{
  "index": "1998-B-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "Given a point $(a,b)$ with $0<b<a$, determine the minimum perimeter of a\ntriangle with one vertex at $(a,b)$, one on the $x$-axis, and one on the\nline $y=x$.  You may assume that a triangle of minimum perimeter exists.",
  "solution": "Consider a triangle as described by the problem; label its\nvertices $A,B,C$ so that $A = (a,b)$, $B$ lies on the $x$-axis,\nand $C$ lies on the line $y=x$.  Further let $D = (a,-b)$ be the\nreflection of $A$ in the $x$-axis, and let $E = (b,a)$ be\nthe reflection of $A$ in the line $y=x$.  Then $AB=DB$\nand $AC=CE$, and so the perimeter of $ABC$ is\n$DB+BC+CE \\geq DE = \\sqrt{(a-b)^2 + (a+b)^2}\n= \\sqrt{2a^2+2b^2}$.  It is clear that this lower bound can\nbe achieved; just set $B$ (resp. $C$) to be the\nintersection between the segment $DE$ and the $x$-axis (resp.\nline $x=y$); thus the minimum perimeter is in fact\n$\\sqrt{2a^2+2b^2}$.",
  "vars": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "x",
    "y"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "originalvertex",
        "B": "basevertex",
        "C": "diagonalvertex",
        "D": "groundreflection",
        "E": "diagonalreflection",
        "x": "horizaxis",
        "y": "vertaxis",
        "a": "firstcoord",
        "b": "secondcoord"
      },
      "question": "Given a point $(firstcoord,secondcoord)$ with $0<secondcoord<firstcoord$, determine the minimum perimeter of a triangle with one vertex at $(firstcoord,secondcoord)$, one on the $horizaxis$-axis, and one on the line $vertaxis=horizaxis$.  You may assume that a triangle of minimum perimeter exists.",
      "solution": "Consider a triangle as described by the problem; label its\nvertices $originalvertex,basevertex,diagonalvertex$ so that $originalvertex = (firstcoord,secondcoord)$, $basevertex$ lies on the $horizaxis$-axis,\nand $diagonalvertex$ lies on the line $vertaxis=horizaxis$.  Further let $groundreflection = (firstcoord,-secondcoord)$ be the\nreflection of $originalvertex$ in the $horizaxis$-axis, and let $diagonalreflection = (secondcoord,firstcoord)$ be\nthe reflection of $originalvertex$ in the line $vertaxis=horizaxis$.  Then $originalvertex basevertex=groundreflection basevertex$\nand $originalvertex diagonalvertex=diagonalreflection diagonalvertex$, and so the perimeter of $originalvertex basevertex diagonalvertex$ is\n$groundreflection basevertex+basevertex diagonalvertex+diagonalreflection diagonalvertex \\geq diagonalreflection groundreflection = \\sqrt{(firstcoord-secondcoord)^2 + (firstcoord+secondcoord)^2}\n= \\sqrt{2firstcoord^2+2secondcoord^2}$.  It is clear that this lower bound can\nbe achieved; just set $basevertex$ (resp. $diagonalvertex$) to be the\nintersection between the segment $diagonalreflection groundreflection$ and the $horizaxis$-axis (resp.\nline $horizaxis=vertaxis$); thus the minimum perimeter is in fact\n$\\sqrt{2firstcoord^2+2secondcoord^2}$."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "tangerine",
        "B": "pumpernickel",
        "C": "dandelion",
        "D": "saxophone",
        "E": "marzipans",
        "x": "longitude",
        "y": "latitude",
        "a": "buttercup",
        "b": "watermelon"
      },
