5,\ncontradicting (1−p)(1−r)<2/5.  Hence not all three α,β,γ can be <2/5, so max{α,β,γ}≥2/5.\n\nCombining (a) and (b) completes the proof.",
      "_meta": {
        "core_steps": [
          "Algebraic expansion to write A,B,C and α,β,γ as explicit functions of p,r",
          "Part (a): maximize p², 2p(1−p), (1−p)² on [0,1] to see one ≥ 4⁄9",
          "Part (b): assume all three < 4⁄9, giving pr<4⁄9 and (1−p)(1−r)<4⁄9",
          "Shift to p' = p−½, r' = r−½ so β = ½−2p'r'; β<4⁄9 ⇒ p'r' > 1⁄36",
          "Compare non-overlapping hyperbolic regions to get a contradiction; hence max≥4⁄9"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Chosen lower-bound constant for the coefficients; any value below the true maximum (½) works with the same hyperbola argument",
            "original": "4/9"
          },
          "slot2": {
            "description": "Parameter range for p and r; any symmetric closed interval [a,b] would keep the expansion and ‘midpoint-shift’ argument intact after rescaling",
            "original": "[0,1]"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}