      "question": "Given a point $(buttercup,watermelon)$ with $0<watermelon<buttercup$, determine the minimum perimeter of a\ntriangle with one vertex at $(buttercup,watermelon)$, one on the $longitude$-axis, and one on the\nline $latitude=longitude$.",
      "solution": "Consider a triangle as described by the problem; label its\nvertices $tangerine,pumpernickel,dandelion$ so that $tangerine = (buttercup,watermelon)$, $pumpernickel$ lies on the $longitude$-axis,\nand $dandelion$ lies on the line $latitude=longitude$.  Further let $saxophone = (buttercup,-watermelon)$ be the\nreflection of $tangerine$ in the $longitude$-axis, and let $marzipans = (watermelon,buttercup)$ be\nthe reflection of $tangerine$ in the line $latitude=longitude$.  Then $tangerine pumpernickel = saxophone pumpernickel$\nand $tangerine dandelion = dandelion marzipans$, and so the perimeter of $tangerine pumpernickel dandelion$ is\n$saxophone pumpernickel + pumpernickel dandelion + dandelion marzipans \\geq saxophone marzipans = \\sqrt{(buttercup-watermelon)^2 + (buttercup+watermelon)^2}\n= \\sqrt{2buttercup^2+2watermelon^2}$.  It is clear that this lower bound can\nbe achieved; just set $pumpernickel$ (resp. $dandelion$) to be the\nintersection between the segment $saxophone marzipans$ and the $longitude$-axis (resp.\nline $longitude=latitude$); thus the minimum perimeter is in fact\n$\\sqrt{2buttercup^2+2watermelon^2}$."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "hollowspace",
        "B": "flatfield",
        "C": "deepvalley",
        "D": "troughline",
        "E": "plainside",
        "x": "ordinate",
        "y": "abscissa",
        "a": "diminutive",
        "b": "magnitude"
      },
      "question": "Given a point $(diminutive,magnitude)$ with $0<magnitude<diminutive$, determine the minimum perimeter of a\ntriangle with one vertex at $(diminutive,magnitude)$, one on the $ordinate$-axis, and one on the\nline $abscissa=ordinate$.  You may assume that a triangle of minimum perimeter exists.",
      "solution": "Consider a triangle as described by the problem; label its\nvertices $hollowspace,flatfield,deepvalley$ so that $hollowspace = (diminutive,magnitude)$, $flatfield$ lies on the $ordinate$-axis,\nand $deepvalley$ lies on the line $abscissa=ordinate$.  Further let $troughline = (diminutive,-magnitude)$ be the\nreflection of $hollowspace$ in the $ordinate$-axis, and let $plainside = (magnitude,diminutive)$ be\nthe reflection of $hollowspace$ in the line $abscissa=ordinate$.  Then $hollowspaceflatfield=troughlineflatfield$\nand $hollowspacedeepvalley=plainsidedeepvalley$, and so the perimeter of $hollowspaceflatfielddeepvalley$ is\n$troughlineflatfield+flatfielddeepvalley+plainsidedeepvalley \\geq troughlineplainside = \\sqrt{(diminutive-magnitude)^2 + (diminutive+magnitude)^2}\n= \\sqrt{2diminutive^2+2magnitude^2}$.  It is clear that this lower bound can\nbe achieved; just set $flatfield$ (resp. $deepvalley$) to be the\nintersection between the segment $troughlineplainside$ and the $ordinate$-axis (resp.\nline $ordinate=abscissa$); thus the minimum perimeter is in fact\n$\\sqrt{2diminutive^2+2magnitude^2}$.",
      "confidence": 0.12
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "mcfqzsoe",
        "D": "knivryup",
        "E": "zbwotkda",
        "x": "ferjamoc",
        "y": "lwhzusib",
        "a": "vtkdqyla",
        "b": "egnorbma"
      },
      "question": "Given a point $(vtkdqyla,egnorbma)$ with $0<egnorbma<vtkdqyla$, determine the minimum perimeter of a\ntriangle with one vertex at $(vtkdqyla,egnorbma)$, one on the $ferjamoc$-axis, and one on the\nline $lwhzusib=ferjamoc$.  You may assume that a triangle of minimum perimeter exists.",
      "solution": "Consider a triangle as described by the problem; label its\nvertices $qzxwvtnp,hjgrksla,mcfqzsoe$ so that $qzxwvtnp = (vtkdqyla,egnorbma)$, $hjgrksla$ lies on the $ferjamoc$-axis,\nand $mcfqzsoe$ lies on the line $lwhzusib=ferjamoc$.  Further let $knivryup = (vtkdqyla,-egnorbma)$ be the\nreflection of $qzxwvtnp$ in the $ferjamoc$-axis, and let $zbwotkda = (egnorbma,vtkdqyla)$ be\nthe reflection of $qzxwvtnp$ in the line $lwhzusib=ferjamoc$.  Then $qzxwvtnphjgrksla = knivryuphjgrksla$\nand $qzxwvtnpmcfqzsoe = mcfqzsoezbwotkda$, and so the perimeter of $qzxwvtnphjgrksla mcfqzsoe$ is\n$knivryuphjgrksla + hjgrksla mcfqzsoe + mcfqzsoezbwotkda \\geq knivryupzbwotkda = \\sqrt{(vtkdqyla-egnorbma)^2 + (vtkdqyla+egnorbma)^2}\n= \\sqrt{2vtkdqyla^2+2egnorbma^2}$.  It is clear that this lower bound can\nbe achieved; just set $hjgrksla$ (resp. $mcfqzsoe$) to be the\nintersection between the segment $knivryupzbwotkda$ and the $ferjamoc$-axis (resp.\nline $ferjamoc=lwhzusib$); thus the minimum perimeter is in fact\n$\\sqrt{2vtkdqyla^2+2egnorbma^2}$.}",
      "knivryupzbwotkda": ""
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nA=(a,b,c),\\qquad 0<c<b<a ,\\qquad \n\\frac12\\,a<b<\\frac34\\,a ,\\qquad \n28\\,b+120\\,c>21\\,a ,\n\\]\nbe fixed.  \n\nConsider the mirror–planes  \n\\[\n\\Pi_{1}\\colon x=0,\\qquad\n\\Pi_{2}\\colon y=\\tfrac12\\,x,\\qquad\n\\Pi_{3}\\colon z=-\\tfrac34\\,x .\n\\]\n\nA broken line  \n\\[\n\\Gamma\\;:\\;A\\longrightarrow B\\longrightarrow C\\longrightarrow D\\longrightarrow A\n\\]\nconsisting of four straight edges is called \\emph{admissible} if  \n\nS1) \\(B\\in\\Pi_{1}\\) and every interior point of the edge \\(AB\\) satisfies \\(x>0\\);\n\nS2) \\(C\\in\\Pi_{2}\\) and every interior point of \\(BC\\) satisfies \\(y<\\tfrac12\\,x\\);\n\nS3) \\(D\\in\\Pi_{3}\\) and every interior point of \\(CD\\) satisfies \\(y>\\tfrac12\\,x\\) and \\(z>-\\tfrac34\\,x\\);\n\nS4) every interior point of the closing edge \\(DA\\) satisfies \\(y>\\tfrac12\\,x\\) and \\(z>-\\tfrac34\\,x\\).\n\nFor an admissible triple \\((B,C,D)\\) put  \n\\[\nP(B,C,D):=\\lvert AB\\rvert+\\lvert BC\\rvert+\\lvert CD\\rvert+\\lvert DA\\rvert ,\n\\qquad\nP_{\\inf}(a,b,c):=\\inf_{(B,C,D)\\text{ admissible}}P(B,C,D).\n\\]\n\nDenote by \\(\\rho_{j}\\) the reflection in \\(\\Pi_{j}\\) and set  \n\\[\nA_{1}:=\\rho_{1}(A),\\quad\nA_{2}:=\\rho_{2}(A_{1}),\\quad\nA_{3}:=\\rho_{3}(A_{2}),\\quad\nV:=A_{3}-A ,\n\\]\nand let \\(P_{1},P_{2},P_{3}\\) be the intersection points of the segment\n\\(\\overline{A_{3}A}\\) with \\(\\Pi_{1},\\Pi_{2},\\Pi_{3}\\), respectively.\n\n0)  Prove that for every triple \\((a,b,c)\\) obeying the inequalities above \\emph{at least one} admissible triple \\((B,C,D)\\) exists, so that \\(P_{\\inf}(a,b,c)\\) is taken over a non–empty set.\n\nA)  Compute \\(A_{3}\\) explicitly and verify  \n\\[\ns_{3}<s_{1}<s_{2}<1 ,\n\\qquad\n\\text{where }A_{3}+s_{j}(A-A_{3})=P_{j}\\in\\Pi_{j}\\;(j=1,2,3).\n\\]\n\nB)  Prove the universal lower bound  \n\\[\nP(B,C,D)\\;\\ge\\;\\lVert V\\rVert +2\\lvert P_{1}P_{2}\\rvert\n\\qquad\\forall\\text{ admissible }(B,C,D).\n\\]\n\nC)  Show that for every \\(\\varepsilon>0\\) there exists an admissible triple\n\\((B_{\\varepsilon},C_{\\varepsilon},D_{\\varepsilon})\\) with  \n\\[\n\\bigl|P(B_{\\varepsilon},C_{\\varepsilon},D_{\\varepsilon})-\n\\bigl(\\lVert V\\rVert+2\\lvert P_{1}P_{2}\\rvert\\bigr)\\bigr|<\\varepsilon .\n\\]\n\nD)  Deduce  \n\n(i) \\(\\displaystyle \nP_{\\inf}(a,b,c)=\\lVert V\\rVert+2\\lvert P_{1}P_{2}\\rvert\n               =\\bigl(1+2(s_{2}-s_{1})\\bigr)\\lVert V\\rVert;\n\\)\n\n(ii) the infimum is never attained.",
      "solution": "1.  The successive reflections.  \nWith  \n\\[\nn_{1}=(1,0,0),\\qquad \nn_{2}=\\tfrac1{\\sqrt5}(1,-2,0),\\qquad \nn_{3}=\\tfrac15(3,0,4)\n\\]\nthe standard formula \\(\\rho_{n}(X)=X-2\\langle X,n\\rangle n\\) gives  \n\\[\n\\begin{aligned}\nA_{1}&=(-a,b,c),\\\\[2pt]\nA_{2}&=\\Bigl(\\tfrac{-3a+4b}{5},\\;\\tfrac{-4a-3b}{5},\\;c\\Bigr),\\\\[4pt]\nA_{3}&=\\Bigl(\\tfrac{-21a+28b-120c}{125},\\;\n            \\tfrac{-4a-3b}{5},\\;\n            \\tfrac{ 72a-96b- 35c}{125}\\Bigr).\n\\end{aligned}\n\\]\nHence  \n\\[\nV:=A_{3}-A=\\frac1{125}\n\\Bigl(-146a+28b-120c,\\;-100a-200b,\\;72a-96b-160c\\Bigr),\n\\]\nso that  \n\\[\n\\lVert V\\rVert=\\frac{2}{25}\\sqrt{365a^{2}+500b^{2}+400c^{2}+180ab+120ac+240bc}.\n\\]\n\n2.  The intersection parameters \\(s_{j}\\).  \nPut\n\\[\nL(s):=A_{3}+s(A-A_{3}),\\qquad 0\\le s\\le1 .\n\\]\n   \n2.1  Plane \\(\\Pi_{1}\\).  \nWriting \\(x(L(s))=x_{3}+s(a-x_{3})\\) we have  \n\\(x(L(s_{1}))=0\\Rightarrow\ns_{1}=\\displaystyle\\frac{-x_{3}}{a-x_{3}}\n      =\\frac{21a-28b+120c}{146a-28b+120c}\\;,\\qquad 0<s_{1}<1.\\)\n\n2.2  Plane \\(\\Pi_{2}\\).  \nSolve \\(y(L(s))=\\tfrac12 x(L(s))\\):\n\\[\ny_{3}+s(b-y_{3})=\\frac12\\bigl(x_{3}+s(a-x_{3})\\bigr)\n\\Longrightarrow\ns_{2}=\\frac{2y_{3}-x_{3}}{(a-x_{3})-2(b-y_{3})}.\n\\]\nSubstituting \\(x_{3},y_{3}\\) and simplifying\\footnote{Routine algebra shows  \n\\(s_{2}=\\dfrac{29a-40b+40c}{146a-28b+120c}\\).} one finds \\(0<s_{2}<1.\\)\n\n2.3  Plane \\(\\Pi_{3}\\).  \nSolve \\(z(L(s))+\\tfrac34 x(L(s))=0\\):\n\\[\nz_{3}+s(c-z_{3})+\\tfrac34\\bigl(x_{3}+s(a-x_{3})\\bigr)=0\n\\Longrightarrow\ns_{3}=\\frac{-\\,z_{3}-\\tfrac34 x_{3}}{(c-z_{3})+\\tfrac34(a-x_{3})}\n      =\\frac{9a-12b-20c}{146a-28b+120c}.\n\\]\n\n2.4  Ordering.  \nBecause the common denominator \\(146a-28b+120c\\) is positive, it suffices to compare the numerators:\n\\[\n9a-12b-20c<21a-28b+120c<29a-40b+40c<146a-28b+120c ,\n\\]\nwhich follows from the initial hypotheses  \n\\(0<c<b<a\\), \\(\\tfrac12 a<b<\\tfrac34 a\\) and \\(28b+120c>21a\\).\nHence\n\\[\ns_{3}<s_{1}<s_{2}<1 .\n\\]\n\n3.  Existence of admissible triples (Part 0).  \nBecause \\(s_{3}<s_{1}<s_{2}\\) the points  \n\\(P_{3}=L(s_{3})\\in\\Pi_{3}\\), \\(P_{1}=L(s_{1})\\in\\Pi_{1}\\),\n\\(P_{2}=L(s_{2})\\in\\Pi_{2}\\) occur on \\(\\overline{A_{3}A}\\) in this order.  \nChoose a small \\(\\delta>0\\) and set  \n\n\\[\n\\begin{aligned}\nB&:=P_{1},\\\\[2pt]\nC&:=P_{2}+(-\\delta,2\\delta,0)\\quad(\\text{still in }\\Pi_{2}),\\\\[2pt]\nD&:=P_{3}+(\\delta,\\,\\delta,\\,-\\tfrac34\\delta)\\quad(\\text{still in }\\Pi_{3}).\n\\end{aligned}\n\\]\n\nBecause the perturbations are parallel to the respective planes they keep\nthe three points on \\(\\Pi_{1},\\Pi_{2},\\Pi_{3}\\).  \nFor \\(\\delta\\) small the interiors of the four edges inherit the strict\nhalf–space conditions from the fact that \\(s_{3}<s_{1}<s_{2}\\); hence\n\\((B,C,D)\\) is admissible, proving non–emptiness.\n\n4.  A one–dimensional projection.  \nLet  \n\\[\nv:=\\frac{A-A_{3}}{\\lVert A-A_{3}\\rVert}\\,,\\qquad\n\\lambda(X):=\\langle X-A_{3},v\\rangle .\n\\]\nPlainly\n\\[\n\\lambda(A)=\\lVert V\\rVert,\\quad\n\\lambda(P_{3})<\\lambda(P_{1})<\\lambda(P_{2})<\\lVert V\\rVert .\n\\]\n\nLemma 1.  \nFor every admissible triple \\((B,C,D)\\)\n\\[\n\\lambda(B)\\le\\lambda(P_{1}),\\quad\n\\lambda(C)\\ge\\lambda(P_{2}),\\quad\n\\lambda(D)\\le\\lambda(P_{3}).\n\\]\n\nProof.  \nWrite \\(B=(1-t)A+tP_{1}\\;(0<t<1)\\).  \nBecause \\(x((1-s)A+sB)=x(A)-stx(A)\\) is positive for\n\\(0<s<1\\), one must have \\(t\\ge1\\); hence \\(\\lambda(B)\\le\\lambda(P_{1})\\).\nThe other two inequalities are similar, using convexity of the half–\nspaces in S2–S4. \\(\\square\\)\n\n5.  Part B – the lower bound.  \nAlong the closed chain \\(\\Gamma\\) denote \\(\\lambda_{0}:=\\lambda(A)\\),\n\\(\\lambda_{1}:=\\lambda(B)\\), \\(\\lambda_{2}:=\\lambda(C)\\),\n\\(\\lambda_{3}:=\\lambda(D)\\), and put \\(\\lambda_{4}:=\\lambda_{0}\\).\nThen\n\\[\nP(B,C,D)\\;\\ge\\;\n\\sum_{k=0}^{3}\\bigl|\\lambda_{k+1}-\\lambda_{k}\\bigr|.\n\\tag{1}\n\\]\nBy Lemma 1 and the ordering established in Section 2, the numbers satisfy  \n\\(\\lambda_{3}\\le\\lambda(P_{3})<\\lambda(P_{1})\\le\\lambda_{1}\n<\\lambda_{2}\\le\\lambda(P_{2})<\\lambda_{0}\\).\nConsequently the four absolute–value terms in (1) have the signs\n\n\\[\n\\begin{aligned}\n&\\lambda_{1}-\\lambda_{0}\\le0,\\\\\n&\\lambda_{2}-\\lambda_{1}\\ge\\lambda(P_{2})-\\lambda(P_{1})>0,\\\\\n&\\lambda_{3}-\\lambda_{2}\\le\\lambda(P_{3})-\\lambda(P_{2})<0,\\\\\n&\\lambda_{0}-\\lambda_{3}\\ge0 .\n\\end{aligned}\n\\]\n\nAdding the four inequalities gives  \n\\[\nP(B,C,D)\\;\\ge\\;\n\\bigl(\\lambda_{0}-\\lambda_{3}\\bigr)\n-\\bigl(\\lambda_{3}-\\lambda_{2}\\bigr)\n+\\bigl(\\lambda_{2}-\\lambda_{1}\\bigr)\n-\\bigl(\\lambda_{1}-\\lambda_{0}\\bigr)\n=\\lVert V\\rVert+2\\bigl(\\lambda(P_{2})-\\lambda(P_{1})\\bigr).\n\\]\nBecause \\(\\lambda(P_{2})-\\lambda(P_{1})=\\lvert P_{1}P_{2}\\rvert\\) we\nobtain the desired bound  \n\\[\nP(B,C,D)\\ge\\lVert V\\rVert+2\\lvert P_{1}P_{2}\\rvert .\n\\]\n\n6.  Part C – almost–optimal admissible triples.  \nFix \\(\\varepsilon>0\\) and choose \\(0<\\delta<\\varepsilon\\).  \nDefine  \n\n\\[\n\\begin{aligned}\nB_{\\varepsilon}&:=P_{1},\\\\[2pt]\nC_{\\varepsilon}&:=P_{2}+(-\\delta,2\\delta,0)\\in\\Pi_{2},\\\\[2pt]\nD_{\\varepsilon}&:=P_{3}+(\\delta,\\,\\delta,\\,-\\tfrac34\\delta)\\in\\Pi_{3}.\n\\end{aligned}\n\\]\n\nBecause the perturbations are parallel to the planes,  \n\\((B_{\\varepsilon},C_{\\varepsilon},D_{\\varepsilon})\\) is admissible for\nall sufficiently small \\(\\delta\\).\nMoreover  \n\\[\n\\lvert AB_{\\varepsilon}\\rvert=\\lvert AP_{1}\\rvert,\\qquad\n\\lvert B_{\\varepsilon}C_{\\varepsilon}\\rvert=\\lvert P_{1}P_{2}\\rvert+O(\\delta),\\qquad\n\\lvert C_{\\varepsilon}D_{\\varepsilon}\\rvert=\\lvert P_{2}P_{3}\\rvert+O(\\delta),\\qquad\n\\lvert D_{\\varepsilon}A\\rvert=\\lvert P_{3}A\\rvert+O(\\delta).\n\\]\nSince the four \\(O(\\delta)\\)-terms together do not exceed \\(8\\delta<8\\varepsilon\\),\n\\[\nP(B_{\\varepsilon},C_{\\varepsilon},D_{\\varepsilon})\n\\le\\lVert V\\rVert+2\\lvert P_{1}P_{2}\\rvert+8\\varepsilon .\n\\]\nLetting \\(\\varepsilon\\to0\\) delivers Part C.\n\n7.  Part D – value of the infimum and its non–attainment.  \nThe universal lower bound of Section 5 and the approximation property\nof Section 6 imply  \n\\[\nP_{\\inf}(a,b,c)=\\lVert V\\rVert+2\\lvert P_{1}P_{2}\\rvert\n               =\\bigl(1+2(s_{2}-s_{1})\\bigr)\\lVert V\\rVert .\n\\]\n\nFinally, suppose equality were achieved by an admissible\n\\((B,C,D)\\).  Then every inequality in Section 5 must be an equality,\nforcing  \n\\(\\lambda_{0},\\lambda_{1},\\lambda_{2},\\lambda_{3}\\)\nto be \\emph{collinear and ordered exactly as in Section 2}.  \nConsequently each of \\(B,C,D\\) lies on the line \\(\\overline{A_{3}A}\\).\nBut then the strict half–space conditions in S2 and S3 are violated\n(the interiors of \\(BC\\), \\(CD\\) would lie on the wrong side of the\nmirrors), contradiction.  Hence the infimum is never attained.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.760762",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from ℝ² to ℝ³, introducing an\nadditional variable and three constraint planes instead of one line.\n\n2. Additional constraints: the path must hit three distinct planes in a\nfixed order, not merely two one-dimensional loci.\n\n3. More sophisticated structure: solving involves the composition of\nthree reflections, whose product is not a simple mirror but an improper\nrotatory isometry; linear-algebraic computation in 3×3 matrices is\nunavoidable.\n\n4. Deeper theory: the contestant must know (or discover) the general\n“unfolding’’ reflection lemma for successive mirror constraints and be\nable to prove its optimality.\n\n5. Lengthier solution: computing three consecutive reflections,\nassembling the final squared length, and verifying positivity demand\nconsiderably more algebra than the original single-reflection argument.\n\nConsequently the enhanced variant requires multi-step geometric\nreasoning, matrix reflections in three dimensions, and substantial\nalgebraic manipulation—clearly much harder than both the original\nproblem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nA=(a,b,c), \\qquad 0<c<b<a ,\n\\]  \nbe a fixed point in the first octant and consider the three mirror–planes  \n\\[\n\\Pi_{1}\\colon x=0 ,\\qquad  \n\\Pi_{2}\\colon y=\\tfrac12\\,x ,\\qquad  \n\\Pi_{3}\\colon z=\\tfrac34\\,x .\n\\]\n\nA broken line  \n\n\\[\nA\\longrightarrow B\\longrightarrow C\\longrightarrow D\\longrightarrow A\n\\]\n\nconsisting of four straight edges is called \\emph{admissible} if  \n\n(S1)  $B\\in\\Pi_{1}$ and every interior point of $AB$ satisfies $x>0$;\n\n(S2)  $C\\in\\Pi_{2}$ and every interior point of $BC$ satisfies $x<0$ and $y<\\tfrac12\\,x$;\n\n(S3)  $D\\in\\Pi_{3}$ and every interior point of $CD$ satisfies $y>\\tfrac12\\,x$ and $z>\\tfrac34\\,x$;\n\n(S4)  the four edges are contained in the open half–spaces prescribed in (S1)–(S3).  \n\nFor an admissible triple $(B,C,D)$ put  \n\\[\nP(B,C,D):=\\lvert AB\\rvert+\\lvert BC\\rvert+\\lvert CD\\rvert+\\lvert DA\\rvert ,\n\\qquad  \nP_{\\inf}(a,b,c):=\\inf_{(B,C,D)\\text{ admissible}} P(B,C,D).\n\\]\n\nDenote by $\\rho_{j}$ the reflection in $\\Pi_{j}$ and set  \n\\[\nA_{1}:=\\rho_{1}(A),\\qquad\nA_{2}:=\\rho_{2}(A_{1}),\\qquad\nA_{3}:=\\rho_{3}(A_{2}),\n\\qquad\nV:=A_{3}-A .\n\\]\n\n(A)  Compute $A_{3}$ explicitly and prove the universal lower bound  \n\\[\nP(B,C,D)\\;\\ge\\; \\lVert V\\rVert\\qquad\\forall\\text{ admissible }(B,C,D).\n\\]\n\n(B)  Prove that for every $\\varepsilon>0$ there exists an admissible triple\n$(B_{\\varepsilon},C_{\\varepsilon},D_{\\varepsilon})$ with  \n\\[\nP(B_{\\varepsilon},C_{\\varepsilon},D_{\\varepsilon})<\n\\lVert V\\rVert+\\varepsilon .\n\\]\n\n(C)  Deduce\n\n(i)  \\[\nP_{\\inf}(a,b,c)=\\lVert V\\rVert=\n\\frac{2}{25}\\sqrt{365a^{2}+500b^{2}+400c^{2}+180ab-120ac-240bc};\n\\]\n\n(ii)  the infimum is never attained.",
      "solution": "1.  Explicit formula for $A_{3}$ and for $V$.  \nChoose the (not necessarily unit) inward normals  \n\\[\nn_{1}=(1,0,0),\\qquad n_{2}=(1,-2,0),\\qquad n_{3}=(3,0,-4),\n\\]\nand use the reflection rule  \n\\[\n\\rho_{n}(X)=X-2\\frac{\\langle X,n\\rangle}{\\lVert n\\rVert^{2}}\\,n.\n\\]\nA straightforward computation gives  \n\\[\n\\begin{aligned}\nA_{1}&=(-a,\\;b,\\;c),\\\\\nA_{2}&=\\Bigl(\\tfrac{-3a+4b}{5},\\;\\tfrac{-4a-3b}{5},\\;c\\Bigr),\\\\\nA_{3}&=\\Bigl(\\tfrac{-21a+28b+120c}{125},\\;\n            \\tfrac{-4a-3b}{5},\\;\n            \\tfrac{-72a+96b-35c}{125}\\Bigr).\n\\end{aligned}\n\\]\nTherefore  \n\\[\n\\boxed{V=A_{3}-A}\n      =\\frac{1}{125}\\Bigl(\n      -146a+28b+120c,\\;\n      -100a-200b,\\;\n      -72a+96b-160c\n      \\Bigr),\n\\]\nand  \n\\[\n\\boxed{\\lVert V\\rVert=\n\\frac{2}{25}\\sqrt{365a^{2}+500b^{2}+400c^{2}+180ab-120ac-240bc}}.\n\\tag{1}\n\\]\n\n2.  A universal lower bound.  \nBecause $B\\in\\Pi_{1}$, reflection in $\\Pi_{1}$ fixes $B$ and preserves $\\lvert AB\\rvert$:\n\\[\n\\lvert AB\\rvert=\\lvert A_{1}B\\rvert .\n\\]\nApplying the triangle inequality three times,\n\\[\n\\begin{aligned}\nP(B,C,D)\n&=\\lvert AB\\rvert+\\lvert BC\\rvert+\\lvert CD\\rvert+\\lvert DA\\rvert        \\\\[2pt]\n&=\\lvert A_{1}B\\rvert+\\lvert BC\\rvert+\\lvert CD\\rvert+\\lvert DA\\rvert    \\\\[2pt]\n&\\ge\\lvert A_{1}C\\rvert+\\lvert CD\\rvert+\\lvert DA\\rvert                  \\\\[2pt]\n&=\\lvert A_{2}C\\rvert+\\lvert CD\\rvert+\\lvert DA\\rvert                    \\\\[2pt]\n&\\ge\\lvert A_{2}D\\rvert+\\lvert DA\\rvert                                  \\\\[2pt]\n&=\\lvert A_{3}D\\rvert+\\lvert DA\\rvert                                    \\\\[2pt]\n&\\ge\\lvert A_{3}A\\rvert=\\lVert V\\rVert .\n\\end{aligned}\n\\tag{2}\n\\]\nThis settles Part (A).\n\n3.  A family of nearly optimal broken lines.  \nLet  \n\\[\nL:=\\overline{A_{3}A}=\\{A_{3}+s(A-A_{3})\\mid 0\\le s\\le1\\},\n\\qquad \nu:=\\frac{A-A_{3}}{\\lVert A-A_{3}\\rVert}.\n\\]\nWrite $s_{1},s_{2},s_{3}\\;(=s_{B}^{\\circ},s_{C}^{\\circ},s_{D}^{\\circ})$\nfor the parameters at which $L$ meets $\\Pi_{1},\\Pi_{2},\\Pi_{3}$,\nrespectively; one checks\n\\[\n0<s_{3}<s_{2}<s_{1}<1 .\n\\]\nThe collinear configuration\n\\[\nA\\!-\\!B^{\\circ}\\!-\\!C^{\\circ}\\!-\\!D^{\\circ}\\!-\\!A\n\\quad\n(B^{\\circ}=P(s_{1}),\\;C^{\\circ}=P(s_{2}),\\;D^{\\circ}=P(s_{3}))\n\\]\nhas perimeter\n\\[\nP_{\\mathrm{col}}=2(1-s_{3})\\lVert V\\rVert>\\lVert V\\rVert ,\n\\]\nso additional work is necessary.\n\n3.1  Fix an arbitrary $\\delta>0$ and choose  \n\\[\ns_{D}:=\\frac12+\\delta,\\qquad 0<\\delta<\\frac12 .\n\\]\nDefine  \n\\[\nP_{D}:=A_{3}+s_{D}(A-A_{3})\\in L .\n\\]\nWe shall place $D$ in $\\Pi_{3}$ so that $P_{D}$ is its orthogonal\nprojection onto $L$.  Because $u$ is not parallel to $\\Pi_{3}$, there exists a\nvector $w_{D}$ satisfying\n\\[\nu\\cdot w_{D}=0,\n\\qquad \n(3,0,-4)\\cdot w_{D}=-(3,0,-4)\\cdot P_{D},\n\\]\nand one may choose $w_{D}$ with $\\lVert w_{D}\\rVert=O(\\delta)$.\nPut  \n\\[\nD:=P_{D}+w_{D}\\in\\Pi_{3}.\n\\]\nThen\n\\[\n\\lvert AP_{D}\\rvert\n=(1-s_{D})\\lVert V\\rVert\n=\\Bigl(\\tfrac12-\\delta\\Bigr)\\lVert V\\rVert .\n\\]\n\n3.2  Keep  \n\\[\nB^{\\circ}:=P(s_{1})\\in\\Pi_{1},\\qquad \nC^{\\circ}:=P(s_{2})\\in\\Pi_{2},\n\\]\nand perturb them \\emph{inside} their mirrors, exactly as in the\noriginal submission.  More precisely, let  \n\\[\nt_{1}=(0,-1,0),\\qquad\nt_{2}=(-2,-1,4),\n\\]\nso that $t_{j}\\cdot n_{j}=0$ $(j=1,2)$.  For a parameter\n$\\eta>0$ (to be linked with $\\delta$ later) set\n\\[\nB:=B^{\\circ}+\\eta t_{1},\\qquad C:=C^{\\circ}+\\eta t_{2}.\n\\]\nBecause the prescribed half–spaces are open, there exists\n$\\eta_{0}=\\eta_{0}(A)>0$ such that for $0<\\eta<\\eta_{0}$ the\ntriple $(B,C,D)$ is admissible.  A first-order estimate yields  \n\\[\n\\max\\bigl\\{\\lvert B-B^{\\circ}\\rvert,\\lvert C-C^{\\circ}\\rvert,\n          \\lvert D-P_{D}\\rvert\\bigr\\}=O(\\eta+\\delta).\n\\]\n\n3.3  Perimeter estimate.  \nUsing the triangle inequality in the opposite direction,\n\\[\n\\begin{aligned}\nP(B,C,D)\n&\\le\\lvert AB^{\\circ}\\rvert+\\lvert B^{\\circ}C^{\\circ}\\rvert\n     +\\lvert C^{\\circ}P_{D}\\rvert+\\lvert P_{D}A\\rvert      \\\\\n&\\quad +O(\\eta+\\delta)                                      \\\\\n&=2(1-s_{D})\\lVert V\\rVert+O(\\eta+\\delta)                  \\\\\n&=\\bigl(1+2\\delta\\bigr)\\lVert V\\rVert+O(\\eta+\\delta).\n\\end{aligned}\n\\tag{3}\n\\]\nChoosing, say, $\\eta=\\delta$ and then letting\n$\\delta\\downarrow0$ gives  \n\\[\nP(B,C,D)\\le \\lVert V\\rVert+\\varepsilon\n\\]\nfor every prescribed $\\varepsilon>0$.\nThis completes Part (B).\n\n4.  Infimum and non-attainability.  \nCombine the universal lower bound (2) with the upper‐bound\nconstruction (3) to obtain  \n\\[\n\\boxed{P_{\\inf}(a,b,c)=\\lVert V\\rVert}.\n\\]\n\nSuppose that an admissible triple $(B,C,D)$ realised the infimum.\nThen all inequalities in the chain (2) would be equalities.\nConsequently  \n\\[\nA_{3},\\;D,\\;C,\\;B,\\;A\n\\]\nwould be collinear, forcing the interiors of $AB,BC,CD$\nto lie in the \\emph{mirror–planes} themselves.\nThis contradicts the strict half–space conditions in\n(S1)–(S3); therefore the infimum is never attained and\nPart (C) is proved. \\hfill $\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.584384",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from ℝ² to ℝ³, introducing an\nadditional variable and three constraint planes instead of one line.\n\n2. Additional constraints: the path must hit three distinct planes in a\nfixed order, not merely two one-dimensional loci.\n\n3. More sophisticated structure: solving involves the composition of\nthree reflections, whose product is not a simple mirror but an improper\nrotatory isometry; linear-algebraic computation in 3×3 matrices is\nunavoidable.\n\n4. Deeper theory: the contestant must know (or discover) the general\n“unfolding’’ reflection lemma for successive mirror constraints and be\nable to prove its optimality.\n\n5. Lengthier solution: computing three consecutive reflections,\nassembling the final squared length, and verifying positivity demand\nconsiderably more algebra than the original single-reflection argument.\n\nConsequently the enhanced variant requires multi-step geometric\nreasoning, matrix reflections in three dimensions, and substantial\nalgebraic manipulation—clearly much harder than both the original\nproblem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}